Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2017 Comptt (Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2017 Comptt (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.
\(\dagger\) Deleted from Syllabus.
SET I
\(\dagger\) Question 1.
A metallic element crystallises into a lattice having a pattern of AB AB … and packing of spheres leaves out voids in the lattice. What type of structure is formed by this arrangement? [1]
Answer:
Tetrahedral void is formed in AB AB … pattern. The hexagonal close packing (hep) is formed in this arrangement.
Question 2.
Which of the following is most effective in coagulating negatively charged hydrated ferric oxide sol? [1]
(i) NaNO
3
(ii) MgSO
4
(iii) AlCl
3
Answer:
AlCl
3
(Aluminium chloride) is most effective in coagulating negatively charged hydrated ferric oxide sol.
Question 3.
Why are low spin tetrahedral complexes not formed? [1]
Answer:
Low spin tetrahedral complexes are rarely observed because orbital splitting energies for tetrahedral complexes are not sufficiently large for forcing pairing.
Question 4.
Write IUPAC name of the following compound : [1]
Answer:
2, 3-Dinitro phenol.
Question 5.
What type of aldehydes undergo cannizaro reaction? [1]
Answer:
Aldehydes which do not contain α-hydrogen atom undergo Cannizzaro reaction e.g. Formaldehyde (HCHO) and Benzaldehyde (C
6
H
5
CHO).
Question 6.
Explain why on addition of 1 mol of glucose to 1 litre of water, the boiling point of water increases. [2]
Answer:
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.
Question 7.
For a chemical reaction R → P, variation in in[R] vs time (f) plot is given below :
For this reaction :
(i) Predict the order of reaction
(ii) What is the unit of rate constant (k)? [2]
Answer:
(i) It is a zero order reaction.
(ii) The unit of rate constant (k) is mol L
-1
S
-1
.
\(\dagger\) Question 8.
“Orthophosphoric add (H
3
PO
4
) is non-reducing whereas hypophosphorus acid (H
3
PO
2
) is a strong reducing agent.” Explain and justify the above statement with 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
Or
\(\dagger\)(a) What is the covalence of nitrogen in N
2
O
5
?
\(\dagger\)(b) BiH
3
is a stronger reducing agent than SbH
3
, why?
Answer:
(a) The covalance of nitrogen in N
2
O
5
is 4 because each nitrogen atom has four shared pairs of electrons
(b) BiH
3
: Because it is a stronger reducing agent as its tendency to liberate H is maximum.
Question 9.
Account for the following : [2]
(i) The two oxygen-oxygen bond lengths in ozone molecule are identical.
(ii) Most of the reactions of fluorine are exothermic.
Answer:
(i) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e., 128 pm
(ii) Due to much higher electrode potential, high electro-negativity and low bond dissociation enthalpy of F
2
.
Question 10.
Which alkyl halide from the following pair is
(i) Chiral and
(ii) undergoes S
N
2 reaction faster? [2]
Answer:
2-bromobutane
is a chiral compound and 1 Bromo Butane undergoes S
N
2 reaction faster.
\(\dagger\)Question 11.
An element crystallises in b.c.c. lattice with cell edge of 400 pm. Calculate its density if 500 g of this element contains 2.5 × 10
24
atoms. [3]
Answer:
Given : a = 400 pm = 400 × 10
-10
cm
Z = 2 (for bcc) M = ? d = ?
Using formula, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^3 \times \mathrm{N}_{\mathrm{A}}}\)
∵ 2.5 × 10
24
atoms of an element have mass = 500 g
∴ 6.022 × 20
23
atoms of an element have mass = \(\frac{500 \times 6.022 \times 10^{23}}{2.5 \times 10^{24}}\)
∴ M = 120.44 g
Substituting all values in the formula :
Question 12.
The vapour pressure of pure liquids A and B at 400 K are 450 and 700 mmHg respectively. Find out the composition of liquid mixture if total vapour pressure at this temperature is 600 mmHg. [3]
Answer:
The composition of the liquid mixture will be
x
A
= 0.40, x
B
= 1 – 0.40 = 0.60
∴ P
A
= x
A
× \(\mathrm{P}_{\mathrm{A}}^{\circ}\) = 0.40 × 450 mm ∴ P
A
= 180 mm
P
B
= x
B
× \(\mathrm{P}_{\mathrm{B}}^{\circ}\) = 0.60 × 700 mm ∴ P
B
= 420 mm
Mole fraction of B in the vapour phase = 1 – 0.30 = 0.70
Question 13.
