Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2014 Comptt (Outside Delhi) with Solutions
Time allowed : 3 hours
Maximum marks : 70
General Instructions:
- All questions are compulsory.
- Questions number 1 to 8 are very short answer questions and carry 1 mark each.
- Questions number 9 to 18 are short answer questions and carry 2 marks each.
- Questions number 19 to 27 are also short answer questions and carry 3 marks each.
- Questions number 28 to 30 are long answer questions and carry 5 marks each.
-
Use log tables if necessary. Use of calculators is not allowed.
Deleted from Syllabus.
SET I
Question 1.
Why are crystalline solids anisotropic?
Answer:
Crystalline solids show different values of their some properties like electrical conductivity, refractive index, thermal expansion etc. in different directions.
Question 2.
What are emulsions? Name an emulsion in which water is a dispersed phase.
Answer:
Emulsions : An emulsion is a colloidal dispersion in which both the dispersed phase and dispersion medium are liquids.
Water in oil → Butter, cold creams.
Question 3.
What are the collectors used in froth floatation process? Name a substance that can be used as such.
Answer:
Collectors : Pine oils, fatty acids etc. enhance non-wettability of the mineral particles.
Question 4.
Why is F
2
a stronger oxidising agent than Cl
2
?
Answer:
Due to low bond dissociation enthalpy and high electronegativity of Fluorine, it has strong tendency to accept electrons and thus get reduced.
F + e
–
→ F
–
Therefore F
2
acts as strong oxidising agent, while Cl
2
is weak oxidising agent due to low electronegativity.
Question 5.
Name the alcohol that is used to make the following ester :
Answer:
Alcohol used : Propan-2ol
Question 6.
Give a test to distinguish between propan-2-one and pentan-3-one.
Answer:
Propan-2-one and pentan-3-one can be distinguished by Iodoform test.
Question 7.
How does a homopolymer differ from a copolymer?
Answer:
Homopolymers : Polymers whose repeating structural units are derived from only one type of monomer units are called homopolymers.
Example : Polyethene, PVC, teflon.
Copolymers : Polymers whose repeating structural units are derived from two or more types of monomer molecules are copolymers.
Example : Buna-S, Nylon 66
Question 8.
Define a ‘Peptide linkage’.
Answer:
Peptide linkage : It is an amide linkage formed between -COOH group of one a-amino acid and NH
2
, group of the other α-amino acid by loss of a molecule of water. -CO – NH – bond is called Peptide linkage.
Question 9.
Set up Neverest equation for the standard dry cell. Using this equation show that the voltage of a dry cell has to decrease with use.
Answer:
Cell reaction of a dry cell can be represented as
Zn + Hg
2+
→ Zn
2+
+ Hg (n = 2)
Nernst equation E
cell
= \(\mathrm{E}_{\text {cell }}^{\circ}\) – \(\frac{0.0591}{2}\) log \(\frac{\left[\mathrm{Zn}^{2+}\right]}{\mathrm{Hg}^{2+}}\)
The voltage of dry cell has to decrease because the concentration of electrolyte decreases in the reaction.
Question 10.
How does a change in temperature affect the rate of a reaction? How can this effect on the rate constant of a reaction be represented quantitatively?
Sol.
The rate constant of a reaction increases with increase of temperature and becomes nearly double for ever, 10°C rise in temperature.
The effect can be represented quantitatively by Arrhenius equation
K = Ae
-Ea/RT
where Ea = Activation energy of the reaction
A = Frequency factor
Question 11.
Describe the underlying principle of each of the following processes :
(i) Recovery of silver from the solution obtained by leaching silver ore with a solution of NaCN.
(ii) Electrolytic refining of a crude metal.
Answer:
(i) NaCN acts as a leaching agent or oxidising agent, thus oxidises Ag to Ag
+
which then combines with CN
–
ions to form respective soluble complex
(ii) Electrolytic refining : Here the impure metal is made to act as anode and a strip of the same metal in pure form is used as cathode. When they both are put in suitable electrolyte containing soluble salt of same metal, the more basic metal remains in the solution and the less basic ones go to the anode mud.
