CH 1 SOME BASIC CONCEPTS OF CHEMISTRY - rjvision.org · Molecular formula = molar mass 14 84...
Transcript of CH 1 SOME BASIC CONCEPTS OF CHEMISTRY - rjvision.org · Molecular formula = molar mass 14 84...
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CH – 1 SOME BASIC CONCEPTS OF CHEMISTRY
1. (b)
Gaseous mixture formed is CO and CO2 when it is passed through KOH, only CO2 is absorbed. So the remaining gas is
CO.
So, weight of remaining gaseous product CO is g8.22820
2
So, the correct option is (b).
2. (a)
(1) Mass of water = 18 x 1 = 18 g
Molecules of water = mole x NA = A
N18
18
(2) Moles of water = 4
104.22
00224.0
Molecules of water = mole x NA = 10-4
NA
(3) Molecules of water = mole x NA = A
N18
18.0
=10—2
NA
(4) Molecules of water = mole x NA = 10-3
NA
3. (a)
Temperature dependent unit is molarity.
4. (c)
Mole ratio of (i) carbon = 1.712
7.85 (ii) Hydrogen = 3.14
1
3.14
Simplest ratio of carbon and hydrogen = 1 : 2
So, empirical formula is CH2
Molecular formula = molar mass 14
84
massformulaempirical
massmolar= 6
So, molecular formula = 126 HC
Note : molecular mass is obtained from following data 42 mg of the compound contains 3.01 x 1020
molecules.
5. (c)
16.9 g AgNO3 is present in 100 mL solution.
∴ 8.45 g AgNO3 is present in 50 mL solution
5.8 g NaCl is present in 100 mL solution
∴ 2.9 g NaCl is present in 50 mL solution.
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mass of AgCl precipitated
= 0.049 × 143.5 g
= 7g AgCl
6. (b)
∵ mass of 1 mol (6.022 × 1023
atoms) of carbon = 12 g
If Avogadro Number (NA) is changed
then mass of 1 mol (6.022 × 1023
atoms) of carbon
23
20
106.022
106.02212
g1012
3
7. (d)
∵ 1 mole water = 6.02 × 1023
molecules
∴ 18 mole water = 18 × 6.02 × 1023
molecules
So, 18 mole water has maximum number of molecules.
8. (d)
MgCO3(s) → MgO(s) + CO2(g)
moles of MgCO3 = 84
20= 0.238 mol
From above equation
1 mole MgCO3 gives 1 mole MgO
∴ 0.238 mole MgCO3 will give 0.238 mole MgO
= 0.238 × 40 g = 9.523 g MgO
Practical yield of MgO = 8 g MgO
∴ % purity = 84%1009.523
8
9. (b)
10. (a)
MgOO2
1Mg 2
32
56.0
24
0.1
4
07.0
12
5.0
2
x
4
0.07x -
12
0.5
Oxygen is limiting reagent so 02
x
4
0.07
2
0.07x
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Excess Mg = mole4
0.07 -
12
0.5
Mass of Mg is = 1 0.7 12 = 0.16 gram.
11. (a)
H2 + CI2 2HCI
22.4 It 11.2 It
= 1 mole = 0.5 mole
Limiting reagent is CI2. So, 1 mole HCI is formed.
12. (c)
Millimoles of solution of chloride = 0.05 × 10 = 0.5
Millimoles of AgNO3 solution = 10 × 0.1 = 1
So, the millimoles of AgNO3 are double than the chloride solution.
∴ XCl2 + 2AgNO3 → 2AgCl + X(NO3)2.
13. (a)
M 0.01106.02
106.02
206.02100
1000106.02M
23
21
23
20
14. (c)
No. of molecules
Moles of CO2 = 44
44 = 1 NA
Moles of O3 = 48
48 = 1 NA
Moles of H2 = 2
8 = 4 4NA
Moles of SO2 = 64
64 = 1 NA
15. (b)
The number of atoms in 0.1 moles of a triatomic gas
= 0.1 × 3 × 6.023 × 1023
= 1.806 × 1023
16. (d)
No. of milli equivalent of HCl = 20 × 0.05 = 1.0
No. of milli equivalent of Br(OH)2 = 30 × 0.1 × 2 = 6.0
After neutralization, no. of milli equivalents in 50 ml. of solution = (6 – 1) = 5
Total volume of the solution = 20 + 30 = 50 ml
∴ No. of milli equivalent of OH– is 5 in 50 ml
M 0.11050
10005][OH
3
17. (b)
OHO2
1H
2
64g
210g
2
(2mol)(5mol)
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In this reaction oxygen is the limiting agent. Hence amount of H2O produced depends on the amount of O2 taken
∵ 0.5 mole of O2 gives H2O = 1 mol
∴ 2 mole of O2 gives H2O = 4 mol
18. (d)
Writing the equation for the reaction, we get
OHPbCl2HClPbO
2
278g71207
2
73g36.52
223g16207
From this equation we find 223 g of PbO reacts with 73 g of HCl to form 278 g of PbCl2. If we carry out the reaction
between 3.2 g HCl and 6.5 g PbO.
Amount of PbO that reacts with 3.32 g HCl
= g 9.773.273
223
Since amount of PbO present is only 6.5 g so PbO is the limiting reagent.
Amount of PbCl2 formed by 6.5 g of PbO
g 6.5223
278
Number of moles of PbCl2 formed
278
6.5
223
278 moles = 0.029 moles
19. (a)
Element % At. wt. Relative number Ratio
C 38.71 12 3.2312
38.71 1
3.23
3.23
H 9.67 1 9.671
9.67 3
3.23
9.67
O 100 – (38.71 + 9.67) = 51.62 16 3.2316
51.62 1
3.23
3.23
Thus, empirical formula is CH3O.
20. (c)
Writing the equation of combustion of propane (C3H8), we get
O4H3CO5OHC 22
L 5 vol5
2
L 1 vol1
83
From the above equation we find that we need 5 L of oxygen at NTP to completely burn 1 L of propane at N. T.P.
If we change the conditions for both the gases from N.T.P. to same conditions of temperature and pressure. The same
results are obtained. i.e. 5 L is the correct answer.
21. (a)
C8H10CO2Mn 22
2
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Fe2+
ions will also oxidise to Fe3+
.
From above equation 2 moles MnO4 required to oxide 5 moles of oxalate.
Thus number of moles of MnO4 required to oxidise one mole of oxalate = 2/5 = 0.4.
And number of moles of MnO4 required to oxidise one mole of Fe
2+ =1/5 = 0.2
Total = 0.6
22. (c)
Volume
MassDensity
3
3
cm
gram 1cm gram 1
Density
MassVolume
3
3cm 1
cm gram 1
gram 1
∴ Volume occupied by 1 gram water = 1 cm3 or volume occupied by
18
106.02323
molecules of water = 1 cm
3
[∴ 1 g water = 18
1 moles of water]
Thus volume occupied by 1 molecules of water
23
106.023
181
cm
3 = 3.0 × 10
–23 cm
3.
i. e. the correct answer is option (c).
23. (b)
The balance chemical equation is:
34 5SO6H2MnO → O3H5SO2Mn 24
From the equation it is clear that
Moles of
4MnO require to oxidize 5 moles of
3SO are 2/5.
24. (d)
Average isotopic mass of
2890
2202819990200X
100
404189218000
100
19996
= 199.96 amu
25. (c)
Molarity of H2SO4 solution.
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18.41.8410098
100098
Suppose V ml of this H2SO4 is used to prepare 1 lit. of 0.1 M H2SO4
∴ V × 18.02 = 1000 × 0.1
Or 18.02
0.11000V
= 5.55 ml.
26. (a)
4
6
2
OH
4
7
nOMKnOMK
Change in oxidation number of Mn in basic medium is 1. Hence mole of KI is equal to mole of KMnO 4.
27. (c)
2Al2O3 + 3C → Al + 3CO2
Gram equivalent of Al2O3 ≡ gm equivalent of C
Now equivalent weight of Al = 93
27
Equivalent weight of C = )OCC(gm34
122
40
No. of gram equivalent of Al = 9
102703
= 30 × 10
3
Hence,
No. of gram equivalent of C = 30 × 103
Again,
No. of gram equivalent of C = weight equivalent gram
gram in mass
⇒ 30 × 103 =
3
mass
⇒ mass = 90 × 103 g = 90 kg
28. (a)
No. of molecules in different cases
(a) ∵ 22.4 litre at STP contains = 6.023 × 1023
molecule of H2
∴ 15 litre at STP contains = 23
106.02322.4
15
(b) ∵ 22.4 litre at STP contains = 6.023 × 1023
molecule of N2
∵ 5 litre at STP contains = 23
106.02322.4
5
(c) ∵ 2 gm of H2 = 6.023 × 1023
molecules of H2
∵ 0.5 gm of H2 = 23
106.0232
0.5
(d) Similarly 10 g of O2 gas 23
106.02332
10 molecules
Thus, (a) will have maximum number of molecules.
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29. (c)
litre 20 vol.2
3
litre 30 vol.3
2
litre 10 vol.1
2 2NH3HN
It is given that only 50% of the expected product is formed hence only 10 litre of NH3 is formed.
N2 used = 5 litres, left = 30 – 5 = 25 litres
H2 used = 15 litres, left = 30 – 15 = 15 litres
30. (b)
2 g of H2 means one mole of H2, hence contains 6.023 × 1023
molecules. Others have less than one mole, so have less
no. of molecules.
31. (d)
Suppose the mol. wt. of enzyme = x
Given 100 g of enzyme wt of Se = 0.5 gm
∴ In x g of enzyme wt. of Se = x100
0.5
Hence, 100
0.578.4
x
∴ x = 15680 = 1.568 × 104
32. (d)
Specific volume (volume of 14 gm) of cylindrical virus particle = 6.02 × 10–2
cc/gm
Radius of virus (r) = 7 Å = 7 × 10–8
cm
Length of virus = 10 × 10–8
cm
Volume of virus
8282 1010)10(7
7
22r lπ
= 154 × 10–23
cc
Wt. of one virus particle = volumesecific
volume
∴ Mol. wt. of virus = Wt. of NA particle
23
2
23
106.02106.02
10154
= 15400 g/mol = 15.4 kg/mole
33. (c)
gm 197
3BaCO → BaO + CO2
∵ 197 gm of BaCO3 released carbon dioxide = 22.4 litre at STP
∴ 1 gm of BaCO3 released carbon dioxide = 197
22.4 litre
∴ 9.85 gm of BaCO3 released carbon dioxide = 9.85197
22.4 = 1.12 litre
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34. (c)
Element % At. wt. Relative number Ratio
C 40 12 12
403.33
3.33
3.33 1
H 6.66 1 1
6.66 6.66
3.33
6.66 2
O 53.34 16 16
53.34 3.33
3.33
3.33 1
Empirical formula of organic compound = CH2O.
35. (c)
According to Stoichiometry they should react as follow
mole 1moles 5
2
moles 0.8moles 4
3 5O4NH →
moles 1.2moles 6
2
mole 0.8moles 4
O6H4NO
Thus for 1 mole of O2 only 0.8 mole of NH3 is consumed. Hence O2 is consumed completely.
36. (c)
There is 0.334 g of Fe per 100 g of Hb.
∴ 100 g Hb ⟶ 0.334 g Fe.
∴ 67200 g Hb ⟶ 100
672000.334 = 224.448 g
56 g Fe ⟶ 1 Fe atom
224.448 ⟶
56
1224.4 4 atoms of He
37. (d)
We know that all non-zero digits are significant and the zeros at the beginning of a number are not significant.
Therefore number 161 cm, 0.161 cm and 0.0161 cm have 3, 3 and 3 significant figures respectively.
38. (c)
As the sum of the percentage of C, H & N is 100. Thus it does not contains O atom.
Table for empirical formula
Element % At. wt. Rel. Number Ratio
C 40.00 12 12
403.66
3.33
3.66 1.09 ~ 1
H 13.33 1 1
13.33 13.33
3.33
13.33 4
N 46.67 14 14
46.67 3.33
3.33
3.33 1
Hence, empirical formula = CH4N
39. (d)
15(32)
22(78)
66 (g)15OH2C → 12CO2(g) + 6H2O(g)
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∴ 156 gm of benzene required oxygen = 15 × 22.4 litre
∴ 1 gm of benzene required oxygen = 156
22.415 litre
∴ 39 gm of Benzene required oxygen = 84.0156
3922.415
litre
40. (d)
Molecular weight of ZnSO4.7H2O = 65 + 32 + (4 × 16) + 7(2 × 1 + 16) = 287.
∴ Percentage mass of zinc (Zn) = % 22.65100287
65
41. (b)
Molecular weight of C60H122 = (12 × 60) + 122 = 842.
Therefore weight of one molecule number sAvagadro'
HC of weightMolecular 12260
g 101.4106.023
842 21
23
42. (c)
On calculation we find
6
5
101.171.37
)10(1.79 20.2)(29.2
As the least precise number contains 3 significant figures therefore answers should also contains 3 significant figures.
43. (a)
No of moles of nitride ion
0.314
4.2 mol = 0.3 × NA = 0.3 NA nitride ions
Valence electrons = 8 × 0.3 NA = 2.4 NA
(5 + 3 due to charge) One N3–
ion contains 8 valence electrons.
44. (a)
5 M H2SO4 = 10 N H2SO4, (∵ Basicity of H2SO4 = 2)
N1V1 = N2V2,
10 × 1 = N2 × 10 or N2 = 1 N
45. (c)
According to Avogadro’s law ‚equal volumes of all gases contain equal numbers of molecules under similar conditions
of temperature and pressure‛. Thus if 1 L of one gas contains N molecules, 2 L of any gas under the same conditions
will contain 2N molecules.
46. (a)
Average atomic mass = 100
mass] Atomic2 of Al [Relativemass] Atomic1 of Abundance [Relative 2
100
11811019 = 10.81
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47. (a)
4.4 g CO2 = 0.144
4.4 mol CO2 (mol. wt. of CO2 = 44)
= 6 × 1022
molecules
= 2 × 6 × 1022
atoms of O.
48. (b)
6.02 × 1023
molecules of CO = 1 mole of CO
6.02 × 1024
CO molecules = 10 moles CO
Which is equal to 10 gm atoms of O and 5 gm molecules of O2
49. (c)
C2H4 + 3 O2 → 2CO2 + 2H2O
28kg 96 kg
∵ 28 kg of C2H4 undergo complete combustion by = 96 kg of O2
∴ 2.8 kg of C2H4 undergo complete combustion by = 9.6 g of O2.
50. (a)
Cp / Cv = 1.4 shows that the gas is diatomic.
22.4 litre at NTP ≡ 6.02 × 1023
molecules
11.2 L at NTP = 3.01 × 1023
molecules
= 3.01 × 1023
× 2 atoms = 6.02 × 1023
atoms
51. (d)
The reaction may given as
Z2O3 + 3H2 → 2Z + 3H2O
0.1596 g of Z2O3 react with H2 = 6 mg = 0.006 g
∴ 1 g of H2 react with 0.006
0.1596 = 26.6 g of Z2O3
∴ Eq. wt. of Z2O3 = 26.6 (from the definition of eq. wt.)
Eq. wt. of Z + Eq. wt. of O = E + 8 = 26.6
⇒ Eq. wt. of Z = 26.6 – 8 = 18.6
Valency of metal in Z2O3 = 3
valency
wt. Atomicmetal of wt. Eq.
∴ At. wt. of Z = 18.6 × 6 = 55.8
52. (d)
At NTP 22400 cc of N2O = 6.02 × 1023
molecule
∴ 1 cc N2O = 22400
106.0223
molecules
22400
106.02323
atoms
22
10224
1.8 atoms
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No. of electrons in a molecule of N2O = 7 + 7 + 8 = 22
Hence no. of electrons 2222400
106.0223
electrons = 22400
101.3223
53. (a)
Mg++
+ Na2CO3 → MgCO3 + 2Na+
1g eq. 1g eq.
Mili eq. of Mg2+
= 112
12
wt. atomic
10wt(g)3
∴ 12 mg Mg++
= 1 milli eq. Na2CO3
54. (a)
Mol wt. of CCl4 = 154 g = 22.4 L at STP
∴ Density = 22.4
154gL
–1 = 6.875 gL
–1
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1. (b)
According to Hund’s Rule of maximum multiplicity, the correct electronic configuration of N -atom is
Option (b) violates Hund’s Rule.
2. (c)
In H–like atom energy of 2s = 2p. orbital.
3. (d)
42 = 16
4. (d)
Ti(22) = 1s22s
22p
63s
23p
64s
23d
2
5. (a)
6. (d)
18
9
834
104.41045
103106.63
λ
hcE
7. (a)
It is 3P orbital with magnetic Q.N = 0
So, it should be 3PZ
8. (d)
AN
hcE
λ
λ
231027106.02103106.62
λ
8101.19
Hence, none of the given option is correct.
9. (c)
Energy of an electron at infinite distance from the nucleus is zero. As an electron approaches the nucleus, the electron
attraction increases and hence the energy of electron decreases and thus becomes negative. Thus as the value of n
decreases, i.e., lower the orbit is, more negative is the energy of the electron in it.
10. (c)
n = 3 → 3rd
shell
l = 1 → p sub shell.
m = –1 is possible for two electrons present in an orbital.
11. (b)
C = v
nm 50106
103
v
C15
17
λ
CH – 2 STRUCTURE OF ATOM
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12. (a)
Orbital angular momentum
1)(2
h
π
For p orbital = 1
So, ππ 2
h2
2
h
13. (c)
Electronic configuration of Rb = [Kr] 5s–1
Set of quantum number, n = 5
= 0, ∴ s orbital
m = 0, s = + 1/2
14. (a)
(n = 4, l = 3) ⇒ 4f subshell
Since, maximum no. of electrons in a subshell = 2(2l + 1)
So, total no. of electron in 4f subshell
= 2 (2 × 3 + 1) = 14 electrons.
15. (c)
MOT configuration of O2 and
2O :
2O : (σ1s)2 (σ*1s)
2 (σ2s)
2 (σ*2s)
2 (σ2pz)
2
)2p2p( 2
y
2
x ππ )2p2p( 1
yx *π*π 1
Number of unpaired electrons = 2, so paramagnetic.
O2: (σ1s)2 (σ*1s)
2 (σ2s)
2 (σ*2s)
2 (σ2pz)
2
)2p2p( 2
y
2
x ππ )2p2p( 0
yx *π*π 1
Number of unpaired electrons = 1, so paramagnetic.
16. (c)
Energy of photon obtained from the transition n = 6 to n = 5 will have least energy.
2
2
2
1
2
n
1
n
113.6ZE Δ
17. (a)
ns → (n – 2)f → (n – 1)d → np [n = 6]
18. (b)
Given E1 = 25eV E2 = 50 eV
1
1
hcE
λ
2
2
hcE
λ
1
2
2
1
E
E
λ
λ
2
1
50
25
1
2 λ
λ ∴ 1 = 22
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19. (a)
O2 16 KK(σ2s)2 (σ*2s)
2 (σ2pz)
2 (π2px)
2 (π2py)
2 2
2O 15 Remove one electron from π*2py from O2 2.5
2O 17 KK(σ2s)2 (σ*2s)
2 (σ2pz)
2 (π2px)
2 (π2py)
2 1.5
2
2O 18 KK(σ2s)2 (σ*2s)
2 (σ2pz)
2 (π2px)
2 (π2py)
2 1
Since, the bond length decreases as the bond order increases, hence,
2O have least bond length.
20. (b)
Total no. of atomic orbital in a shell = n2.
Given n = 4; Hence number of atomic orbitals in 4th
shell will be 16.
21. (d)
m1011000.66
106.6
mv
h 3534
λ
22. (b)
Bond order of Be2 = 0, hence Be2 cannot exis t.
23. (a)
Molecular orbital configuration of
1
z
1
y2
x2
z
2
y22222
22p*
2p* σ2p
2p
2p2s*σ σ2s1s*σ σ1sN
π
π
π
π
Bond order = 22
610
0
z
1
y2
x2
z
2
y2222
22p*
2p* 2p
2p
2p2s* 2s1s* 1sN
π
πσ
π
πσσσσ
Bond order = 2.52
510
2p 2p
2p2s* 2s1s* 1sN 2
x2
z
2
y2222
2 σπ
πσσσσ
Bond order = 32
410
∴ The correct order is = 22
2
2 NNN
24. (b)
m = 2l + 1, thus for l = 2, m = 5, hence values of m will be –2, –1, 0, +1, +2.
Therefore, for l = 2, m cannot have the value – 3.
25. (d)
The number of sub shell is (2 + 1). The maximum number of electrons in the sub shell is 2 (2 + 1) = (4 + 2)
26. (d)
K.E per atom
2
104.0104.41919
20
19
102.02
100.4
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27. (a)
∆P = m∆v
Substituting the given values of ∆x and m, we get
1 × 10–18
gcms–1
= 9 × 10–28
g × ∆v
or 28
18
109
101v
Δ
= 1.1 × 109 cms
–1 ≃ 1 × 10
9 cm s
–1
i.e. option (a) is correct.
