Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2017 Comptt (Outside Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2017 Comptt (Outside Delhi) with Solutions
Time allowed: 3 hours
Maximum marks : 70
General Instructions
- All questions are compulsory.
- Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
- Questions number 6 to 10 are short-answer questions and carry 2 marks each.
- Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
- Question number 23 is a value based question and carries 4 marks.[*]
- Questions number 24 to 26 are long-answer questions and carry 5 marks each.
-
Lise log tables, if necessary, Use of calculators is not allowed.
† Deleted from Syllabus.
SET I
Question 1.
What type of stoichiometric defect is shown by ZnS? [1]
Answer:
ZnS shows Frenkel defect.
Question 2.
What are emulsions? Give an example. [1]
Answer:
An emulsion is a colloidal dispersion in which both the dispersed phase and dispersion medium are liquids. For example, milk, cream.
Question 3.
Write IUPAC name of the complex : [CoCl
2
(en)
2
]
+
. [1]
Answer:
[CoCl
2
(en)
2
]
+
: Dichloridobis(ethylenediamine)Cobalt(III)ion.
Question 4.
What happens when phenol is oxidized by Na
2
Cr
2
O
7
/H
2
SO
2
? [1]
Answer:
Phenol forms benzoquinone on oxidation with Na
2
Cr
2
O
7
/H
2
SO
4
.
Question 5.
Write IUPAC name of the following compound : [1]
Answer:
IUPAC name : N, N-Dimethylbutanamine.
Question 6.
Following reactions can occur at cathode during the electrolysis of aqueous silver nitrate solution using Pt electrodes : [2]
On the basis of their standard electrode potential values, which reaction is feasible are cathode and why?
Answer:
As the standard electrode potential of silver is greater than that of the other hydrogen electrode, so reduction of silver takes place and reaction (i) will be feasible
i.e., Ag
+
aq
+ e
–
→ Ag
(s)
Question 7.
“Orthophosphoric acid (H
3
PO
4
) is not a reducing agent whereas hypophosphorus acid (H
3
PO
2
) is a strong reducing agent.” Explain and justify the above statement with the help of a suitable example. [2]
Answer:
Orthophosphoric acid (H
3
PO
4
) is not a reducing agent because it doesn’t contain any P-H bond whereas hypophosphorus acid (H
3
PO
2
) is a strong reducing agent as it contains two P-H bonds. H
3
PO
2
can reduce silver nitrate (AgNO
3
) into metallic silver which H
3
PO
4
cannot.
4AgNO
3
+ H
3
PO
2
+ 2H
2
O → 4Ag ↓ + H
3
PO
4
+ 4HNO
3
Question 8.
(a) Explain why H
2
and O
2
do not react at room temperature.
(b) Write the rate equation for the reaction A
2
+ 3B
2
→ 2C, if the overall order of the reaction is zero. [2]
Answer:
(a) H
2
and O
2
do not react at room temperature because they do not have enough activation energy to overcome the exceptionally high activation energy barrier.
(b) A
2
+ 3B
2
→ 2C
Rate = \(\left(\frac{d x}{d t}\right)\) = K[A]
0
[B]
0
= K (rate constant)
Or
Derive integrated rate equation for rate constant of a first order reaction.
Answer:
In a first order reaction, the rate of reaction, is directly proportional to the concentration of the reactant.
Let us consider the reaction,
A → Products
The instantaneous reaction rate can be expressed as :
\(\frac{-d[\mathrm{~A}]}{d t}\) = K[A] [K = rate constant)
\(\frac{-d[\mathrm{~A}]}{[\mathrm{A}]}\) = Kdt ….. (i)
On integrating equation (i)
\(-\int \frac{d[\mathrm{~A}]}{[\mathrm{A}]}\) = K∫dt
or —ln[A] = Kt + I ……. (ii)
where, I = Integration constant
If t = 0 and [A] = [A]
0
, where [A]
0
is the initial concentration of the reactant.