Example : In refining of Cu
At anode : (oxidation)
Cu → Cu
2+
+ 2e
–
At cathode : (reduction)
Cu
2+
+ 2e
–
→ Cu
Or
Describe the principle involved in each of the following processes
(i) Zone refining of a metal
(ii) Vapour phase refining of metals
Answer:
(i) Zone refining of metals : It is based on the principle that the impurities are more soluble in melt than in the solid state of the metal.
(ii) Vapour phase refining :
Vapour phase refining of inetals Here the metal is converted into its volatile compound and collected elsewhere which then decomposed to give pure metal. The requirements are the metal should form a volatile compound with an available reagent. the volatile compound should be easily decomposable so that recovery is easy.
Example :
Question 12.
Complete the following chemical equations :
(i) SO
2
+ MnO
–
4
+ H
2
O →
(ii) F
2+
(g) + H
2
O (l) →
Answer:
(i) 5SO
2
+ 2MnO
–
4
+ 2H
2
O → 5SO
4
2-
+ 2Mn
2+
+ 4H
+
(ii) Fluorine oxidises H
2
O to O
2
and O
3
2F
2
(g) + 2H
2
O (l) → 4H
+
(aq) + 4F
–
(aq) + O
2
(g)
3F
2
(g) + 3H
2
O (l) → 6H
+
(aq) + 6F
–
(aq) + O
3
(g)
Question 13.
Assign reasons for the following :
(i) Copper(I) ion is not known to exist in aqueous solutions.
(ti) Both O
2
and F
2
stabilize high oxidation states of transition metals but the ability of oxygen to do so exceeds that of fluorine.
Answer:
(z) See Q. 23 (i), 2014 Comptt. (II Outside Delhi). [Page 78
(i) Cu
2+
(aq) is much more stable than Cu
2+
(aq). This is because although second ionization
enthalpy of copper is large but ∆
hyd
H for Cu
2+
(aq) is much more negative than that for Cu
+
(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows:
2Cu
+
→ Cu
2+
+ Cu
(ii) Oxygen stabilizes the highest oxidation state even more than fluorine.
Example : Highest fluoride of Mn is MnF
4
whereas highest oxide is Mn
4
O
7
. It is due to ability of oxygen to form multiple bonds with the metal atoms.
Question 14.
Write the IUPAC names of the following compounds :
(i) CH
2
= CHCH
4
Br
(ii) (CCl
3
)
3
CCl
Answer:
(i) CH
2
= CHCH
4
– Br
IUPAC name : 3-Bromopropene
(ii) (CCl
3
)
3
– C – Cl
IUPAC name : 2-(Trichloromethyl) – 1, 1, 1, 2, 3, 3, 3-heptachloropropane
Question 15.
What are ambident nucleophiles? Explain with an example.
Answer:
Ambident nucleophile : A nucleophile that can form new bonds at two or more spots in its structure, usually due to resonance contributors.
Example : S = C = N
–
can act as a nucleophile with either the S or N attacking.
Question 16.
(i) Arrange the following compounds in an increasing order of basic strength :
C
6
H
5
NH
2
, C
6
H
5
N(CH
3
)
2
, (C
2
H
5
)
2
NH and CH
3
NH
2
(ii) Arrange the following compounds in a decreasing order of pK
b
values :
C
2
H
5
NH
2
, C
6
H
5
NHCH
3
, (C
2
H
5
)
2
NH and C
6
H
5
NH
2
Answer:
(i) Increasing order of basic strength :
C
6
H
5
NH
2
< C
6
H
5
N(CH
3
) < CH
3
NH
2
< (C
2
H
5
)
2
NH (More +I effect)
(ii) Decreasing order of pK
b
values :
C
6
H
5
NH
2
< C
6
H
5
NHCH
3
< C
2
H
5
NH
2
< (C
2
H
5
)
2
NH
Question 17.
Give a chemical test to distinguish between each of the following pairs of compounds
(i) Ethylamine and Aniline
(ii) Aniline and Benzylamine
Answer:
(i) Ethylamine and aniline :
By Azodye test: It involves the reaction of any aromatic primary amine with HNO
2
(NaNO
2
+ dil. HCl) at 273-278 K followed by treatment with an alkaline solution of 2-napthol when a brilliant yellow, orange or red coloured dye is obtained.
Aliphatic 1° amines under these conditions give a brisk evolution of N
2
gas with the formation of 1° alcohol i.e., solution remains clear.
