Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2013 Outside Delhi to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2013 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.
\(\dagger\) Deleted from Syllabus.
Set I
Question 1.
Of physisorption or chemisorption, which has a higher enthalpy of adsorption? [1]
Answer:
Chemisorption has higher enthalpy of adsorption than Physisorption due to chemical bond formation.
Question 2.
Name the method used for refining of copper metal. [1]
Answer:
Electrolytic refining of copper metal.
Question 3.
Name two poisonous gases which can be prepared from chlorine gas. [1]
Answer:
- Phosgene gas (COCl2) and
- Chloropicrin or tear gas (CCl 2 NO 2 ).
Question 4.
Write the IUPAC name of the following compound: [1]
Answer:
IUPAC name : 2-Chloro-3, 3-dimethylbutane.
Question 5.
Rearrange the following compounds in the increasing order of their boiling points: [1]
CH
3
-CHO, CH
3
-CH
2
-OH
2
, CH
3
– CH
2
– CH
3
Answer:
Question 6.
Write the structure of N-methylethanmine. [1]
Answer:
Structure of n-methylethamamine : H
3
C — H
2
C — NH — CH
3
Question 7.
What are the products of hydrolysis of sugar? [1]
Answer:
Invert sugar, An equimolar mixture of glucose and fructose is obtained by hydrolysis of sucrose in presence of an acid such as dil. HQ or the enzyme invertase or sucrose and is called invert sugar.
Question 8.
Is
n a homopolymer or a copolymer?
Answer:
is a homopolymer as it is made up of same monomer units.
Question 9.
Account for the following: [2]
(i) Schottky defects lower the density of related solids.
(ii) The conductivity of silicon increases on doping it with phosphorus.
Answer:
(i) Schottky defect is produced due to missing of equal number of cation and anion from lattice as a result of which the density of the lattice solid decreases.
(ii) The conductivity of silicon increases due to negatively charged extra electron of doped pentavalent phosphorus.
Question 10.
Aluminium crystallizes in an FCC structure. The atomic radius of the metal is 125 pm, what is the length of the side of the unit cell of the metal? [2]
Answer:
For fcc, Formula : r = \(\frac{a}{2 \sqrt{2}}\)
Given: r = 125 pm
∴ a = 2\(\sqrt{2}\) r ⇒ a = 2\(\sqrt{2}\) × 125
⇒ a = 2 × 1.414 × 125 = 353.5 pm
Question 11.
The standard electrode potential (E°) for Daniell cell is +1.1 V. Calculate ∆G° for the reaction: [2]
Zn (s) + Cu
2+
(aq) → Zn
2+
(aq) + Cu(s)
(1F = 96500 C/mol)
Answer:
We know, ∆G° = -nF E°
cell
Given : E°
cell
= 1.1 volt
∴ ∆G° = -2 × 96500 C mol
-1
× 1.1 volt
= -212300 CV mol
-1
= -212300 J mol
-1
= -212.3 KJ mol
-1
Question 12.
For a reaction A + B → P, the rate law is given by [2]
r = k [A]
1/2
[B]
2
(a) What is the order of this reaction?
(b) A first-order reaction is found to have a rate constant k = 5.5 x 10
-14
s
-1
. Find the half life of the reaction.
Answer:
(a) According to the formula : r = k[A]
1/2
[B]
2
Order w.r.t A = \(\frac{1}{2}\) Order w.r.t B = 2
∴ Overall order = \(\frac{1}{2}\) + \(\frac{2}{1}\) = \(\frac{5}{2}\)
(b) For first order reaction, =
Given: k = 5.5 × 10
-14
s
-1
Thus, t
1/2
= \(\frac{0.693}{5.5 \times 10^{-14} \mathrm{~s}^{-1}}\) Hence t
1/2
= 1.26 × 10
13
s
Question 13.
(a) Name the method used for removing gangue from sulphide ores. [2]
(b) How is wrought iron different from steel?
Answer:
(a) Froth Floatation method is used for removing gangue from sulphide ores.