Then equation (ii) becomes
—ln[A]
0
= I …….. (iii)
Substitute the value of I in equation (ii)
-ln[A] = Kt – ln[A]
ln[A]
0
– ln[A] = Kt
ln\(\frac{[\mathrm{A}]_0}{[\mathrm{~A}]}\) = Kt
Or K = \(\frac{1}{t} \ln \frac{[\mathrm{A}]_0}{[\mathrm{~A}]}\)
K = \(\frac{2.303}{t} \log \frac{[\mathrm{A}]_0}{[\mathrm{~A}]}\)
This is called integrated rate equation for the first order reaction.
Question 9.
Explain the following observations: [2]
(i) Copper atom has completely filled d orbitais (3d
10
) in its ground state, yet it is
regarded as a transition element.
(ii) Cr
2+
is a stronger reducing agent than Fe
2+
in aqueous solutions.
Answer:
(i) Copper atom has completely filled d orbitals (3d
10
) in its ground state, yet it is regarded as a transition element due to incompletely filled d-orbital in its ionic states i.e. Cu (3dì.
(ii) The highest oxidation state for Cr is +6, therefore it can loose 3 more electrons, whereas
Fe needs to loose only 1 eletron to achieve its highest oxidation state of +3. Thus, Cr
3+
is more reducing than Fe
2+
.
Question 10.
How will you carry out the following conversions: [2]
(i) 2-Bromopropane to 1-bromopropane
(ii) Benzene to p-chloronitrobenzene
Answer:
\(\dagger\) Question 11.
An element exists in bcc lattice with a cell edge of 288 pm. Calculate its molar mass if its density is 7.2 g/cm
3
. [3]
Answer:
Given : Cell edge, a = 288 pm = 288 × 10
-10
cm
Density, d = 7.2 g/cm
3
For bcc formula units per cell Z = 2, M = ?
Using formula and substituting values,
M = \(\frac{d \times \mathrm{N}_{\mathrm{A}} \times a^3}{\mathrm{Z}}\)
= \(\frac{7.2 \mathrm{~g} / \mathrm{cm}^3 \times\left(6.022 \times 10^{23}\right) \times\left(288 \times 10^{-10} \mathrm{~cm}\right)^3}{2}\)
= \(\frac{7.2 \mathrm{~g} / \mathrm{cm}^3 \times 6.022 \times 10^{23} \times 2.39 \times 10^{-23} \mathrm{~cm}^3}{2}\) = \(\frac{103.626}{2}\)
∴ M = 51.8 g mol
-1
Question 12.
Calculate ∆
r
G° and log K
c
for the following reaction at 29 K.
2C
(s)
+ \(3 \mathrm{Cd}_{(\mathrm{aq})}^{2+}\) → \(2 \mathrm{Cr}_{\text {(aq) }}^{3+}\) + 3Cd
(s)
[Given: \(\mathrm{E}_{\text {cell }}^0\) = +0.34 V, IF = 96500 C mol
-1
]
Answer:
Question 13.
For a first order reaction, show that time required for 99% completion is twice the time required for completion of 90% reaction. [3]
Answer:
Question 14.
Explain the following phenomenon giving reasons: [3]
(i) Tyndall effect
(ii) Brownian movement
(iii) Physical adsorption decreases with increase in temperature.
Answer:
(i) Tyndall effect: When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called Tyndall cone and the phenomenon is called Tyndall effect. The phenomenon is due to scattering of light by the colloidal particles.
(ii) Brownian movement is the continuous and random zig-zag motion of the colloidal
= \(\frac{7.2 \mathrm{~g} / \mathrm{cm}^3 \times 6.022 \times 10^{23} \times 2.39 \times 10^{-23} \mathrm{~cm}^3}{2}\) = \(\frac{103.626}{2}\)
∴ M = 51.8 g mol
-1
Question 12.
Calculate ∆
r
G° and log K
c
for the following reaction at 298 K.
[Given : \(\mathrm{E}_{\text {cell }}^0\) = +0.34, IF = 9650 C mol
-1
]
Answer:
Question 13.
For a first order reaction, show that time required for 99% completion is twice the time required for completion of 90% reaction. [3]
Answer:
Question 14.