28. (a)
We know ∆p.∆x π4
h
Or m.∆v.∆x = π4
h [∴ ∆p = m∆v+
Since ∆p = ∆x (given)
∴ ∆p.∆p = π4
h or m∆v m ∆v =
π4
h
Or (∆v)2 =
2 4 mπ
h
Or ∆v =
h
mmπ
h
2
1
42
Thus, option (a) is the correct option.
29. (b)
(ii) is not possible for any value of because varies from 0 to (n – 1) thus for n = 2, can be only 0, 1, 2.
(iv) is not possible because for = 0, m = 0.
(v) is not possible because for = 2, m varies from –2 to +2.
30. (b)
Magnetic quantum no. represents the orientation of atomic orbitals in an atom. For example p x, py & pz have
orientation along X-axis, Y-axis & Z-axis.
31. (d)
We know that ∆x . ∆p ≥ π4
h
∆x.m∆v > π4
h
∆v > xm
h
4
3110
34
109.11100.14
106.626v
πΔ
= 7
1094
66
π
= 5.79 × 10
6 m/sec
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32. (b)
We know that
21
nn
zEE
Given 1
2
12
mol kJ 3282
EE
11
2
24
mol kJ 82mol kJ4
328
2
EE
33. (c)
2
2
2
1n
1
n
1IE
h
1v
115
34
18
s 103.0816
1
1
1
106.625
102.18
34. (c)
Amongst isoelectric species, ionic radii of anion is more than that of cations. Further size of anion increase with
increase in-ve change and size of cation decrease with increase in +ve charge. Hence, ionic radii decreases from O– to
Al+++
.
35. (d)
v
cor
hchvE λ
λ
m 103.75108
103 8
15
8
λ
In nanometer = 3.75 × 10
Which is closest to 4 × 101.
36. (a)
Suppose the nucleus of hydrogen atom have charge of one proton i.e., The electron revolves in radius of r around it.
Therefore the centripetal force is supplied by electrostatic force of attraction i.e.
2
22
r
ze
r
mv
Or r
ze
r
mv 22
Or K.Er
ze
2
1mv
2
1 22
now total energy (En) = K.E + P.E in first excited state
r
zemv
2
1E
22
r
ze
r
ze
2
122
–3.4 eV = r
ze
2
1 2
∴ K.E = r
ze
2
12
= + 3.4 eV
37. (c)
Both CN– and CO have 14 electrons.
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30 YEAR’S BOOK CHEMISTRY PAGE: 93
38. (a)
Quantum number n = 3, l = 2, m = +2 represent an orbital with
2
1s 22
yxxy 3dor 3d
Which is possible only for one e–.
39. (c)
The molecule which contains same number of electrons are called isoelectronic. Eg.,
N2 = CO = CO– = 2
2O = 14e
40. (c)
The energy of photon,
19
103.03hv
E
λ
Or 19
834
103.03
103106.626
λ
Or 7
103.03
19.878 λ
= 6.56 × 10
–7 m = 656 nm
41. (c)
Energy of electron in 2nd orbit of Li+2
2
2
n
z13.6
eV 30.6(2)
(3)13.62
2
Energy required = 0 – (–30.6) = 30.6 eV
42. (d)
By Heisenberg uncertainty Principle 4π
hΔpΔx (which is constant)
As ∆x for electron and helium atom is same thus momentum of electron and helium will also be same therefore the
momentum of helium atom is equal to 5 × 10–26
kg. m.s–1
.
43. (d)
Given: Radius of hydrogen atom = 0.530Å, Number of excited state (n) = 2 and atomic number of hydrogen atom (Z) =
1. We know that the Bohr radius.
Z
n(r)
2
Radius of atom 0.5301
(2)2
= 4 × 0.530 = 2.12 Å
44. (a)
We know that ions which have the same number of electrons are called isoelectronic. We also know that both CO and
CN– have 14 electrons, therefore these are isoelectronic.
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30 YEAR’S BOOK CHEMISTRY PAGE: 94
45. (c)
According to deBroglie
17
34
10
106.626hmv
mv
h
λ
λ
⇒ p = 6.626 × 10–17
kg m/s
46. (d)
The orbitals which have same energy are called degenerate orbitals eg. p x, py, pz.
47. (a)
Given mass of an electron(m) = 9.1 × 10–28
g;
Velocity of electron (v) = 3 × 104 cm/s;
Accuracy in velocity = 0.001% = ;100
0.001
Actual velocity of the electron
(∆v) = 3 × 104 ×
100
0.001= 0.3 cm/s.
Planck’s constant (h) = 6.626 × 10–27
erg-sec.
∴ Uncertainty in the position of the electron
(∆x) = vm
h
4 0.3)10(9.1224
7106.62628
27
= 1.93 cm
48. (a)
State of hydrogen atom (n) = 1 (due to ground state)
Radius of hydrogen atom (r) = 0.53 Å.
Atomic number of Li(Z) = 3.
Radius of Li2+
ion 0.173
(1)0.53
Z
nr
22
1
49. (d)
as m = (2 + 1), hence m = –1 means,
–1 = 2 + 1. = 1
i.e., least value of = 1. So it cannot be present in s-orbital. Because for s orbital = 0.
50. (c)
K.E. of emitted electrons = hʋ - hʋ (i.e. smaller than hʋ)
51. (b)
P.E. = work done = r
zedr
r
ze 2r
2
2
52. (b)
Wave nature of electron was shown by Davisson and Germer. Davisson and germer demonstrated the physical reality
of the wave nature of electrons by showing that a beam of electrons could also diffracted by crystals just like light of x-
rays.
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30 YEAR’S BOOK CHEMISTRY PAGE: 95
53. (b)
Cathode rays are never electromagnetic waves.
54. (b)
The sub-shell are 3d, 4d, 4p and 4s, 4d has highest energy as n + value is maximum for this.
55. (a)
The ionization energy of any hydrogen like species (having one electron only) is given by the equation
2
422
h
e mZ 2I.E
π or I.E ∝ Z
2
Since the atomic number of H is 1 and that of He is 2, therefore, the I.E. of He+ is four times (2
2) the I.E. of H i.e., 13.6 ×
4 = 54.4 eV
56. (c)
Outer electronic configuration of Cl = 1
z
2
y
2
x
2 3p3p3p3s
Outer electronic configuration of Cl– =
2
z
2
y
2
x
2 3p3p3p3s
Hence Cl– contain four lone pairs of electron
57. (c)
Energy of an electron in Bohr’s orbit is given by the relationship. eVn
13.6E
2n
58. (b)
The sub-shell with lowest value of (n + ) is filled up first. When two or more sub-shells have same (n + ) value of
subshell with lowest value of ‘n’ is filled up first therefore the correct order is
Orbital 4s 3d 4p 5s 4d
n + 4 + 0 3 + 2 4 + 5 + 4 + 2
Value = 4 5 5 5 6
59. (d)
3 means f-subhsell. Maximum no. of electrons = 4 + 2 = 4 × 3 + 2 = 14
60. (d)
This is as per the definition of Pauli’s Exclusion principle.
61. (a)
Cu+ = 29 – 1 = 28e
–
Thus the electronic configuration of Cu+ is Cu
+(28) = 1s
2 2s
2 2p
6
18e
1062 3d 3p 3s
62. (c)
n = 2, 1 = 1 means 2 p-orbital. Electrons that can be accommodated = 6 as p sub-shell has 3 orbital and each orbital
contains 2 electrons.
63. (b)
No. of orbitals in a sub-shell = 12
⇒ No. of electrons = 241)2(2
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30 YEAR’S BOOK CHEMISTRY PAGE: 96
64. (c)
N(7) = 1s22s
22p
2
N2+(s)
= 1s2, 2s
2 1
x2p
Unpaired electrons = 1.
65. (d)
Is uncertainty principle which was given by Hiesenberg and not Bohr’s postulate
66. (a)
Radius of nth
orbit = r1 n2. (for H-atom)
67. (a)
No. of radial nodes in 3p-orbital
= (n – – 1)
[for p orbital = 1]
= 3 – 1 – 1 = 1
68. (b)
Both He and Li+ contain 2 electrons each therefore their spectrum will be similar.
RJ VISION PVT. LTD.
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1. (d)
Elements B Ga Al In Tl
Atomic radii (pm) 85 135 143 167 170
2. (d)
Element (X) electronic configuration 1s22s
22p
3
So, valency of X will be 3.
Valency of Mg is 2.
Formula of compound formed by Mg and X will be Mg3X2.
3. (a)
Z = 114 [Rn]86
7s2 5f
14 6d
10 7p
2 14
th gp. (carbon family)
4. (a)
5. (c)
6. (b)
Be2 = 1s
2 = Li
+
7. Bonus
F = 133 pm
O2 = 140 pm
Na+ = 102 pm
There is no correct option.
8. (a)
As the nuclear charge increases, the force of attraction between the nucleus and the incoming electron increases and
hence the electron gain enthalpy becomes more negative, hence the correct order is:
Ca < Al < C < O < F
9. (a)
As the positive charge increases on metal cation, radius decreases. This is due to the fact that nuclear charge in the
case of a cation is acting on lesser number of electrons and pulls them closer.
10. (a)
IE1 of Na = – Electron gain enthalpy of Na+
= – 5.1 volt
11. (c)
12Mg 15P 17Cl 20Ca
160p 110 99 197 (pm)
Cl < P < Mg < Ca
12. (b)
O < S < F < CI
Electron gain enthalpy – 141 – 200 – 333 – 349 kJ mol–1
13. (c)
Among the isoelectronic species, size increases with the increase in negative charge. Thus S2–
has the highest negative
charge and hence largest in size followed by Cl–, K
+ and Ca.
CH – 3 CLASSIFICATION OF ELEMENTS AND PERIODICITY
IN PROPERTIES
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30 YEAR’S BOOK CHEMISTRY PAGE: 98
14. (d)
The smaller the atomic size, larger is the value of ionisation potential. Further the atoms having half filled or fully fille d
orbitals are comparatively more stable, hence more energy is required to remove the electron from such ato ms.
15. (d)
The stability of + 1 oxidation state increases from aluminium to thallium i.e.,
Al < Ga < In < Tl
16. (a)
Proton affinity decreases in moving across the period from left to right due to increase in charge, within a group the
proton affinities decreases from to p to bottom.
Nitrogen family > Oxygen family > Halogens
17. (b)
18. (a)
For isoelectronic species, size of anion increase as negative charge increases whereas size of cation decreases with
increase in positive charge. Further ionic radii of anions is more than that of cations. Thus the correct order is
Ca++
< K+ < Ar < Cl
– < S
– –
19. (d)
The amount of energy released when an electron is added to an isolated gaseous atom produce a monovalent anion is
called electron gain enthalpy.
Electron affinity value generally increase on moving from left to right in a period however there are exceptions of this
rule in the case of those atoms which have stable configuration. These atoms resist the addition of extra electron,
therefore the low value of electron affinity
3.83.62.01.48ClFSO
On the other hand Cl because of its compariti very bigger size than F, allow the addition of an extra electron more
easily.
20. (d)
Due to odd number of electrons in ClO2.
21. (a)
Ionic radii are inversely proportional to effective nuclear charge.
Ionic radii in the nth
orbit is given as
Z
1ror
Z
anr n
0
2
n
When n = principal quantum number
Z-effective nuclear change.
22. (a)
Among the given options, only Fe shows variable oxidation states so it can form two chlorides, viz. FeCl2 and FeCl3.
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23. (b)
This is because of inter-electronic repulsions between lone pairs.
B.E.: F – F Cl – Cl
(kJ mol–1
) 158.8 242.6
24. (a)
The electronic configuration clearly suggest that it is a d-block element (having configuration (n – 1)d1 – 10
ns 0 – 2
) which
starts from III B and goes till II B. Hence with d3 configuration it would be classified in the group.
25. (b)
Along the period, I.P. generally increases but not regularly. Be and B are exceptions. First I.P. increases in moving from
left to right in a period, but I.P. of B is lower than Be.
26. (a)
Be – 1s22s
2; B – 1s
22s
22p
1; C – 1s
22s
22p
2; N – 1s
22s
22p
3; O – 1s
22s
22p
4. IP increases along the period. But IP of Be > B.
Further IP of O < N because atoms with fully or partly filled orbitals are most stable and hence have high ionisation
energy.
27. (a)
Mg = 1s22s
22p
63s
2
After removing of 2 electron, the magnesium acquired noble gas configuration hence removing of 3rd electron will
require large amount of energy.
28. (d)
First ionisation potential of Be of greater than boron due to following configuration
4Be = 1s2, 2s
2 5B = 1s
2, 2s
22p
1
Order of attraction of electrons towards nucleus 2s > 2p, so more amount of energy is required to remove the electron
with 2s-orbital in comparison to 2p orbital.
29. (a)
In a period on moving from left to right ionic radii decreases.
(a) So order of cationic radii is
Cr2+
> Mn2+
> Fe2+
> Ni2+
and
(b) Sc > Ti > Cr > Mn (correct order of atomic radii)
(c) For unpaired electrons
Mn2+
(Five) > Ni2+
(Two) < Co2+
(Three) < Fe2+
(Four)
(d) For unpaired electrons
Fe2+
(Four) > Co2+
(Three) < Ni2+
(Two) < Cu2+
(One)
30. (b)
Greater is the positive charge on atom, large is effective nuclear charge. Hence , smaller is the size.
31. (b)
Electronic configuration of element with atomic number 118 will be [Rn] 5f14
6d10
7s27p
6. Since its electronic
configuration in the outer most orbit (ns2np
6) resemble with that of inert or noble gases, therefore it will be noble gas
element.
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32. (a)
Atomic number of the given element is 15 and it belongs to 5th group. Therefore atomic number of the element below
the above element = 15 + 18 = 33.
33. (d)
In general non-metals are more electronegative then metals.
34. (d)
Abnormally high difference between 2nd and 3rd ionization energy means that the element has two valence electrons,
i.e., configuration (d).
35. (d)
Atomic volume is the volume occupied by one gram of an element. Within a period from left to right, atomic volume
first decreases and then increases due to increases of nuclear charge and increase in the number of electrons in the
valence shell.
36. (c)
Element with Z = 33
(1s2 2s
2p
63s
2p
6d
104s2p
3) lies in fifth (or 15th) group.
37. (c)
Amongst isoelectronic ions, the size of the cation decreases as the magnitude of the charge increases.
38. (c)
Proton (H+) being very small in size would have very large hydration energy.
39. (c)
N, O and F (p-block elements) are highly electronegative non metals and will have the strongest tendency to form
anions by gaining electrons from metal atoms.
40. (c)
Electron affinity values are high in case of halogen because halogens have seven electrons (ns2np
5) in the valence shell,
they have a strong tendency to acquire the nearest inert gas configuration by gaining an electron from the metallic
atom and form halide ions easily.
41. (a)
Metallic character decreases in a period and increases in a group.
42. (c)
Elements (a), (b) and (d) belong to the same group since each one of them has two electrons in the s sub shell. In
contrast, element (c) has seven electrons in the valence shell and hence does not lie in the same group in which
elements (a), (b) and (c) lie.
43. (a)
Pauling scale of electronegativity was helpful in predicting
(i) Nature of bond between two atoms
(ii) Stability of bond
By calculating the difference in electronegativities polarity of bond can be calculated.
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30 YEAR’S BOOK CHEMISTRY PAGE: 101
1. (c)
The structure of CiF3 is
The number of lone pair of electrons on central Cl is 2.
2. (c)
NO : (1s)2, (
*1s)
2, (2s)
2,(
*2s)
2,( 2pz)
2,(2px)
2 = (2py)
2,(
*2px)
1=(
*2py)
0
BO = 5.22
510
CN- : (1s)
2,(
*1s)
2,(2s)
2,(
*2s)
2,(2px)
2=(2py)
2,(2pz)
2
32
4-10 BO
CN : (1s)2, (
*1s)
2,(2s)
2,(
*2s)
2,(2px)
2=(2py)
2,(2pz)
1
5.22
4-9 BO
CN+ : (1s)
2, (
*1s)
2, (2s)
2,(
*2s)
2, (2px)
2=(2py)
2
22
4-8 BO
Hence, option (c) should be the right answer.
3. (b)
Total no. of electrons in CN. is 14.
Total no. of electrons in CO is also 14.
Hence B.O. of both CN. & CO is 3.
4. (c)
5. (d)
243 ])([ NHCu - dsp
2 hybidisaiton
6. (a)
Neutral structure is most stable than charged.
7. (d)
(i) Structure of SiCl4 is tetrahedral
Structure of
4PCl is tetrahedral
CH – 4 CHEMICAL BONDING AND MOLECULAR STRUCTURE
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30 YEAR’S BOOK CHEMISTRY PAGE: 102
(ii) Diamond & SiC both are isostructural because both have tetrahedral arrangement and central atom is sp3
hybridised.
(iii) Hybridiation of NH3 [σ = 3, lp = 1]
sp3 geometry : tetrahedral
(iv) Structure of XeF4 is square planar.
Structure of XeO4 is tetrahedral
So, XeF4 and XeO4 are not isostructural.
8. (a)
Given species: O2, 1
2O , 1
2O , -2
2O
Total number of electrons
O2 → 16e–
1
2O → 17e–
1
2O → 15e–
2
2O → 18e–
1
2O O2 1
2O 2
2O
Bond order 2.5 2 1.5 1
* Stability order [ 1
2O > O2 > 1
2O > 2
2O ]
9. (d)
Species 2
2O
2O
2O
Bond order 3 2.5 1.5
10. (c)
In
2NO , the bond angle is 180.
11. (b)
Species
2O 2O
2O
Bond order 2.5 2 1.5
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12. (d)
Species 2
3SO
3ClO
Shape Pyramidal Pyramidal
13. (b)
Species 4XeO
Shape Tetrahedral
14. (b)
Sp2 (triangular planer)
15. (c)
CO2 CH4 NH3 NF3
= 0 = 0 = 1.47D = 0.23D
16. (d)
17. (d)
(a) N2 →
2N
B.O 3 2.5
Bond energy decreases
Magnetic behaviour changes from diamagnetic to paramagnetic.
(b) O2 →
2O
B.O. 2 2.5
Bond energy increases
Magnetic behaviour does not change
(c) C2 →
2C
B.O. 2 2.5
Bond energy decreases
Magnetic behaviour changes from diamagnetic to paramagnetic.
(d) NO → NO+
B.O. 2 2.5
Bond energy increases
Magnetic behaviour changes from paramagnetic to diamagnetic.
18. (b)
No. of electrons in CO = 6 + 8 = 14
No. of electrons in NO+ = 7 + 8 – 1 = 14
∴ Co and NO+ are isoelectronic species.
Isoelectronic species have identical bond order.
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30 YEAR’S BOOK CHEMISTRY PAGE: 104
19. (c)
Applying VSEPR theory, both NF3 and H2O are sp3 hybridized.
20. (a)
21. (a)
Molecular orbital configuration of
2O is
2O (17) = σ1s2, σ*1s
2, σ2s
2, σ*2s
2,
1
y
2
x
2
y
2
x
2
z 2p*2p*,2p2p,σ2p ππππ
22. (a)
23. (b)
(BH3)2 or (B2H6)
It contains two 3 centre – 2 electron bonds and present above and below the plane of molecules compounds which do
not have sufficient number of electrons to form normal covalent bonds are called electron deficient molecules.
24. (d)
Bond order = 2
NN ab
2He = σ(1s)2 σ*(1s) B.O. = 0.5
2O = KK σ(2s)2 σ*(2s)
2 σ(2pz)
2 π(2px)
2 π(2py)
2 π*(2px)
2 π*(2py)
1 B.O. = 1.5
NO = KK σ(2s)2 σ*(2s)
2 π(2px)
2 π(2py)
2 σ*(2pz)
2 π*(2px)
1 B.O. = 2.5
2
2C = KK σ(2s)2 σ*(2s)
2 π(2px)
2 π(2py)
2 σ(2pz)
2 B.O. = 3.0
25. (a)
O2 = KK (σ2s)2(σ*2s)
2(σ2pz)
2(π2px)
2(π2py)
2(π*2px)
1(π*2py)
1
2O = KK (σ2s)2(σ*2s)
2(σ2pz)
2(π2px)
2(π2py)
2(π*2px)
2(π*2py)
1
26. (a)
Both 2
2O and B2 has bond order equal to 1
] π πσσσσ 1
z
1
y
2222
2 2p2p2s* 2s 1s* 1s[(10)B
Bond order = 12
2
2
46
2
NN ab
B2 is known in the gas phase
2222
2 2s*σ σ2s 1s*σ σ1sO 2y
2
x
2
y
2
x
2
z 2p*2p*2p2pσ2p ππππ
Bond order = 16)(82
1
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30 YEAR’S BOOK CHEMISTRY PAGE: 105
27. (d)
PCl3
28. (b)
(O2) = σ1s2 σ*1s
2 σ2s
2 σ*2s
2 2
x
2
z 2pσ2p π1
y
1
x
2
y 2p*2p*2p π ππ
Bond order = 22
4
2
610
2
aNNb
ionO2
= σ1s2 σ*1s
2 σ2s
2 σ*2s
2
1
x
2
y
2
x
2
z 2p* 2p2p σ2p πππ
Bond order = 2
12
2
5
2
510
2
NN ab
2O = σ1s2 σ*1s
2 σ2s
2 σ*2s
2
1
y
2
x
2
y
2
x
2
z 2p*2p*2p2pσ2p π ππ π
Bond order = 2
11
2
3
2
710
2
NN ab
2
2O = σ1s2 σ*1s
2 σ2s
2 σ*2s
2
2
y
2
x
2
y
2
x
2
z 2p*2p* 2p2pσ2p πππ π
Bond order = 12
2
2
810
2
NN ab
29. (d)
4BF hybridisation sp3, tetrahedral structure.