(ii) Distinction between Aniline and Benzylamine :
By Nitrous acid test : Benzylamine reacts with HNO
2
to form a diazonium salt which being unstable even at low temperature, decomposes with evolution of N
2
gas
Aniline on the other hand, reacts with HNO
2
, to form benzenediazonium chloride which is stable at 273-278 K and hence does not decompose to evolve N
2
gas.
Question 18.
Write the names and structures of monomers used for getting the following polymers :
(i) Buna-S
(ii) Nylon-6, 6
Answer:
(i) Buna-S :
(ii) Nylon-6, 6 : It has two repeating monomers
Question 19.
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm
-3
. Use this information to calculate Avogadro’s number. [3]
(Gram atomic mass of Fe = 55.84 g mol
-1
).
Answer:
Given :
a = 286.65 pm = 286.65 × 10
-10
, d = 7.87 g cm
-3
, M = 56 g mol
-1
z = 2, N
A
= ?
Using formula : d = \(\frac{zM}{a^3 \mathrm{~N}_{A}}\) or N
A
= \(\\frac{z \mathrm{M}}{a^3 d}\)
Question 20.
The resistance of 0.01 M NaCl Solution at 25° C is 200 Ω. The cell constant of the conductivity cell used is unity. Calculate the molar, conductivity of the solution.
Solution:
For 0.01 M NaCl solution,
R = 200 Ω, cell constant is unity.
Conductivity (K) = \(\frac{\text { Cell constant }}{\text { Resistance }}\) = \(\frac{1}{200}\) = 0.005 Sm
-1
Concentration of solution = 0.01 M = 0.01 mol L
-1
= 0.01 × 10
3
mol m
-3
= 10 mol m
-3
Molar conductivity = \(\frac{K}{C_m}\) = \(\frac{0.005}{10}\) = 5 × 10 -4 Sm 2 mol -1
Question 21.
For a decomposition reaction, the values of k at two different temperatures are given below:
k
1
= 2.15 × 10
-8
L/(mol.s) at 650 K
k
2
= 2.39 × 10
-7
L/(mol.s) at 700 K
Calculate the value of Ea for the reaction.
(Log 11.11 = 1.046) (R = 8.314 J K
-1
mol
-1
)
Solution:
Given :
k
1
= 2.15 × 10
-8
L mol
-1
S
-1
, T
1
= 650 K
k
2
= 2.39 × 10
-7
L mol
-1
S
-1
, T
2
= 700 K
R = 8.314 J K
-1
mol
-1
, E
a
= ?
Question 22.
Giving appropriate examples, explain how the two types of processes of adsorption (physisorption and chemisorption) are influenced by the prevailing temperature, the surface area of adsorbent and the activation energy of the process?
Answer:
| Physisorption | Chemisorption |
| 1. Temperature: It decreases with increase in temperature. | 1. It increases with increase in temperature initially after adsorption it is decreasing |
| 2. Surface area of adsorbent: The adsorption increases with increase in surface area of adsorbent. | 2. It also increases with increase in surface area of adsorbent. |
| 3. Activation energy: It requires very low or no activation energy. | 3. It requires high activation energy. |
Or
Explain clearly how the phenomenon of adsorption finds application in (i) production of vacuum in a vessel
(ii) heterogeneous catalysis
(iii) froth floatation process in metallurgy.
Answer:
(i) Production of vacuum : In Dewar flasks, activated charcoal is placed between the walls of the flask so that any gas which enters into annular space either due to glass imperfection or diffusion through glass is adsorbed and create a vacuum.
(ii) Heterogeneous catalysis : If the catalyst is present in a different phase than that of the reactants, it is called a heterogeneous catalyst and this type of catalysis is called Heterogeneous catalysis.
Example: Manufacture of NH
3
from N, and H
2
by Haber’s process using iron as catalyst
Reactants are gaseous whereas catalyst is solid.
(iii) Froth Floatation process : This method is used for removing gangue from suiphide ores.
In this powdered ore is mixed with collectors (e.g. pine oils, fatty acids etc.) and froth stabilisers (e.g. cresols, aniline) which enhance non-wettability of the mineral particles and froth stabilisation respectively. As a result of which ore comes with froth and gangue remain in the solution.
Question 23.