Explain the following phenomenon giving reasons: [3]
(i) Tyndall effect
(ii) Brownian movement
(iii) Physical adsorption decreases with increase in temperature.
Answer:
(i) Tyndall effect: When a beam of light is passed through a colloidal solution and viewed perpendicular to the path of the incident light, the path of light becomes visible as a bright streak. The illuminated path is called Tyndall cone and the phenomenon is called Tyndall effect. The phenomenon is due to scattering of light by the colloidal particles.
(ii) Brownian movement is the continuous and random zig-zag motion of the colloidal particles. This motion is due to kinetic motion of striking colloidal particles which produces a resultant force to cause motion.
(iii) Physical adsorption decreases with increase in temperature due to Le-Chatelier’s principle which shifts the equilibrium in forward direction to complete the reaction. As the adsorption is an exothermic process, it decreass with increase in temperature.
Question 15.
(a) Write the principle involved in the following:
(i) Zone refining of metals
(ii) Electrolytic refining
(b) Name the metal refined by each of the following processes:
(i) Mond process
(ii) van Arkel Method [3]
Answer:
(a) (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) 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 anode: (oxidation)
Cu
2+
+ 2e
–
→ Cu
Question 16.
A mixed oxide of iron and chromium is fused with sodium carbonate in free access of air to form a yellow coloured compound (A). On acidification the compound (A) forms an orange coloured compound (B), which is a strong oxidizing agenL. Identify compound (A) and (B). Write chemical reactions involved. [3]
Answer:
The mixed oxide of iron and chromium is chromite or chrome ion i.e. FeO.Cr
2
O
3
Or
(a) Give reasons for the following:
(i) Compounds of transition elements are generally coloured.
(ii) MnO is basic while Mn
2
O
7
is acidic.
(b) Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is 26.
Answer:
(a) (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. The colour of transition metal ions is due to d-d transition. When electrons jump from one orbital to another in their partially filled d-orbitals, another light is emitted due to which the compounds of transition elements seem to be coloured.
(ii) MnO is basic while Mn
2
O
7
is acidic because the basic nature decreases as the oxidation state or number of oxygen atoms increases i.e. MnO (+4) and Mn
2
O
7
(+7)
(b) Divalent ion with atomic number 26 is Fe
2+
Fe
2+
= [Ar] 3d
6
µ = \(\sqrt{n(n+2)}\) BM. ⇒ µ = \(\sqrt{4(4+2)}\) B.M. ⇒ µ = 4.90B.M.
Question 17.
For the complex ion [Fe(en)
2
Cl
2
]
+
write the hybridizations type and magnetic behaviour. Draw one of the geometrical isomer of the complex ion which is optically active. [Atomic No. : Fe = 26] [3]
Answer:
In the complex of
26
Fe
3+
= 3d
5
4s
0
4p
0
- Due to presence of one unpaired electron in d orbital the complex is paramagnetic.
-
In the complex of [Fe(en)
2
Cl
2
]Cl, only cis-isomer shows optical isomerism.t
Question 18.
(a) Account for the following:
(i) Electrophilic substitution reactions in haloarenes occur slowly.
(ii) Haloalkanes, though polar, are insoluble in water.
(b) Arrange the following compounds in increasing order of reactivity towards S
N
2 displacement:
2-Bromo-2-Methylbutane, 1-Bromopentane, 2-Bromopentane [3]
Answer:
(a) (i) Due to —I effect of halogen atom, it withdraws electrons from the henzene ring and thus ring gets deactivated.
(ii) They fail to form hydrogen bonds with water. More energy is required to break hydrogen bonds in water and less energy is released when new attractions are set up.
(b) 2-Bromo-2-Methylbutane < 2-Bromopentane < 1-Bromopentane
Question 19.
(a) Why phenol is more acidic than ethanol?
(b) Write the mechanism of acid dehydration of ethanol to yield ether:
[3]
Answer:
(a) Phenol on losing H+ ion forms phenoxide ion, and ethanol on losing H+ ion forms ethoxide ion. Phenoxide ion is more stable than ethoxide ion as phenoxide ion exists in resonance structure. Due to this phenol is more acidic than ethanol.