The following data were obtained during the first order thermal decomposition of SO
2
Cl
2
at a constant volume:
Calculate the rate constant (k).
[Given : log 2 = 0.3010; log 4 = 0.6021]
Answer:
Question 14.
Define the following terms : [3]
(i) Desorption
(ii) Critical micelle concentration
(iii) Shape selective catalysis
Answer:
(i) Desorption is a phenomenon whereby an adsorbed substance is removed from or through a surface.
(ii) Critical micelle concentration: The formation of micelles takes place only above a particular temperature i.e., Kraft temperature (T
k
) and above a particular concentration.
(iii) Shape selective catalysis : The catalyst reaction in which small sized molecules are absorbed in the pores and cavities of selective adsorbents like zeolites is known as shape-selective catalysis.
Question 15.
(a) Write the principle involved in the ‘vapour phase refining’ of metals.
(b) Write the name of the metal refined by each of file following processes :
(i) Mond process
(ii) van Arkel method
(c) What is the role of depressant in froth floatation process? [3]
Answer:
(a) Vapour phase refining of metals : 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 :
(b)
(i) Ni
(ii) Ti/Zr
(c) Depressants prevent the formation of froth with air bubbles of other sulphide ores.
Question 16.
(a) Arrange the hydrides of group 16 in increasing order of their acidic character. Justify your answer.
(b) Draw structure of XeOF
4
. [3]
Answer:
(a) H
2
O < H
2
S < H
2
Se < H
2
Te
As we move down the group the bond dissociation enthalpy decreases due to increase in bond length and size of the central atom.
(b) XeOF
4
Or
(a) Account for the following :
\(\dagger\)(i) PCl
5
is more covalent than PCl
3
.
(ii) Iron on reaction with HCl forms FeCl
2
and not FeCl
3
.
(b) Draw structure of XeO
3
. [1]
Answer:
(a) (i) In PCl
5
, phosphorus has +5 oxidation state and has less tendency to lose electrons than in +3 of PCl
3
. Therefore, in PCl
5
has more tendency to share e
-1
s than PCl
3
.
(ii) Because HCl on reaction with Iron liberates H
2
gas which prevents the formation of ferric chloride.
(b)
Question 17.
For the complex ion [CoF
6
]
3-
write the hybridization type, magnetic character and spin nature. [Atomic number : Co = 27] [3]
Answer:
It is paramagnetic due to presence of 4 unpaired electrons and form high spin complex.
Question 18.
(a) Write the structural formula of A, B, C and D in the following sequence of reaction :
(b) Illustrate Sandmeyer’s reaction with the help of a suitable example. [3]
Answer:
(a)
(b) Sandmeyer’s reaction : The substitution of diazo group of benzene diazonium chloride by Chloro, Bromo and Cyano group with the help of solution of CuCl dissolved in HCl, CuBr/HBr and CuCN/KCN respectively is known as Sandmeyer’s reaction.
Question 19.
(a) What happens when CH
3
—O—CH
3
is heated with HI?
(b) Explain mechanism for hydration of acid catalyzed ethene :
Answer:
(a) Methyl Iodide (CH
3
I) and Methanol (CH
3
OH) are formed when CH
3
—O—CH
3
is heated with HI.
(b) Acid catalysed hydration : Alkenes react with water in the presence of acid as catalyst to form alcohols.
Mechanism : It involves the following three steps :
Question 20.
Identify A, B and C in the following reactions :
Answer:
Question 21.
(a) What type of linkage is present in disaccharides?
(b) Write one source and deficiency disease of vitamin B
12
.
(c) Write the difference between DNA and RNA. [3]
Answer:
(a) Glycosidic linkage is present in disaccharides.
(b) Eggs are the source of Vitamin B12 and its deficiency causes pernicious anaemia.
(c) DNA is a double strand while RNA is a single strand molecule.
Question 22.
Write the therapeutic action of following on human body and mention the class of drugs to which each of these belong:
(i) Ranitidine
(ii) Morphine
(iii) Aspirin [3]
Answer:
(i) Ranitidine belongs to antacids and it neutralizes the excess acid and raises the pH to an appropriate level in stomach.
(ii) Morphine belongs to narcotic analgesics and it relieves pain and produces sleep even when taken in small dose.
(iii) Aspirin belongs to non-narcotic analgesics and it inhibits the synthesis of compounds which stimulate inflammation in the tissues and cause pain. Aspirin relieves pain and reduces fever.
Question 23.
