CBSE/12th Class/2010/CHEMISTRY - Poornima University · Write chemical equations for the involved...

CBSE/12th Class/2010/CHEMISTRY

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S.No Questions Answers Q.1 Write a feature which will distinguish a metallic solid from an ionic solid.

Ans.1 Metallic solids Metallic solids are conductors of electricity in solid state as well as in molten state. Ionic solids Ionic solids are insulators in solid state but conductors in molten state and in aqueous solutions.

Q.2 Define 'order of a reaction'.

Ans.2 In chemical kinetics, the order of reaction with respect to a given substance (such as reactant, catalyst or product) is defined as the index, or exponent, to which its concentration term in the rate equation is raised.

Q.3 What is an emulsion?

Ans.3 A fine dispersion of minute droplets of one liquid in another in which it is not soluble or miscible.

Q.4 Why does NO2 dimerise?

Ans.4 In the NO2 molecule the nitrogen atom has a single unpaired electron. The molecule will tend to dimerise so that this unpaired electron can be paired with another to form the N2O4 molecule.

Q.5 Give an example of linkage isomerism.

Ans.5

Q.6 A solution of KOH hydrolyses CH3CH(Cl)CH2CH3 and

CH2CH2CH2CH2Cl.Which one of these is more easily hydrolysed? Ans.6 CH2CH2CH2CH2Cl can easily hydrolysed by SN2 mechanism because it is less sterically hindered as compared to CH3CH(Cl)CH2CH3.

Q.7 Draw the structural formula of 1-phenylpropan-1-one molecule.

Ans.7 1-phenylpropan-1-one.

Q.8 Give the IUPAC name of H2N - CH2 - CH2 - CH = CH2

Ans.8 H2N – CH2 – CH2 – CH = CH2

But-3-en-1-amine Q.9 Non-ideal solutions exhibit either positive or negative deviations from Raoult's

law. What are these deviations and why are they caused? Explain with one example of each type.

Ans.9 The vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law If it is higher, the solution exhibits positive deviation and if it is lower, it exhibits negative deviation from Raoult’s law. Eg. For positive deviation Mixtures of ethanol and acetone. Eg of negative deviation a mixture of chloroform and acetone.

Q.10 A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when: (i) The concentration of B alone is increased to three times (ii)The concentrations of A as well as B are doubled?

Ans. 10 It is given that a reaction is first order in reactant A and second order in reactant B R=K[A][B]2 In 1st case: R=K[A][3B]2 =9 K[A][B]2, hence rate of reaction increases 9 times.


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In case 2nd R=K[2A][2B]2 =8K[A][B]2 Hence rate of reaction increases 8 times.

Q.11 The rate constant for a reaction of zero order in A is 0.0030 mol L-1 s-1. How long willit take for the initial concentration of A to fall from 0.10 M to 0.075 M?

Ans.11 Rate constant,k= 0.0030mol L-1 s-1 Initial concentration, [R0 ] = 0.10M Concentration after timet, [R]= 0.075M For Zero otder reaction , K= [R0]-[R]/t Hence, t= [R0]-[R]/K = 0.10-0.075/0.003 = 8.33 S

Q.12 Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?

Ans.12

White Phosphorus

White phosphorus is less stable and therefore, more reactive than the red phosphorus under normal conditions because of angular strain in the P4 molecule where the angles are 60° only.

Q.13 Explain the following observations: (i) Generally there is an increase in density of elements from titanium (Z=22) to copper (Z=29) in the first series of transition elements. (ii) Transition elements and their compounds are generally found to be good catalysts in chemical reactions.

Ans.13 (i) The decrease in metallic radius coupled with increase in atomic mass results in a general increase in the density of these elements. Thus from titanium (Z=22) to copper (Z=29) the increase in the density is significant. (ii) The transition metals and their compounds show catalytic activity in a variety of chemical processes. This activity is ascribed to their ability to exhibit multiple oxidation states and to form complexes, alloys and interstitial compounds.

Q.14 Name the following coordination compounds according to IUPAC system of nomenclature:

(i) [Co(NH3)4(H2O)Cl]Cl2 (ii) [CrCl2(en)2]Cl, (en=ethane - 1,2-diamine)

Ans.14 (i) [Co(NH3)4(H2O)Cl]Cl2 Tetraammineaquachloridocobalt(III) chloride (ii) [CrCl2(en)2]Cl, (en=ethane - 1,2-diamine)

Dichloridobis(ethane-1,2-diamine)chromium(III) chloride

Q.15 Illustrate the following reactions giving a chemical equation for each: (i) Kolbe's reaction. (ii) Williamson synthesis

Ans.15 (i) Kolbe's reaction.