(b) Wrought iron is the purest form of commercial iron which contains about 0.2 – 0.5% carbon while steel contains about (0.5 – 1.5)% carbon.
Question 14.
Draw the structures of the following molecules:
(i) XeOF
4
(ii) H
3
PO
3
[2]
Answer:
(i)
XeOF
4
(ii) H
3
PO
3
(phosphorus acid)
Question 15.
How are interhalogen compounds formed? What general compositions can be assigned to them? [2]
Answer:
Interhalogen compounds : Halogens reacts with each other to form a number of compounds called interhalogen compounds, whose general formula is XX’n.
where [X = less electronegative atom (have larger size)
X’ = more electronegative atom (have smaller size)
n = no. of more electronegative atoms]
They are of four hjpes:
XX’ = CIF, BrF, IF, BrCl, ICl, IBr
XX’
3
= ClF
3
, BrF
3
, IF
3
, ICl
3
XX
5
= Cl
5
, BrF
5
, IF
5
XX’
7
= IF
7
Naming: The halogen with positive O.S. named first and with negative OS. named after first with suffix ‘ide’.
Example :
BrCl
3
→ Bromine trichioride
IF
7
= Iodine heptafluoride
Preparation of Interhalogen Compounds:
By direct combination:
Example:
Question 16.
Explain the mechanism of the following reaction: [2]
Answer:
Question 17.
Write the equations involved in the following reactions: [2]
(i) Reimer-Tiemann reaction
(ii) Williamson ether synthesis
Answer:
(i) Reimer-Tiemann reaction :
(ii) Williamson’s ether synthesis : Alkyl Halide reacts with Alkoxide
Question 18.
Define thermoplastic and thermosetting polymers. Give one example of each. [2]
OR
What is a biodegradable polymer? Give an example of biodegradable aliphatic polyester.
Answer:
Thermoplastic polymers : I.inear polymers in which the intermolecular forces of attraction are in between those of elastomers and fibres and can be melted again and again on heating followed by moulding to give desired shape.
Example : Polyethene, Polyvinyl chloride (PVC) etc.
Thermosetting polymers : These are semifluid substances with low molecular masses which when heated in a mould, undergo change in chemical composition to give a hard, infusible and insoluble mass. These cannot be re-melted.
Example : Bakelite, Melamine etc.
Or
Bio-degradable polymer : All those biopolymers which disintegrate by themselves in biological systems during certain period of time by enzymatic hydrolysis are called biodegradable polymers.
Example : Poly-β-Hydroxybutyrate-Co-β-Hydroxyvalerate (PHBV)
Uses : These are used
- in packaging
- in orthopaedic devices
- in controlled drug release
- in bacterial degradation
Question 19.
The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E
a
) of the reaction assuming that it does not change with temperature. [3]
[R = 8.314 J/K mol
-1
, log 4 = 0.6021
Answer:
Question 20.
What are the characteristics of the following colloids? Give one example of each. [3]
(i) Multimolecular colloids
(ii) Lyophobic sols
(iii) Emulsions.
Answer:
(i) Multimolecular colloids :
-
They are formed by the aggregation of a large number of atoms or molecules which generally have diameter less than 1 run.
Example : sols of gold, sulphur etc. - Their molecular masses are not very high.
- Their atoms or molecules are held together by weak Vander Waals’ forces.
(ii) Lyophobic sols : Liquid hating colloids in which there is no affinity between disperse phase and dispersion medium
Example: As
2
S
3
sol. Fe(OH)
3
sol.
(iii) Emulsions : The colloidal solution in which both dispersed phase and medium are in liquid state.
Example ; Milk, cream.
Question 21.
Give reasons for the following: [3]
(i) When R is an alkyl group R
3
P = 0 exist but R
3
N = 0 doesn’t.
(ii) PbCl
4
is more covalent than PbCl
2
.
(iii) N
2
is much less reactive at room temperature.