4NH hybridisation sp3, tetrahedral structure.
30. (a)
Hybridisation 2
1 [No. of valence electrons of central atom + no. of monovalent atoms attached to it–
2NO 2
1H [5 + 0 + 1 – 0] = 3 = sp
2
3NO 2
1H [5 + 0 + 1 – 0] = 3 = sp
2
2NH
2
1H [5 + 2 + 1 + 0] = 4 = sp
3
4
NH 2
1H [5 + 4 + 0 – 1] = 4 = sp
3
SCN– = sp
i.e.,
2NO and
3NO have same hybridisation.
31. (b)
32. (c)
(a) :NH4
sp
3 hybridisation
(b) CH4 : sp3 hybridisation
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30 YEAR’S BOOK CHEMISTRY PAGE: 106
(c) SF4 : sp3d hybridisation
(d) : BF4
sp3 hybridisation
∴ Correct choice : (c)
33. (b)
In ,23NO nitrogen is in sp2
Hybridisation, thus planar in shape. In H3O+, oxygen is in sp
3 hybridisation, thus tetrahedral in shape.
∴ Correct choice : (b)
34. (c)
62
255 :SbCl
2
5
means sp3d
2
43 SF ,I and PCl5; all have sp
3d hybridisation.
35. (b)
32
33 : BF3
means sp
2
32
15 : NO 2
means sp
2
36. (c)
For neutral molecules:
No. of electron pairs = No. of atoms bonded to it 2
1 [Group number of central atom – valency of the central atom].
For ions:
No. of electron pairs = No. of atoms bonded to it 2
1 [Group no. of central atom- valency of the central atom ± no. of
electron]
On calculating no. of electron pairs in given molecules. We find that in the given molecule hybridisation is:
BF3 → sp2
2NO → sp2
2NH → sp3
H2O → sp3
37. (d)
Due to intermolecular hydrogen bonding in methanol, it exist at associated molecule.
38. (d)
Calculating the bond order of various species
:O2
kkσ2s
2 σ*2s
2
2
z2pσ
2
y
2
x
2
y
2
x 2p*2p*2p2p π π ππ
2
bondingnon in electrons ofNumber bonding in electrons ofNumber B.O.
1.5or
2
58
NO : 2
x
22 2p2s*σ σ2s kk π 1
x
2
z
2
y 2p*σ2p 2p ππ
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2.5or 2
38
2
NNB.O.
ab
2
x
222
2 2p 2s* 2s kk:C πσσ 2
z
2
y 2p 2p σπ
3or 2
38
2
NNB.O.
ab
122
2 1s*σ1sHe σ
0.5or 2
12
2
NNB.O.
ab
From these values we find the correct order of increasing bond order is
2
22
2
2 CNOOHe
39. (b)
From the structure of three species we can determine the number of lone pair electrons on central atom (i.e., N atom)
and thus the bond angle.
We know that higher the number of lone pair of electron on central atom, greater is the lp – lp repulsion between
Nitrogen and oxygen atoms. Thus, smaller is bond angle.
The correct order of bond angle is:
222 ONNONO
i.e. option (b) is correct.
40. (d)
The shape of ozone molecule is
In it we find 2σ and 1π bond, i.e., option (d) is correct.
41. (c)
All these structures exhibits resonance and can be represented by the following resonating structures.
More is the single bond character. More will be the bond length. Hence, the correct order is:
CO < CO2 <
3CO
RJ VISION PVT. LTD.
30 YEAR’S BOOK CHEMISTRY PAGE: 108
42. (c)
Hybridisation of Br is
3BrO
Total valence electrons = 7 + 3 × 6 = 25
Charge = –1 hence
Total = 25 + 1 = 26
52(R)3(Q)8
26
Hybridisation = dsp3
Hybridization of Xe in XeO3
Total valence electrons
= 8 + 3 × 6 = 26 = 8
26 = 3(Q) + 2(R) = 5
Hybridization = dsp3
In both cases, the structure is trigonal pyramidal.
43. (d)
In SnCl4 there is sp3 hybridisation so the structure is tetrahedral. In
4
3
4
2
4 NH,PO,SO the structure is tetrahedral because
in all hybridisation is sp3. But in SCl4, sp
3d hybridization present so shape is different which is see saw.
44. (b)
2NO will have linear shape as it will have sp hybridisation.
45. (a)
In NH3 the atomic dipole and bond dipole are in the same direction whereas in NF3 these are in opposite direction so
in the former case they are added up whereas in the latter case net result is reduction of dipole moment. It has been
shown in the following figure:
46. (d)
In BF3, AlF3 & NF3 all fluoride atoms are symmetrically oriented with respect to central metal atom but in ClF3 three
fluorine atoms are arranged as follows:
Here two bonds are in equatorial plane & one bond is in axial plane.
47. (d)
Electronic configuration of the molecule according to molecular orbital theory, is
σ 1s2 σ*1s
2 σ2s
2 σ*2s
2 σ2pz
2 (π2px
2 = σ2py
2)
(π*2px2 = π2py
2). σ*2pz
2 σ3s
2 σ*3s
2 σ3pz
2
(π3px2 = π3py
2) (π*3px
1 = 3py
1)
Last two electrons are unpaired. So no. of unpaired electron is 2.
RJ VISION PVT. LTD.
30 YEAR’S BOOK CHEMISTRY PAGE: 109
48. (b)
Statement (a), (c), (d) are correct. Statement (b) is incorrect statement.
AB5 may have two structures as follows:
49. (b)
SF4 has permanent dipole moment.
SF4 has sp3d hybridization and see saw shape (irregular geometry)
Whereas XeF4 shows square planar geometry SiF4 has tetrahedral shape and BF3 has Trigonal planar shape. All these
are symmetric molecule. Hence μ ≠ 0.
50. (a)
BF3 is sp2 hybridised. So, it is trigonal planner. NH3, PCl3 has sp
3 hybridisation hence has trigonal bipyramidal shape, IF3,
has sp3d hybridization and has linear shape.
51. (c)
As difference of electronegativity increases % ionic character increases and covalent character decreases i.e., negativity
difference decreases covalent character increase.
Further greater the charge on the cation more will be its covalent character. Be has maximum (+2) charge.
52. (b)
The compound, of which central atom is octetless known as electron deficient compound. Hence B2H6 is electron
deficient compound.
53. (c)
In BrF3, both bond pairs as well as lone pairs of electrons are present. Due to the presence of lone pairs of electrons
(1p) in the valence shell, the bond angle is contracted and the molecule takes the T-shape. This is due to greater
repulsion between two lone pairs or between a lone pair and a bond pair than between the two bond pa irs.
RJ VISION PVT. LTD.
30 YEAR’S BOOK CHEMISTRY PAGE: 110
54. (a)
Only those d orbitals whose lobes are directed along X, Y and Z directions hybridise with s and p orbital. In other three
d orbitals namely dxy, dyz and dxz, the lobes are at an angle of 45° from both axis, hence the extent of their overlap with
s and p orbitals is much lesser than 22 yxd
and 2z
d orbitals.
55. (d)
In a linear symmetrical molecule like BeF2, the bond angle between three atoms is 180°, hence the polarity due to one
bond is cancelled by the equal polarity due to other bond, while it is not so in angular molecules, like H2O.
56. (a)
Thus here bond angles between
X4 – M – X2 = 180°
X1 – M – X3 = 180°
X5 – M – X6 = 180°
57. (a)
Shape (structure) of a species can be ascertained on the basis of hybridisation of its central atom, which in turn can be
known by determining the number of hybrid bonds
(H). Thus A)CX(V2
1H
For SiF4 : 4;0)04(42
1H sp
3 (Tetrahedral structure)
For SF4: 5;0)04(62
1H sp
3d (See saw structure)
Hence the two compounds have different structures.
58. (c)
As sigma bond is stronger than the π(pi) bond, so it must be having higher bond energy than π(pi) bond.
59. (d)
To form ,NO3
nitrogen uses one p-electron for π-bond formation two p-electrons for σ-bond formation. 2s electrons
are used for coordinate bond formation. Thus there is no lone pair on nitrogen but four bond pairs.
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30 YEAR’S BOOK CHEMISTRY PAGE: 111
60. (b)
In 2
3SO
S is sp3 hybridised, so
16S = 1s2, 2s
2 2p
6,
ionhybridisat sp
1
z
1
y
1
x
2
3
3p3p3p3s
unhybride
1
xy3d
In ‘S’ unhybride d- orbit is present, which is involved in π bond formation
8O = 1s2, 2s
2
1
z
1
y
2
x 2p2p2p
In oxygen two unpaired p- orbital is present in these one is involved in σ bond formation while other is used in π bond
formation.
Thus in ,SO2
3
pπ and dπ orbitals are involved for pπ – dπ bonding.
61. (a)
In X — H - - - - Y, X and Y both are electronegative elements (i.e., attracts the electron pair) then electron density on X
will increase and on H will decrease.
62. (a)
For π-overlap the lobes of the atomic orbitals are perpendicular to the line joining the nuclei.
63. (c)
In XeF2 and
2IF . Both XeF2 and
2IF are sp3d hybridized and have planar shape.
64. (c)
We know that bond angles of NH3 = 107°,
4NH = 109.5°, PCl3 = 100°. Therefore bond angle of
4NH is maximum.
65. (c)
In P–O bond, π bond is formed by the sidewise overlapping of d-orbital of P and p-orbital of oxygen. Hence it is
formed by pπ and dπ overlapping.
66. (a)
We know that the electrostatic force that binds the oppositely charged ions which are formed by transfer of electron
from one atom to another is called ionic bond. We also know that cation and anion are oppositely charged particles
therefore they form ionic bond in crystal.
67. (c)
Dissociation energy of any molecules depends upon bond order. Bond order in N2 molecule is 3 while bond order in
2N is 2.5.
Further we know that more the Bond order, more is the stability and more is the BDE.
68. (a)
Boron in BCl3 has 6 electrons in outer most shell. Hence BCl3 is a electron pair deficient compound.
69. (d)
HF form linear polymer structure due to hydrogen bonding.
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30 YEAR’S BOOK CHEMISTRY PAGE: 112
70. (d)
PF3 has pyramidal shape
Phosphorus exist in sp3 hybridisation state hence it exist in tetrahedral shape. But due to presence of lone pair its
shape is pyramidal.
F–H----F—H----F—H----F
71. (b)
NO+ = σ1s
2 σ*1s
2 σ2s
2 σ*2s
2 σ2px
2
π2py2 = π2pz
2
Bond order of NO+ = )N(N
2
1ab
362
14)(10
2
1
Similarly, Bond order of NO 5)(102
1
2.5(5)2
1
Bond order of NO– = 2(4)
2
16)(10
2
1
Bond order of 7)(102
1O2
1.5(3)2
1
By above calculation, we get decreasing bond order
NO+ > NO > NO
– >
2O
72. (c)
As dipole moment = electric charge × bond length
D. M. of AB molecules
= 4.8 × 10–10
× 2.82 × 10–8
= 13.53 D
D. M. of CD molecules
= 4.8 × 10–10
× 2.67 × 10–10
= 12.81 D
Now % ionic character
compound ionic pure of moment Dipole
bond the of moment dipole Actual then % ionic character in
76.94%10013.53
10.41AB
% ionic character in
80.23%10012.81
10.27CD
73. (b)
XeF4 hybridisation is = 2
1(V + X – C + A) hence V = 8 (no. of valence v
–)
RJ VISION PVT. LTD.
30 YEAR’S BOOK CHEMISTRY PAGE: 113
X = 4 (no of monovalent atom) 60)04(82
1 sp
3d
2
C = 0 charge on cation
A = 0 (charge on anion). The shape is square planar shape.
74. (c)
Bond order between P – O 1.254
5
structures resonating of no. total
direction possible all in bonds of no.
or formal charge on oxygen = 0.754
3
(Formal charge = No. of electrons in valence shell
electrons free of No,electrons bonding of No.
2
1
75. (a)
Paramagnetic character is based upon presence of unpaired electron.
17Cl– = 1s
2, 2s
22p
6, 3s
2 2
z
2
y
2
x 3p3p3p
4Be = 1s2, 2s
1
10Ne2+
= 1s2, 2s
2
1
z
1
y
2
x 2p2p2p
33As+ = 1s
2, 2s
2 2p
6, 3s
2 3p
6 3d
10
4s2
0
z
1
y
1
x 4p4p4p
Hence, only Cl– do not have unpaired electrons.
76. (d)
Total no. of electrons in 2
2O = 16 + 2 = 18. Distribution of electrons in molecular orbital
σ1s2, σ*1s
2, σ2s
2, σ*2s
2, ,2p2
xσ2
y2pσ
2
y
2
x
2
z 2p*2p* ,2p πππ
Anti bonding electron = 8 (4 pairs)
77. (c)
The electronic configuration of As is
As = 1s2, 2s
2, 2p
6,
tiondhybridisa sp
11
z
1
y
1
x
1
3
3d3p3p3p3s
Thus, the hyrbidisation involved in the AsF5 molecule is trigonal bipyramidal. So, the hybrid orbitals used are s, p x, py,
pz, 2
zd
78. (b)
We know that in O2 bond, the order is 2 and in
2O bond, the order is 1.5. Therefore the wrong statement is (b).
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30 YEAR’S BOOK CHEMISTRY PAGE: 114
79. (d)
In alkynes the hybridisation is sp i.e. each carbon atom undergoes sp hybridisation to form two sp -hybrid orbitals. The
two 2p-orbitals remain unhybridised. Hybrid orbitals form one sigma and two unhybrisdised orbitals form π-bonds.
Hence two π bond and one sigma bond between C – C lead to cylindrical shape.
80. (a)
We know that due to polar nature, water molecules are held together by intermolecular hydrogen bonds. The structure
of ice is open with large number of vacant spaces, therefore the density of ice is less than water.
81. (a)
The bond represented by dots form the 3 -centred electron pair bond. The idea of three centred electron pair bond B–
H–B bridges is necessitated because diborane does not have sufficient electrons to form normal covalen t bonds. It has
only 12 electrons instead of 14 required to give simple ethane like structure for diborane.
82. (c)
The N – O bond length decreases in the order.
83. (d)
In this configuration, there are four completely filled bonding molecular orbitals and one completely filled antibonding
molecular orbital. So that Nb = 8 and Na = 2.
32)(82
1)N(N
2
1order Bond ab
84. (d)
The bond length of O – O in O2 is 1.21 Å, in H2O2 it is 1.48 Å and in O3 it is 1.28 Å.
∴ Correct order of bond length is H2O2 > O3 > O2.
85. (d)
As there is no lone pair on boron in BCl3 therefore no repulsion takes place. But there is a lone pair on nitrogen in NCl3.
Therefore repulsion takes place. Thus BCl3 is planar molecule but NCl3 is a pyramidal molecule.
86. (b)
Paramagnetism is caused by the presence of atoms, ions or molecules with unpaired electrons. In NO the presence of
unpaired electron is clear. Therefore it is paramagnetic.
87. (c)
The C – C bond distance decreases as the multiplicity of the bond increases. Thus, bond distance decreases in the
order: butane (1.54 Å) > benzene (1.39 Å) > ethene (1.36 Å) > ethyne (1.20 Å) . Thus in butane, C – C bond distance is
the largest.
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88. (c)
The b.p. of p-nitrophenol is higher than that of o-nitrophenol because in p-nitrophenol there is intermolecular H-
bonding but in o-nitrophenol it is intermolecular H-bonding.
89. (c)
Chemical bonds.
90. (b)
The angle between the bonds formed by px and py orbitals is the minimum i.e., 90°
91. (b)
The stability of the ionic bond depends upon the lattice energy which is expected to be more between Mg and F due
to +2 charge on Mg atom.
92. (b)
BH3 has sp2 hybridization and hence does not have tetrahedral structure while all others have tetrahedral structures.
93. (d)
H-bond is the weakest
94. (b)
The removal of an electron from a diatomic molecule may increases the bond order as in the conservation O 2(2) →
2O (2.5) or decrease the bond order as the conversion, N2(3.0) →
2N (2.5). As a result, the bond energy may increase or
decrease, thus, statement (b) is incorrect.
95. (a)
According to Fajan rules, ionic character increases with increase in size of the cation and decrease in size of the anion.
Thus, CsF has higher ionic character than NaCl and hence bond in CsF is stronger than in NaCl.
96. (c)
The strength of the interactions follows the order Vander Waal’s < hydrogen – bonding < dipole-dipole < covalent.
97. (b)
For compounds containing ions of same charge, lattice energy increases as the size the ions decrease. Thus, NaF has
highest lattice energy. The size of cation is in the order Na+ < K
+ < Rb
+ < Cs
+
98. (b)
Sigma bond is stronger than π-bond. The electrons in the π bond are loosely held. The bond is easily broken and is
more reactive than σ-bond. Energy released during sigma bond formation is always more than π bond because of
greater extent of overlapping.
99. (c)
H – F shows strongest H-bonds.
100. (b)
CO2 has sp-hybridization and is linear. SO2 and 2
3CO are planar (sp2) while
2
4SO is tetrahedral (sp3).
101. (c)
The bond length decreases in the order sp3 > sp
2 > sp.
Because of the triple bond, the carbon-carbon bond distance in ethyne is shortest.
102. (c)
Polarity of the bond depends upon the electronegativity difference of the two atoms forming the bond. Greater the
electronegativity difference more is the polarity of the bond.
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N – Cl O – F N – F N – N
3.0–3.0 3.5–4.0 3.0–4.0 3.0–3.0
As the electronegativity difference between N and F is maximum hence this bond is most polar.
103. (b)
CCl4 has sp3 hybridisation, tetrahedral geometry and all bond angles of 109° 28’.
104. (d)
The star marked carbon has a valency of 5 and hence this formula is not cor rect.
105. (a)
Linear combination of two hybridized orbitals leads to the formation of sigma bond.
106. (d)
With the increase of electronegativity and decrease in size of the atom to which hydrogen is covalently linked, the
strength of hydrogen bond increases. As F is most electronegative thus HF shows maximum strength of hydrogen
bond.
107. (a)
A σ-bond is stronger than a π-bond hence option (a) is not correct. Sigma (σ) bonds are formed by head on overlap of
unhybridized s-s, p-p or s-p orbitals and hybridised orbitals (sp, sp2, sp
3, sp
3d and sp
3d
2) hence σ bonds are strong
bonds. Where as Pi (π)-bonds are formed by side ways overlap of unhybridised p- and d orbitals-hence π bonds are
weak bonds.
108. (c)
As we move in period form Li → Be → B → C, the electronegativity (EN) increases and hence the EN difference between
the element and Cl decreases and accordingly the covalent character increases Thus option (c) i.e.,
LiCl < BeCl2 < BCl3 < CCl4 is correct.
109. (b)
BeF2 is linear and hence has zero dipole moment while H2O, being a bent molecule, has a finite or non-zero dipole
moment.
110. (b)
BF3 involves sp2-hybridization
111. (d)
In metallic bonds each ion is surrounded by equal no. of oppositely charged ions hence have electrostatic attraction on
all sides and hence do not have directional characteristics.
112. (a)
For linear arrangement of atoms the hybridisation should be sp (linear shape, 180° angle). Only H 2S has sp3-
hybridization and hence has angular shape while C2H2, BeH2 and CO2 all involve sp - hybridisation and hence have
linear arrangement of atoms.
113. (b)
Equilateral or triangular planar shape involves sp2 hybridization.
114. (a)
The overlap between s- and p-orbitals occurs along internuclear axis and hence and angle is 180°.
RJ VISION PVT. LTD.
30 YEAR’S BOOK CHEMISTRY PAGE: 117
1. (a)
Van der waal constant ‘a’, signifies intermolecular forces of attraction.
