Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2017 Delhi to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2017 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.
-
Use log tables, if necessary, Use of calculators is not allowed.
† Deleted from Syllabus.
Question 1.
Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation
Answer:
Permanganate ion, i.e., Mn04~ with oxidation number +7.
Question 2.
Write IUPAC name of the following compound: [1]
(CH
3
CH
2
)
2
NCH
3
Answer:
N-Ethyl-N-methylhexanamine.
Question 3.
For a reaction R → P, half-life (t
1/2
) is observed to be independent of the initial concentration of reactants. What is the order of reaction? [1]
Answer:
The t
1/2
of a first order reaction is independent of initial concentration of reactants
Question 4.
Write the structure of l-Bromo-4-chlorobut-2- ene. [1]
Answer:
Question 5.
Write one similarity between physisorption and Chemisorption. [1]
Answer:
Physisorption and chemisorption both are the surface phenomenon and both increases the surface area during the process of adsorption.
Question 6.
Complete the following reactions:
(i) NH
3
+ 3Cl
2
(excess) →
(ii) XeF
6
+ 2H
2
O →
Answer:
OR
What happens when
(i) (NH
4
)2Cr
2
O
7
is heated?
(ii) H
3
PO
3
is heated?
Write the equations. [2]
Answer:
Question 7.
Define the following terms:
(i) Colligative properties
(ii) Molality (m) [2]
Answer:
(i) Colligative properties. All those properties which depend on the number of solute particles irrespective of the nature of solute’ are called as colligative properties.
(ii) Molality (m). Number of moles of solute dissolved per kg of the solvent.
Question 8.
Draw the structures of the following:
(i) H
2
S
2
O
7
(ii) XeF
6
Answer:
Question 9.
Calculate the degree of dissociation (α) of acetic acid if it’s molar conductivity (Λ
m
) is 39.05 S cm
2
mol
-1
.
Given: Λ
0
(H
+
) = 349.6 S cm
2
mol
-1
and Λ° (CH
3
COO
–
)= 40.9 S cm
2
mol
-1
. [2]
Answer:
Question 10.
Write the equations involved in the following reactions:
(i) Wolff-Kishner reduction
(ii) Etard reaction. [2]
Answer:
Question 11.
A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15K.
[Given: (Molar mass of sucrose = 342 g mol
-1
), (Molar mass of glucose = 180 g mol
-1
)] [3]
Answer:
Molality (m) = \(\frac{a}{2 \sqrt{2}}\)
Given:
Molar mass of sucrose = C
12
H
22
O
11
= 12 × 12 + 22 + 11 × 16 = 342
10% solution (by mass) of sucrose in water means 10 g of sucrose is present in (100 – 10) = 90 g of water
10% solution of sucrose means, w = 10 g
Mass of water, W = 90 g
Question 12.
(a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO
3
for 15 minutes.
Given: Molar mass of Ag = 108 g mol
-1
, 1F = 96500 C mol
-1
)
(b) Define fuel cell [3]
Answer:
(a) Q = I × t
= 2 × 15 × 60 = 1800 C
∵ 96500 C deposit Ag = 108 g
∴ 1800 C deposit Ag = \(\frac{a}{2 \sqrt{2}}\) × 1800 = 2.0145 g
(b) Cells that convert the energy of combustion of fuels like hydrogen, methanol, methane, etc. directly into electrical energy are called fuel cells.
Question 13.
(i) What type of isomerism is shown by the complex [Co(NH
3
)
6
]Cr(CN)
6
]?
(ii) Why a solution of [Ni(H
2
O)
6
]
2+
is green while a solution of [Ni(CN)
4
]
2-
is colourless? (At no. of Ni = 28)
(iii) Write the IUPAC name of the following complex: [3]
[CO(NH
3
)
5
(CO
3
)]Cl.
Answer :
(i) Coordination isomerism
(ii) [Ni(H 2 O) 6 ] 2+ is an outer orbital complex due to weak field ligand H20 and the presence of unpaired electrons undergoes d – d transition by absorbing red light and shows green colour while [Ni(CN) 4 ] 2- is an inner orbital complex and has no unpaired electrons hence colourless.