The Kolbe reaction is formally a decarboxylative dimerisation of two carboxylic


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acids (or carboxylate ions) The overall general reaction is:

Eg,

(ii) ) Williamson synthesis involves the reaction of an alkoxide ion with a primary alkyl halide via an SN2 reaction. This reaction is important in the history of organic chemistry because it helped prove the structure of ethers.

The general reaction mechanism is as follows:

Q.16 How are the following conversions carried out?

(i) Benzyl chloride to benzyl alcohol, (ii) Methyl magnesium bromide to 2-methylpropan-2-ol.

Ans.16 (i) ) Benzyl chloride to benzyl alcohol


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(ii) Methyl magnesium bromide to 2-methylpropan-2-ol.

Q.17 Explain the following terms :

(i) Invert sugar (ii) Polypeptides

OR Name the products of hydrolysis of sucrose. Why is sucrose not a reducing sugar?

Ans.17 (i) Invert sugar-It is a mixture of glucose and fructose; it is obtained by splitting sucrose into these two components. Compared with its precursor, sucrose, inverted sugar is sweeter and its products tend to retain moisture and are less prone to crystallization. Inverted sugar is therefore valued by bakers. (ii) Polypeptides Polypeptides are chains of amino acids. Proteins are made up of one or more polypeptide molecules. The amino acids are linked covalently by peptide bonds. The graphic on the right shows how three amino acids are linked by peptide bonds into a tripeptide. One end of every polypeptide, called the amino terminal or N-terminal, has a free amino group. The other end, with its free carboxyl group, is called the carboxyl terminal or C-terminal.

OR Hydrolysis breaks the glycosidic bond converting sucrose into glucose and fructose. The alpha-1, beta-2 acetal is really part of a double acetal, since the two monosaccharides are joined at the hemiacetal (alpha-1) of glucose and the hemiketal (beta-2) of the fructose. There are no hemiacetals remaining in the sucrose and therefore sucrose is a non-reducing sugar.

Q.18 What are essential and non-essential amino acids in human food? Give one example of each type.

Ans.18 Essential amino acids cannot be made by the body. As a result, they must come from food. The nine essential amino acids are: histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine. Non- essential amino acids are usually not essential, except in times of illness and stress.


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They include: arginine, cysteine, glutamine, tyrosine, glycine, ornithine, proline, and serine.

Q.19 The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral, there are 4 Ca2+ ions and 8 ions and that Ca2+ ions are arranged in fcc lattice. The Fions fill all the tetrahedral holes in the face centred cubic lattice of Ca2+ions. The edge of the unit cell is 5.46x10-8 cm in length. The density of the solid is 3.18 g cm-3. Use this information to calculate Avogadro's number (Molar mass of CaF2 = 78.08 g mol-1)

Ans.19 Given: Edge of unit cell=5.46 x 10-8cm Density of solid, d=3.18 g cm-3

Molar mass of CaF ,M=78.08 gmol-1

Since lattice is fcc type,z=4 Density of unit cell,d=zM/a3Na Na=zM/a3 x d =4x78.08/(5.46 x10-8 ) x 3.18 =6.03 x1023mol-1

Q.20 A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.310C. Determine molar mass of this compound. (B.P. of pure benzene = 80.10 °C and Kb for benzene = 2.530C kg mol-1.)

Ans.20 Elevation temperature∆ Tb=Kb x1000 x w2/M2 x w1 Where w2=wt. of solute(1.25gm),w1=wt. of solvent(99.0gm) B.P of solution Tb= 80.310C ,solvent Tb0= 80.10 0C Therefore∆ T=Tb –Tb0 =80.31- 80.10=0.210C M2=Kb x w2 /∆Tb x w1 =2.53 x 103 x 1.25 /0.21 x 99.0 =152.11 gm mol-1

Q.21 What is the difference between multimolecular and macromolecular colloids? Give one example of each type. How are associated colloids different from these two types of colloids?