Give reasons for the following :
(i) Transition metals exhibit a wide range of oxidation states.
(ii) Cobalt(II) is very stable in aqueous solutions but gets easily oxidised in the presence of strong ligands.
(iii) Actinoids exhibit a greater range of oxidation states than lanthanoids.
Answer:
(i) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals as ns, and (n – 1) d electrons have approximate equal energies.
(ii) Co
2+
ion is easily oxidised to Co
3+
ion in presence of a strong ligand because of its higher crystal field energy which causes pairing of electrons to give inner orbital complexes (d
2
sp
3
).
(iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states.
Question 24.
Write the IUPAC name and draw the structure of each of the following complex entities :
(i) \(\left[\mathrm{Co}\left(\begin{array}{c}
\mathrm{COO} \\
| \\
\mathrm{COO}
\end{array}\right)_3\right]^{3-}\)
(ii) [Cr(CO)
6
]
(iii) [PtCl
3
(C
2
H
4
)]
(At. nos. Cr = 25, Co = 27, Pt = 78) [3]
Answer:
(i) IUPAC name : Trioxalato chromate (III) ion
(ii) IUPAC name : Hexa carbonyl chromium (o)
(iii) IUPAC name : Trichlorido ethylene platinum (II)
Question 25.
Explain the following with an example for each :
(i) Kolbe’s reaction
(ii) Reimer-Tiemann reaction
(iii) Williamson ether synthesis [3]
Solution:
(i) Kolbe’s reaction : Phenol reacts with CO
2
in presence of sodium hydroxide (NaOH) at 4-7 Atm and 390 – 410 K giving salicytic acid
(ii) Reimer-Tiemann reaction :
(iii) Williamson synthesis of an ether : AlkvI Halide reacts with Alkoxide
Question 26.
What are essential and non-essential amino acids? Give two examples of each. 3
Answer:
Essential amino acids : Amino acids which the body cannot synthesize are called essential amino acids.
Example : Valine, leucine etc. Therefore they must be supplied in diet.
Non-essential amino acids : Amino acids which the body can synthesize are called non¬essential amino acids. Therefore, they may or may not be present in diet.
Example : Glycine, alanine etc.
Question 27.
What is meant by the following terms? Explain with an example for each.
(i) Target molecules as used in medicinal chemistry
(ii) Food preservatives
(iii) Non-ionic detergents [3]
Answer:
(i) Drugs interact with macromolecules like proteins, carbohydrates, lipids etc. and are called as target molecules.
(ii) Food preservatives : They are used to prevent spoilage of food due to microbial growth. Example ; Table salt, vegetable oils, sodium benzoate etc.
(iii) Non-ionic detergents : These are esters of high molecular mass alcohols obtained by reaction between polyethylene glycol and steric acid.
Example :
Question 28.
(a) What is van’t Hoff factor? What types of values can it have if in forming the solution, the solute molecules undergo
(i) Dissociation?
(ii) Association?
(b) How many mL of a 0.1 M HCl solution are required to react completely with 1 g of a mixture of Na
2
CO
3
and NaHCO
3
containing equimolar amounts of both?
(Molar mass : Na
2
CO
3
= 106 g, NaHCO
3
= 84 g) [3]
Answer:
(i) Van’t Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property
i = \(\frac{\text { Experimental value }}{\text { Calculated value }}\)
If there is dissociation of the solute in the solution, the Van’t Hoff factor T will be greater than one i.e. i > 1.
It means observed colligative property will be greater than calculated value.
(ii) Association : If there is association of solute in the solution, the Van’t Hoff factor T’ will be less than one i.e. i < 1. Thus, observed colligative property will be less than the calculated value.