Answer:
(i) Due to presence of d-orbitals in P, it can form pπ-dπ: bonds and can extend its covalency beyond 4 while N cannot do so due to absence of d-orbitals.
(ii) According to Fajan’s rule, highly charged Pb
4+
can polarise the anion i.e., Cl
–
more effectively than Pb
2+
and hence PbCl
4
becomes more covalent than PbCl
2
.
(iii) Due to presence of triple bonds between 2 N atoms, their bond length decreases and hence bond dissociation energy increases which makes N
2
lesser reactive. While in phosphorus due to presence of single bond, more bond length, bond dissociation energy is low, hence very reactive.
Question 22.
For the complex [NiCl
4
]
2-
, write [3]
(i) the IUPAC name
(ii) the hybridization type
(iii) the shape of the complex (Atomic no. of Ni = 28)
OR
‘What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configuration of d
4
in terms of t
2g
and e
g
in an octahedral field when
(i) ∆
0
> P
(ii) ∆
0
< P
Answer:
(i) IUPAC name : [NiCl
4
]
2-
Tetrachloridonickelate (II) ion
(ii) Hybridization type : The above complex shows sp
3
hybridization
(iii) Shape : The above complex shows tetrahedral shape due to sp
3
hybridization.
Or
Crystal field splitting energy: When ligands approach the central metal ion, the degenerate d-orbitals split into two sets, one with lower energy (t
2g
) and the other with higher energy.
(eg). The difference of energy between these two sets of orbitals is called crystal field splitting energy. (∆
0
for octahedral complexes).
The magnitude of ∆
0
decides the actual configuration of d-orbitals by the help of mean pairing energy.
- if P > ∆ 0 then pairing of electrons does not occur and electrons enter in the higher energy e.g. orbitals and thus form high spin complexes due to weak field ligands.
- If P < ∆ 0 then pairing of electrons occurs within the same set and form low spin complexes due to strong field ligands.
Question 23.
Give reasons for the following:
(i) Ethyl iodide undergoes S
N
2 reaction faster than ethyl bromide.
(ii) (±) 2-Butanol is optically inactive.
(iii) C-X bond length in halobenzene is smaller than C-X bond length in CH
3
-X.
Answer:
(i) I, is better leaving group/C—I bond is weaker than C—Br bond.
(ii) (±)-Butan-2-ol is optically inactive because in racemic mix one type of rotation is cancelled by other.
(iii) In Halogen Benzene, halogen atom is attached to the sp
2
hybrid carbon atom while in CH
3
-X halogen atom is attached to sp
3
hybrid carbon atom. Hence C-X bond length in halobenzene is smaller than CH
3
—X.
Question 24.
Complete the following reactions: [3]
Answer:
Question 25.
(i) What class of drug is Ranitidine? [3]
(ii) If the water contains dissolved Ca
2+
ions, out of soaps and synthetic detergents, which will you use for cleaning clothes?
(iii) Which of the following is an antiseptic?
0.2% phenol, 1% phenol
Answer:
(i) Ranitidine is an Antacid.
(ii) We will use synthetic detergents because they can produce lather with the hard water containing Ca
2+
ions.
(iii) 0.2% phenol acts as an antiseptic.
Question 26.
Calculate the emf of the following cell at 25°C: [3]
Ag(s) | Ag
+
(10
-3
M) || Cu
2+
(10
-1
M) | Cu(s)
Given E
cell
° = +0.46V & log
10
n = n.
Given cell notation is incorrect correct cell formula is
Cu
2+
(10
-1
M) | Cu(s) || Ag
+
(10
-3
M) | Ag(s)
Answer:
Given: Cell notation is incorrect. Correct cell formula is
Cu
2+
(10
-1
m) | Cu
(s)
|| Ag
+
(10
-3
M) | Ag
(s)
According to Nernst equation
Question 27.