Ans.21 Multimolecular Colloids: When a large number of atoms or small molecules (having diameters of less than 1nm) of a substance combine together in a dispersion medium to form aggregates having size in the colloidal range, the colloidal solutions thus formed are called multimolecular colloids. The species (atoms or molecules) constituting the dispersed particles in multimolecular colloids are held together by Vander Waals’ forces. The gold sol, sulphur sol etc. are some examples of multimolecular colloids. A gold sol may contain particles of various size composed of several atoms of gold. Similarly, sulphur sol consists of particles containing about a thousand of S8molecules. Macromolecular Colloids: Certain substances form large molecules whose dimensions are comparable to those of colloidal particles. Such molecules have very high molecular masses and are termed as macromolecules. When such substances are dispersed in suitable dispersion medium, the resulting colloidal solutions are known as macromolecular colloids. Thus, in macromolecular colloids, the dispersed particles are themselves large molecules having very high molecular masses. Most of the lyophilic sols are macromolecular colloids. For example colloidal dispersion of naturally occurring macromolecules such as starch, proteins, gelatin, cellulose, nucleic acids etc. are macromolecular colloids.Associated colloids are those colloids which behave as normal strong electrolytes at low concentrations but exhibit colloidal properties at higher concentrations due to the formation of aggregated particles. The aggregated particles thus formed are called micelles. The formationof micelles takes place only above a particular temperature called Kraft temperature and above a particular concentration called critical


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micelle concentration. On dilution, these colloids revert back to individual ions. Q.22 Describe how the following changes are brought about:

(i) Pig iron into steel. (ii) Zinc oxide into metallic zinc. (iii) Impure titanium into pure titanium.

OR Describe the role of: (i) NaCN in the extraction of gold from gold ore. (ii) SiO2 in the extraction of copper from copper matte. (iii) Iodine in the refining of zirconium. Write chemical equations for the involved reactions.

Ans.22 (i) Pig iron into steel Pig iron is the basic building block of all ferrous metals. It is formed when iron ore is refined in a furnace in the presence of charcoal, limestone and air. It is the basic form of iron used to make decorative wrought iron items. When the pig iron is further refined and a minimum amount of carbon is added, the crystalline structure of the metal changes and steel is produced. Though possible to create at home, the majority of steel is made in enormous batches using heavy machinery. (ii) Zinc oxide into metallic zinc Zinc oxide is converted to metallic zinc by reacting it with coke at 673 K. Coke,673K ZnO + C Zn + CO (iii)The ore rutile (impure titanium (IV) oxide) is heated with chlorine and coke at a temperature of about 900°C. TiO2+2Cl2+2C⟶TCl4+2CO Other metal chlorides are formed as well because of other metal compounds in the ore. Very pure liquid titanium (IV) chloride can be separated from the other chlorides by fractional distillation under an argon or nitrogen atmosphere. Titanium(IV) chloride reacts violently with water. Handling it therefore needs care and is stored in totally dry tanks.

OR (i)NaCN in the extraction of gold from gold ore: In the metallurgy of gold, gold metal is leached with a dilute solution of NaCN in the presence of air (for O2). The gold metal is then obtained from the product by displacement reaction. 4Au(s) + 8CN- (aq) + 2H2 O(aq) + O2 (g) 4[Au(CN)2 ]- (aq) + 4OH- (aq)

2[Au(CN)2 ]- (aq) + Zn(s) 2Au(s) + [Zn(CN) ]-2 (aq) (ii)SiO2 in the extraction of copper from copper matte.

The roasted ore is mixed with coke and silica (sand) SiO2 and is introduced in to a blast furnace. The hot air is blasted and FeO is converted in to ferrous silicate (FeSiO3). FeO + SiO2 FeSiO3 Cu2O + FeS Cu2S + FeO FeSiO3 (slag) floats over the molten matte of copper. (iii)Iodine is heated with Zirconium to form a volatile compound which on further heating decompose to give pure zirconium as shown: Zr2(impure) + 2I2 ZrI4


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ZrI4 Zr2 (pure) + 2I2

Q.23 How would you account for the following? (i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series. (ii) The E0 value for the Mn3/Mn2+ couple is much more positive than that for Cr3/Cr2+ couple or Fe3+/Fe2+ couple. (iii)The highest oxidation state of a metal is exhibited in its oxide or fluoride.

Ans.23 (i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) because of lanthanide contraction which is Lanthanide contraction is a term used in chemistry to describe the greater than expected decrease in ionic radii of the elements in the lanthanide series from atomic number 57, lanthanum, to 71, lutetium, which results in smaller than otherwise expected ionic radii for the subsequent elements starting with 72, hafnium. (ii)The E0 value for the Mn3/Mn2+ couple is much more positive than that for Cr3/Cr2+ couple or Fe3+/Fe2+ couple.Because Mn3+ has the outer electronic configuration of 3d4 and Mn2+ has the outer electronic configuration of 3d5. Thus, the conversion of Mn3+ to Mn2+ will be a favourable reaction since 3d5 is a very stable configuration as it is half filled configuration. Hence, Eo value for Mn3+ / Mn2+ couple is positive. Cr3+ to Cr2+ undergoes a change in outer electronic configuration from 3d3 to 3d4. Fe3+ to Fe2+ undergoes a change in outer electronic configuration from 3d5 to 3d6. Both these configurations of the resultant are not stable and hence have a lower Eo value. (iii)The highest oxidation state of a metal is exhibited in its oxide or fluoride.Because Oxigen and florine are the most electro negative elements in the whole 7hem.. These elements attract the electrons present in the outermost orbital the most. Thus the highest oxidation state of a metal is exhibited in its oxide or floride.

Q.24 (i) State one use each of DDT and iodoform. (ii) Which compound in the following couples will react faster in SN2 displacement and why? (a) 1-Bromopentane or 2-bromopentane (b) 1-Bromo-2-methylbutane or 2-bromo-2methylbutane.

Ans.25 (i) DDT- It is used as an insecticide Iodoform- It is used as a mild antiseptic. (ii) 1-Bromopentane will undergo faster SN2 displacement reaction than 2-bromopentanebecause 1- bromopentane has less steric hindrance than 2 -bromopentane. This is because1- bromopentane is a primary alkyl halide whereas 2-bromopentane is a secondary alkyhalide. (iii) 1- Bromo-2-methylbutane will undergo SN2 reaction faster than 2- Bromo-2-methylbutane because 1- Bromo-2-methylbutane has less steric hindrance than 2- Bromo-2-methylbutane. This is because 1- Bromo-2-methylbutane is a primary alkyl halidewhereas 2- Bromo-2-methylbutane is a tertiary alkyl halide.

Q.25 In the following cases, rearrange the compounds as directed: (i) In an increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3 NH2 (ii) In decreasing order of basic strength: Aniline, p-nitroaniline and p-toluidine (iii) In an increasing order of pkb values: C2H5NH2, C6H5 NHCH3, (C2H5)2NH and C6H5NH2

Ans.25 (i) In aromatic ammine there is delocalization of electrons of the N-atom over the benzene ring.In aliphatic amines the +I effect of two C2H5 groups in (C2H5)2NH is more than that of CH3NH2. Therefore (C2H5)2NH is more basic than CH3NH2.In C6H5NHCH3 and C6H5NH2 , C6H5NH2 is more basic than C6H5NHCH3 due to delocalization of electrons on the N-atom.So the increasing order of basic strength of the amines are: C6H5NH2 > C6H5NH(CH3 )2> CH3NH2 >( C2H5)2NH2. (ii) In decreasing order of basic strength: Aniline, p-nitroaniline and p-toluidine p-toluidine >Aniline> p-nitroaniline


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This is because –NO2 group has an electron withdrawing inductive effect or –I effect and – CH3 group has electron releasing inductive effect or +I effect. Groups with –I effect decreases the electron density on the nitrogen of amino group and hence decreases the basic strength. Groups with +I effect increases the electron density on the nitrogen of amino group and hence increases the basic strength. (iii) (i) The order of increasing basicity of the given compounds is as follows: C6H5NH2 < C6H5NHCH3 < C2H5NH2 < (C2H5)2NH We know that the higher the basic strength, the lower is the pKb values.C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH due to –C6H5 group has an electron withdrawing inductive effect or –I effect and – C2H5 group has electron releasing inductive effect. Groups with –I effect decreases the electron density on the nitrogen of amino group and hence decreases the basic strength. Groups with +I effect increases the electron density on the nitrogen of amino group and hence increases the basic strength. Greater the basic strength, the smaller is the pKb value.

Q.26 Give one example each of (i) Addition polymers, (ii) Condensation polymers, (iii) Copolymers

Ans.26 (i) Addition polymers,

(ii) Condensation polymers

(iii) Copolymers


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Q.27 What are analgesic medicines? How are they classified and when are they

commonly recommended for use?

Ans.27 Analgesics include paracetamol (known in the US as acetaminophen or simply APAP), the non-steroidal anti-inflammatory drugs (NSAIDs) such as the salicylates, and opioid drugs such as morphine and oxycodone. These are classified as follows: (i)Non-narcotic (non-addictive) analgesics (ii) Narcotic drugs Non-narcotic analgesics drugs are effective in relieving skeletal pain such as that due to arthritis and preventing platelet coagulation. Narcotic drugs analgesics are chiefly used for the relief of severe pain like postoperative pain, cardiac pain and pains of terminal cancer, and in child birth.

Q.28 (a) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law. (b) Calculate �0 m for acetic acid. Given that �0m (HCl) = 426 S cm2 mol-1 �0m (NaCl) = 126 S cm2 mol-1 �0m (CH3COONa) = 91 S cm2 mol-1

OR (a) Write the anode and cathode reactions and the overall reaction occurring in a lead storage battery. (b) A copper-silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell. (Given E0 Ag+/Ag = +0.80 V, E0 Cu2+/Cu=+0.34V)

Ans.28 (a) Kohlrausch’s law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If a salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.

Where �m=Molar conductivity, �0m=limiting molar conductivity,C= concentration So for acetic acid- �0mCH3COOH=λ0

CH3COO- + λ0H+

(b) �0m(CH3COOH)=? �0m(HCl)=426S cm2mol-1 �0m(NaCl)=126S cm2mol-1 �0m(CH3COONa)=91S cm2mol-1 therefore �0m(CH3COOH)= �0m (HCl) +�0m (CH3COONa) -�0m(NaCl) =426 +91 -126 =391Scm2mol-1

OR (a) The anode and cathode reactions and the overall reaction occurring in a lead storage battery are as follows- Anode Reaction: PbSO4(s) + 2e- PbSO4(s)+ 2e-

Cathode Reaction: PbO2(s) + 4H+(aq) + SO4

2-(aq)+ 2e- PbSO4(s) + 2H2O(l)

Overall Reaction: Pb(s) + PbO2(s) + 4H+(aq) + 2SO4

2-(aq) 2PbSO4(s) + 2H2O(l)

(b) Cu(s) Cu+2(aq) + 2e-


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2Ag+(aq) + 2e- 2Ag(s) So overall reaction- Cu(s) + 2Ag+(aq) Cu+2(aq) + 2Ag(s) Cell reaction for this- E0cell= E0Ag+/Ag - E0Cu+/Cu (where given E0Ag+/Ag=+0.80V, E0Cu+/Cu=0.34V) =+0.80 - 0.34 =0.46V If n=2 then Ecell=E0cell - 0.059/2 log [Cu2+] /[Ag+]2 0.422V =0.46V -0.059/2 log 0.1/[Ag+]2 So -0.038V = -0.059/2 log 0.1/[Ag+]2 Or log 0.1/[Ag+]2 =0.038V x 2 /0.059 =1.2881 0.1 / [Ag+]2 =Antilog (1.2881) = 1.941 x 10-7 [Ag+]2 = 0.1 / 1.941 x10 [Ag+]2 =0.00515 [Ag+] =0.071 mol L-1

Q.29 (a) Complete the following chemical equations: (i) NaOH(aq)+Cl2(g) (Hot and conc.) (ii) XeF6(s) + H2O(l) (b) How would you account for the following? (i) The value of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen. (ii) NF3 is an exothermic compound but NCl3 is endothermic compound. (iii) ClF3 molecule has a T-shaped structure and not a trigonal planar one.

OR (a) Complete the following chemical reaction equations: (i) P4+ SO2Cl2 (ii) XeF4 + H2O (b) Explain the following observations giving appropriate reasons: (i) The stability of +5 oxidation state decreases down the group in group 15 of the periodic table. (ii) Solid phosphorus pentachloride behaves as an ionic compound.

(iii) Halogens are strong oxidizing agents.

Ans.29 (a)

(i)6NaOH(aq)+3Cl2(g) 5 NaCl + NaClO3 + 3H2O (Hot and conc.) (ii)XeF6(s) + 3H2O(l) XeO3 + 6HF (b)(i) It is the same reason that oxygen has less negative electron gain enthalpy than sulphur which is the succeeding element down a group. When an electron is added to oxygen or fluorine (electron gain enthalpy) the added electron goes into the lower energy (n=2) level, where it suffers sufficient repulsion from the other electrons that are already present in the 2p sub-shell. However in chlorine or sulphur, the added electron goes into 3p sub-shell and therefore occupies a larger space, thus electron-electron repulsion is reduced and thus electron gain enthalpy is more -ve. (more energy is released) (ii) N≡N + 3X2 → 3NX3 ΔHrxn +ve(endothermic) or -ve(exothermic) Approximately ΔHrxn = 3ΔHbond(NX) - ΔHbond(N2) - 3ΔHbond(X2) The N-N bond energy is huge (-ve) and formation will only be thermodynamically allowed if X-X is low -ve (the F-F bond strength is anomalously small) and the N-X bond energy is high (-ve). The N-F covalent bond is much stronger than the N-Cl bond due to better 2p-2p overlap. (iii)