Higher is the value of ’a’, easier will be the liquefaction of gas.
2. (d)
In real gas equation, nRTnbVV
anP
)(
2
2
Van der Waal’s constant, ‘a’ signifies intermolecular forces of attraction.
3. (d)
)s(SrCO3 )g(CO)s(SrO 2
2COp PK
Maximum pressure of CO2 = 1.6 atm
P1V1 = P2V2
0.4 20 = 1.6 V2
V2 = 5L
4. (a)
Real gases show ideal gas behaviour at high temperatures and low pressures.
5. (c)
H2
32
16
O2
32
1
CH4
32
2 mole ratio
6. (d)
Density = g/ml 3.415000.0821
285
RT
PM
7. (b)
This type of attractive force operates between the polar molecules having permanent dipole and the molecules lacking
permanent dipole.
HCl is polar (μ ≠ 0) and He is non polar (μ = 0), thus gives dipole-induced dipole interaction.
8. (c)
Higher the critical temperature more easily will be the gas liquify. Now since most easily liquefiable gas show larger
deviation, NH3 will show maximum deviation from ideal behavior.
9. (d)
‚Molar mass increase, ‘a’ increases
size of molecule increase, ‘b’ increase
b(L/mol) a(bar. L2/mol
2)
H2 → 0.02661 CH4 → 2.25
He → 0.0237 O2 → 1.36
O2 → 0.03183 H2 → 0.244
CO2 → 0.04267
CH – 5 STATES OF MATTER
RJ VISION PVT. LTD.
30 YEAR’S BOOK CHEMISTRY PAGE: 118
10. (b)
M
1r
1
2
2
1
M
M
r
r
4
M
r
3r 2
1
1
4
M9 2
M2 = 36 g/mole
11. (c)
⇒ AM
36
150
200 ⇒
AM
36
3
4
⇒
AM
36
9
16
⇒ 20.254
81MA
So, most approximate answer is 32.
12. (a)
Given
P1 = 1.5 bar T1 = 273 + 15 = 255 K V1 = V
P2 = 1.0 bar T1 = 273 + 25 = 298 K V2 = ?
2
22
1
11
T
VP
T
VP
298
V1
288
V1.5 2
V2 = 1.55 V i.e., volume of bubble will be almost 1.6 time to initial volume of bubble.
13. (c)
For free expansion of an ideal gas under adiabatic condition
q = 0, ∆T = 0, w = 0
14. (a)
Given nCO = 2Nn
PCO + 2NP = 1 atm
Partial pressure of a gas = mole fraction of gas × total pressure
1nn
nP
2
2
2
NCO
N
N
12n
n
2
2
N
N
atm 0.52
1
15. (b)
A
B
B
A
M
M
r
r
49
M
2
1
49
M
10
V20
V
BB
12.25494
1MB
A
B
M
M
B
B
t
V
A
A
t
V
RJ VISION PVT. LTD.
30 YEAR’S BOOK CHEMISTRY PAGE: 119
16. (d)
Average velocity = M
8RT
π
i.e., v ∝ T
1.41T
2T
V
V
1
2
17. (a)
Ideal gas during spontaneous expansion into vacuum does not do any external work.
18. (d)
Pa 416480.03
4028.314
16.02
6
V
nRTP
19. (c)
At any constant temperature the K.E. of gaseous molecules remains same. (K.E. ∝ T).
Thus, options (c) is correct answer.
20. (c)
Due to intermolecular H-bonding the surface tension of H2O is more than other liquid. One H2O molecule is joined
with 4 another H2O molecules through H–bond.
Hydrogen bonding is in order H2O > C2H5OH > CH3OH.
21. (a)
At higher temperature and low pressure real gas acts as an ideal gas .
22. (c)
Average molar kinetic energy = kT2
3
As temperature is same hence average kinetic energy of CO and N2 is same.
23. (d)
(Given)1
8
O of Moles
H of Moles
2
2
1
8
O of weight
H of weight .
H of M.W.
O of M.W.
2
2
2
2
2
1
132
28
O of weight
H of weight
2
2
24. (a)
Applying Boyle’s law P1V1 = P2V2 for both gases
3
200P3P400
1000
500
3
400P'3P'
1000
666.6600
⇒ PT = P + P'
= tor 2003
600
3
400
3
200
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30 YEAR’S BOOK CHEMISTRY PAGE: 120
25. (a)
Rate of diffusion depend upon molecular weight
2
1
2
1
M
M
r
r
⇒ r1 \ r2 if M1 = M2
Hence, compounds are N2O and CO2 as both have same molar mass i.e. 22
26. (b)
At low pressure and high temperature real gas nearly behave like ideal gas. Hence deviation is minimum from ideal
behaviour.
27. (d)
By Ideal gas equation
P1 V = n1 RT
n1 ∝ P1 and n2 ∝ P2
2
1
2
1
P
P
n
n ⇒ 0.30
570
170
n
n
2
1
28. (a)
Given initial volume (V1) = 600 C, C: Initial pressure (P1) = 750 mm and final volume (V2) = 500 c.c. according to Boyle’s
law.
P1V1 = P2V2
Or 750 × 600 = P2 × 500
Or 900mm500
600750P2
Therefore increases in pressure = (900 – 750) = 150 mm.
29. (b)
Given initial volume (V1) = 500 ml; Initial temperature (T1) = 27°C = 300 K and final temperature (T2) = –5°C = 268 K.
From Charle’s law:
2
2
1
1
T
V
T
V or
268
V
300
500 2
Where V2 = New volume of gas
ml 446.66268300
500V2
30. (d)
A substance exists as a liquid above its melting point and below its Boiling point.
31. (a)
By definition of Nernst distribution law.
When a solute is shaken with two immiscible liquids, having solubility in both, the solute distributes itself between the
two liquids in such a way that the ratio of its concentrations in two liquids is constant at a given temperature, provided
the molecular state of the solute remains the same in both the liquids.
32. (d)
u α T or u1/u2 = 21 TT
273927
27327
=
1200
300 =
2
1
u2 = 2u1
33. (b)
At low temperature and high pressure.
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30 YEAR’S BOOK CHEMISTRY PAGE: 121
34. (d)
Most probable velocity (α) = M
2RT
Mean velocity M
8RT)v(
π
Root mean square velocity (u) = M
3RT
∴ α : v : u = M
2RT :
M
8RT
π :
M
3RT
= 3:8
:2π
35. (a)
PV = 3
1 mnu
2 =
3
1Mu
2
= 2
1
3
2. Mu
2 =
3
2E per unit vol.
36. (c)
PV = nRT or P = V
n RT = CRT.
Hence 1 = 1 × 0.082 × T is T = 0.082
1 = 12 K
37. (a)
Average kinetic energy depends only on temperature
KT
2
3K.E.
38. (c)
In the equation PV = nRT, n moles of gas have volume V.
39. (d)
In the ideal gas, the intermolecular forces of attraction are negligible and hence it cannot be liquefied.
40. (c)
R = 0.082 litre atm K–1
mole–1
.
41. (c)
Molecules in an ideal gas move with different speeds.
42. (b)
Velocity and hence the K.E is maximum in the gaseous state.
43. (b)
PV = 3
1mnu
2 =
3
1Mu
2 or u = 3PV/M (∵ n = NA)
At STP u ∝ M
1
i.e., higher will be the molar mass lower will be the value of urms.
Molecular masses of H2, N2, O2 and HBr are 2, 28, 32 and 81. Hence the correct order is HBr < O2 < N2 < H2.
44. (d)
According to Boyle’s law at constant temperature, V ∝ P
1 or PV = constant
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30 YEAR’S BOOK CHEMISTRY PAGE: 122
45. (c)
Here volume is constant. Again the mass of H2 is fixed so the number of moles of the gas do not change. As
temperature increases the pressure also increases. The rate of collis ion among the gas molecules and their energy also
increase.
46. (d)
At low pressure and high temperature: At low pressure volume correction for 1 mole of a gas in negligible, i.e., b = 0
thus the gas equation becomes
2V
aP V =RT
or RTV
a1
RT
PVZ
m
m
At higher pressure, the pressure correct for 1 mole of gas in negligible i.e., 0V
a2
Or (P + 0) (V – b) = RT
Or P(Vm – b) = RT
Or PVm = RT + Pb
Or RT
Pb1
RT
PVZ m
47. (c)
Vander Waal’s equation for 1 mole:
2V
aP (V – b) = RT
Here,
2V
aP represent the intermolecular forces. (V – b) is the corrected volume.
48. (a)
Absolute zero is the temperature at which kinetic energy of gas molecules becomes zero i.e., all molecular motio n
ceases.
49. (c)
According to kinetic gas equation 3
1PV mNu
2rms, u = root mean square velocity
⇒ u2 =
mN
3PV or u ∝
m
1 i.e. u ∝ 2
1
m
50. (b)
T
PVconstant or
2
22
1
11
T
VP
T
VP
2
1
22
11
T
T
VP
VP
51. (a)
Charle’s Law – The volume of the given mass of a gas increases or decreases by 273
1 of its volume at 0°C for each
degree rise or fall of temperature at constant pressure.
Vt = V0
273
t1 at const. pressure
52. (d)
PV = nRT = M
mRT
Or PM = V
mRT = dRT ⇒ d =
RT
PM
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30 YEAR’S BOOK CHEMISTRY PAGE: 123
1. (b)
The reaction for ∆fHo(XY)
)()(2
1)(X
2
122
gXYgYg
Bond energies of X2, Y2 and XY are X, XX
,2
respectively
∆H= 20042
X
XX
On solving, we get
20042
XX
X = 800 kJ/mole
2. (b)
Work done in irreversible process
W = –Pext V
= – 2.5 [4.5 – 2.5] = – 5 L atm
= – 5 × 101.3J = – 505J
Since system is well insulated q = 0
By FLOT E = q + W
E = W = – 505 J.
3. (a)
G = H – TS
For equilibrium G = 0
H = TS
KS
HT
eq425
6.83
10005.35.
Since, the reaction is endothermic it will be spontaneous at T > 425 K.
4. (c)
Entropy of a molecule is directly proportional to number of bond.
5. (b)
Non-expansion work done by the system.
6. (a)
Formation of CO2 from carbon and dioxygen gas can be represented as
C(s) + O2(g) → CO2(g); ∆fH = – 393.5 kJ mol–1
(1 mole = 44 g)
Heat released on formation of 44 g CO2
= – 393.5 kJ mol–1
g 35.244g
mol kJ 393.51
= – 315 kJ
7. (c)
CH – 6 THERMODYNAMICS
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30 YEAR’S BOOK CHEMISTRY PAGE: 124
8. (b)
H = U + ng RT
3.32.11.21000
300221.2
G = H Ts
calK 7.263.31000
203003.3
9. (b)
Bonus According to AIPMT Answer Key 2014.
G = 2.303 RT log Ksp
63.3 103 = 2.303 8.31 298 log Ksp
11.09 = log Ksp
8 1012
= Ksp
10. (c)
Applying Hess’s law, equation (i) can be obtained by adding equations (ii) and (iii).
∴ x = y + z
11. (c)
O2HCO2OCH 222x
2x
4
O4H3CO5OHC 22x)5(52
x)(583
2x + 5(5 – x) = 16
⇒ x= 3L
∴ Heat released
317222022.4
2890
22.4
3
12. (c)
∆G = –nFE°
For the reaction 4323
2n
960 × 103 = –4 × 96500 × E°
E° = –2.5 volt
13. (d)
1.673
5
R2
3
R2
5
C
C
V
P
14. (a)
H2O )( ⇌ H2O(g) + Q
∆E = 37558 J/mol
∆E = 37.56 kJ mol–1
15. (c)
273
101.435
T
HS
3
ΔΔ
= 5.260 cal/mol – K
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16. (a)
Since, in the first reaction gaseous products are forming from solid carbon hence en tropy will increase i.e. ∆s = + ve.
C(gr.) + 2
1O2(g) → CO(g); ∆S° = + ve
Since, ∆G° = ∆H° – T∆S hence the value of ∆G decrease on increasing temperature.
17. (b)
Given ∆H
A2
1 ⟶ B + 150 …(1)
3B ⟶ 2C + D –125 …(2)
E + A ⟶ 2D +350 …(3)
To calculate ∆H operate
2 × eq. (1) + eq. (2) – eq. (3)
∆H = 300 – 125 – 350 = – 175
18. (c)
4H(g) → 2H2(g) ; ∆H = –869.6 kJ
Or 2H2(g) → 4H(g) ; ∆H = 869.6 kJ
H2(g) → 2H(g) ; ∆H = 2
869.6 = 434.8 kJ
19. (d)
Given ∆H = 30 kJ mol–1
T = 273 + 27 = 300 K
1
4
T
T mol J300
103
T
ΔHΔS
= 100 J mol–1
K–1
20. (a)
Fe2O3 + CO(g) → 2FeO(s) + CO2(g)
∆H = –26.8 + 33.0 = +6.2 kJ
21. (d)
(1) Kp > 0 (iv) Spontaneous
(2) ∆G < R T In Q (i) Non spontaneous
(3) Kp = 0 (ii) Equilibrium
(4) ΔS
ΔHT (iii) Spontaneous and endothermic
22. (d)
H2O )( H2O(g)
∆H = 40630 J mol–1
∆S = 108.8 JK–1
mol–1
∆G = ∆H – T ∆ S When ∆G = 0, ∆G – T ∆ S = 0
K 373.4mol J108.8
mol J40630
S
HT
1
1
Δ
Δ
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23. (a)
∆S for the reaction 22 Y
2
3X
2
1 ⇌ XY3
∆S = 50 – (30 + 60) = –40 J
For equilibrium ∆G = 0 = 0 = ∆H – T ∆ S
K 7540
30000
S
HT 0
Δ
Δ
24. (b)
Enthalpy of reaction
= B. E(Reactant) – B.E(Product)
= [B.E(C=C) + 4 B.E(C–H) + B.E(H–H)] – [B.E(C–C) + 6B.E.(C–H)]
= [606.1 + (4 × 410.5) + 431.37)] – [336.49 + (6 × 410.5)]
= –120.0 kJ mol–1
25. (b)
∆G = ∆H – T ∆ S
At equilibrium, ∆G = 0
⇒ 0 = (170 × 103J) – T (170 JK
–1)
⇒ T = 1000 K
For spontaneity, ∆G is –ve, which is possible only if T > 1000 K.
26. (c)
The reaction for formation of HCl can be written as
H2 + Cl2 → 2HCl
H – H + Cl – Cl → 2(H – Cl)
Substituting the given values, we get enthalpy of formation of
2HCl = –(862 – 676) = –186 kJ.
∴ Enthalpy of formation of
93kJ.kJ2
186HCl
27. (b)
For the reaction
PCl5(g) ⇌ PCl3(g) + Cl2(g)
The reaction given is an example of decomposition reaction and we know that decomposition reactions are
endothermic in nature, i.e., ∆H > 0.
Further
∆n = (1 + 1) – 1 = +1
Hence, more number of molecules are present in products which shows more randomness i.e.
∆S > 0 (∆S is positive)
28. (d)
We know that q (heat) and work (w) are not state functions but (q + w) is a state functions. H – TS (i.e. G) is also a state
functions. Thus II and III are not state functions so the correct answer is option (d).
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29. (b)
HClCl2
1H
2
122
∆HHCl = reactant of B.E. – products of B.E.
2402
1430
2
190 B.E. of HCl
∴ B.E. of HCl = 215 + 120 + 90 = 425 kJ mol–1
30. (d)
This reaction show the formation of H2O, and the X2 represents the enthalpy of formation of H2O because as the
definition suggests that the enthalpy of formation is the heat evolved or absorbed when one mole of substance is
formed from its constituent atoms.
31. (d)
We know that
∆H = ∆E + P∆V
In the reactions, H2 + Br2 → 2HBr this is no change in volume or ∆V = 0
So, ∆H = ∆E for this reaction
32. (a)
If ∆Gsystem = 0 the system has attained equilibrium is right choice.
In it alternative (d) is most confusing as when ∆G > 0, the process may be spontaneous when it is coupled with a
reaction which has ∆G < 0 and total ∆G is negative, so right answer is (a) .
33. (d)
∆G = ∆H – T∆S
When the reaction is in equilibrium, ∆G = 0
0 = ∆H – T∆S ⇒ T = ΔS
ΔH
K 285.7105
100030T
34. (a)
= – 358.5 kJ
The resonance energy provides extra stability to the benzene molecule so it has to be overcome for hydrogenation to
take place.
So ∆H = –358.5 – (–150.4) = –208.1 kJ
35. (a)
Measure of disorder of a system is nothing but Entropy. For a spontaneous reaction,
∆G < 0. As per Gibbs Helmnoltz equation, ∆G = ∆H – T∆S
Thus ∆G is –ve only
When ∆H = –ve (exothermic)
and ∆S = +ve increasing disorder)
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36. (c)
For a spontaneous reaction
∆G(–ve), which is possible if ∆S = +ve, ∆H = +ve
and T∆S > ∆H *As ∆G = ∆H – T∆S+
37. (c)
As MgO is a oxide of weak base hence some energy is lost to break MgO(s). Hence enthalpy is less than –57.33 kJmol–1
38. (b)
W = –p∆V
= –3(6 – 4) = –6 litre atmosphere
= –6 × 101.32 = –608 J
39. (a)
For a spontaneous process, ∆Stotal is always positive
40. (b)
∆G = ∆H – T∆S
= –382.64 – (298 × (-145.6) × 10–3
)
= –339.3 kJ mol–1
41. (d)
H2(g) + Br2(g) → 2HBr(g)
∆H°= (B E)reactant – (B E)product
= (433 + 192) – (2 × 364)
= 625 – 728 = –103 kJ
42. (c)
∆H = ∆E + nRT
∆n = 3 – (1 + 5)= 3 – 6 = –3
∆H – ∆E = (–3RT)
43. (d)
XeF4 sublimes at room temp., while N2O3 does not exist at room temp. It decomposes even at –30°C.
44. (b)
∆G = –P∆V = Work done
∆V is the change in molar volume in the conversion of graphite to diamond.
L102.25
12
3.31
12ΔV 3
= –1.91 × 10
–3 L
Work done = – (–1.91 × 10–3
) × P × 101.3 J
9794atm101.3101.91
mol J1895P
3
1
1 atm = 105 × 1.013 Pa
⇒ P = 9.92 × 108 Pa
45. (a)
T
HS
ΔΔ
∆S (per mole) = 273
6000
T
moleper H
Δ= 21.98 JK
–1 mol
–1
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46. (c)
Given Cp = 75 JK–1
mol–1
18
100n mole. Q = 1000 J ∆T = ?
Q = nCp∆T
K 2.475100
181000T
Δ
47. (a)
T
qΔS
q → required heat per mole
T → constant absolute temperature
Unit of entropy is JK–1
mol–1
48. (a)
C(s) + O2(g) → CO2(g) ∆H = –94 kCal/mole …(i)
H2(g) + 2
1O2 → H2O(g) ∆H = –68 kCal/mole …(ii)
CH4 + 2O2 → CO2 + 2H2O, ∆H = –213 kCal/mole …(iii)
C(s) + 2H2 → CH4(g), ∆H = ? …(iv)
Eqn. (iv) can be obtained by eq. (i) + eq. (ii) × 2 – eq. (iii)
C(s) + O2 → CO2(g)
2H2 + O2 → 2H2O(g)
CO2 + 2H2O → CH4(g) + 2O2
C(s) + 2H2(g) → CH4(g)
So, 4CHΔH = –94 + 2 (–68) – (–213)
= –94 –136 + 213 = –17 k Cal/mole
49. (d)
For isothermal reversible expansion
w = q = nRT × 2.303 log
1
2
v
v
= 2RT × 2.303 log2
20
= 2 × 2 × T × 2.303 × 1 = 9.2 T
Entropy change, cal. 9.2T
9.2T
T
qΔS
50. (a)
Internal energy is dependent upon temperature and according to firs t law of thermodynamics total energy of an
isolated system remains same, i.e., in a system of constant mass, energy can neither be created nor destroyed by any
physical or chemical change but can be transformed from one form to another
∆E = q + w
For closed insulated container, q = 0, so, ∆E = +w, as work is done by the system
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51. (a)
2H2O2(l) → 2H2O(l) + O2(g) ∆H = ?
∆H = *2 × (∆Hf of H2O(l)) + (∆Hf of O2) – (2 × ∆Hf of H2O2(l))]
= [(2 × –286) + (0) – (2 × –188)]
= [–572 + 376] = –196 kJ/mole
52. (b)
As volume is constant hence work done in this process is zero hence heat supplied is equal to change in internal
energy.
53. (b)
CH4(g) + 2
1O2(g) → CH3OH(l) ∆H = ?