(iii) Pentaamminecarbonatocobalt (III) Chloride
Question 14.
Write one difference in each of the following:
(i) Lyophobic sol and Lyophilic sol.
(ii) Solution and Colloid
(iii) Homogeneous catalysis and Heterogeneous catalysis. [3]
Answer:
(i) Lyophobic sol and Lyophilic sol. Lyophobic solutions are liquid (dispersion medium) – hating and lyophilic solutions are liquid (dispersion medium) – loving colloids.
(ii) Solution and Colloid. Solution is a homogenous solutibn whose particle size is less than 10 -9 m while colloid is a heterogenous solution whose particle size is in between 10 -9 to 10 -6 m.
(iii) Homogeneous catalysis and Heterogeneous catalysis. Homogeneous catalysis is the phenomenon of changing the rate of reaction when catalyst has same phase as the reactants while in heterogeneous catalysis, the catalyst has different phase than that of the reactants.
Question 15.
Following data are obtained for reaction:
N
2
O
5
→ 2NO
2
+ \(\frac { 1 }{ 2 }\) O
2
(a) Show that it follows first order reaction.
(b) Calculate the half-life.
(Given log 2 = 0.3010, log 4 = 0.6021) [3]
Answer:
Since the value of k comes to nearly constant therefore it follows the first order reaction.
(b) t 1/2 = \(\frac{0.693}{k}\) = \(\frac{0.693}{0.0023106}\) = 299.9 ≈ 300 s
Question 16.
Following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane
(i) Write the compound which is most reactive towards SN
2
reaction.
(ii) Write the compound which is optically active.
(iii) Write the compound which is most reactive towards β-elimination reaction. [3]
Answer:
(i) 1-Bromopentane is most reactive towards SN
2
reaction as it follows the order 1° > 2° > 3°.
(ii) 2-Bromopentane is optically active.
(iii) 2-Bromo-2-methyl butane is most reactive towards β-elimination reaction.
Question 17.
(a) Write the principle of the method used for the refining of germanium.
(b) Out of PbS and PbCO
3
(ores of lead), which one is concentrated by froth floatation process preferably?
(c) What is the significance of leaching in the extraction of aluminium? [3]
Answer:
(a) Zone refining method is used for the refining of germanium and it is based on the principle that the impurities are more soluble in the melt than in the solid-state of the metal.
(b) PbS, Sulphide ore has more tendency to stick to the oil which comes on the surface being lighter and easily skimmed off so PbS is concentrated by froth floatation method.
(c) Leaching of alumina is done to remove the impurities like SiO
2
by using NaOH solution and pure alumina is obtained.
Question 18.
Write structures of compounds A, B and C in each of the following reactions: [3]
Answer:
OR
Do the following conversions in not more than two steps:
(a) Benzoic acid to Benzaldehyde
(b) Ethylbenzene to Benzoic acid
(c) Propanone to Propene
Answer:
Question 19.
Write the structure of the monomers used for getting the following polymers: [1×3=3]
(i) Dacron
(ii) Melamine-formaldehyde polymer
(iii) Buna-N
Answer:
(ii) Monomers of melamine-formaldehyde polymer are melamine and formaldehyde.
Question 20.
Define the following:
(i) Anionic detergents
(ii) Broad spectrum antibiotics
(iii) Antiseptic
Answer:
(a) Anionic detergents. Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example : Sodium alkyl suiphates
These are obtained from long straight chain alcohols containing 12-48 carbon atoms by treatment with conc. H
2
SO
4
followed by neutralization with NaOH.
Example: Sodium lauryl sulphate.
(ii) Broad spectrum antibiotics: Antibiotics which kill or inhibit a wide range of Gram¬positive and Gram-negative bacteria are called broad spectrum antibiotics.
Example: Chloramphenicol
(iii) Antiseptics are the chemicals which either kill or prevent growth of microbes on living tissues.
Question 21.
Give reasons:
(i) Thermal stability decreases from H
2
O to H
2
Te.