Once there was a heavy downpour for about three hours in the early morning. Irfan and his family were finding it difficult to carry out their morning chores as the sewer water was flowing back into the toilets, the road outside was flooded with water and they could not move out. On this serious problem, Irfan called a meeting of the residents and said that we are using too much polythene bags and other plastic items which we throw here and there, which finally move into the drains and sewer lines which get choked. As these are non-biodegradable, they remain as such for long time. So we should use bags made up of cloth and jute which are biodegradable.
After reading the above passage, answer the following questions : [4]
(i) Name a polymer which is biodegradable. Write the structures of its monomers and the repeating unit.
(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) When a bright silver object is placed in the solution of gold chloride, it acquires a golden tinge but nothing happens when it is placed in a solution of copper chloride. Explain this behaviour of silver.
[Given : \(\mathbf{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}\) = +0.34V, \(\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}\) = + 0.80V, \(\mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}\) = + 1.40V]
(b) Consider the figure given and answer the following questions :
(i) What is the direction of flow of electrons?
(ii) Which is anode and which is cathode?
(iii) What will happen if the salt bridge is removed?
(iv) How will concentration of Zn
2+
and Ag
+
ions be affected when the cell functions?
(v) How will concentration of these ions be affected when the cell becomes dead? [5]
Answer:
(a) The standard electrode potential, E° for silver is 0.80 V and that of gold is 1.5 V, hence silver can displace gold from its solution. The replaced gold is deposited on silver object due to which golden tinge is obtained. On the other hand E° for Cu is 0.34 V which is lower than that of silver, thus silver cannot replace copper from its solution.
(b) (i) Electrons flow from anode (Zinc plate) to cathode (Silver plate).
(ii) Zinc plate where oxidation occurs acts as anode and silver plate where reduction occurs acts as cathode.
(iii) If the salt bridge is removed then electrons from zinc electrode will flow to the silver electrode where they will neutralize some of Ag
+
ions and the S\(\mathrm{O}_4^{2-}\) ions will be left and the solution will acquire a negative charge. Secondly the Zn
2+
ions from zinc plate will enter into ZnSO
4
solution producing positive charge. Thus due to accumulation of charges in two solutions, further flow of electrons will stop and hence the current stops flowing and the cell will stop functioning.
(iv) As silver from silver sulphate solution is deposited on the silver electrode and sulphate ions migrate to the other side, the concentration of AgSO
4
solution decreases and of ZnSO
4
solution increases as the cell operates.
(v) When the cell becomes dead, the concentration of these ions become equal due to attainment of equilibrium and zero EMF.
Or
(a) What is limiting molar conductivity? Why there is steep rise in the molar conductivity of weak electrolyte on dilution?
(b) Calculate the emf of the following cell at 298 K :
Mg (s) | Mg
2+
(0.1 M) || Cu
2+
(1.0 × 10
-3
M) | Cu (s)
[Given = E°
cell
= 2.71 V].
Answer:
(a) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol \(\Lambda_m^{\circ}\).
There is steep rise in the molar conductivity of weak electrolyte on dilution because as the concentration of the weak electrolyte is reduced, more of it ionizes and thus increase in the number of ions in the solution.
(b) Mg (s) | Mg
2+
(0.1 M) || Cu
2+
(1.0 × 10
-3
M) | Cu (s)
Given : E°
cell
= 2.71 V
Applying Nernst equation
Question 25.
When chromite ore is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (A) is obtained. On acidifying the yellow solution with sulphuric acid, compound (8) is crystallised out. When compound (B) is treated with KC1, orange crystals of compound (C) crystallise out. Identify (A), (B) and (C) and write the reactions involved. [5]
Answer:
Fusion of chromite ore with sodium carbonate :
Or
(a) (i) Which transition element in 3d series has positive \(\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{\circ}\) value and why?
(ii) Name a member of lanthanoid series which is well know to exhibit +4 oxidation state and why?
(b) Account for the following :
(i) The highest oxidation state is exhibited in oxoanions of transition metals.
(ii) HCl is not used to acidify KMnO
4
solution.
(iii) Transition metals have high enthalpy of atomisation.
Answer:
(a) (i) Copper has positive \(\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^0\) value because the sum of enthalpies of sublimation and ionization is not balanced by hydration enthalpy.
(ii) Cerium shows +4 oxidation state because it acquires stable empty orbital configuration and therefore Ce+4 is also used as a good analytical reagent and good oxidising agent.
(b) (i) The highest oxidation state shown in oxoanions of transition metals is \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) (Cr shows +6) and it is due to the ability of oxygen to form multiple bonds with the metal atoms.