∴ ∆H = *(∆H of combustion of CH3OH) – (∆H of combustion of CH4)]
= [(–y) – (–x)] = –[–y + x) = x – y
Given ∆H = –ve
∴ x – y < 0
Hence x < y.
54. (c)
Heat of reaction
∆H = Eforward – Ebackward
= 19 – 9 = 10 kJ/mol
55. (c)
322 SOO
2
1SO
o
)f(SO
o
)f(SO 23HHH ΔΔΔ
= –98.2 + 298.2 = 200 kJ/mole
56. (b)
For the reaction
2 ZnS → 2 Zn + S2; ∆G1° = 293 kJ …(1)
2 Zn + O2 → ZnO; ∆G2° = –480 kJ …(2)
S2 + 2O2 → 2SO2; ∆G3° = –544 kJ …(3)
∆G° for the reaction
2 ZnS + 3 O2 → 2 ZnO + 2 SO2
Can be obtained by adding eqn. (1), (2) and (3)
⇒ ∆G° = 293 – 480 – 544 = –731 kJ
57. (a)
T
H
point Melting
fusion of heat LatentS fΔ
Δ
11mol K J
300
2930
= 9.77 J K–1
mol–1
58. (c)
As all reactant and product are liquid
∆n(g) = 0
∆H = ∆E - ∆nRT ∆H = ∆E (∵ ∆n = 0)
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59. (a)
∆E = ∆Q – W
For adiabatic expansion, ∆Q = 0
⇒ ∆E = –W
The negative sign shows decrease in Internal energy, which is equal to the work done on the system by the
surroundings.
60. (d)
For a cyclic process
∆E = 0, ∆H = 0 & ∆G = 0. As all depend upon final state and initial state, w doesn’t depend initial and final state.
61. (b)
For an isothermal process ∆E = 0
62. (a)
We know from the third law of thermodynamics, the entropy of a perfectly crystalline substance at absolute zero
temperature is taken to be zero.
63. (b)
Given C + O2 = CO2, ∆H° = – x kJ …(i)
2CO2 = 2CO + O2, ∆H° = +y kJ …(ii)
Or CO2 = CO + 1/2O2, ∆H° = +y/2 kJ …(iii)
From eq. no (1) and (3)
CO,O2
1C 2 ∆H° = y/2 – x =
2
2xy kJ
64. (b)
Hydration energy of Cl+ is very less than H
+ hence it doesn’t from Cl
+ (aq) ions.
65. (d)
Entropy change
∆S = ∆Sproduct – ∆Sreactant
= 2 (186.7) – (223 + 130.6)
= 373.4 – 353.6
= 19.8 JK–1
mol–1
66. (b)
Entropy states the randomness or disorderness of the system. At absolune zero, the movement of molecules of the
system or randomness of the system is zero, hence entropy is also zero.
67. (a)
The relation between free energy change and equilibrium constant is given by Nernst equation
Q lnnF
RTEE o
At equilibrium, E = 0 and Q = KC
K lnnF
RTEo …(i)
Again ∆G = –nFE° …(ii)
Put in (i)
;K lnnF
RT
nF
ΔGo
∆G° = –RT ln K
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68. (b)
Enthalpy of formation of C2H4, CO2 and H2O are 52, –394 and –286 kJ/mol respective. (Given)
The reaction is
C2H4 + 3O2 → 2CO2 + 2H2O
Change in enthalpy,
(∆H) = ∆Hproducts – ∆Hreactants
= 2 × (–394) + 2 × (–286) – (52 + 0)
= –1412 kJ/mol
69. (a)
If more trans-2-pentene is added, then its concentration in right hand side will increase. But in order to maintain the K
constant, concentration of cis-2-pentene will also increase. Therefore more cis-2-pentene will be formed.
70. (b)
This is combustion reaction, which is always exothermic hence
∆H = –ve
As the no. of gaseous molecules are increasing hence entropy increase
Now ∆G = ∆H – T∆S
For a spontaneous reaction
∆G = –ve
Which is possible in this case as ∆H = –ve and ∆S = +ve.
71. (d)
∆G is negative for a spontaneous process.
72. (c)
During isothermal expansion of ideal gas,
∆T = 0. Now H = E + PV
∵ ∆H = ∆E + ∆(PV)
∴ ∆H = ∆E + ∆(nRT);
Thus if ∆T = 0., ∆H = ∆E
i.e., remain unaffected
73. (b)
∆ng = 2 – 4 = –2, ∆H = ∆E – 2RT.
74. (b)
1 M H2SO4 = 2g eq. of H2SO4
Hence y = 2x or y2
1x
75. (d)
As ∆H = ∆E + ∆ngRT
If np < nr; ∆ng = np – nr = –ve
Hence ∆H < ∆E
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1. (d)
Meq of HCl = 1515
175
Meq of NAOH = 515
125
Meq of HCl in resulting solution = 10
Molarity of [H+] in resulting mixture = 10
1
100
10
pH = -log[H+] = -log 0.110
1
2. (a)
Solubility of BaSO4, )L (mol233
1042.2 1-3
s )L (mol1004.1
-15
)(4 sBaSO ⇌ )()(2
4
2aqSOaqBa
ss
22
4
2]][[ sSOBaKsp
= (1.04 x 10
—5)2
= 1.08 x 10-10
mol2 L
-2
3. (a)
)()( 22 gBgA ⇌ xkJ);(2 HgX
On increasing pressure equilibrium shifts in a direction where pressure decreases i.e. forward direction .
On decreasing temperature, equilibrium shifts in exothermic direction i.e., forward direction.
So, high pressure and low temperature favours maximum formation of product.
4. (a)
22 H3N K NH2 13 → (1)
22 ON K NO2 2 → (2)
22 O2
1H
K O H 32 → (3)
For reaction 23 O2
5NH2 OH3NO2 2 (4)
Equation (4)
= equation (2) + 3 equation (3) – equation (1)
1
332
K
K.KK , Option (a).
5. (a)
Equilibrium constant is not affected by presence of catalyst hence statement (a) is incorrect.
CH – 7 EQUILIBRIUM
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6. (c)
422 OCAg Ag2 + 242OC
2.2 10–4
M 1.1 10–4
M
]OC[]Ag[K2
422
sp
]101.1.[]102.2[ 424
12sp 103.5K
7. (d)
n
pc RTKK )( = 825 1056.3)29808314.0(108.5
8. (a)
Catalyst do not affect the value of equilibrium.
9. (b)
n
pc RTKK )(
nc
P
RTK
K
)(
1
10. (a)
N1V1 – N2V2 = N.V.
0.1 × 1 – 0.01 × 1 = N × 2
[OH–] = N 0.045
2
0.09NR
pOH = –log (0.045) = 1.35
∴ pH = 14 – pOH = 14 – 1.35 = 12.65
11. (a)
HClO4 and NaClO4 cannot act as an acidic buffer.
12. (a)
N2(g) + O2(g) ⇌ 2NO(g); K
(g)O2
1(g)N
2
122 ⇌ NO(g); K'
when a reaction is multiplied by 1/2 then
K' = (K)1/2
13. (d)
14. (c)
15. (a)
21
0
1
2
T
1
T
1
2.303R
ΔH
K
Klog
T2> T1 So Kp<Kp’ (exothermic reaction)
(Assuming T2> T1, although it is not mentioned, which temperature is higher )
If T1> T2 then Kp>K’p then answer should be (2).
16. (d)
According to Le-Chatalier principle.
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17. (c)
;0.10
K;0.037
c
Kα aa
Ka = (0.037)2 × 0.10 = 1.37 × 10
–4
18. (c)
pH = pKa + log [Acid]
[Salt]
5 = 4 + log [Acid]
[Salt]
∵ pKa = – log Ka
Given Ka = 1 × 10–4
∴ pKa = – log (1 × 10–4
)
= 4
Now, from Handerson equation
pH = pKa + log[Acid]
[Salt]
putting the values
5 = 4 + log [Acid]
[Salt]
log[Acid]
[Salt] = 5 – 4 = 1
taking antilog
[Salt]/[Acid] = 10 = 1
10
19. (a)
Kw at 25°C = 1 × 10–14
At 25°C
Kw = [H+] [OH
–] = 10
–14
At 100°C (given)
Kw = [H+] [OH
–] = 55 × 10
–14
∵ for a neutral solution
[H+] = [OH
–]
∴ [H+]2 = 55 × 10
–14
or [H+] = (55 × 10
–14)1/2
∵ pH = – log [H+]
On taking log on both side
– log [H+] = – log (55 × 10
–14)1/2
pH = 2
1 log 55 + 7 log 10
pH = – 0.87 + 7
= 6.13
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20. (a)
CaCO3 → x
2
3x
2 COCa
CaC2O4 → y
2
42y
2 OCCa
∴ [Ca2+
] = x + y
Now, Ksp (CaCO3) = [Ca2+
] ][CO2
3
Or 4.7 × 10–9
= (x + y)x
Similarly, Ksp (CaC2O4) = [Ca2+
] ]O[C 2
42
Or 1.3 × 10–9
= (x + y)y
On solving, we get
[Ca2+
] = 7.746 × 10–5
M
21. (b)
BF3 is Lewis acid (e– pair acceptor)
22. (c)
Solubility of alkali metal is maximum among the following. Among ZnS (1.7 × 10–5
) & Cus (8 × 10–37
), ZnS has higher
value of Ksp.
23. (c)
A2 + B2 ⇌ 2AB ]][B[A
[AB]K
22
2
c
33
23
c104.2103
)10(2.8K
0.624.23
(2.8) 2
24. (c)
2SO2 + O2 ⇌ 2SO3 K = 278 (given)
SO3 ⇌ SO2 + 2
1O2
K
1K' 24 1061035.97
278
1
25. (a)
Lets take an example of an acidic buffer CH3COOH and CH3COONa.
CH3COOH ⇌ CH3COO– + H
+; CH3COONa ⇌ CH3COO
– + Na
+
When few drops of HCl are added to this buffer, the H+ of HCl immediately combine with CH3COO
– ions to form
undissociated acetic acid molecules. Thus there will be no with CH3COO– ions to form undissociated acetic acid
molecules. Thus there will be no appreciable change in its pH value. Likewise if few drops of NaOH are added, the OH–
ions will combine with H+ ions to form unionized water molecule. Thus pH of solution will remain constant.
26. (a)
(AlCl3, LiCl & BeCl2) all these solutions are acidic due to cationic hydrolysis, where as BaCl2, is salt of strong base and
strong acid, hence its solution will almost neutral i.e., pH ≈ 7.
27. (b)
Given pH = 12
Or [H+] = 10
–12
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30 YEAR’S BOOK CHEMISTRY PAGE: 137
Since, [H+][OH
–] = 10
–14
∴ 2
12
14
1010
10][OH
Ba(OH)2 ⇌ 2ss
2 2OHBa
[OH–] = 10
–2
2s = 10
–2
2
10s
2
Ksp = 4s3
32
2
104
= 5 × 10–7
28. (c)
Ksp = [Ag+] [Cl
–]
1.8 × 10–10
= [Ag+] [0.1]
[Ag+] = 1.8 × 10
–9 M
Ksp = [Pb+2
] [Cl–]2
1.7 × 10–5
= [Pb+2
] [0.1]2
[Pb+2
] = 1.7 × 10–3
M
29. (c)
BF3 behaves as lewis acid
30. (b)
Given [NH3] = 0.3 M, [NH4+] = 0.2 M, Kb = 1.8 × 10
–5.
[base]
[salt]logpKpOH b [pKb = –log Kb; pKb = –log 1.8 × 10
–5]
∴ pKb = 4.74
4.560.47710.30104.740.3
0.2log4.74
pH = 14 – 4.56 = 9.436
31. (c)
2A(g) + B(g) ⇌ 3C(g) + D(g)
Mole ratio 2 1 3 1
Molar concentration at t = 0 1 1 0 0
Equilibrium molar concentration 0.50 0.75 0.75 0.25
(0.75)(0.50)
(0.25)(0.75)
[B][A]
[D][C]K
2
3
2
3
c
32. (c)
Boron in B2H6 is electron deficient.
33. (d)
Kb = 10–10
; Ka = 10–4
or pKa = 4
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30 YEAR’S BOOK CHEMISTRY PAGE: 138
For the buffer solution containing equal concentration of B– and HB
pH = pKa + log ][
][
Acid
Salt= pKa + log
x
x= pKa + log 1 = pKa
pH = pKa = 4
The octahedral complex ion [CoCl2(NH3)4]+ i.e., tetra amminedichloro cobalt (III) ion exis ts as cis and trans isomers.
34. (d)
2C(s) + O2(g) ⇌ 2CO2(g)
∆n = 2 – 1 = 1
∴ Kc and Kp are not equal.
35. (d)
Acid
Salt logpKpH a
log [H+] = log Ka – log
Acid
Salt
log [H+] = log Ka + log
Salt
Acid
Salt
AcidK][H a
= 1.8 × 10–5
× 0.2
0.1 = 9 × 10
–6 M
36. (d)
Ba(OH)2(s) → (aq)
2
(aq) 2OHBa
pH = 12 or pOH = 2
[OH–] = 10
–2 M
Ba(OH)2 → 2
2
100.5Ba
+ 210
2OH
[∴ Concentration of Ba2+
is half of OH–]
Ksp = [Ba2+
] [OH–]2
= [0.5 × 10–2
] [1 × 10–2
]2
= 0.5 × 10–6
= 5 × 10–7
M3
37. (d)
Ammonium chloride is a salt of weak base and strong acid. In this case hydrolysis constant K h can be calculated as
10-
5
14
b
wh 105.65
101.77
101
K
KK
38. (d)
(CH3)3 B is electron deficient, thus behave as a lewis acid.
39. (c)
Given, CH3COOH ⇌ CH3COO– + H
+;
5
a 101.5K1
…(i)
HCN ⇌ H+ + CN
–;
01
a 104.5K2
or H+ + CN
– ⇌ HCN;
10
a
'
a104.5
1
K
1K
2
2 …(ii)
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30 YEAR’S BOOK CHEMISTRY PAGE: 139
∴ From (i) and (ii), we find the equilibrium constant (Ka) for the reaction,
CN– + CH3COOH ⇌ CH3COO
– + HCN, is
'
aaa 21KKK
45
10105.6510
3
1
104.5
101.5
5
40. (b)
Given: Equilibrium constant (K1) for the reaction
HI(g) 8;K(g);I2
1(g)H
2
1122 …(i)
To find equilibrium constant for the following reaction
H2(g) + I2(g) ⇌ 2HI(g); K2 = ? …(ii)
For this multiply (i) by 2, we get
2HI(g) ⇌ H2(g) + I2(g); K1 = 82 = 64 …(iii)
[Note: When the equation for an equilibrium is multiplied by a factor, the equilibrium constant must be raised to the
power equal to the factor]
Now reverse equation (iii), we get
H2(g) + I2(g) ⇌ 2HI(g); 64
1K …(iv)
[Note: For reversible reaction, the equilibrium constant of the backward reaction is inverse of the equilibrium constant
for the forward reaction.]
Equation (iv) is the same as the required equation (ii), thus K2 for equation (ii) is 64
1
i.e., Option (b) is correct.
41. (c)
Given reaction are
X ⇌ Y + Z …(i)
And A ⇌ 2B …(ii)
Let the total pressure for reaction (i) and (ii) be P1 and P2 respectively, then
1
9
KP
K
2
P1 (given)
After dissociation,
X ⇌ Y + Z
At equilibrium (1 – α) α α
[Let mole of X dissociation with α as degree of dissociation]
Total number of moles = 1 – α + α + α
= (1 + α)
Thus 1Z1Y1X .P1
P;P1
P;.P1
1P
α
α
α
α
α
α
111P .P1
1/.P
)(1.P
1K
1
α
α
α
α
α
α …(i)
Similarly for A ⇌ 2B
At equilibrium (1 – α) 2α
1K
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30 YEAR’S BOOK CHEMISTRY PAGE: 140
We have,
2
2
2P P
1
1/
1
P2K
2
α
α
α
α …(ii)
Dividing (i) by (ii), we get
2
1
P2
P1
2
2
1
2
P2
P1
P
P.
4
1
K
Kor
.P4
.P
K
K
α
α
Or
2
1
P
P.
4
19
1
9
K
K
2P
P1
1
36
P
P or
2
1 or P1 : P2 = 36 : 1
42. (b)
For the reaction
At equilibrium x)2(12 (g)2AB
⇌ x2
2x
(g)B2AB(g)
2
2
2
2
c][AB
][B[AB]K or
33
4
4x
x
2
2
cx)}{2(1
x(2x)K
= x3[(1 – x) can be neglected in denominator (1 – x) = 1]
The partial pressure at equilibrium is calculated on the basis of total number of moles at equilibrium.
Total number of moles
= 2 (1 – x) + 2x + x = (2 + x)
∴ Px)(2
x)2(1P
2AB
where P is the total pressure.
,Px)(2
2xPAB
P
x)(2
xP
2B
Since x is very small so can be neglected in denominator
Thus, we get
Px)(1P2AB
PxPAB
P2
xP
2B
Now,
2
2
AB
B
2
AB
PP
PPK
22
22
Px)(12
x.PP(x)
2
33
P12
.Px
[∴ 1 – x ≃ 1]
3
1
Pp33
P
2K xor
P
2.K xor
2
.Px
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30 YEAR’S BOOK CHEMISTRY PAGE: 141
43. (b)
The highest pH will be recorded by the most basic solution. The basic nature of hydroxides of alkaline earth metals
increase as we move from Mg to Ba and thus the solution of BaCl2 in water will be most basic and so it will have
highest pH.
44. (c)
For this reaction Keq. Is given by
3
33
Fe(OH)
OHFeK
= (Fe3+
) (OH–)3 [∴ [solid] = 1].
If (OH–) is decreased by
4
1 times then for reaction equilibrium constant to remain constant, we have to increase the
concentration of [Fe3+
] by a factor of 43 i.e., 4 × 4 × 4 = 64. Thus option (c) is correct answer.
45. (b)
[H3O]+ for a solution having pH = 3 is given by
[H3O]+ = 1 × 10
–3 moles/litre
[∴ [H3O]+ = 10
–pH]
Similarly for solution having pH = 4,
[H3O]+ = 1 × 10
–4 moles/litre and for pH = 5
[H3O]+ = 1 × 10
–5 moles/litre
Let the volume of each solution in mixture be IL, then total volume of mixture solution L = (1 + 1 + 1 ) L = 3L
Total [H3O]+ ion present in mixture solution = (10
–3 + 10
–4 + 10
–5) moles
Then [H3O]+ ion concentration of mixture solution
M3
0.0011M
3
101010 543
= 0.00037 M = 3.7 × 10–4
M.
46. (a)
Given [H3O+] = 1 × 10
–10 M
at 25° [H3O+] [OH
–] = 10
–14
4
10
14
1010
10][OH
Now, [OH–] =
OHp10 = 10
–4 =
OHp10
∴ pOH = 4.
47. (d)
Given,
N2 + 3H2 ⇌ 2NH3; K1 …(i)
N2 + O2 ⇌ 2NO; K2 …(ii)
H2 + 2O2
1 ⇌ H2O; K3 …(iii)
We have to calculate
4NH3 + 5O2 → 4NO + 6H2O; K = ?
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30 YEAR’S BOOK CHEMISTRY PAGE: 142
or 2NH3 + 2O
2
5 → 2NO + 3H2O
for this equation, 5/2
2
2
3
3
2
2
][O][NH
O][H[NO]K
but ,3
22
2
3
1]][H[N
][NHK
]][O[N
[NO]K
22
2
2
& 2
1
][O][H
O][HK
2
3
2
2
3 or 3/2
2
3
2
3
2
3][O][H
O][HK
Now operate,
1
3
32
K
.KK
2
3
3
22
3/2
2
3
2
3
2
22
2
][NH
]][H[N
][O][H
O][H
]][O[N
[NO]
K][O][NH
O][H[NO]5/2
2
2
3
3
2
2
1
3
32
K
KKK
48. (a)
Given Ka = 1.00 × 10–5
, C = 0.100 mol
For a weak electrolyte,
Degree of dissocation
1%100.100
101
C
Kaα) ( 2
5
49. (d)
HNO2 is a weak acid and NaNO2 is salt of that weak acid and strong base (NaOH).