(b)
Question 20.
Identify A, B and C in the following reactions:
Answer:
Question 21.
(a) Why water soluble vitamins must be supplied gu1arly in the diet? Give one example of it.
(b) Differentiate between the following:
(i) Essential and non-essential amino acids.
(ii) Fibrous and globular proteins. [3]
Answer:
(a) Water soluble vitamins must be supplied regularly in the diet because they are regularly excreted in urine and cannot be stored in our body. For eg., Vitamin C, Vitamin B, etc.
(b) (i) Essential amino acids: Amino acids which the body cannot synthesize are called essential amino acids.
Examples : Valine, leucine etc. Therefore they must be supplied in diet.
Non-essential diet: Amino acids which the body can synthesize are called non-essential amino acids. Therefore, they may or may not be present in diet.
Examples : Glycine, alanine etc.
(ii)
| Globular Proteins | Fibrous Proteins |
| (i) Globular proteins have almost spheroidal shape due to folding of the polypeptide chain. | Polypeptide chains of fibrous proteins consist of thread like molecules which tend to lie side by side to form fibres. |
| (ii) Globular proteins are soluble in water. | Fibrous proteins are insoluble in water. |
| (iii) Globular proteins are sensitive to small changes of temperature and pH. Therefore they undergo denaturation on heating or on treatment with acids/bases. | Fibrous proteins are stable to moderate changes of temperature and pH. |
|
(iv) They possess biological activity that’s why they act as enzymes.
Example: Maltase, invertase etc. harmones (insulin) antibodies, transport agents (haemoglobin), etc. |
They do not have any biological activity but serve as chief structural material of animal tissues.
Example: Keratin in skin, hair, nails and wool, collagen in tendons, fibroin in silk etc. |
Question 22.
(i) Name a substance which can be used as an antiseptic as well as disinfectant.
(ii) Name an artificial sweetener whose use is limited to cold foods and drinks.
(iii) What are cationic detergents? 3
Answer:
(i) Phenol
(ii) Aspartame
(iii) Cationic detergents : They are quatemary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.
Example : Cetyltrimethyl ammonium bromide
Question 23.
Once there was a heavy downpour for about 3 hours in the early morning. Irfan and his family were finding it difficult to cariy out their daily morning chores as the sewer water was flowing back into the toilets, the road in front of their house was flooded with water and they could not move out. On this very serious problem, Irfan called a meeting of all the residents. In the meeting Irfan discussed the problem and said that we are using too much polythene bags and other plastic items which we throw here and there. All these move into the drains and sewer lines which get choked and do not allow flow of water. As these are non-biodegradable, they remain as such for a long time. So to overcome this problem, we should use bags made up of cloth or jute which are biodegradable.
Answer the following questions: [4]
(i) Name the polymer which is biodegradable. Write the structures of monomers and the repeating unit.
Answer:
(ii) Write two uses of this polymer.
*(iii) Write any two values shown by Irfan.
Answer:
(i) Polymer: Poly-β-hydroxybutyrate-co-β-hydroxyvalerate (PHBV)
Monomers of this are:
(ii) PHBV is used in packaging, orthopaedic devices and in controlled drug disease.
*(iii) As per latest CBSE curriculum, Value Based Questions will not be asked in the examination.
Question 24.
(a) Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases.
(b) Henry’s law constant for CO
2
in water is 1.67 × 10
8
Pa at 298 K. Calculate the number of moles of CO
2
in 500 ml of soda water when packed under 2.53 × 10
5
Pa at the same temperature.
Answer:
(a) Glucose is a non-volatile solute, therefore, addition of glucose to water lowers the vapour pressure of water as a result of which boiling point of water increases.
(b) Given: K
H
= 1.67 × 10
8
Pa
pCO
2
= 2.53 × 10
5
Pa
Using Henry’s law
pCO
2
= K
H
× xCO
2
Or
(a) Define the following terms:
(i) Ideal solution
(ii) Osmotic pressure
(b) Calculate the boiling point elevation for a solution prepared by adding 10 g CaCl
2
to 200 g of water, assuming that CaCl
2
is completely dissociated.