(b) Calculation of no. of moles of the components in the mixture
Suppose Na
2
CO
3
in the mixture = x g
∴ NaHCO
3
in the mixture = (1 – x) g
Molar mass of Na
2
CO
3
= 106 g mol
-1
Molar mass of NaHCO
3
= 84 g mol
-1
∴ Moles of Na
2
CO
3
= \(\frac{x}{106}\)
and Moles of NaHCO
3
= \(\frac{1-x}{84}\)
As the mixture contains equimolar amounts of both
No. of moles of HCl required
Na
2
CO
3
+ 2HCl → 2NaCl + H
2
O + CO
2
NaHCO
3
+ HCl → NaCl + H
2
0 + CO
2
1 mole of Na
2
CO
3
requires HCl = 2 moles
∴ 0.00526 mole of Na
2
C0
3
requires HCl = 0.00526 × 2
= 0.01052 mole
Similarly, 0.00526 mole of NaHC0
3
requires HCl = 0.00526
∴ Total HC1 required = 0.01052 + 0.00526 = 0.01578 moles
Thus 0.1 mole of HCl is present in 1000 mL of HCl
∴ 0.01578 mole of HCl present in 1000 mL = \(\frac{1}{2}\) × 0.01578
= 157.8 mL
Or
(a) Define
(i) Mole fraction (ii) Molality (iii) Raoult’s law
(b) Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00 g of Glauber’s salt, Na
2
SO
-1.10H
2
O in 0.100 kg of water. (K
f
for water = 1.86 K kg mol
2
, Atomic masses : Na = 23, S = 32, O = 16, H = 1) [3,2]
Answer:
(i) Mole fraction : Mole fraction of a constituent is the fraction obtained by dividing number of moles of that constituent by the total number of moles of all the constituents present in the solution. It is denoted by ‘x’.
Example: x
1
= \(\frac{\text { No. of moles of } x_1}{\text { Total no. of moles }}\) = \(\frac{n x_1}{n x_1+n x_2}\)
(ii) Molality: Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent. ¡t is denoted by ‘m’.
zi’ 1000
m = \(\frac{w×1000}{M×W}\)
where r = ‘eight of solute in grams
M = Molecular mass of solute
W = Weight of solvent in grams
(iii) Raoult’s law: For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
P = P°x
Non-ideal solution shows positive and negative deviations from Raoult’s law.
→ is greater than the corresponding ideal solution of same composition. Such behaviour is called positive deviation.
Example : Mixtures of ethanol + cyclohexane
Mixture of acetone + carbon disulphide
→ Negative deviation from Raoult’s law : When the total vapour pressure will be less than corresponding vapour pressure, then it is termed as negative deviation.
Example : Chloroform + Acetone
Chloroform + Diethylether
(b) Since Na
2
SO
4
.10H
2
O is an ionic compound, so undergoes complete dissociation.
Question 29.
(a) Write the formula and describe the structure of a noble gas species which is isostructural with
(i) IBr
–
2
(ii) BrO
–
3
(b) Assign reasons for the following :
(i) SF
6
is kinetically inert.
(ii) NF
3
is an exothermic compound whereas NCl
3
is not.
(iii) HCl is a stronger acid than HF though flourine is more electronegative than chlorine. 2,3
Answer:
(a) (i) IBr
–
2
: It has 2 bond pairs and 3 lone pairs. Therefore according to VSEPR theory it should have Linear structure.
IBr
–
2
has (7 + 2 × 7 + 1) i.e. 22 valence electrons.
No noble gas has 22 electrons
∴ The isoelectronic species for IBr
–
2
is XeF
2
(ii) IBr
–
3
: It has 3 bond pairs and 1 lone pair of electrons. Therefore according to VSEPR theory it has pyramidal structure.
It has 26 valence electrons i.e. (7+ 3 × 6 + 1 = 26).
A noble gas having 26 valence electrons is XeO
3
(8 + 3 × 6 = 26). Thus XeO
3
also has pyramidal structure.
(b) (i) Because SF
6
is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom.
(ii) Due to smaller size of F as compared to Cl, the N – F bond is much stronger than N-Cl bond while bond dissociation energy of F
2
is much lower than that of Cl
2
. Therefore, energy released during the formationlof NF
3
molecule is more than the energy needed to break N
2
and F
2
molecules into individual atoms. In other words, formation of NF
3
is an exothermic reaction.
The energy released during the formation of NCl
3
molecule is less than the energy needed to break N
2
and Cl
2
molecule into individual atoms. Thus formation of NCl is an endothermic reaction.
(iii) Because HCl has large size and less bond strength than HF which makes easier liberation of H + .
Or
+(a) How is ammonia prepared on a large scale? Name the process and mention the optimum conditions for the production of ammonia by this process.
(b) Assign reasons for the following :
(i) H
2
S is more acidic than H
2
O.
t(ii) NH
3
is more basic than PH
3
.