50. (a)
For a solution of 10–8
M HCl [H+] = 10
–8
[H+] of water = 10
–7
Total [H+] = 10
–7 + 10
–8 = 10 × 10
–8 + 10
–8
10–8
(10 + 1) = 11 × 10–8
51. (a)
First option is incorrect as the value of Kp given is wrong. It should have been
2
OCH
CO
P][PP
PK
24
2
52. (d)
Given Kb = 1.0 × 10–12
[BOH] = 0.01 M [OH] = ?
x)c(1eq ct
BOH0t
⇌
cxcx00
OHB
x)(1
cx
x)c(1
xcK
222
b
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30 YEAR’S BOOK CHEMISTRY PAGE: 143
x)0.01(1
0.01x101.0
212
On calculation, we get, x = 1.0 × 10–5
Now [OH
–] = cx
= 0.01 × 10–5
= 1 × 10–7
mol L–1
53. (a)
The solution formed from isomolar solutions of sodium oxide, sodium sulphide, sodium selenide H2O, H2S, H2Se &
H2Te respectively. As the acidic strength increases from H2O to H2Te thus pH decreases and hence the correct of pHs is
pH1 > pH2 > pH3 > pH4
54. (a)
IVth
group needs higher S2–
ion concentration. In presence of HCl, the dissociation of H2S decreases hence produces
less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H2S which is sufficient to
precipitate IInd
group radicals.
55. (d)
For an acid-base indicator
HIn ⇌ H+ + In
–
[HIn]
]][In[HK ln
or [H+] =
][In
[HIn]K In
or log H+ = log KIn + log
][In
[HIn]
Taking negative on both sides
–log[H+] = –log KIn – log
][In
[HIn]
or log [HIn]
][In
= pH – pKIn
56. (c)
For AX2; Ksp = 4s3
∴ 3.2 × 10–11
= 4s3
or 4
103.13s
11 = 2 × 10
–4
57. (c)
Ksp for AgI = 1 × 10–16
In solution of KI, I– would be due to the both AgI and KI, 10
–4 solution KI would provide = 10
–4 I
–
AgI would provide, say = x I– (x is solubility of AgI)
Total I– = (10
–4 + x), Ksp of AgI = (10
–4 + x)x
⇒ Ksp = 10–4
x + x2
as x is very small ∴ x2 can be ignored
∴ 10–4
x = 10–16
or )(mol1010
10x 112
4
16
y)(solubilit
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30 YEAR’S BOOK CHEMISTRY PAGE: 144
58. (b)
B(OH)3 does not provide H+ ions in water instead it accepts OH
– ion and hence it is Lewis acid.
B(OH)3 + H2O ⇌ [B(OH)4]– + H
+
59. (c)
The higher is the tendency to donate proton, stronger is the acid. Thus the correct order is
R – COOH > HOH > R – OH > CH ≡ CH depending upon the rate of donation of proton.
60. (d)
For reaction to proceed from right to left rate forward
crate backward
KQ i.e. the reaction will be fast in backward direction.
i.e. rb > rf.
61. (b)
pOH = pKb + log [Base]
[Salt]
or pKb = pOH – log [Base]
[Salt]
but pOH + pH = 14 or pOH = 14 – pH
∴ 14 – pH – log bpK
0.1
0.1
62. (b)
Na2CO3 is a salt of weak acid H2CO3 and strong base NaOH, therefore, its aqueous solution will be basic hence has pH
more than 7.
Na2CO3 + 2H2O → acid weak
32base strong
CO2H2NaOH
63. (d)
Given s = 0.5 × 10–4
moles/lit
[MX2 ⇌ M2+
+ 2X–]
∵ For MX2, Ksp = s × (2s)2 = 4s
3
Ksp = 4 × (0.5 × 10–4
)3 = 4 × 0.125 × 10
–12
= 0.5 × 10–12
= 5 × 10–13
64. (c)
For the reaction
BaO2(s) ⇌ BaO(s) + O2(g); ∆H = +ve
At equilibrium Kp = 2OP
[For solid and liquids concentration term is taken as unity]
Hence, the value of equilibrium constant depends only upon partial pressure O 2. Further on increasing temperature
formation of O2 increases as this is an endothermic reaction.
65. (c)
Strong base has higher tendency to accept the proton. Increasing order of base and hence the order of accepting
tendency of proton is
I– < HS
– < NH3 < RNH2
66. (c)
CH3COOH ⇌ CH3COO– + H
+
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30 YEAR’S BOOK CHEMISTRY PAGE: 145
COOH][CH
]][HCOO[CHK
3
3a
Given that,
[CH3COO–] = [H
+] = 3.4 × 10
–4 M
Ka for CH3COOH = 1.7 × 10–5
CH3COOH is weak acid, so in it [CH3COOH] is equal to initial concentration. Hence
COOH][CH
)10(3.4 )10(3.4101.7
3
445
5
44
3101.7
103.4103.4COOH][CH
= 6.8 × 10–3
M
67. (a)
M2S ⇌
x2xS2M
Solubility product = (2x)2(x) = 4x
3
= 4 (3.5 × 10–6
)3 = 1.7 × 10
–16
68. (b)
CH3COOH is weak acid while NaOH is strong base, so one equivalent of NaOH cannot be neutralized with one
equivalent of CH3COOH. Hence the solution of one equivalent of each does not have pH value as 7. Its pH will be
towards basic side as NaOH is a strong base hence conc. of OH– will be more than the conc. of H
+.
69. (b)
In polyprotic acids the loss of second proton occurs much less readily than the first. Usually the Ka values for successive
loss of protons from these acids differ by at least a factor of 10–3
i.e., 21 aa KK
H2X ⇌ H+ + HX
– (
1aK )
HX– ⇌ H
+ + X
2– (
2aK )
70. (b)
Because NH3 after losing a proton (H+) gives
2NH
NH3 + H2O ⇌ OHNH 32
(Conjugate acid-base pair differ only by a proton)
71. (a)
MgCO3(s) → MgO(s) + CO2(g)
MgO & MgCO3 are solid and they donot exert any pressure and hence only pressure exerted is by CO 2.
Therefore Kp = 2COP
72. (a)
Given: Hydroxyl ion concentration [OH–] = 0.05 mol L
–1.
We know that the [H+][OH
–] = 1 × 10
–14
or [H+] =
1314
1020.05
101
mol L–1
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30 YEAR’S BOOK CHEMISTRY PAGE: 146
We also know that
Ph = –log[H+] = –log[2 × 10
–13]
= –log2 – log10–13
= –log2 –(–13) log10
= –0.3010 + 13.0000 = 12.6990
Since the value of pH > 7, therefore the solution is basic.
73. (d)
2xx
2 2ABBA
Solubility product = [x] [2x]2 = 4x
3
4 × 10–12
= 4x3 or 3
12
4
104x
∴ x = 10–4
74. (d)
No. of moles of NaOH = 0.140
4
[Molecular weight of NaOH = 40]
No. of moles of OH– = 0.1
Concentration of OH– =
litre 1
0.1 = 0.1 Mole/L
As we know that, [H+] [OH
–] = 10
–14
∴ [H+] = 10
–13 (∵ OH
– = 10
–1)
75. (c)
Lewis acid is that compound which have electron deficiency. e.g. BF3, SnCl2.
76. (d)
For the reaction
XeF6(g) + H2O(g) ⇌ XeOF4(g) + 2HF(g)
O]][H[XeF
][HF][XeOFK
26
2
41 …(1)
and for the reaction
XeO4(g) + XeF6(g) ⇌ XeOF4(g) + XeO3F2(g)
]][XeF[XeO
]F][XeO[XeOFK
64
2342 …(2)
For reaction:
XeO4(g) + 2HF(g) → XeO3F2(g) + H2O(g)
2
4
223
][HF][XeO
O]][HF[XeOK
∴ From eq. no. (1) and (2)
K = K2 / K1
77. (d)
A2(g) + B2(g) ⇌ 3C(g) ⇌ D(g)
step-1 step-2
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78. (d)
pH = pKa + log10 [Acid]
[Salt]
For small concentration of buffering agent and for maximum buffer capacity [Acid]
[Salt]
≅ 1
∴ pH = pKa
79. (d)
CuS ⇌
2
S
2
SSuC
Ksp = S2 or S =
spK
For Binary salts like CuS & HgS, solubility, S = spK
⇒ SCuS = ,1031
SHgS =45
10
For Ag2S → s
2
2s
S2Ag
Ksp = 4s3 or 3
44
3 sp
SAg4
10
4
KS
2
∴ The order is CuS > Ag2S > HgS
80. (b)
A2 ⇌ 2A Equilibrium constant is given by
Kc = reactants of terms conc.
product of terms conc.
][A
[A]
2
2
Since the value given is very small, hence conc. of products is less. It means the reaction is slow.
81. (b)
According to equation
2HI ⇌ H2 + I2
At t = 0 (2 moles) 0 0
At equilibrium (2 - 2α) moles α mole α mole
Total moles at equilibrium = 2 - 2𝛂 + α + α = 2 mole
82. (d)
The buffer system present in serum is H2CO3 + NaHCO3 and as we know that a buffer solution resist the change in pH
therefore pH value of blood does not change by a small addition of an acid or a base.
83. (d)
Rate constant of forward reaction (Kf) = 1.1 × 10–2
and rate constant of backward reaction (Kb) = 1.5 × 10–3
per minute.
Equilibrium constant (Kc) = 7.33101.5
101.1
K
K3
2
b
f
84. (a)
Molarity (M) = 10 M. HCl is a strong acid and it is completely dissociated in aqueous solutions as:
HCl ⟶ H+ + Cl
–
10 M 10 M
So, for every moles of HCl, there is one H+.
Therefore [H+] = HCl or [H
+] = 10.
pH = –log[H+] = –log[10] = –1
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85. (b)
For Bi2S3. Ksp– = (2s)2 . (3s)
3
= 4s2 . 27s
3 = 108s
5
or 5
70sp
108
101
108
K5s
For MnS. Ksp = s2
or 16
sp 107Ks
for CuS 73
sp 108Ks
for Ag2S Ksp = 2s2.s = 4s
3
or 3 sp
4
Ks
Thus MnS has maximum solubility.
86. (c)
In presence of the given salts, the solubility of AgCl decreases due to common ion effect.
87. (d)
Solid ⇌ Liquid
It is an endothermic process. So when temperature is raised, more liquid is formed. Hence adding heat will shift the
equilibrium in the forward direction.
88. (b)
AgBr has the highest solubility in 10–3
M NH4OH. All other solvents will dissolve AgBr poorly. Moreover bromides of
Ag+, 2
2Hg and 2
2Cu are water insoluble.
89. (d)
According to Le-chatelier’s principle whenever a constraint is applied to a system tends to readjust so as to nullify the
effect of the constraint.
90. (a)
Sodium borate is a salt of strong base (NaOH) and weak acid (H3BO3). Hence its aqueous solution will be basic.
91. (b)
An aqueous solution of acetic acid dissociates as
CH3COOH + H2O ⇌ CH3COO– + H3O
+
92. (a)
For reaction (1)
]][O[N
[NO]K
22
2
1
and for reaction (2)
[NO]
][O][NK
21
21
222
Therefore 2
2
1K
1K
93. (c)
NaCl is a salt of strong acid and strong base hence its aqueous solution will be neutral ie pH = 7. NaHCO 3 is an acidic
salt hence pH < 7. Na2CO3 is a salt of weak acid and strong base. Hence its aqueous solution will be strongly basic i.e.,
pH > 7.
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1. (d)
V 59.1E ,o
HBrO/Br
01
2
rBrOBH
,3
51
rOBrOBH V 5.1Eo
/BrO-3
HBrO
o
cellE for the disproportionation of HBrO,
o
cellE = o
/BrO
o
HBrO/Br -32
EEHBrO
= 1.59 -1.5
=0.095 V = + ve
Hence, option (d) is correct answer.
2. (c)
n-factor of 54
MnO
n-factor of 22
42
OC
Ratio of n-factor of
4MnO and 2
42OC is 5 : 2
So, molar ratio in balanced reaction is 2 : 5
The balanced equation is
OHCOMnHOCMnO 22
22
424 81021652
3. (a)
4. (c)
FeCI2 and SnCI2 (both are reducing agent and have lower oxidation no.)
5. (a)
Higher the value of reduction potential higher will be the oxidizing power whereas the lower the value of reduction
potential higher will be the reducing power.
6. (c)
OHCOKClSOKSOHOCHKClO 22
16
42
6
42422
5
3
i.e. maximum change in oxidation number is observed in Cl (+5 to –1)
7. (b)
On reaction with hot and concentrated alkali a mixture of chloride and chlorate is formed
3Cl2 + 3NaOH(excess) Hot O3HNaClO5NaCl 23
51
8. (d)
3
4PO = x + 4 (–2) = –3; x – 8 = –3; x = +5
2
4SO = x + 4 (–2) = –2; x – 8 = –2; x = +6
2
72OCr = 2x + 7 (–2) = –2; 2x – 14 = –2; 2x = 12; x = +6
CH – 8 REDOX REACTIONS
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9. (b)
2
3SO → S is in +4 oxidation state
2
42OS → S is in +3 oxidation state
2
62OS → S is in +5 oxidation state
10. (d)
Zinc gives H2 gas with dil. H2SO4/HCl but not with HNO3 because in HNO3,
3NO ion is reduced and give NH4NO3,
N2O, NO and NO2 (based upon the concentration of HNO3)
[Zn + 6%)(nearly
32HNO → Zn(NO3)2 + 2H] × 4
HNO3 + 8H → NH3 + 3H2O
NH3 + HNO3 → NH4NO3
4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + 3H2O
Zn is on the top position of hydrogen in electrochemical series. So Zn displaces H2 from dilute H2SO4 and HCl with
liberation of H2.
Zn + H2SO4 → ZnSO4 + H2
11. (b)
Oxidation number of a compound must be 0. Using the values for A, B and C in the four options we find that A3(BC4)2
is the answer.
12. (d)
Zn → Zn+2
+ 2e– …(1)
8e– + 10H+ + O3HNHNO 243 …(2)
operate eq. (1) × 4 + eq. (2) × 1
4Zn + 10H+ +
3NO → 4Zn+2
+
4NH + 3H2O
13. (d)
Pyrophosphoric acid H4P2O7
Let oxidation state of phosphorus is x
(4 × 1 + (–2) × 7 + 2x) = 0
∴ 2x = 10 or x = 5
14. (c)
2NaOH + H2SO4 → ation)(neutraliz
242 O2HSONa (not redox)
3
0Light
2
0
O2O3 (not redox reaction)
22
Light
02
02 ON2ON
(redox reaction)
Here oxidation of N2 & reduction of O2 is taking place
H2O(l) ΔH2O(g) (not redox reaction)
15. (c)
Carbon has the maximum oxidation state of +4, therefore carbon dioxide (CO2) cannot act as a reducing agent.
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16. (a)
Losing of electron is called oxidation.
17. (a)
Let x = oxidation no. of Cr in K2Cr2O7.
∴ (2 × 1) + (2 × x) + 7 (–2) = 0
Or 2 + 2x – 14 = 0 or x = +6.
18. (c)
2
4
61
2
1
2
2
OSHOBa
→ 1
2
1
2
2
4
62
OHOSBa
In this reaction, none of the elements undergoes a change in oxidation number of valency.
19. (a)
O.N. of P in H3PO3 (phosphorous acid)
3 × 1 + x + 3 × (–2) = 0 or x = +3
In orthophosphoric acid (H3PO4) O.N. of P is +5, in hypophosphorous acid (H3PO2) it is +1 while in metaphosphoric
acid (HPO3), it is +5.
20. (b)
The element undergo oxidation itself and reduces other is known as reducing agent. In this reaction O N of Ni Changes
from 0 to +2 and hence Ni acts as a reducing agent.
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1. (c)
O3 is reduced into O2 ion and
Ag2O is reduced to Ag so
H2O2 is reducing agent in both (a) and (b).
2. (b)
∴ Correct choice: (b)
3. (b)
H2O2 + [O] OxidationH2O + O2↑
4. (a)
O2H2Na 2A'' →
B''2
'C'H2NaOH
B''222
C''D''HZnONa2NaOHZn
B''2442
D''HZnSOSOdil.HZn
Na produces golden yellow colour with smokeless flame of Bunsen burner.
5. (c)
Fe + dil.H2SO4 → FeSO4 + H2↑
3Fe + Steam
2O4H → Fe3O4 + 4H2↑
Cu + dil. HCl → No reaction
Copper does not evolve H2 from acid as it is below hydrogen in electro chemical series.
2Na +C2H5OH → 2C2H5ONa + H2↑
6. (b)
2 acid
22 base1 acid
21 base
(g)H(aq)OHO(l)H(aq)H
In this reaction H– acts as bronsted base as it accepts one proton (H
+) from H2O and for H2.
7. (c)
Volume strength = 5.6 × Normality
= 5.6 × 1.5 = 8.4 L
8. (b)
Temporary hardness is due to presence of bicarbonates of calcium and magnesium and permanent hardness is due
to the sulphates or chlorides of bond of calcium and magnesium.
9. (d)
O – O – H bond angle in H2O2 is 97°.
10. (b)
Electrostatic forces of attraction are reduced to 1/80th
in water.
CH – 9 HYDROGEN
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11. (b)
In the structure of ice each molecule of H2O is surrounded by three H2O molecules in hexagonal honey comb
manner. On the other in water, each molecule is surrounded by four neighbouring molecules randomly which results
an open cage like s tructure. As a result there are a number of ‘hole’ or open species. In such a structure lesser
number of molecules are packed per ml. When ice melts a large no. of hydrogen bonds are broken. The molecule s
therefore move into the holes or open spaces and come closer to each other than they were in solid state. this result
sharp increase in the density. Therefore ice has lower density than water.
12. (c)
H(g) → H+(g) + e
–.
13. (a)
The complex salt of metaphosphoric acid sodium hexametaphosphate (NaPO3)6, is known as calgon. It is represented
as Na2[Na4(PO3)6]
14. (b)
is the structure of H2O2.
15. (d)
In this reaction H2S is oxidized to sulphur and H2O2 is reduced to H2O, hence this reaction show oxidation-reduction
both i.e., redox reaction.
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1. (c)
So, the most acidic should be BeO. In fact, BeO is amphoteric oxide while other given oxides are basic.
2. (a)
For 2nd
group hydrides, on moving down the group metallic character of metals increases so ionic character of metal
hydride increases. Hence the option (a) should be correct option.
3. (c)
ion hy drated of size
1 mobility Ionic
Smaller size hydrated ion in aq. soln – Rb
+(aq)
Larger size hydrated ion in aq. soln – Li
+(aq)
Lowest ionic mobility in aq. soln ⟶ Li
+(aq) due to high hydration.
4. (c)
All the above compounds absorb CO2, but only superoxide ( 2KO ) releases oxygen.
5. (c)
Thermal stability order
K2CO3 > Na2CO3 > CaCO3 > MgCO3
Therefore MgCO3 release CO2 most easily
MgCO3 Δ MgO + CO2
6. (a)
7. (c)
8. (c)
Na2CO3 is basic due to hydrolysis of 2
3CO ion
OHCO 2
23
⇌ OHHCO3
9. (d)
At the cathode, since the discharge potential of Na+ ions is lower than that of H
+ ions on the mercury, cathode, so
Na+ ions are discharged while H
+ ions remain in the solution.
Na+ + e
– → Na
2Na + xHg → Amalgam Sodium
2HgxNa
10. (c)
Tertiary halide can show ionic reaction with MF so, MF should be moist ionic for reaction to proceed forward. Hence
‘M’ should be ‘Rb’.
11. (b)
All alkali metal salts are ionic (except Lithium) and soluble in water due to the fact that cations get hydrated by water
molecules. The degree of hydration depends upon the size of the cation. Smaller the size o f a cation, greater is its
hydration energy.
CH – 10 THE s-BLOCK ELEMENTS
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Relative ionic radii:
Cs+ > Rb
+ > K
+ > Na
+ > Li
+
Relative ionic radii in water or relative degree of hydration
Li+ > Na
+ > K
+ > Rb
+ > Cs
+
12. (c)
All the a lkali metals when heated with oxygen form different types of oxides for example lithium forms lithium oxide
(Li2O), sodium forms sodium peroxide (Na2O2), while K, Rb and Cs form their respective superoxide.
OLiO2
12Li 22
13. (b)
(A) Plaster of paris = OH2
1CaSO 24 (B) Epsomite = MgSO4.7H2O
(C) Kieserite = MgSO4.H2O (D) Gypsum = CaSO4.2H2O
14. (c)
2Al(s) + 2NaOH(aq) + 2H2O(l) → aluminate meta sod.
22 3H2NaAlO
15. (b)
Active ingredient in bleaching powder for bleaching action is Ca(OCl)2
16. (c)
Melting point of halides decreases as the size of the halogen increases. The correct order is
CaF2 > CaCl2 > CaBr2 > CaI2.
17. (b)
(s)colourless
2(g)3(s) CaOCOCaCOA
Δ
2(aq)B
2(s) OH)(Ca2COCaO
C2(aq)322 )Ca(HCO2COCa(OH)
O(g)HCOCaCO)Ca(HCO 22(g)3(s)2(s)3A
∴ Correct choice: (b)
18. (a)
Lattice energy decreases more rapidly than hydration energy for alkaline earth metal hydroxides.
19. (b)
Be2+
is very small, hence its hydration enthalpy is greater that its lattice enthalpy.