(ii) HCl is not used to the acidify KMnO
4
solution because KMnO
4
is a very strong oxidizing agent and it can oxidize HCl to liberate chlorine gas.
(iii) Due to strong interatomic bonding or metal-metal bonding, the enthalpies of atomization of transition elements are quite high.
Question 26.
(a) How will you convert:
(i) Benzene to acetophenone
(ii) Propanone to 2-Methylpropan-2-ol
(b) Give reasons :
(i) Electrophilic substitution in benzoic acid takes place at meta position.
(ii) Carboxylic acids are higher boiling liquids than aldehydes, ketones and alcohols of comparable molecular masses.
(iii) Propanal is more reactive than propanone in nucleophilic addition reactions. [5]
Answer:
(a) (i) Benzene to acetophenone
(ii) Propane to 2-methylpropan-2-ol
(b) (i) Because -COOH group is electron withdrawing group and deactivates the
benzene ring. As a result of this ortho and para position acquires positive charge but only meta does not, so electrophile can attack on meta position.
(ii) Because -COOH group of carboxylic acids is capable to do intermolecular hydrogen bonding forming a dimer while alcohols, aldehydes and ketones cannot.
(iii) Because of smaller +1 effect of one alkyl group in propanal as compared to larger + I effect of 2 alkyl groups of propanone, the magnitude of positive charge on the carbonyl carbon is more in propanal than propanone.
Or
(a) Write the products of the following reactions :
(b) Write simple chemical tests to distinguish between the following pairs of compounds
(i) Propanal and propanone
(ii) Benzaldehyde and Benzoic acid
Answer:
(a)
(b)
(i) Propanal and propanone : Propanal gives positive test with Fehling solution in which a red ppt. of cuprous oxide is obtained while propanone does not respond to test.
(ii) Distinction between Benzaldehyde and Benzoic acid: Benzaldehyde has no alpha hydrogen atom. Therefore, it undergoes Cannizaro’s reaction as follows :
Whereas, Benzoic acid does not undergo Cannizaro’s reaction.
SET II
Note: Except for the following questions, all the remaining questions have been asked in Set-I.
Question 3.
What type of aldehydes undergo Aldol condensation? [1]
Answer:
Aldehydes with a-hydrogen atom undergo Aldol condensation.
Question 4.
Which of the following is most effective in coagulating positively charged hydrated ferric oxide sol? [1]
(i) NaNO
3
(ii) Na
2
SO
4
(iii) (NH
4
)
3
PO
4
Answer:
Ammonium phosphate (NH
4
)
3
PO
4
is most effective in coagulating positively charged hydrated ferric oxide sol.
\(\dagger\)Question 5. A metallic element crystallises into a lattice having a ABC ABC … pattern and packing of spheres leaves out voids in the lattice. What type of structure is formed by this arrangement? [1]
Answer:
Octahedral voids are formed in ABC ABC … pattern. The cubic close packing (ccp) is formed in this arrangement.
Question 7.
Which alkyl halide from the following pair is
(i) Chiral and
(ii) undergoes S
N
1 reaction faster? [2]
(a) (CH
3
)
3
CBr
(b) CH
3
CH
2
CHBrGH
3
[2]
Answer:
(i) 2-Bromobutane is chiral as the central C atom has all 4-different groups.
(ii) 3° Alkyl bromide i.e. (CH
3
)
3
CBr undergoes S
N
1 reaction faster due to more stability of 3° carbocation.
Question 10.
Account for the following : [2]
(i) Two S-O bond lengths in SO
2
are equal.
(ii) Fluorine shows only-1 oxidation state in its compounds.
Answer:
(i) Due to resonance in SO
2
the double bond (π) electrons are distributed equally in both resonating structures as a result of which the bond length of two S-O becomes equal.
(ii) Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.
Question 11.
(a) What type of linkage is present in proteins?
(b) Give one example each of water soluble and fat soluble vitamins.
(c) Draw pyranose structure of glucose. [3]
Answer:
(a) Peptide linkage is present in proteins.
(b) Vitamin C is water soluble and Vitamin D is fat soluble vitamin.
(c) Pyranose structure of glucose
Question 12.
Define the following terms : [3]
(i) Kraft temperature
(ii) Peptization
(iii) Electrokinetic potential
Answer:
(i) Kraft temperature : The formation of micelles takes place only above a particular temperature known as Kraft temperature (T
k
).