(iii) Sulphur has a greater tendency for catenation than oxygen. [2,3]
Answer:
(a) According to Le Chatelier’s principle the favourable conditions for the production of NH
3
by Haber’s process are
- pressure of 200 atmosphere
- temperature of ~700 K
-
use of catalyst like Fe
2
O
3
with small amount of K
2
O and AL
2
O
3
N 2 + 3H 2 → 2NH3
(b) (i) H
2
S is more acidic than H
2
O because bond dissociation enthalpy of H – S bond in
H
2
S is less than that of H – O bond in H
2
O.
(ii) Since both P and N contains lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH
3
, the electron density is much higher than PH
3
, therefore it can easily donate electrons and acts as strong Lewis base than PH
3
.
(iii) The greater catenation tendency of sulphur is due to two reasons :
- The lone pair of electrons feels more repulsion in O – O bond than S – S bond due to its small size and thus S – S forms strong bond.
- As the size of atom increases down the group from O to P, the strength of bond increases and therefore catenation tendency also increases.
Question 30.
(a) Write the IUPAC names of the following compounds :
(i) CH
3
CO(CH
2
)
4
CH
3
(ii) Ph – CH = CH – CHO
(b) Describe the following conversions in not more than two steps :
(i) Ethanol to 3-Hydroxybutanal
(ii) Benzoic acid to m-Nitrobenzyl alcohol
(iii) Propanone to Propene [2,3]
Answer:
(a) (i) CH
3
CO(CH
2
)
4
CH
3
:
IUPAC name : Heptan-2-one
(ii) Ph – CH = CH – CHO :
IUPAC name : 3-phenylprop-2-en-1-al
(b) (i) Ethanol to 3-hydroxybutanal :
(ii) Benzoic acid to m-Nitrobenzyl alcohol
(iii) Propanone to Propene
Or
(a) Draw the structures of the following compounds :
(i) 4-Chloropentan-2-one
(ii) p-Nitropropiophenone
(b) Give tests to distinguish between the following pairs of compounds
(i) Ethanal and Fropanal
(ii) Phenol and Benzoic acid
(iii) Benzaldehyde and Acetophenone
Answer:
(a) (i) 4-Chloropentane-2-one
(ii) p-Nitropropiophenone
(b) (i) Ethanal and Propanal: Ethanal and propanal can be distinguished by iodoform test. Warm each compound with iodine and sodium hydroxide solution in water. Ethanal gives yellow crystal of iodoform while propanal does not respond to iodoform test.
(ii) Phenol and Benzoic acid: On addition of NaHCO3 to both solutions carbondioxide gas is evolved with benzcic acid while phenol does not form CO
2
.
(iii) Benzaldehyde and Acetophenone :
By Iodoform test: Acetophenone being a methyl ketone on treatment with I
2
/NaOH (NaOI) undergoes iodoform test to give yellow ppt. of iodoform but benzaldehyde does not.
SET II
Note : Except for the following questions, all the remaining questions have been asked in Set-I.
Question 1.
What is meant by ‘antiferromagnetism’? [1]
Answer:
Antiferromagnetism : These substances possess zero Viet magnetic moment because of presence of equal number of electrons with opposite spins.
Question 2.
Define dialysis. [1]
Answer:
Dialysis : The process of separating the particles of colloids from those of crystalloids by diffusion of the mixture through a parchment or an animal membrane is known as dialysis.
Question 3.
What is the role of CO
2
in the extractive metallurgy of aluminium from its ore? [1]
Answer:
The solution of sodium meta-aluminate is neutralized by passing CO
2
when hydrated alumina separates out while sodium silicate remains in the solution.
Question 4.
Why is nitrogen gas very unreactive? [1]
Answer:
Nitrogen gas is not particularly reactive due to presence of triple bond between two N-atoms. The bond length is very small and hence the bond dissociation energy is very large. As a result of which nitrogen gas is unreactive.
Question 9.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with change in temperature. [2]
Answer:
Conductivity : Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. It is represented by K.
Its unit is S cm
-1
Molar conductivity : Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm
3
of the solution when the electrodes are 1 cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by Λ
m
.
Its unit is S cm
2
mol
-1
Conductivity and molar conductivity of electrolytes increase with increasing temperature.