(K
b
for water = 0.512 K kg mol
-1
; Molar mass of CaCl
2
= 111 g mol
-1
)
Answer:
(a) (i) Ideal solution : The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.
(ii) The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent sito the solution through the semipermeable membrane is called the osmotic pressure.
(b) Given w
2
= 10 g, w
1
= 200 g,
K
b
= 0.512 K kg mol
-1
, M
2
= 111 g mol
-1
Using formula:
\(\dagger\)Question 25.
(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling the gas (A) changed into a colourless solid (B). Identify (A) and (B). Write chemicai reactions involved.
(b) Draw structure of XeOF
4
. [5]
Answer:
(a) The salt is sodium nitrate which on heating with conc. H
2
SO
4
evolves a brown gas i.e. NO
2
which gets intensified when Cu turnings are added.
On cooling, NO
2
condenses as a brown liquid which turns paler on cooling and eventually becomes a colourless solid due to formation of dimerized NO
2
i.e. N
2
O
4
(Dinitrogen tetraoxide).
A : Nitrogen dioxide (NO
2
)
B : Dinitrogen tetraoxide (Na
2
O
4
)
(b) Structure of XeOF
4
:
(a) Account for the following:
(i) Reducing character decreases from SO
2
to TeO
2
.
(ii) HClO
3
is a stronger acid than HClO.
(iii) Xenon forms compounds with fluorine and oxygen only.
(b) Complete the following equations:
(i) 4NaCl + MnO
2
+ 4H
2
SO
4
→ 4NaHSO
4
+ MnCl
2
+ 2H
2
O + Cl
2
(ii) 6XeF
4
+ 12H
2
O → 4Xe + 2XeO
3
+ 3O
2
Answer:
(a) (i) Reducing character decreases from SO
2
to TeO
2
because the pπ-pπ bonds in them become weaker with increase in size and bond length along the group.
(ii) HClO
3
is a stronger acid than HClO because with increase in oxidation state and
oxidation number the acidic character increases i.e. HClO
3
(+5) and HClO (+1)
(iii) Xenon forms compounds with fluorine and oxygen only due to their high electronegativity and reactivity. The first ionisation energy of it is fairly close to that of O
2
and F
2
.
(b) (i) 4NaCl + MnO
2
+ 4H
2
SO
4
→
4NaHSO + MnCl
2
+ 2H
2
O + Cl
2
(ii) 6XeF
4
+ 12H
2
O → 4Xe + 2XeO
3
+ 24HF + 3O
2
Question 26.
(a) Account for the following:
(i) Propanal is more reactive than propanone towards nucleophilic reagents.
(ii) Electrophilic substitution in benzoic acid takes place at meta position.
(iii) Carboxylic acids do not give characteristic reactions of carbonyl group.
(b) Give simple chemical test to distinguish between the following pairs of compounds:
(i) Acetophenone and benzaldehyde
(ii) Benzoic acid and ethylbenzoate. [5]
Answer:
(a) (i) Due to steric and + I effect of two methyl groups in propanone.
(Ii) The benzene ring of henzoic acid undergoes electrophilic substitution reaction such as nitration, suiphonation etc. Since the —COOH group in henzene is an electron withdrawing group, therefore it is mcta directing group.
(iii) The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible resonance structure.
(b) (i) 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.
(ii) Benzoic acid and Ethyl benzoate:
By lodoform test: Ethyl benzoate on boiling with excess of NaOH solution gives ethyl alcohol which on heating with iodine gives yellow ppt. of iodoform.
Or
(a) Write structures of A, B, C and D in the following reaction sequence:
(b) Arrange the following compounds in the increasing order of their boiling points:
CH
3
CHO, CH
3
CH
2
OH, CH
3
OCH
3
, CH
3
COOH.
Answer:
(a)
(i)
(b) CH
3
OCH
3
< CH
3
CHO < CH
3
CH
2
OH < CHCOOH
SET II
Note: Except for the following questions, all the remaining questions have been asked in Set-I.