Shanti, a domestic helper of Mrs. Anuradha, fainted while mopping the floor. Mrs. Anuradha immediately took her to the nearby hospital where she was diagnosed to be severely ‘anaemic’. The doctor prescribed an iron-rich diet and multivitamins supplement to her. Mrs. Anuradha supported her financially to get the niedkines. After a month, Shanti was diagnosed to be normal. 3
After reading the above passage, answer the following questions:
(i) Name the vitamin whosé deficiency causes ‘pernicious anaemia’.
(ii) Give an example ofa water soluble vitamin.
Answer:
(i) Deficiency of Vitamin B
12
causes ‘pernicious anaemia.
(ii) Water soluble vitamin- : Vitamin B
2
(Riboflavin).
Question 28.
(a) State Raoult’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law?
(b) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (K
f
for benzene = 5.12 K kg/ mol) [5]
OR
(a) Define the following terms:
(i) Ideal solution
(ii) Azeotrope
(iii) Osmotic pressure.
(b) A solution of glucose (C
6
H
12
O
6
) in water is labelled as 10% by weight. What would be the molality of the solution?
(Molar mass of glucose = 180 g/mol)
Answer:
(a) 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 + Dieth1ether
According to Raoult’s law P
A
= \(\mathrm{P}_{\mathrm{A}}^0 \times x_{\mathrm{A}}\)
According to Henry’s law P
A
= K
H
× x
A
Thus both laws are identical and differ by their proportionality constants.
(b) We know that M
2
= \(\frac{1000 \mathrm{~K}_f w_2}{w_1 \Delta \mathrm{T}_f}\)
Given : w
2
= 1.0 g, w
1
= 50 g, ∆T
f
= 0.40 K, K
f
= 5.12 K kg mol
-1
∴ M
2
= \(\frac{1000 \mathrm{~g} \mathrm{~kg}^{-1} \times 5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 1.0 \mathrm{~g}}{50 \mathrm{~g} \times 0.40 \mathrm{~K}}\) = 256 g mol
-1
Or
(a) (i) Ideal solution: An ideal solution is that which obeys Raoult’s law and in which the intermolecular interactions between the different components are of same magnitude as that is found in pure components.
(ii) Azeotrope : It is a type of liquid mixture having a definite composition and boiling like a pure liquid (distills without change in compositions).
(iii) Osmotic pressure : The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through semi-permeable membrane is called osmotic pressure.
(b) 10% of glucose means 10 g of solute in 100 g of solvent
∴ Mass of solute = 10 g
Mass of solvent = 100 – 10 = \(\frac{90}{1000}\) g = kg
Molar mass of glucose = 180 g mol
-1
No. of moles = \(\frac{10}{180}\) = \(\frac{1}{18}\) mole
∴ Molarity \(=\frac{\text { No. of moles of solute }}{\text { mass of solvent in } \mathrm{kg}}\)
= \(\frac{1}{18} \times \frac{1000}{90}\) = \(\frac{100}{162}\) = 0.617 mol kg
-1
Question 29.
(a) Give reasons for the following: [5]
(i) Mn
3+
is a good oxidising agent.
(ii) E°M
2+
/M values are not regular for first-row transition metals (3d series).
(iii) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF
4
, whereas the highest oxide is Mn
2
O
7
.
(b) Complete the following equations:
OR
(a) Why do transition elements show variable oxidation states?
(i) Name the element showing the maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).
(ii) Name the element which shows only +3 oxidation state.
(b) What is lanthanoid contraction? Name an important alloy which contains some of the lanthanoid metals.
Answer:
(a) (i) Mn
3+
has electronic configuration 3d
sup>4
4s
0
. On reduction it gains one electron to become 3d
5
4s
0
which is half filled stable configuration. Hence it is a good oxidising agent.
(ii) \(E^o{ }_{\mathrm{M}}{ }^{2+} / \mathrm{M}\) values are not regular for first row transition metals due to abnormalities and irregularities in their ionization enthalpies (IE
1
+ IE
2
) and sublimation enthalpies.
(iii) Because oxygen stabilizes the highest oxidation state even more than fluorine and has ability to form multiple bonds with metal atoms.