(At. Nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)
20. (a)
NaOH is a strong alkali. It combines with acidic and amphoteric oxides to form salts. Since CaO is a basic oxide hence
does not reacts with NaOH.
21. (d)
The stability of alkali metal hydrides decreases from Li to Cs. It is due to the fact that M–H bonds becomes weaker
with increase in size of alkali metals as we move down the group from Li and Cs. Thus the order of stability of
hydrides is
LiH > NaH > KH > RbH > CsH
i.e., option (d) is correct answer.
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22. (b)
Smaller the ion more is its ionic mobility in aqueous solution. Ionic radii of the given a lkali metals in the order Na+ <
K+ < Rb
+ < Cs
+ and thus expected ionic mobility will be in the order Cs
+ < Rb
+ < K
+ < Na
+. However due to high
degree of solvation (or hydration) because of lower size or high charge density, the hydrate ion size follows the order
Cs+ < Rb
+ < K
+ < Na
+ and thus conductivity order is Cs
+ > Rb
+ > K
+ > Na
+ i.e., option (b) is correct answer.
23. (a)
The solubility of sulphates of alkaline earth metals decreases as we move down the group from Be to Ba due to the
reason that ionic size increases down the group. The lattice energy remains constant because sulphate ion is so large,
so that small change in cationic sizes do not make any difference. Thus the order:
BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4
24. (a)
As the basicity of metal hydroxides increases down the group from Be to Ba, the thermal stability of their carbonates
also increase in the same order. Further group 1 compounds are more thermally stable than group 2 because their
hydroxide are much basic than group 2 hydroxides therefore, the order of thermal stability is:
BeCO3 < MgCO3 < CaCO3 < K2CO3
25. (c)
Ionic radii of alkali metals in water follows the order Li+ > Na
+ > K
+ > Rb
+ > Cs
+
Thus in aqueous solution due to larger ionic radius Li+ has lowest mobility and hence the correct order of ionic
mobility is
Li+ < Na
+ < K
+ < Rb
+
26. (b)
The given properties coincide with CaCO3
27. (a)
In Castner process, for production of (Na) Sodium metal, Sodium hydroxide (NaOH) is electrolysed at temperature
330°C.
28. (c)
As Cs+ ion has larger size than Li
+ and I
– has larger size than F
–, therefore maximum distance between centres of
cations and anions is in CsI.
29. (b)
Calcium is obtained by electrolysis of a fused mass consisting six parts calcium chloride and one part ca lcium fluoride
at about 700°C in an electrolytic cell.
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30. (c)
MgSO4 is the only alkaline earth metal sulphate which is soluble in water and for solubility hydration energy should
be greater than lattice energy i.e. hydration energy > lattice energy
31. (c)
2Na2CO3 + NO + 3NO2 → 4NaNO2 + Co
32. (d)
Gypsum is CaSO4.2H2O and plaster of Paris is (CaSO4)2.H2O. Therefore, gypsum contain a lower percentage of calcium
than plaster of Paris.
33. (d)
Sodium is obtained by electrolytic reduction of its chloride. Melting point of chloride of sodium is high (803°C) so in
order to lower its melting point (600°C), calcium chloride is added to it.
34. (d)
Be(OH)2 is amphoteric, but the hydroxides of other alkaline earth metals are basic. The basic strength increases
gradually.
35. (d)
Ca2+
ions is an essential element for the contraction of muscles.
36. (b)
MgO, N2O5 is strongly acidic, ZnO and Al2O3 are amphoteric, therefore, MgO is most basic.
37. (c)
Ca and CaH2 both react with H2O to form H2 gas,
Ca + 2H2O → Ca(OH)2 + H2
CaH2 + 2H2O → Ca(OH)2 + 2H2
Whereas K gives H2 while KO2 gives O2 and H2O2
2K + 2H2O ⟶ 2KOH + H2
2KO2 + 2H2O ⟶ 2KOH + O2 + H2O2
Similarly, Na gives H2 while Na2 O2 gives H2O2
2Na + 2H2O → 2NaOH + H2
Na2O2 + 2H2O → 2NaOH + H2O2
Likewise Ba gives H2 while BaO2 gives H2O2
Ba + 2H2O → Ba(OH)2 + H2
BaO2 + 2H2O → Ba(OH)2 + H2O2
38. (c)
Mixture of K2CO3 and Na2CO3 is called fusion mixture.
39. (c)
A cation is always much smaller than the corresponding atom, whereas an anion is always larger than the
corresponding atom, hence the size decreases in the order
Na– > Na > Na
+
40. (c)
Atomic size of K+ > Ca
2+ > Mg
2+ and that of Cl
– > F
–. therefore, Mg
2+/Cl
– ratio has the minimum value.
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41. (b)
20Ca = 1s22s
22p
63d
23p
6 = [Ar]4s
2
42. (b)
Ca(20) = 1s22s
23p
63s
23p
64s
2 = [Ar]4s
2
43. (b)
Washing soda is Na2 CO3. 10H2O
44. (d)
Because of larger size and smaller nuclear charge, alkali metals have low ionization potential relative to alkaline ear th
metals.
45. (c)
Within a period, the atomic size decreases from left to right. Fur ther atomic size increases down the group. Hence the
correct order is i.e., Na > Mg > Li > Be.
46. (a)
Within a group, ionic radius increases with increase in atomic number. the melting points decrease down the group
due to weakening of metallic bond. The electronegativity and the 1st ionization energy also decreases down the group.
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1. (b)
‘B’ has no vacant d-orbitals in its valence shell, so it can’t extend its covalency beyond 4, i.e. ‘B’ cannot form the ion
like MF63(-)
i.e. BF63(-)
Hence, the correct option is (b).
2. (b)
Due to inert pair effect.
3. (a)
Strong reducing behaviour of H3PO2
All oxy-acid of phosphorus which contain P.H bond act as reductant.
Presence of one –OH group and two P–H bonds.
4. (b)
Stability of +1 oxidation state due to inert pair effect Tl > In > Ga > Al
5. (b)
Due to strong H-bonding in HF molecule, boiling point is highest for HF
HF > HI > HBr > HI
6. (d)
(i) No. of electron in ONF = 24
No. of electron in
2NO = 24
both are isoelectronic
(ii) OF2 is a fluoride of oxygen not oxide of fluorine because EN of fluorine is more than oxygen
OF2 = oxygen difluoride
(iii) Cl2O7 is an anhydride of perchloric acid
2HClO4 OH
Δ
2 Cl2O7
(iv) O3 molecule is bent bent shape
7. (d)
8. (a)
Feldspars are aluminosilicates.
9. (a)
4
4SiO is basic structural unit of silicates.
10. (b)
Since Me3SiCl contains only one Cl, therefore it can’t form high molecular mass silicon polymer. It can fo rm only
dimer.
CH – 11 THE p-BLOCK ELEMENTS
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11. (d)
Boron nitride (BN) is known as inorganic graphite. The most stable form is hexagonal one. It has layered structure
similar to graphite.
12. (c)
Fused alumina (Al2O3) is a bad conductor of electricity. Therefore, cryolite (Na3AlF6) and fluorspar (CaF2) are added to
purified alumina which not only make alumina a good conductor of electricity but also reduce the melting point of
the mixture to around 1140 K.
13. (c)
Pyrosilicate [Si2O7]6–
14. (b)
p-p overlap between B and F is maximum due to identical size and energy of p-orbitals, so electron deficiency in
boron of BF3 is neutralized partially to the maximum extent by back donation.
Hence, BF3 is least acidic.
15. (c)
Hydrolysis of substituted chlorosilanes yield corresponding silanols which undergo polymerization.
Polymerisation of dealkyl silandiol yields linear thermoplastic polymer.
16. (b)
Al2O3 can be converted to anhydrous AlCl3 by heating a mixture of Al2O3 and carbon in dry Cl2.
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17. (c)
Zeolite are microporous crystalline solid with well defined structure. They contain silicon, aluminium and oxygen in
their framework and cations, water & other molecules within their pores. The general formula of zeolite is
Mx/y[(AlO2)x(SiO2)y]. mH2O.
18. (b)
Na2B4O7. 10H2O O210H
Δ
anhydride Boric
32metaborate sod.
2anhydrous
74a2 OB2NaBOOBN Δ
CuO + B2O3 → beeda) (Blue metaborate cupric
22)Cu(BO
19. (b)
6HF + SiO2 → H2SiF6 + 2H2O
20. (d)
In graphite, each carbon is sp2-hybridized and the single occupied unhybridized p-orbitals of C-atoms over lap side
wise to give π-electron cloud which is delocalized and thus the electrons are spread out between the structures.
21. (a)
H3BO3 is a weak monobasic acid.
22. (c)
Aluminium can be extracted by electrolysis of pure alumina (Al2O3). Alumina ionizes as
32OAl ⇌ anode
33cathode
3 AlOAl
Al3+
+ 3e → Al (at cathode)
3
34AlO → 2Al2O3 + 3O2 + 12e (at anode)
Aluminium of 99.8% purity is obtained from this process.
23. (a)
Water gas is made by blowing steam through the layer of incandescent coal.
28kcalCOHCOHgasWater
2hot RedSteam2
24. (b)
In diamond each carbon atom is sp3 hybridized and thus forms covalent bonds with four other carbon atoms lying at
the corners of a regular tetrahedron.
25. (c)
Glass is a super cooled liquid.
26. (a)
SiCl4 gets hydrolysed in moist air and gives white fumes which are used as a smoke screen in warfare.
27. (d)
Potash Alum, K2SO4 . Al2(SO4). 24H2O is a double salt.
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1. (b)
-NO2 group exhibit –l effect and it decreases with increase in distance. In option (b) positive charge present on C-atom
at maximum distance so –l effect reaching to it is minimum and stability is maximum.
2. (c)
-l effect increases on increasing electronegativity of atom. So, correct order of –l effects is –NH2 < -OR < - F.
*Most appropriate Answer is option (a); however option (c) may also be correct answer.
3. (c)
The ortho and para isomers can be separated by steam distillation o-Nitrophenol is steam volatile due to
intramolecular hydrogen bonding while p-nitro phenol is less volatile due to intermolecular hydrogen bonding which
cause association of molecule.
4. (c)
Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electron
from a nucleophile.
5. (c)
6. (a)
Reason: Nucleophiles are electron rich species so act as Lewis base.
7. (a)
8. (b) 9. (a)
10. (a) 11. (c)
12. (c)
13. (a)
Volume of 1M H2SO4 = 10 m mol
Volume of NH3 consumed = 20 m mol
Weight of 0.280gg1000
2014N
37.33%1000.75
0.280%N
CH – 12 ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES
AND TECHNIQUES
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14. (b)
In Ethyne (CHCH) both carbon atoms are sp
hybrid as the hybridization of combustion
15. (c)
The blue colour is of Fe4[Fe(CN)6]3
16. (b)
In the carbonium ion the carbon atom carrying the positive charge is sp2 hybridized.
17. (b)
Stability depends on number of hyperconjugative structure.
18. (c)
Conformers are form of stereoisomers in which isomers can be interconverted by rotations about single bonds. I and
II are staggered and eclipsed conformers respectively.
19. (b)
Stability depends on number of hyperconjugative structures.
20. (d)
21. (b)
–NO2 is a powerful electron withdrawing group. Its presence on ring makes the ring less active
22. (c)
Decreasing order of deactivating effect of the given m-directing group is
> NO2 > –CN > –SO3H > –COOH
—NO2 group is most deactivating group due to strong – E, – I and – M effects.
23. (a)
IUPAC name of the structure is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-yonic acid.
24. (a)
It shows geometrical isomerism but does not show optical isomerism.
25. (a)
The correct name is 3-Bromoprop-1-ene
26. (c)
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Electrophilic rate order
Substitution hence presence of electron releasing group like CH3 in the nucleus facilitates nitration.
27. (a)
Conc. HNO3 decomposes NaCN and Na2S to avoid their interference.
NaCN + HNO3 → NaNO3 + HCN↑
Na2S + 2HNO3 → 2NaNO3 + H2S↑
28. (b)
Given wt of compound taken (w) = 0.35 g
Volume of nitrogen collected (V) = 55 ml
Room temperature (t K) = 300 K
Atmospheric pressure (P) = 715 mm
Aq. Tension (ρ) = 15 mm
Calculation-
Volume of N2 at NTP
ml760273
tV)(P
ρ
ml 46.098260273
3005515)(715
% of nitrogen
compound organic of wt22400100NTP at N of vol.28 2
16.46%0.3522400
10046.09828
29. (a)
Correct IUPAC name of above compound is trans-2-chloro-3-iodo-2-pentene
30. (a)
In SN Ar reactions, a carbanion is formed as an intermediate, so any substituent that increases the stability of
carbanion and hence the transition state leading to its formation will enhance the SNAr reactions. To compare the
rates of substitution in chlorobenzene, chlorobenzene having electron-withdrawing group, and chlorobenzene having
electron releasing group, we compare the structures carbanion I (from chlorobenzene). II (from chlorobenzene
containing electron-withdrawing group) and III (from chlorobenzene containing electron-releasing group).
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G withdraws electrons, neutralizes (disperses) –ve charge of the ring, stabilizes carbanion, facilitates SN reaction
(activation effect)
G release electrons. Intensifies –ve charge, destabilizes carbanion, retards SN reaction (deactivation)
NO2 is activating group and CH3 and OCH3 are deactiving group.
Hence, the correct order of nucleophilic substitution reactions.
31. (c)
Because of high electronegativities of the halogen atom, the carbon halogen (C – X) is highly polarised covalent
bond. Thus, the carbon atom of the C – X bond becomes a good site for attack by nucleophiles (electron rich
species). Nucleophilic substitution reactions are the most common reactions of alkyl halides.
32. (a)
Bond length order is
A1.54A1.40A1.34A1.10CCOCCCHC
33. (a)
The given compound is
4 ethyl, 3 proxyl hex – 1 – eme
34. (b)
Due to +M effect of –OH group and hyperconjugation of –CH3 group
35. (d)
Has a lone pair of electrons on oxygen atom, thus not an electrophile. Also octet is complete.
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36. (a)
Due to hydrogen bonding between the two OH groups, gauche conformation of ethylene glycol (a) is the most stable
conformation.
37. (b)
The bulky methyl groups are maximum away from each other.
38. (d)
Among –OH, –CH2OH, –NHCOCH3 and –OCH3, methoxy group has the highest +M effect.
39. (d)
Cl– is the weakest base and hence better leaving group.
40. (a)
Tertiary alkyl halide is most reactive towards nucleophilic substitution because the corresponding carbocation (3°) is
most stable. Aryl halide is least reactive due to partial double bond character of the C – Cl bond.
Presence of —NO2 groups in ortho and para positions increases the reactivity of the –Cl towards nucleophiles.
(CH3)3 – C – X > (CH3)2 – CH – X >
Or I < II < IV < III
41. (a)
Explanation: Presence of three —NO2 groups in o–, p– positions to phenolic groups (in III) makes phenol strongly
acidic because its corresponding phenate ion (conjugate base) is highly stabilized due to resonance. Conjugate base
of CH3COOH, II (i.e. CH3COO–) is resonance hybrid of two equivalent structures. The conjugate base of phenol, IV is
stabilized due to resonance note that here all resonating structures are not equivalent). The conjugate base of
cyclohexanol, I does not exhibit resonance, hence not formed.
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42. (d)
Alkenes with double bonds cannot undergo free rotation and can have different geometrical shapes with two
different groups on each and of the double bond.
43. (d)
44. (b)
C – 1 is sp hybridized (C ≡ C)
C – 3 is sp3 hybridized (C – C)
C – 5 is sp2 hybridized (C = C)
Thus the correct sequence is sp, sp3, sp
2.
45. (d)
The stability of carbanions is affected due to resonance, inductive effect and s -character of orbitals. Greater the
number of groups having +I group (alkyl group) lesser stable would be the carbanion.
Further stability of carbanion decreases with decrease in s-character. Benzene carbanions are stabilized due to
resonance, hence the correct order is
The correct order of stability of given carbanion is in order a > c > b > d.
46. (c)
A strong base can abstract and α-hydrogen from a ketone.
47. (a)
In the molecule
the number of stereoisomers is given by sum of geometrical isomers (because of presence of C = C) and optical
isomers (because of presence of chiral carbon atom). Number of geometrical isomers = 2 (one C = C is present).
Number of optical isomers = 2 (one chiral carbon atom).
Total number of stereoisomers = 2 + 2 = 4
48. (d)
The amount of s-character in various hybrid orbitals is as follows.
sp = 50%, sp2 = 33% and sp
3 = 25%
Therefore character of the C – H bond in acetylene (sp) is greater than that of the C – H bond in alkene (sp2
hybridized) which in turn has greater s character of the C – H bond in alkanes. Thus owing to a high s character of the
C – H bond in alkynes, the electrons constituting this bond are such a carbon atom can be easily removed as proton.
The acidic nature of three types of C – H bonds follows the following order
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≡ C – H > = C – H > –C – H
Further, as we know that conjugate base of a strong acid is a weak base, hence the correct order of basicity is
3
)()(
2
)(CHCHCHCHCCH
49. (c)
Out of the given compounds the most reactive towards nucleophilic attac k is
Phenoxide ion is stable due to resonance. i.e., the correct answer is option (c).
50. (c)
Electrophiles have high affinity for electrons. They attack at the site where electron density is highest. Electron
donating groups increases the electron density. The electron donating tendency decreases in the order:
–OH > –CH3 > –H > –Cl
Therefore, the correct order of reactivity towards electrophile is:
C6H5OH > C6H5CH3 > C6H6 > C6H5Cl
51. (d)
Nucleophilicity increases down the periodic table. I– > Br
– > Cl
– > F
–
52. (b)
53. (a)
Compounds which do not show optical activity inspite of the presence of chiral carbon atoms are called meso-
compounds. The absence of optical activity in these compound is due to the presence of a plane of symmetry in their
molecules. E.g. mesotartaric acid is optically inactive.
54. (c)
The degree of hydrolysis increases as the magnitude of positive charge on carbonyl group increases. Electron
withdrawing group increases the positive charge and electron releasing group decreases the positive charge. Among
these NO2 & CHO are electron withdrawing group from which NO2 has more –I effect than –CHO. On the other hand
CH3 is a electron releasing group therefore the order of reactivity towards hydrolysis is:
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55. (b)
*marked are chiral carbons.
56. (a)
Amond the three given hybrid orbitals, sp hybrid orbital is most electronegative. Contribution of s in sp hybrid orbital
is maximum (50%) so this orbitals is closer to nucleus. Naturally it will have greater tendency to pull electron towards
it. Hence it becomes more electronegative and sp3 becomes least electronegative as it have only 25% S character.
57. (c)
It is 2, 3 dimethyl pentanoyl chloride.
58. (c)
General molecular formula of alkanols is C2H2n+2O ∙ *CnH2n+1OH)
59. (b)
Among the given compounds naphthalene is volatile but benzoic acid is non-volatile (it forms a dimer). So, the best
method for their separation is sublimation, which is applicable to compounds which can be converted directly into
the vapour phase from its solid state on heating and back to the solid state on cooling. Hence it is the most
appropriate method.
60. (a)
Correct IUPAC name of
is 4-Ethyl-3-methyl heptanes
61. (c)
Phenols are more acidic than alcohol as they are resonance stabilized whereas alcohols are not.
Further nitro is an electron withdrawing which increases acidic character and facilitates release of proton, whereas –
CH3 is an electron donating group. Which decreases acidic character, thus removal of H+ becomes very difficult.
[Chiral]
[Not chiral]
[chiral]
[chiral]
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62. (d)
More the number of alkyl groups, the greater the dispersal of positive charge and therefore more the stability of
carbocation hence.
63. (d)
1NS reaction is favoured by heavy groups on the carbon atom attached to halogen i.e. Benzyl > allyl > tertiary >
primary > secondary > primary > alkyl halides.
Obtained from SN1 path
This molecule is resonance stabilized.
64. (b)
Optical and geometrical isomerism pair up to exhibit stereoisomerism. This is because the isomers differ only in their
orientation in space.
65. (a)
Clockwise rotation.
Hence configuration is R.
If the eye travel in a clockwise direction, the configuration is specified as the order of priority is Br > Cl > CH3 > H.
66. (b)
In diphenylmethane monochlorination at following positions will produce structured isomers.
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67. (b)
H2C = CHCl is capable of showing resonance which develops a partial double bond, character to C–Cl bond, thereby
making it less reactive toward nucleophilic substitution.
68. (a)
Carbon atoms from C1 to C5 are chiral.
69. (b)
Compound which are mirror image of each other and are not super imposable are termed as enantiomers.
are anantiomers.
70. (b)
In the presence of UV rays or energy, by boiling chlorine, free radical is generated which attack the methyl carbon
atom of the toluene.
71. (a)
Amino group is ring activating while nitro group is deactivating. Hence, correct order is aniline > benzene >
nitrobenzene.