(ii) Fluoride ion has higher hydration enthalpy than chloride ion.
(iii) Nitrogen does not form pentahalide. [3]
Answer:
(i) Thermal stability decreases from H
2
O to H
2
Te due to weakening of bond between hydrogen and the atom from O to Te as size is increasing down the group.
(ii) Fluoride ion has higher hydration enthalpy than chloride ion due to stronger attractions of smaller in size fluoride ion.
(iii) Nitrogen does not contain’d’ orbitals.
Question 22.
Give reasons: 1×3=3
(i) Acetylation of aniline reduces its activation effect.
(ii) CH
3
NH
2
is more basic than C
6
H
5
NH
2
.
(iii) Although – NH
2
is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline.
Answer:
(i) Acetylation of aniline reduces its activation effect because acetyl group being electron withdrawing group attracts the lone pair of electrons of the N-atom towards carboxyl group and the lone pair of electrons on N is less available for donation to benzene ring by resonance.
(ii) CH 3 NH 2 is more basic than aniline due to availability of lone pair of electrons for donation while in aniline lone pair of electrons on the nitrogen atom is delocalised over benzene ring and thus unavailable for donation.
(iii) Because of nitration in an acidic medium, aniline gets protonated to give anilinium ion which is m-directing.
Question 23.
After watching a programme on TV about the presence of carcinogens (cancer causing agents) Potassium bromate and Potassium iodate in bread and other bakery products, Ritu a class XII student decided to aware others about the adverse effects of these carcinogens in foods. She consulted the school principal and requested him to instruct canteen contractor to stop selling sandwiches, pizza, burgers and other bakery products to the students. Principal took an immediate action and instructed the canteen contractor to replace the bakery products with some proteins and vitamins rich food like fruits, salads, sprouts, etc. The decision was welcomed by the parents and students. After reading the above passage, answer the following questions:
*(t) What are the values (at least two) displayed by Ritu?
(ii) Which polysaccharide component of carbohydrates is commonly present in bread?
(iii) Write the two types of secondary structure of proteins.
(iv) Give two examples of water soluble vitamins. [4]
Answer:
*(i) As per latest CBSE curriculum, Value Based Questions will not be asked in the examination.
(ii) Starch is present in bread.
(iii) a-helix and P-pleated sheet structures
(iv) Vitamin-B and Vitamin-C are water soluble.
Question 24.
(a) Account for the following:
(i) Transition metals form a large number of complex compounds.
(ii) The lowest oxide of transition metal is basic whereas the highest oxide is amphoteric or acidic.
(iii) E
0
value for the Mn
3+
/Mn
2+
couple is highly positive (+1.57 V) as compare to Cr
3+
/Cr
2+
.
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements. [5]
Answer:
(a) (i) Transition metals form large number of complexes because:
small size of metal ion;
high ionic charge; and
availability of empty d-orbitals.
(ii) The lowest oxide of transition metal is basic because of low oxidation state some of valence electrons are not involved in bonding and acts as base by donating electron. However in higher oxides due to high oxidation state, it cannot donate electrons but can accept electrons due to high effective nuclear charge. Hence, they are acidic in nature.
(iii) Mn 2+ exists in half-filled d 5 state which is very stable while Mn 3+ is d 4 which is not so stable. Mn 3+ can be easily reduced to Mn 2+ . Conversion from d 4 to d 5 will be quick and have negative ∆G value. Hence, because of the stability factor the E° value is high for this process. While Cr 3+ is d 3 is half-filled (t 2g 3) is stable in nature and Cr 2+ is d 4 , has one extra electron which it would like to donate to attain the stable half-filled (t 2g 3) configuration. Hence for the process Cr 3+ to Cr 2+ , the value of E° is less.
(b) Similarity : Both lanthanoids and actinoids show contraction in size and irregularity in their electronic configuration.
Difference : Actinoids show wide range of oxidation states but lanthanoids do not.
OR
(a) (i) How is the variability in oxidation states of transition metals different from that of the p-block elements?