(b)
(i) \(2 \mathrm{CrO}_4^{2-}\) + 2H
+
→ \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) + H
2
O
(ii)
Or
(a) Because the energy difference between (n-1) d-orbitais and ns-orbitals is very less.
(i) It is Manganese (Mn) which shows oxidation states from +2 to +7
(ii) Scandium (Sc) shows only +3 oxidation state.
(b) The steady decrease in the ionic radius from La
3+
to Lu
3+
is termed as lanthanoid contraction.
Consequences:
(i) Similar physical chemical propertiés among lanthanoids.
(ii) Zr and Hf exist together.
Important alloy: Mischmetal alloy which contains 95% lanthanoids and 5% Fe.
Question 30.
(a) How will you convert the following: [5]
(i) Propanone to Propan-2-ol
(ii) Ethanal to 2-hydroxy propanoic acid
(iii) Toluene to benzoic acid
(b) Give a simple chemical test to distinguish between:
(i) Pentan-2 -one and Pentan-3 -one
(ii) Ethanal and Propanal
OR
(a) Write the products of the following reactions:
(b) Which acid of each pair shown here would you expect to be stronger?
(i)
F-CH
2
-COOH or Cl-CH
2
-COOH
(ii)
Answer:
(a) (i) Propanone to Propan-2-ol
(ii) Ethanal to 2-hydroxy propanoic acid
(iii) Toluene to benzoic acid
(b)
(i) Pentan-2-one and Pentan-3-one
By lodoform test
(ii) Ethanal and Propanal : Add I 2 and NaOH in both the solutions. Ethanal gives yellow coloured precipitate but propanal does not.
Or
(a)
(b) (i) Due to much stronger I-effect of F over Cl, the FCH
2
COO
–
ion is much more stable than ClCH
2
COO
–
ion and hence FCH
2
COOH is a stronger acid than ClCH
2
COOH.
(ii) Ethanoic acid is more stronger acid than phenol due to less pKa than that of phenol and the carboxylate ion is much more resonance stabilized than phenoxide ion.
Set II
Note: Except for the following questions, all the remaining questions have been asked in the previous set.
Question 1.
Write the structure of 2-amino toluene. [1]
Answer:
Question 2.
Which aerosol depletes ozone layer? [1]
Answer:
Aerosols like foams, sprays etc. contain Freons which are responsible for depletion of ozone layer.
Question 4.
Ethanal is soluble in water. Why? [1]
Answer:
Ethanal is soluble in water due to H-bonding between the polar carbonyl group and water molecules.
Question 5.
Write the IUPAC name of the following compound: [1]
Answer:
IUPAC name : 2-Bromo-4-chloropentane
Question 7.
Write the name of the linkage joining two amino acids. [1]
Answer:
Peptide linkage joins two amino acids.
Question 8.
Give one example of a condensation polymer. [1]
Answer:
Example: Nylon 6,6
Question 9.
(a) Why does the presence of an excess of lithium make LiCl crystals pink? [2]
(b) A solid with cubic crystal is made of two elements P and Q Atoms of Q are at the comers of the cube and P at the body-centre. What is the formula of the compound?
Answer:
(a) This is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres).
(b) As atoms of Q are present at the 8 centres of the cube, therefore, number of atoms of
Q in the unit cell = \(\frac{1}{8}\) × 8 = 1
The atom P is at the body centre
∴ Number of atoms = 1
Ratio of atoms P: Q = 1 : 1
Hence, the formula of the compound is PQ.
Question 14.
Draw the structures of the following molecules: [2]
(i) XeF
6
(ii) H
2
S
2
O
7
Answer:
(i) XeF
6
(ii) H
2
S
2
O
7
Question 18.
Outline the principles of refining of metals by the following methods: [2]
- Zone refining
- Vapour phase refining
Answer:
(i) Zone refining :
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 métals 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.
Question 19.
Define the following terms giving an example of each: [2]
(i) Associated colloids
(ii) Lyophilic sol
(iii) Adsorption.