Question 10.
Define each of the following :
(i) Specific rate of a reaction
(ii) Energy of activation of a reaction [2]
Answer:
(i) Specific rate of a reaction: Specific rate of reaction is the rate of reaction when the molar concentration of each of the reactants is unity.
(ii) Activation energy of a reaction : The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value, is called activation energy.
Question 12.
Complete the following chemical reaction equations :
Question 14.
Write the structures of the following organic halogen compounds :
(i) 4-tert-Butyl-3-iodoheptane
(ii) 4-Bromo-3-methylpent-2-ene [2]
Solution:
(i) 4-tert-Butyl-3-iodoheptane
(ii) 4-Bromo-3-methylpent-2-ene
Question 23.
Assign reasons for the following : [3]
(i) Cu(I) ion is not known to exist in aqueous solutions.
(ii) Transition metals are much harder than the alkali metals.
(iii) From element to element actinoid contraction is greater than the lanthanoid contraction. .
Answer:
(i) Cu
2+
(aq) is much more stable than Cu
+
(aq). This is because although second ionization enthalpy of copper is large but AhydH for Cu
2+
(aq) is much more negative than that for Cu
+
(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows :
2Cu
+
→ Cu
2+
+ Cu
(ii) In transitional elements, in addition to metallic bonding there is extra covalent bonding due to presence of unpaired electrons in their ‘d’ orbitals, hence they are much harder.
(iii) The actinoid contraction is greater than lanthanoid contraction due to poorer shielding of 5f electrons as they are extended in space beyond 6s and 6p orbitals whereas 4f orbitals are burried deep inside the atom.
Question 24.
Giving one example in each of the following caSes, discuss briefly the role of coordination compounds in
(i) extraction metallurgy of metals
(ii) analytical chemistry [3]
Answer:
(i) Extraction metallurgy of metals : Cold and silver are extracted from their ores through formation of cyanide complexes [ Ag(CN)
2
]
–
and [(Au(CN)
2
]
–
respectively.
Example : Ag
2
S + 4NaCN ⇔ 2Na[Ag(CN)
2
l + Na
2
S
(ii) In analytical chemistry, they are used in qualitative analysis in which basic radicals are determined by converting them into suitable complexes with specific colour.
Example : Ni
2+
is determined by DMG (Dimethyl Glyoxime) in the presence of NH
4
OH and forms a red ppt. of Ni DMG complex.
Similarly cobalt, Fe, Zn are also determined by converting them into complexes.
Question 27.
Answer the following questions :
(i) Why should medicines not be taken without consulting a doctor?
(ii) What is meant by ‘broad spectrum antibiotics’?
(iii) What are the main constituents of Dettol? [3]
Answer:
(i) Because medicines can cause harm to human body if a person does not know its physiological function on body.
(ii) Antibiotics which kill or inhibit a wide range of gram positive and gram negative bacteria, are called broad spectrum antibiotics.
(iii) Dettol is mixture of chloroxylenol and a-terpineol in a suitable solvent.
SET III
Note : Except for the following questions, all the remaining questions have been asked in Set-I and Set-11.
Question 1.
Write a distinguishing feature of a metallic solid compared to an ionic solid. [1]
Answer:
Metallic solid conducts electricity in solid state but ionic solids do so only in molten state or in solution. Metals conduct electricity through electrons while ionic substances conduct through ions. Metallic solids are malleable and ductile while ionic solids are hard and brittle.
Question 2.
What are enzymes? [1]
Answer:
Enzymes are protein molecules which act as catalyst in biochemical reaction.
Question 3.
Name the chief ores of aluminium and zinc. [1]
Answer:
Aluminium ore → Bauxite
Zinc ore → Zinc blende
Question 4.
Draw the structure of PCl
5
(s) molecule.
Solution:
PCl
5
(s)
Question 10.
A reaction is of second order with respect to its reactant. How will its reaction rate be affected if the concentration of the reactant is (i) doubled (ii) reduced to half? [2]
Solution:
Since Rate = K[A]
2
For second order reaction
Let [A] = a then Rate = Ka
2
(i) If [A] = 2a then Rate = K (2a)
2
= 4 Ka
2
∴ Rate of reaction becomes 4 times
(i) If [A] = \(\frac{a}{2}\) then Rate = K (\(\frac{a}{2}\))
2
= 4 \(\frac{Ka^2}{4}\)
∴ Rate of reaction will be \(\frac{1^th}{4}\)
Question 12.