(ii) Out of Cu
+
and Cu
2+
, which ion is unstable in aqueous solution and why?
(iii) The orange colour of Cr
2
O
7
2-
ion changes to yellow when treated with an alkali. Why?
(b) Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons. [3+2*5]
Answer:
(i) In transition elements, the oxidation states can vary from +1 to highest oxidation state by removing all its valence electrons and the oxidation states differ by 1, e.g., Fe
2+
and Fe
3+
while in p-block elements, the oxidation states differ by 2, e.g., +2 and +4 or +3 and +5, etc.
(ii) 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 that 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
(iii) On adding sodium hydroxide the pH of solution increases results in decrease in FI+ concentration which shifts the reaction equilibrium in forward direction producing yellow chromate solution.
(b)
- Lanthanoids show limited oxidation states, i.e., +2, +3, +4 out of which +3 is most common which is due to large energy gap between 4/ and 5d subshells while actinoids show large number of oxidation states due to small energy gap between 5/, 6d and 7s subshells.
- Actinoid contraction is greater than lanthanoid contraction due to poor shielding of 5 electrons.
- Actinoids are radioactive in nature.
Question 25.
(a) An element has atomic mass 93 g mol
-1
and density 11.5 g cm
-3
. If the edge length of its unit cell is 300 pm, identify the type of unit cell.
(b) Write any two differences between amorphous solids and crystalline solids. [5]
Answer:
(a) Given:
M = 93 g mol
-1
ρ = 11.5 g cm
-3
a = 300 pm = 300 × 10
-10
cm = 3 × 10
-8
cm
Using Formula
Z = \(\frac{\rho \times a^3 \times N_A}{M}\)
= \(\frac{11.5 \times\left(3 \times 10^{-8}\right)^3 \times 6.022 \times 10^{23}}{93}\) = 2.01 (approx.)
(b) Tablee 1
Amorphous solids
(i) They are isotropic, i.e., they will show same value of all physical properties in all directions
(ii) They have short range order.
Crystalline solids
(i) They are anisotropic, i.e., value of physical properties will be different when measured along different directions.
(ii) They have long range order.
OR
(a) Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in an f.c.c. structure.
(Atomic mass of Al = 27 g mol
-1
)
(b) Give reasons:
(i) In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect.
(ii) Silicon on doping with phosphorous forms n-type semiconductor.
(iii) Ferrimagnetic substances show better magnetism than antiferromagnetic substances.
Answer:
(a) Given:
Mass of A1 = 8.1,
Atomic mass of A1 = 27 g mol
-1
No. of atoms = η × 6.022 × 10
23
= \(\frac{8.1}{27}\) × 6.022 × 10
23
= 0.3 × × 6.022 × 10
23
= 1.8066 × 10
23
Since one of fcc unit cell has 4 atoms
∴ No of unit cells = \(\frac{1.8066 \times 10^{23}}{4}\)
= 4.5 x 1022 unit cells
(b) (i) Schottky defect is shown by the ionic solids having very small difference in their cationic and anionic radius whereas Frenkel defect is shown by ionic solids having large difference in their cationic and anionic radius. NaCl exhibits Schottky defect because radius of both Na + and Cl – have very small difference.
(ii) Phosphorus is pentavalent that is it has 5 valence electrons, an extra electron results in the formation of n-type semi conductors on doping with Silicon. The conductivity is due to presence of extra electrons.
(iii) To antiferromagnetic substances the magnetic moments of domains are half aligned in one direction and remaining half in opposite direction in the presence of magnetic field so magnetic moment will be zero while in ferrimagnetic substances the magnetic moments of domains are aligned in parallel and anti¬parallel directions in unequal numbers, hence shows some value of magnetic moment.
Question 26.
(a) Write the product(s) in the following reactions:
(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Ethanol and Phenol
(ii) Propanol and 2-methyl propane-2-ol [5]
Answer:
(a)
(b) (i) Ethanol gives a positive Iodoform test.
Phenol gives a negative Iodoform test.