(ii) Peptization : It is a process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of a suitable electrolyte.
(iii) The potential difference between the fixed layer and the diffused layer is known as electrokinetic potential.
\(\dagger\)Question 16.
An element crystallises in fee lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contain 2.5 × 10
24
atoms. [3]
Answer:
Given : a = 400 pm = 400 × 10
-10
cm
Z = 4 (for fcc), M = ?, d = ?
Using formula, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^3 \times \mathrm{N}_{\mathrm{A}}}\)
∵ 2.5 × 10
24
atoms of an element have mass = 250 g
∴ 6.022 × 10
23
atoms of an element have mass = \(\frac{250 \times 6.022 \times 10^{23}}{2.5 \times 10^{24}}\)
∴ M = 60.22 g
Substituting all values in formula:
d = \(\frac{4 \times 60.22}{\left(400 \times 10^{-10}\right)^3 \times 6.022 \times 10^{23}}\) = \(\frac{240.88}{38.5408}\)
∴ d = 6.25 g cm
-3
Set III
Note: Except for the following questions, all the remaining questions have been asked in Set-I and Set II.
\(\dagger\)Question 2.
What type of Stoichiometric defect is shown by AgCl? [1]
Answer:
Frenkel defect.
Question 3.
Which of the following is most effective in coagulating positively charged methylene blue sol? [1]
(i) Na
3
PO
4
(ii) K
4
[Fe(CN)
6
]
(iii) Na
2
SO
4
Answer:
Potassium ferrocyanide K
4
[Fe(CN)
6
].
Question 7.
Account for the following: [2]
\(\dagger\)(i) Bond angle in \(\mathrm{NH}_4^{+}\) is higher than that in NH
3
.
(ii) ICl is more reactive than I
2
.
Answer:
(i) Because in \(\mathrm{NH}_4^{+}\) ion there is no lone pair of electrons which is present in NH
3
due to which lone pair-bond pair repulsion occurs and bond angle decreases from 109°28′ to 107.3°.
(ii) Because I-Cl bond is weaker than I-I bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence more reactive.
\(\dagger\)Question 15.
An element crystallises in bcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contains 2.5 × 10
24
atoms. [3]
Answer:
Given : a = 400 pm = 400 × 10
-10
cm
Z = 2 (for bcc), M = ?, d = ?
Using formula, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^3 \times \mathrm{N}_{\mathrm{A}}}\)
∵ 2.5 × 10
24
atoms of an element have mass = 250 g
∴ 6.022 × 10
23
atoms of an element have mass = \(\frac{250 \times 6.022 \times 10^{23}}{2.5 \times 10^{24}}\)
∴ M = 60.22 g
Substituting all values in formula :
d = \(\frac{2 \times 60.22}{\left(400 \times 10^{-10}\right)^3 \times 6.022 \times 10^{23}}\) = \(\frac{120.44}{38.5408}\)
∴ d = 3.125 g cm
-3
Question 16.
For the first order decomposition of azoisopropane to hexane and nitrogen at 543 K , the following data were obtained :
Calculate the rate constant. The equation for the reaction is :
(CH
3
)
2
CHN = NCH(CH
3
)
2
C
6
H
14
(g) + N
2
(g)
[Given : log 3 = 0.4771; log 5 = 0.6990] [3]
Answer:
Question 19.
For the complex ion [Ni(CN)
4
]
2-
write the hybridization type, magnetic character and spin nature.
[Atomic No. : Ni = 28] [3]
Answer:
∴ Diamagnetic due to paired electrons. Complex is low spin.
Question 20.
Write the therapeutic action of following on human body and mention the class of drugs to which each of the these belong : [3]
(i) Equanil
(ii) Aspirin
(iii) Chloramphenicol
Answer:
(i) Equanil belongs to the class of tranquilizers and it is used in controlling depression and hypertension.
(ii) Aspirin belongs to non-narcotic analgesics and it inhibits the synthesis of compounds which stimulate inflammation in the tissues and cause pain. Aspirin reduces pain and fever.
(iii) Chloramphenicol belongs to antibiotics and it is used for treatment of typhoid. It kills or inhibits the growth of micro-organisms.
Question 22.
(a) What is the principle behind ‘zone refining’ of metals? Name an element which is refined by this method.
(b) Write the name of the metal refined by each of the following processes :
(i) Distillation
(ii) Liquation [3]
Answer:
(a)
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.
Germanium can be refined by this method.
(b) (i) Zinc can be refined by distillation.
(ii) Lead can be refined by liquation.