I > II > III
–NO2 is an electron attracting group hence decrease the electron density on ring, whereas –NH2 group is electron
releasing group hence increases electron density on ring. Benzene is also rich due to delocalization of electrons.
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72. (b)
73. (c)
Geometrical isomers differ in spatial arrangement of atoms.
74. (a)
Both are resonating structures.
75. (b)
The IUPAC name of HCCHCHCHCHC65
2
4
2
32
2
1 is Hex-1-en-5-yne or 1-hexene-5-yne
The lowest number is given to the C = C double bond.
76. (a)
77. (b)
The principle of steam distillation is based on Dalton’s law of partial pressure. Suppose p 1 and p2 be the vapour
pressure of water vapour and the toluene at the distillation temperature. Toluene boils when total pressure is equal
to atmospheric pressure p
i.e., p = p1 + p2
or p2 = p – p1
as a result when toluene boil in the presence of steam. Its partial pressure P2 is less than atmospheric pressure.
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78. (a)
Electron releasing group (–CH3) decreases acidity while electron withdrawing group (–NO2) increases acidity.
79. (a)
A compound is said to exhibit optical isomerism if it atleast contains one chiral carbon atom, which is an atom
bonded to 4 different atoms or groups.
80. (a)
CH3 and NH2 are electron repelling groups, whereas NO2 and OH are electron attracting groups and they leave the
benzene ring deactivated. Due to stronger electron attraction (–I effect) effect of NO2 group C6H5NO2 shows least
reactivity towards electrophilic substitution.
81. (b)
Stability of an alkene depends upon the heat of hydrogenation of an alkene. The lower the heat of hydrogenation of
an alkene higher will be stability.
Order of stability Heat of hydrogenation (kJ/mol)
trans-2-butene 115.5
cis-2-butene 119.6 and
1-butene 126.8 respectively.
82. (a)
Electron withdrawing group (–NO2) increases acidity while electron releasing (–CH3, –H) decreases acidity. Also effect
will be more if functional group present at para position then ortho and meta position.
83. (c)
Chiral molecules are those molecules which have atleast one symmetric carbon atom (a carbon atom attached to 4
different groups). This is true in case of 3-methyl pentanoic acid.
84. (c)
223 sp
2
spspsp
3 CHCCHCH
85. (a)
When similar atoms are on opposite side the compound is in trans-form
86. (c)
The atom or group which has more power to attract electrons in comparison to hydrogen is said to have –I effect.
Thus higher the electronegativity of atom stronger will be the –I effect. As electronegativity of N, O and F follow the
order N < O < F hence based upon electronegative character order of –I effect is –NR2 < –OR < –F.
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87. (c)
The compound is diethyl ether (CH3CH2)2O which is resistant to nucleophilic attack by hydroxyl ion due to absent of
double or triple bond, whereas all other compounds given are unsaturated.
88. (a)
Due to absence of a symmetric (chiral) C-atom.
D—CH2—CH2—CH2Cl
Molecule is not a chiral molecule.
89. (d)
—Cl atom shows o/p-directive influence but deactivate the benzene ring, while [—OH/—CH3] groups show o/p
influence but activate the benzene ring but in these —OH group is more activating than —CH3. Hence order of
electrophilic substitution.
90. (c)
As in this compound the common groups i.e., highly electronegative halogen atoms are on opposite side, hence it is
a trans isomer.
Thus, its name is trans-2-chloro-3-iodo-2-pentene.
91. (d)
–OCH3 activates the benzene ring. –NO2 deactivates the ring. Hence the reaction of the given compounds with
electrophiles is in the order, I > II > III.
92. (c)
Order of stability : staggred anti > gauche > skew boat > eclipsed.
Newmann projection of n butane is given as
The maximum staggered conformation is most stable in which methyl groups are far apart as for as possible, due to
minimum repulsion between methyl groups and is also called anti conformation.
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93. (c)
In arenes, π electrons are delocalized, hence arens do not undergo addition reactions easily. Aromatic compounds
(Arenes) are highly stable and show resonance. e.g., Benzene is the simplest example.
94. (d)
When (–)2 bromo octane is allowed to react with sodium hydroxide under given conditions, where seco nd order
kinetics are followed, the product obtained is (+)2 octanol.
In this reaction Walden Inverson takes place so it is an example of SN 2-reaction.
95. (d)
Tautomerism is exhibited by the oscillation of hydrogen atom between two polyvalent atoms present in the molecule.
As option (d) has α-hydrogen atom. Therefore it shows tautomerism whereas other structures do not.
96. (c)
IR spectroscopy is used for the purification of cyclohexanone from a mixture of benzoic acid, isoamyl alcohol,
cyclohexane and cyclohexanone. Because in this method, each functional group upper at a certain peak. So,
cyclohexanone can be identified by carbonyl peak.
97. (c)
More is the electron - deficiency of the carbonyl carbon, greater will be the reactivity of the carbonyl compounds
towards nucleophilic addition.
98. (b)
Due to presence of four different groups on carbon, (C*) it is chiral.
99. (c)
Due to presence two similar methyl group at same carbon atom, above compound doesn’t show geometrical
isomerism.
100. (d)
Huckel’s rule states that for aromaticity there must be (4n + 2)π electron present in compound where n is an integer.
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101. (b)
When many substituents are present, the numbering is done from the end where the sum of locants is the lowest
(lowest sum rule)
2, 3, 6 TriMethyl Heptane (correct)
102. (b)
The possible isomers of the compound with molecular formula C7H8O is 5. These are (Anisol)
356 ,OCHHCalcohol benzyl
256 OHCHHC
and
103. (c)
The vital force theory suffered first death blow in 1828 when Wohler synthesized the Ist organic compound urea in
the laboratory from inorganic compounds reported below:
NH4CNO leading entrearrangemcharge isomeric to
Urea
22CONHNH
Later on a further blow to vital force theory was given by Kolbe (1845) who prepa red acetic acid, the first organic
compound, in laboratory from its elements.
104. (d)
The boiling point of o-nitrophenol is less than para-nitrophenol due to presence of intramolecular hydrogen
bonding. Since p-nitrophenol is less volatile in than o-nitrophenol due to presence of intermolecular hydrogen
bonding hence they can be separated by steam distillation.
105. (b)
Hydrazine (NH2NH2) does not contain carbon and hence on fusion with Na metal, it cannot form NaCN; consequently
hydrazine does not show Lassaigne’s test for nitrogen.
106. (b)
3-Ethyl-2-2methylpentane
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107. (c)
Chlorination of methane proceeds via free radical mechanism. Conversion of methyl chloride to methyl alcohol
proceeds via nucleophilic substitution. Formation of ethylene from e thyl alcohol proceeds via dehydration reaction.
Nitration of benzene is electrophilic substitution reaction.
108. (d)
Such question can be solved by considering the relative basic character of their conjugated bases which for H 2O,
C2H2, H2CO3 and C6H5OH are:
OHC,HCO,CHCOH,563
More the possibility for the dispersal of the negative charge, weaker will be the base. Thus the relative basic character
of the four bases is:
Thus the acidic character of the four corresponding acids will be
HC ≡ CH < H2O < C6H5OH < H2CO3
109. (a)
Resolution.
110. (c)
Diastereomers since they have different melting points, boiling points, solubilities etc.
111. (d)
Rotation around π bond is not possible. If any attempt is made to rotate one of the carbon atoms, the lo bes of π-
orbital will no longer remain coplanar i.e. no parallel overlap will be possible and thus π-bond will break. This is
known as concept of restricted rotation. In other words the presence of π-bonds makes the position of two carbon
atom.
112. (d)
Resonance structures are separated by a double headed arrow (⟷)
113. (d)
Due to +I-effect of the CH3 group, toluene has much higher electron density in the ring than benzene, nitrobenzene
and benzoic acid as nitro and carboxylic group show-I-effect and hence toluene is most reactive towards nitration.
114. (d)
Angle increases progressively
Sp3 (109°2.8'), sp
2 (120°), sp (180°)
115. (b)
116. (b)
2-Methyl-2-butene
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117. (d)
Carboxylic acids dissolve in NaHCO3 but phenols do not.
RCOOH 3NaHCO RCOONa + H2O + CO2
By evolving CO2 gas
118. (d)
Higher the possibility of delocalization of the positive charge, greater is stability of the species. Thus
22333256 HCCH3HC)(CHC)(CHHCHC
Benzyl carbocation is more stable than tert-butyl due to resonance in the former.
119. (d)
Organic compounds having same molecular formula but differ from each other in physical properties or chemical
properties or both are known as isomers.
120. (d)
spspspsp
2spspsp
3 CHCCHCHCHCHCH23223
121. (a)
sp3 orbital has 1/4 (25%) s-character. & 75% p character.
122. (d)
Shortest C – C distance (1.20 Å) is in acetylene.
As acetylene has sp hybridisation, the bond length increases in the order
)Asp(1.20
oCC <
)A(1.34spo2
CC < )Asp(1.54
oCC
123. (b)
Sodium cyanide (Na + C + N → NaCN). (Lassaigne’s test)
124. (a)
Hence it is homocyclic (as the ring system is made of one type of atoms, i.e. carbon) but not aromatic. As it does not
have (4n + 2)π electron required for aromaticity.
125. (a)
Kjeldal’s method is suitable for estimating nitrogen in those compounds in which nitrogen is linked to carbon and
hydrogen. The method is not used in case of nitro, azo and azoxy compound. This method is basically used for
estimating nitrogen in food fertilizers and agricultural products.
126. (c)
Such isomers, which possess the same molecular and structure formula but differ in the arrangement of atoms
around the double bonded carbon atoms are known as geometrical isomers.
127. (d)
Nitrogen, sulphur and halogens are tested in an organic compound by Lassaigne’s test. The organic compound is
fused with sodium metal as to convert these elements into ionisable inorganic substances,
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Na + C + N → NaCN
2Na + S → Na2S
2Na + X2 → 2NaX
The cyanide, sulphide or halide ions can be confirmed in aqueous solution by usual test.
128. (d)
All the properties mentioned in the question suggest that it is a benzene molecule. Since in benzene all carbons are
sp2–hybridized, therefore, C – C – C angle is 120°
129. (c)
22 sp2
spsp2 CHCCH
130. (d)
Tetrachloroethene being an alkene has sp2-hybridizied C– atoms and hence the angle Cl–C–Cl is 120° while in
tetrachloromethane, carbon is sp3 hybridized, therefore the angle Cl–C–Cl is 109.5°.
131. (c)
The compound containing a chiral carbon atom i.e., (a carbon atom which is attached to four different substituents is
known as a chiral carbon atom) is optically active.
132. (b)
Five chain isomers are possible which are
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1. (c)
HCCCHCHspspspsp
22
2
Number of orbital require in hybridization = Number of -bonds around each carbon atom.
2. (d)
Hence the correct option is (d).
3. (a)
Correct order of acidic strength
CH CH > CH3 – C = CH > CH2 = CH2 > CH3 – CH3
acc. To EN and Inductive effect.
4. (c)
In conformation bond angle and bond length remain same.
5. (d)
Sigma bond can be considered to be cylindrically symmetrical.
6. (b)
7. (b)
8. (c)
CH – 13 HYDROCARBONS
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9. (b)
10. (c)
11. (a)
Huckel rule is not obeyed. It has only four electrons. Further it does not have continuous conjug ation.
12. (d)
13. (d)
Presence of 6p orbitals, each containing one unpaired electron, in a six membered cyclic structure is in accordance
with Huckel rule of aromaticity.
14. (a)
1-Butyne and 2-butyne are distinguish by NaNH2 because 1-Butyne react with NaNH2 due to presence of terminal
hydrogen.
CH3 — CH2 — C ≡ CH + NaNH2 → 1 – Butyne
CH3CH3 — C ≡ CNa + NH3
15. (a)
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16. (b)
In this case dehydration is governed by Saytezeff’s rule according to which hydrogen is preferentially eliminated from
the carbon atom with fewer number of hydrogen atoms i.e., poor becomes poorer. Thus, 2 methyl butane -2 is the
major product.
This reaction is governed by Markownikoff’s rule according to which when an unsymmetrical reagent e.g. HBr adds to
an unsymmetrical alkene, then the negative part of the reagent is added to that carbon atom of the double bond
which bears the least number of hydrogen atom. Thus, in above case. 2-methyl 2-bromo butane will be the major
product.
17. (b)
When both double and triple bonds are present, then double bond is considered as the principal group.
18. (d)
C6H5CH2Br + Mg ether
C6H5CH2MgBr (in ether as solution)
19. (b)
During cracking higher hydrocarbons (liquid) are converted to lower gaseous hydrocarbons.
20. (c)
If both the double and triple bonds are present the compound is regarded as derivative of alkene. Further if double
and triple bonds are at equidistance from either side, the preference is given to double bond.
21. (b)
We know that in case of an unsymmetrical alkene there is the possibility of forming two products. In such cases the
formation of major product is decided on the basis of Markownikoffs rule which is rationalized in terms of stability of
the intermediate carbocation. Also remember that 3° carbocation and 2° carbocation is more stable than 1°
carbocation.
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Of the two possibilities 2° carbocation is more stable so the product of the reaction expected was predominantly one
formed by 2° carbocation i.e.
i.e. 2-Bromo-3-Methylbutane
However some electrophilic addition reaction form products that are clearly not the result of the addition of
electrophile to the sp2 carbon bonded to the most hydrogens and the addition of a nucleophile to the other sp
2
carbon
In the above cases the addition of HBr to 3-methyl-1butene the two products formed are shown below.
In this case the major product is 2-Bromo-2-methylbutane i.e. option (b) is correct answer.
(Note: the unexpected product results form a rearrangement of carbocation intermediate . Please note that all
carbocation do not rearrange.
22. (d)
23. (c)
This reaction occurs according to Markownikoff’s rule which states that when an unsymmetrical alkene undergo
hydrohalogenation, the negative part goes to that C-atom which contain lesser no. of H-atom.
CH3 – CH2 – C ≡ CH + HCl →
24. (b)
CH3 – C ≡ C – CH2 – CH3 3O
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The glyoxal formed as an intermediate is oxidized by H2O2 to give the acids.
25. (a)
Heat of hydrogenation of alkene alkene of Stability
1
Hence, the alkene which will react fastest with H2 will be the least stable. Among the given options the compound
having least number of alkyl groups (R) will be the least stable.
Further the relative rates of hydrogenation decrease with increase of steric hindrance
26. (c)
C6H6 + CH2 = CH2 3AlCl C6H5CH2CH3
27. (d)
In presence of peroxide, HBr adds on alkenes in anti-markovanikov’s way thus
H3C CH = CH2 + HBr PeroxideH3C CH2CH2Br
Kharasch observed that the addition of HBr of unsymmetrical alkene in the presence of organic peroxide follows an
opposite course to that suggested by Markownikoff. This is termed anti-Markownikoff or peroxide effect.
28. (c)
29. (a)
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30. (a)
Temperature
R — CH2 — CH2OHC380350
O Aloo 32
R — CH = CH2 + H2O
While at 220°–250° C it forms ether
31. (c)
32. (d)
We know that
6CH3 – CH = CH2 Cether,0HB
o62
2(CH3CH2CH2)3 OH
2O2H
Propanol33223 BO2HOHCHCH6CH
33. (a)
C2H5MgBr + H2O →
34. (a)
Given
A 2ClB
alc./KOHC
O/HO 23 CH2O
Hydrocarbon
Since hydrocarbon C give only CH2O, on ozonolysis, C should be CH2 = CH2 hence going backward A should be
ethane. Thus the reactions are
(A)
33CHCH /hvCl2
(B)23 ClCHCH
KOH
alc.
(C)22 CHCH
Δ O/HO 23
(D)HCHO
35. (a)
Gasoline (petrol) is a mixture of alkenes, alkenes and aromatic hydrocarbons. The quality of a gasoline is determined
by the amount of branched chain hydrocarbons (2, 2, 4-trimethylpentane, commonly known as iso-octane) present in
it.
36. (a)
Electrolysis of a concentrated aqueous solution of either sodium or potassium salts of saturated non-carboxylic acids
yields higher alkane at anode.
2RCOOK OxidationicElectrolyt
CathodeAnode2K2RCOO
At anode 2RCOO– → 2RCOO + 2e
– → R — R + 2CO2
At cathode 2K+ + 2e
– → 2K
2K + H2O → 2KOH + H2↑
(Kolbe’s method)
37. (c)
We know that
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38. (a)
39. (a)
When 3, 3 dimethyl 2-butanol is heated with H2SO4 the major product obtained is 2, 3 dimethyl 2-butene.
40. (b)
6R – CH = CH2 CEther,0
HB o22
2(RCH2CH2)3
NaOHH 2O2 6RCH2CH2OH + 2H3BO3.
41. (a)
Cl2 ationChaininitihv 2Cl
42. (c)
CH3Cl + AlCl3 → leElectrophi
43 AlClCH
43. (d)
Of all the options listed CH3CH2CH2CH3 has the least number of C-atoms and hence has the lowest b.p.
44. (a)
Peroxide effect is observed only in case of HBr.
Therefore, addition of HCl to propene even in the presence of benzyoyl peroxide occurs according to Markonikov’s
rule‛
CH3 – CH = CH2 2O2CO)5H6(C HCl CH3 – CHCl – CH3
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45. (d)
The reactivity of H-atoms depends upon the stability of free radicals follows the order:
Tertiary > secondary > primary, therefore, reactivity of H-atoms follows the same order, i.e., tertiary > secondary >
primary.
46. (a)
Only C2H2 (acetylene) has acidic H-atoms and hence reacts with NaNH2 to form sodium salt, i.e.,
HC ≡ CH + NaNH2 → HC ≡ Can + NH3.
47. (c)
Reduction of alkynes with Na/liq. NH3 gives trans-alkenes. This reaction is called Birch reduction
48. (c)
On heating ethylene chloride (1, 1 dichloroethane) with alcoholic potash followed by sodamide alkyne is obtained
R – CH2 – CCl2 – R
alc.KOHR – CH = CCl – R
2NaNHR – C ≡ C – R
49. (c)
Benzene do not show addition reaction like other unsaturated hydrocarbons. However it show substitution reactions.
Due to resonance all the C – C bonds have the same nature, which is possible because of the cyclic delocalization of
π-electrons in benzene. Monosubstitution will give only a single product.
50. (d)
Br2 in CCl4(a) Br2 in CH3COOH(b) and alk. KMnO4(c) will react with all unsaturated compounds, i.e., 1, 3 and 4 while
ammonical AgNO3(d) reacts only with terminal alkynes, i.e., 3 and hence compound 3 can be distinguished from 1, 2
and 4 by ammonical AgNO3(d).
51. (a)
The acidity of acetylene or 1—alkynes can be explained on the basis of molecular orbital concept according to which
formation of C—H bond in acetylene involves sp-hybridised carbon atom. Now since a electrons are closer to the
nucleus than p electrons, the electrons present in a bond having more s character will be correspondingly more closer
to the nucleus. Thus owing to high s character of the C—H bond in alkynes (s = 50%), the electrons constituting this
bond are more strongly held by the carbon nucleus (i.e., the acetylenic carbon atom or the sp orbital acts as more
electronegative species than the sp2 and sp
3 with the result the hydrogen present on such a carbon atom (≡C—H) can
be easily removed as a proton.
RJ VISION PVT. LTD.
30 YEAR’S BOOK CHEMISTRY PAGE: 188
1. (a)
Fact
2. (d)
3. (d)
Acid rain is the rain water containing sulphuric acid and nitric acid which are formed from the oxides of sulphur and
nitrogen present in the air as pollutants and rain water a pH of 4-5.
4. (a)
Based on the features given below the gas must be SO2.
5. (d)
The oxidized hydrocarbons and ozone in presence of humidity cause photochemical smog.
Hydrocarbons + O2, NO2, NO, O, O3 → Peroxides, formaldehyde, peroxyacetylnitrate (PAN), acrolein etc.
It is oxidizing in nature and causes irritation to eyes, lungs, nose, asthamatic attack and damage plants.
6. (b)
The ideal value of D.O for growth of fishes is 8 mg/ . 7 mg is desirable range, below this value fishes get
susceptible to disease. A value of 2 mg/ or below is lethal for fishes.
7. (b)
Green chemistry may be defined as the programme of developing new chemical products and chemical processes or
making improvements in the already existing compounds and processes so as to make less harmful to human health
and environment. This means the same as to reduce the use and production of hazardous chemicals.
i.e. correct answer is option (b).
8. (c)
CO and oxides of Nitrogen are poisonous gases present in automobile exhaust gases.
9. (a)
Green house gases such as CO2, ozone, methane, the chlorofluoro carbon compounds and water vapour form a thick
cover around the earth which prevents the IR rays emitted by the earth to escape. It gradually leads to increase in
temperature of atmosphere.
CH – 14 ENVIRONMENTAL CHEMISTRY