(ii) Distinction between propanol and 2-methylpropan-2-ol:
By Lucas Test:
OR
(a) Write the formula of reagents used in the following reactions:
(i) Bromination of phenol to 2, 4, 6-tribromophenol
(ii) Hydroboration of propene and then oxidation to propanol.
(b) Arrange the following compound groups in the increasing order of their property indicated:
(i) p-nitrophenol, ethanol, phenol (acidic character)
(ii) Propanol, propane, Propanal (boiling point)
(c) Write the mechanism (using curved arrow notation) of the following reaction:
Answer:
(a)
(b) (i) p-nitrophenol > Phenol > Ethanol (Acidic character)
(ii) Propanol > Propanal > Propane (Boiling point)
(c)
Set II
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Question 2.
Write the structure of 2, 4-dinitrochlorobenzene. [1]
Answer:
Question 4.
Write IUPAC name of the following compound:
CH
3
NHCH(CH
3
)
2
[1]
Answer:
N-methyl propane-2-amine.
Question 5.
Write the formula of an oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number. [1]
Answer:
Cr
2
O
7
2-
(dichromate ion) in which oxidation state of Cr is +6 which equal to its group number 6.
Question 7.
Draw the structures of the following: [2]
(i) H
3
PO
2
(ii) XeF
4
Answer:
Question 8.
Define the following terms:
(i) Ideal solution
(ii) Molarity (M) [2]
Answer:
(i) (z) Ideal solution. The solution that obeys Raoults Law over the entire range of concentration.
(ii) Molarity is the number of moles of solute dissolved per litre of solution or
\(M=\frac { { W }_{ b }\times 1000 }{ {M}_{ b } \times \text { Volume }(mL)}\)
Question 9.
Complete the following reactions: [2]
(i) Cl
2
+ H
2
O →
(ii) XeF
6
+ 3H
2
O →
Answer:
(i) Cl
2
+ H
2
O → 2HCl + [O]
(ii) XeF
6
+ 3H
2
O → XeO
3
+ 6HF
OR
What happens when
(i) conc. H
2
SO
4
is added to Cu?
(ii) SO
3
is passed through water?
Write the equations.
Answer:
(i) Cu + 2H
2
SO
4
(conc.) → CuSO
4
+ 2H
2
O + SO
2
(ii) SO
3
+ H
2
O → H
2
SO
4
Question 10.
Write the reactions involved in the following:
(i) Hell-Volhard-Zelinsky reaction
(ii) Decarboxylation reaction [2]
Answer:
Question 13.
Write the principles of the following methods:
(i) Vapour phase refining [3]
(ii) Zone refining
(iii) Chromatography
Answer:
(i) 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
(ii) 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.
(iii) 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.
Question 15.
Define the following:
(i) Cationic detergents
(ii) Narrow spectrum antibiotics .
(iii) Disinfectants [3]
Answer:
(i) Cationic detergents. They are quaternary 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
(ii) Narrow spectrum antibiotics. Narrow spectrum antibiotics are those antibiotics which are mainly effective against gram positive or gram negative bacteria.
(iii) Disinfectants. Disinfectants kill or prevent growth of microbes and are applied on inanimate/non living objects. Example: Phenol
Question 19.
Write the structures of the monomers used for getting the following polymers: 1×3=3
(i) Neoprene
(ii) Melamine-formaldehyde polymer
(iii) Buna-S
Answer:
(i) Neoprene
(ii) Monomers of melamine-formaldehyde polymer are melamine and formaldehyde.
(iii) Buna-S
Set III
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Question 1.
What is the effect of catalyst on: [1]
(i) Gibbs energy (∆G) and
(ii) activation energy of a reaction?
Answer:
(i) There will be no effect of catalyst on Gibbs energy.
(ii) The catalyst provides an alternative pathway by decreasing the activation energy of a reaction.
Question 4.
Write the structure of 3-Bromo-2-methylprop-1-ene. [1]
Answer:
3-Bromo-2-methylprop-1-ene
Question 5.
Write IUPAC name of the following compound:
(CH
3
)
2
N-CH
2
CH
3
) [1]
Answer:
N, N-dimethylethanolamine
Question 6.