Answer:
(i) Associated colloids :
-
They are formed by aggregation of a large number of ions in concentrated solution
Example : soap sol - Their molecular masses are generally high.
- Higher is the concentration, greater are the Vander Waals’ forces.
(ii) Lyophilic sol :
Liquid loving colloids in which there is affinity between disperse phase and dispersion medium.
Example : Starch sol, Gum sol, Gelatin sol
(iii) Adsorption : The process of attracting and retaining the molecules of a substance on the surface of a liquid or a solid resulting into higher concentration of the molecules on the surface is called adsorption.
Question 22.
Write the main products of the following reactions: [2]
Answer:
Question 27.
Give reasons for the following: [3]
(i) Oxygen is gas but sulphur is solid.
(ii) O
3
acts as a powerful oxidising agent.
(iii) BiH
3
is the strongest reducing agent amongst all the hydrides of Group 15 elements.
Answer:
(ii) O
3
is a powerful oxidising agent due to its high energy content than oxygen and hence
decomposes to give diatomic oxygen and atomic oxygen
(iii) As we move down the group, the E—H bond length increases and their strength decreases. Bi—H bond is the weakest. It can break easily and evolves H
2
gas which acts as the reducing agent.
Set III
Note: Except for the following questions, all the remaining questions have been asked in previous sets.
Question 1.
What is especially observed when a beam of light is passed through a colloidal solution? [1]
Answer:
Tyndall effect is observed when a beam of light is passed through a colloidal solution.
Question 2.
What is the basicity of H
3
PO
3
and why? [1]
Answer:
The basicity of H
3
PO
3
is 2 because it contains only two ionizable H-atoms which are present as OH groups.
Question 3.
Write the IUPAC name of the following compound: [1]
Answer:
IUPAC name : 2, 5-dichiorotoluene
Question 8.
Write the structure of prop-2-en-l-amine. [1]
Answer:
H
2
C=CH — H
2
C — NH
2
Question 12.
Draw the structures of the following molecules:
(i) N
2
O
5
(ii) XeF
2
[2]
Answer:
(i) Structure of N
2
O
5
:
(ii)
Question 13.
(a) What change occurs when AgCl is doped with CdCl
2
?
(b) What type of semiconductor is produced when silicon is doped with boron? [2]
Answer:
(a) Impurity defect of ionic solids is produced when AgCl is doped with CdCl
2
. Due to this defect vacancies are created that result in higher electrical conductivity of the solid.
(b) p-type semi-conductor is obtained when silicon is doped with boron.
Question 18.
Name the principal ore of aluminium. Explain the significance of leaching in extraction as aluminium. [2]
Answer:
Bauxite (Al
2
O
3
2H
2
O) is the principal ore of aluminium.
The significance of leaching in the extraction of akiminiumis to prepare pure alumina from the bauxite ore in the following steps
After filteration of impurities sod. meta aluminate is neutralized by passing CO
2
Question 19.
Define the following terms with an example in each case: [3]
(i) Macromolecular Sol
(ii) Peptization
(iii) Emulsion.
Answer:
(i) Macromolecular sol :
-
They are molecules of large size
Example: polymers like rubber, nylon, starch proteins etc. - They have high molecular masses.
- Due to long chain, the Vander Waals’ forces holding them are comparatively stronger.
(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) Emulsions : The colloidal solution in which both dispersed phase and medium are in liquid state.
Example: Milk, cream.
Question 21.
Give reasons for the following: [3]
(i) Though nitrogen exhibits a +5 oxidation state, it does not form pentahalide.
(ii) Electron gain enthalpy with a negative sign of fluorine is less than that of chlorine.
(iii) The two oxygen-oxygen bond lengths in ozone molecules are identical.
Answer:
(i) Due to absence of empty d-orbitals, N
2
does not form pentahalide.
(ii) Because of small size of fourme atom and strong electron-electron repulsions in its compact 2p orbitals.
(iii) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm
