Download CBSE/12th Class/2010/CHEMISTRY

CBSE/12th Class/2010/CHEMISTRY
S.No
Q.1
Questions
Write a feature which will distinguish a metallic solid from an ionic solid.
Answers
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.
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.
Ans.3 A fine dispersion of minute droplets of one liquid in another in which it is not
soluble or miscible.
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.
Ans.5
Q.2
Define 'order of a reaction'.
Q.3
What is an emulsion?
Q.4
Why does NO2 dimerise?
Q.5
Give an example of linkage isomerism.
Q.6
Q.7
A
solution
of
KOH
hydrolyses
CH3CH(Cl)CH2CH3
CH2CH2CH2CH2Cl.Which one of these is more easily hydrolysed?
Draw the structural formula of 1-phenylpropan-1-one molecule.
Q.8
Give the IUPAC name of H2N - CH2 - CH2 - CH = CH2
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.
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?
and
Ans.6 CH2CH2CH2CH2Cl can easily hydrolysed by SN2 mechanism because it is less
sterically hindered as compared to CH3CH(Cl)CH2CH3.
Ans.7 1-phenylpropan-1-one.
Ans.8 H2N – CH2 – CH2 – CH = CH2
But-3-en-1-amine
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.
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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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?
Q.12
Draw the structures of white phosphorus and red phosphorus. Which one of
these two types of phosphorus is more reactive and why?
In case 2nd
R=K[2A][2B]2
=8K[A][B]2
Hence rate of reaction increases 8 times.
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
Ans.12
White Phosphorus
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.
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)
Q.15
Illustrate the following reactions giving a chemical equation for each:
(i) Kolbe's reaction.
(ii) Williamson synthesis
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.
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.
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
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?
Q.18
What are essential and non-essential amino acids in human food? Give one
example of each type.
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 Cterminal.
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.
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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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 5.46x108 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)
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.)
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?
They include: arginine, cysteine, glutamine, tyrosine, glycine, ornithine, proline, and
serine.
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
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
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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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.
micelle concentration. On dilution, these colloids revert back to individual ions.
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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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.
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.
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
ZrI4
Zr2 (pure) + 2I2
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.
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 2bromopentanebecause 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-2methylbutane because 1- Bromo-2-methylbutane has less steric hindrance than 2Bromo-2-methylbutane. This is because 1- Bromo-2-methylbutane is a primary alkyl
halidewhereas 2- Bromo-2-methylbutane is a tertiary alkyl halide.
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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Q.26
Give one example each of
(i) Addition polymers,
(ii) Condensation polymers,
(iii) Copolymers
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.
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?
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
0
m (NaCl) = 126 S cm2 mol-1
0
m (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.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.
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 acid0
mCH3COOH=λ0CH3COO- + λ0H+
(b) 0m(CH3COOH)=?
2
-1
0m(HCl)=426S cm mol
2
-1
0m(NaCl)=126S cm mol
2
-1
0m(CH3COONa)=91S cm mol
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 followsAnode Reaction: PbSO4(s) + 2ePbSO4(s)+ 2e+
2Cathode Reaction: PbO2(s) + 4H (aq) + SO4 (aq)+ 2ePbSO4(s) + 2H2O(l)
Overall Reaction: Pb(s) + PbO2(s) + 4H+(aq) + 2SO42-(aq)
2PbSO4(s) + 2H2O(l)
(b) Cu(s)
Cu+2(aq) + 2e-
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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.
2Ag+(aq) + 2e2Ag(s)
So overall reactionCu(s) + 2Ag+(aq)
Cu+2(aq) + 2Ag(s)
Cell reaction for thisE0cell= 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
+ 2
0.1 / [Ag ] =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
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)
(iii) Halogens are strong oxidizing agents.
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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-
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).
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.
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
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