Write the reactions involved in the following reactions:
(i) Clemmensen reduction
(ii) Cannizzaro reaction [1+1=2]
Answer:
(i) Clemmensen reduction. The carbonyl group of aldehyde and ketones is reduced to CH2 group on treatment with zinc amalgam and concentrated hydrochloric acid.
(ii) Cannizzaro reaction. Aldehydes, which do not have an a-hydrogen atom undergo self oxidation and reduction on treatment with cone, alkali and produce alcohol and carboxylic acid salt.
Question 7.
Draw the structures of the following:
(i) H
4
P
2
O
7
(ii) XeOF
4
Answer:
(i) H
4
P
2
O
7
(ii) XeOF
4
Question 8.
Define the following terms:
(i) Abnormal molar mass
(ii) van’t Hoff factor (i) [2]
Answer:
(i) Abnormal molar mass. If the molar mass calculated by using any of colligative properties tends to be different than theoretically expected molar mass, it is called abnormal molar mass.
(ii) van’t Hoff factor (i). Extent of dissociation or association or ratio of the observed colligative property to calculated colligative property.
i = \(\frac{\text { Observed colligative property }}{\text { Theoretical colligative property }}\)
Question 10.
Complete the following chemical equations: [2]
(i) F
2
+ 2Cl
–
→
(ii) 2XeF
2
+ 2H
2
O →
Answer:
(i) F
2
+ 2Cl
–
→ 2F
–
+ Cl
2
(ii) 2XeF
2
+ 2H
2
O → 2Xe + 4HF + O
2
OR
What happens when
(i) HQ is added to MnO
2
?
(ii) PCl
5
is heated?
write the equations involved.
Answer:
(i) MnO
2
+ 4HCl → MnCl
2
+ Cl
2
+ 2H
2
O
(ii) PCl
5
→ PCl
3
+ Cl
2
Question 13.
Define the following: 1×3=3
(i) Anionic detergents
(ii) Limited spectrum antibiotics
(iii) Tranquilizers
Answer:
(i) Those detergents in which large part of their molechles are anions and used in cleansing action, are called anionic detergents.
Example: Sodium alkyl sulphates
(ii) Limited spectrum antibiotics are those which are effective against a single organism or disease.
(iii) Tranquilizers are class of chemicals used for treatment of stress or mild or severe mental diseases.
Question 14.
Write the structures of the monomers used for getting the following polymers:
(i) Nylon-6
Answer:
(i) Monomers of Nylon-6:
Question 19.
Write one difference between each of the following:
(i) Multimolecular colloid and Macromolecular colloid
(ii) Sol and Gel
(iii) O/W emulsion and W/O emulsion [3]
Answer:
(i) In multimolecular colloids, a large number of atoms or smaller molecules of a substance aggregates together to form species having a size in the colloidal range.
Example: Sulphur sol whereas in macro-molecular colloids the colloidal particles are large molecules having colloidal dimensions.
Example: Starch.
(ii) In sol the dispersing phase is solid and dispersing medium is liquid;
Example: paint, gold sol etc., whereas in Gel the dispersing phase is liquid and dispersing medium is solid;
Example: Jelly, butter etc.
(iii) In O/W emulsion, oil is the dispersed phase while water is the dispersion medium
Example: milk, vanishing cream etc whereas in W/O emulsion water is the dispersed phase while oil is the dispersion medium.
Example: Cold cream, butter etc.
Question 20.
(i) What type of isomerism is shown by the complex [Co(en)
3
]Cl
3
?
(ii) Write the hybridisation and magnetic character of [Co(C
2
O
4
)
3
]
3-
(At. no. of Co = 27)
(iii) Write IUP AC name of the following Complex [Cr(NH
3
)
3
Cl
3
]. [3]
Answer:
(i) [Co(en)
3
]Cl
3
shows optical isomerism.
(ii) [Co(C 2 O 4 ) 3 ] 3- shows d 2 sp 3 hybridisation and is diamagnetic in nature.
(iii) IUPAC : Triamminetrichloridochromium(III).