Question 1.
Write the dispersion medium and dispersed phase in milk.
Answer:
Liquid fat is the dispersed phase and water is the dispersion medium.
Question 2.
Write IUPAC name of the complex [Co(NH
3
)
4
Cl(NO
2
)]
+
.
Answer:
Tetra amminechioridonitro cobalt (III) ion
\(\dagger\) Question 4.
What type of stoichiometric defect is shown by NaCl? [1]
Answer:
Schottky defect in shown by NaCl.
Question 5.
What happens when phenol is heated with zinc dust? [1]
Answer:
Benzene is formed when phenol is heated with zinc dist.
Question 7.
Explain the following observations:
(a) Silver atom has completely filled d-orbitais (4d
10
) in its ground state, yet it is
regarded as a transition element.
(b) E° value for Mn
3+
/Mn
2+
couple is much more positive than Cr
3+
/Cr
2+
. [2]
Answer:
(a) Because silver has incomplete d-orbital (4d
9
) in its +2 oxidatìon state, hence it is a transition element.
(b) The large positive E° value for Mn
3+
/Mn
2+
shows that Mn
2+
is much more stable than Mn
+3
due to stable half filled configuration (3d
5
). Therefore the 3
rd
ionisation energy of Mn will be very high and Mn
3+
is unstable and can be easily reduced to Mn
2+
. E° value for Cr
3+
| Cr is positive but šmall i.e. Cr
3+
can also be reduced to Cr
2+
but less easily. Thus Cr
3+
is more stable than Mn
3+
.
Question 10.
Following reactions may occur at cathode during the electrolysis of aqueous silver nitrate solution using silver electrodes:
\(\mathbf{A g}_{(\mathrm{aq})}^{+}\) + e
-1
→ Ag
(s)
; E
0
= 0.80V
\(\mathbf{H}_{(\mathrm{aq})}^{+}\) + e
-1
→ \(\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}\) ; E
0
= 0.00V
On the basis of their standard electrode potential values, which reaction is feasible at cathode and why? [2]
Answer:
The value of E
0
cell
is positive due to higher standard electrode potential (E° = +0.80 V) of
Ag+ than that of H+ so reaction (i) will be feasible at cathode
i.e. \(\mathrm{Ag}_{(\mathrm{aq})}^{+}\) + e
–
→ Ag
(s)
. Silver has higher reduction potential.
Question 15.
(i) Write Reimer-Timann reaction.
(ii) Write the mechanism of acid dehydration of ethanol to yield ethene:
Answer:
(i) Reimer-Timman reaction
(ii) Mechanism of acid dehydration of ethanol into ethene—
Step 1. Formation of protonated ethanol
Step 2. Formation of carbocation—It is the slowest step and hence the rate determination
step of the reaction.
Step 3. Formation of ethene by elimination of a proton
Question 16.
For the complex ion [CoCl
2
(en)
2
]
+
write hybridization type and spin behaviour. Draw one of the geometrical isomers of the complex ion which is optically active.
[Atomic No. : Co = 27] [3]
Answer:
In the complex [CoCl
2
(en)
2
]
+
,
27
Co
3+
= [Ar]3d
6
4s
0
4p
0
It is low spin complex.
Since inner d-orbitals are involved so it is an inner orbital complex.
In this complex only cis-isomer shows optical isomerism
Question 17.
Write the principles involved in the following methods of refining of metals:
(i) Zone refining
(ii) Chromatographic method
(iii) Electrolytic refining [3]
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) Chromatography. This is the method used for the separation and purification of elements. It can also be used for testing the purity of a compound. The principle behind the chromatography is that different components of a mixture are differently adsorbed on an adsorbent.
(iii) 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
Question 22.
Calculate ∆
r
G° and log K
c
for the following reaction at 298 K.