Complete the following chemical equations : [2]
(i) P
4
+ SOCl
2
→
(ii) F
2
(Excess) + Cl
2
\(\rightarrow{300^{\circ} \mathrm{C}}\)
Solution:
(i) P
4
+ 8SOCl
2
→ 4PCl3 + 4SO
2
+ 2S
2
Cl
2
(ii) 3F
2
+ Cl
2
\(\rightarrow{300^{\circ} \mathrm{C}}\) 2ClFsub>3 (excess)
Question 13.
Assign reasons for the following :
(i) Transition metals and many of their compounds act as good catalysts.
(ii) Transition metals generally form coloured compounds. [2]
Answer:
(i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states.
(ii) Because presence of unpaired d electrons, which undergoes d-d transition by absorption of energy from visible region and then the emitted light shows complementary colours.
Question 14.
Write the structures of the following organic halogen compounds :
(i) p-Bromochlorobenzene
(ii) 1-Chloro-4-ethylcyclohexane [2]
Answer:
(i) P-Bromochlorobenzene
(ii) 1-chloro-4-ethylcyclohexane
Question 24.
Write down the IUPAC names of the following complexes and also give stereochemistry and magnetic moment of the complexes :
(i) [Co(NH
3
)
5
Cl]Cl
2
(ii) [CrCl
3
(py)
3
]
(iii) K
4
[Mn(CN)
6
]
(At. nos. Cr = 24, Mn = 25, Co = 27, py = pyridine)
Solution:
(i) [CO(NH
3
)
5
Cl]Cl
2
IUPAC name : Pentaammine chlorido cobalt (III) chloride
C.N. of Co = 6
Shape = octahedral
o.s. of Co : x + 0 – 1 = +2
∴ x = +3
Magnetic moment (μ) = \(\sqrt{0(0 + 2)}\) = 0 BM
(ii) [CrCl
3
(py)
3
]
IUPAC name : Trichloridotripyridine chromium (III)
C.N. of Cr = 6
Shape = octahedral
o.s. of Cr : x – 3 + 0 = 0
∴ x = +3
n = 3,
∴ μ = \(\sqrt{3 (3 + 2)}\) = \(\sqrt{15}\) = 3.87 BM
(iii) K
4
[Mn(CN)
6
]
IUPAC name : Potassium hexacyano manganate (II)
C.N. of Cr = 6
Shape = octahedral
o.s. of Cr : x – 6 = 4
∴ x = +2
n = 1
∴ μ = \(\sqrt{1 (1 + 2)}\) = \(\sqrt{3}\) = 1.73 BM
Question 25.
How are the following conversions carried out?
(i) Propene → Propan-2-ol
(ii) Ethylmagnesium chloride → Propan-1-ol
(iii) Benzyl chloride → Benzyl alcohol [3]
Answer:
(i) Propene to propan-2-ol
(ii) Ethylmagnesium chloride to propan-1-ol
(iii) Benzyl chloride to benzyl alcohol
Question 27.
Answer the following :
(i) Why is the use of aspartame limited to cold foods and drinks?
(ii) How do antiseptics differ from disinfectants?
(iii) Why do soaps not work in hard water?
Answer:
(i) Use of aspartame is limited to cold foods and soft drinks because it is unstable at cooking temperature.
(ii)
| Antiseptic | Disinfectants |
| 1. They are chemical substances which prevent the growth of micro-organisms and may even kill them. | 1. They are chemical substances which kill micro-organisms. |
| 2. They are safe to be applied to the living tissues. | 2. They are not safe to be applied to the living tissues. |
|
3. They are generally applied on wounds, cuts, ulcers and diseased skin surfaces.
Example : Furacin, soframycin, dettol and savlon. |
3. They are used to kill micro-organisms present in the drains, toilets, floors etc.
Example : Phenol (> 1% solution) and chlorine (0.2 to 0.4 ppm). |
(iii) Hard water contains insoluble calcium and magnesium chlorides which forms insoluble ppt (scum) with soap and thus cannot be rinsed off easily.