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ClF3 molecules are T-shaped. The chlorine atom has five electron-pairs in its outer shell; three of these are in Cl-F bonds, two are non-bonding ("lone") pairs. These keep as far apart as possible, minimizing repulsion between each of the negatively charged clouds, by adopting a trigonal bipyramidal arrangement. The two lone pairs occupy equatorial positions at an angle of 120° to each other; this gives the lowest energy arrangement of the electron pairs in the molecule so ClF3 molecule not a show trigonal planar structure.As shown in following structure-

OR

(a)(i) P4 + 10SO2Cl2 4PCl5 + 10SO2 (ii) 6XeF4+ 12H2O 4Xe + 2XeO3 + 24HF + 3O2 (b)(i) All these elements have 5 valence electrons and require three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The remaining elements of this group show a formal oxidation state of −3 in their covalent compounds. In addition to the −3 state, N and P also show −1 and −2 oxidation states. All the elements present in this group show +3 and +5 oxidation states. However, the stability of +5 oxidation state decreases down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect. (ii) In the solid state it can exist as [PCl4]+[PCl6]- . PCl5(s) → N/2[PCl4]+[PCl6]- ΔH = -ve ΔS = probably -ve (more ordered lattice) Briefly, the energy gained by formation of an ionic lattice more than compensates for the energy needed to form [PCl4]+ and [PCl6]- from PCl5. (iii) An oxidizing agent is the substance in redox reaction that gains electrons and whose oxidation number is reduced. Halogens are one electron short of having their shell filled, and they should be especially greedy for more electrons.So Halogens are strong oxidizing agents.

Q.30 (a) Explain the mechanism of a nucleophilic attack on the carbonyl group of an aldehyde or a ketone. (b) An organic compound (A) (molecular formula C8H16O2) was hydrolyzed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (c).

Ans.30 (a)Nucleophile attacks the electrophilic carbon atom of the polar carbonyl group of an aldehyde and a ketone from a direction approximately perpendicular to the plane of sp2 hybridised orbitals of carbonyl carbon. The hybridisation of carbon changes from sp2 to


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Oxidation of (C) with chromic acid also produced (B). On dehydration (C) gives but-1-ene. Write the equations for the reactions involved.

OR (a) Given chemical tests to distinguish between the following pairs of compounds: (i) Ethanal and Propanal (ii) Phenol and Benzoic acid (b) How will you bring about the following conversions? (i) Benzoic acid to benzaldehyde (ii) Ethanal to but-2-enal (iii) Propanone to propene Give complete reaction in each case.

sp3 in this process, and a tetrahedral alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product. The net result is addition of Nu- and H+ across the carbon oxygen double bond.

(b)An organic compound A with molecular formula C8H16O2 gives a carboxylic acid (B) andan alcohol (C) on hydrolysis with dilute sulphuric acid. Thus, compound A must be an ester. Further, alcohol C gives acid B on oxidation with chromic acid. Thus, B and C must contain equal number of carbon atoms.Since compound A contains a total of 8 carbon atoms, each of B and C contain 4 carbon atoms.Again, on dehydration, alcohol C gives but-1-ene. Therefore, C is of straight chain and hence, it is butan-1-ol.On oxidation, Butan-1-ol gives butanoic acid. Hence, acid B is butanoic acid.Hence, the ester with molecular formula C8H16O2 is butylbutanoate.

OR

(a)(i) Propanal and propanone can be distinguished by the following tests. Tollen’s test: Propanal is an aldehyde. Thus, it reduces Tollen’s reagent. But, propanone being a ketone does not reduce Tollen’s reagent.


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(ii) Phenol and benzoic acid can be distinguished by ferric chloride test. Ferric chloride test: Phenol reacts with neutral FeCl3 to form an ironphenol complex giving violet colouration.

(b)(i) Benzoic acid to benzaldehyde

(ii) Ethanal to but-2-enal

(iii) Propanone to propene

CBSE/12th Class/2010/CHEMISTRY - Poornima University · Write chemical equations for the involved...

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