2Cr
(s)
+ 3\(\mathrm{Fe}_{(\mathrm{aq})}^{2+}\) → 2\(\mathrm{Cr}_{(\mathrm{aq})}^{3+}\) + 3Fe
(s)
[(E°
cell
= 0.30 V), 1F = 96500 C mol
-1
] [3]
Answer:
SET III
Note: Except for the following questions, all the remaining questions have been asked in Set-I and Set II.
Question 1.
What happens when phenol is treated with bromine water? [1]
Answer:
2, 4, 6-tribromophenol is formed when phenol is treated with bromine water.
Question 2.
Write IUPAC name of the complex: [CoCl
2
(en
2
)]
+
Answer:
Dichioridobis ethylenediamme cobalt (III) ion
\(\dagger\)Question 3.
Which ionic compound shows both Frenkel and Schottky defects? [1]
Answer:
Silver bromide (AgBr) shows both Schottky and Frenkel defect.
Question 5.
Write the dispersed phase and dispersion medium in butter. [1]
Answer:
Water is the dispersed phase and oil is the dispersion medium in butter.
Question 7.
Explain the following observations:
(i) Zn
2+
salts are colourless.
(ii) Copper has exceptionally positive \(\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^0\) value. [2]
Answer:
(i) Zn
2+
salts are colourless due to absence of unpaired electrons in its ground state and ionic state i.e. Zn
2+
= [Ar] 3d
10
4s
0
4p
0
(ii) The \(E_{\mathrm{M}^{2+} / \mathrm{M}}^{\mathrm{o}}\) for any metal is related to the sum of the enthalpy changes taking place in the following steps:
in the following steps:
M(g) + ∆
a
H → M(g) (∆
a
H = enthalpy of atomization)
M(g) + ∆
1
H → M
2+
(g) (∆
i
H = ionization of atomization)
M
2+
(g) + aq → M
2+
(g) + ∆
hyd
H (∆
hyd
H = hydration atomization)
Copper has high enthalpy of atomization (te. energy absorbed and low enthalpy of hydration (i.e. energy released). Hence \(E_{\mathrm{M}^{2+} / \mathrm{M}}^{\mathrm{O}}\) for copper is positive. The high energy required to transform Cu(s) to Cu
2+
(aq) is not balanced by its hydration enthalpy.
Question 9.
Following reactions may occur at cathode during the electrolysis of aqueous CuCl
2
solution using Pt electrodes:
Answer:
On the basis of their standard electrode potential values, which reaction is feasible at cathode and why? [2]
Answer:
Since the standard electrode potential of Cu
2+
is greater than that of H
+
, so reaction (I) will be feasible at cathode i.e. \(\mathrm{Cu}_{(\mathrm{aq})}^{2+}\) + 2e
–
→ Cu
Cu
2+
has higher reduction potential
Question 10.
(i) What is the order of the reaction whose rate constant has same units as the rate of reaction?
(ii) For a reaction A + H
2
O → B; Rate ∝ [A]. What is the order of this reaction? [2]
Answer:
(i) The reaction whose rate constant has same units as the rate of reaction, will have zero order of reaction.
(ii) The reaction A + H
2
O → B Rate ∝ [A]
The order of this reaction will be pseudo first order reaction as the rate of reaction depends only on concentration of A only.
Question 14.
Calculate ∆
r
G° and log K
c
for the following reaction at 298 K.
Answer:
Question 20.
When a coordination compound CrCl
3
.6H
2
O is mixed with AgNO
3
solution, 3 moles of AgCl are precipitated per mole of the compound. Write:
(i) Structural formula of the complex
(ii) IUPAC name of the complex
(iii) Magnetic and spin behaviour of the complex [3]
Answer:
(i) [Cr(H
2
O)
6
]
3+
\(\mathrm{Cl}_3^{-}\)
(ii) IUPAC name: Hexaaquachromium (III) chloride
(iii) E.C. of Cr
3+
= 3d
3
4s°4p°; unpaired electrons = 3
Inner orbital complex so it is low spin complex.
Since 3 unpaired electrons are present, it is paramagnetic in nature.
Question 21.
Half-life for a first order reaction 693 s. Calculate the time required for 90% completion of this reaction. [3]
Answer:
