Addition of salt KY (having a common ion Y^-) decreases the solubility of MY and NY₃ due to common ion effect.
Option (a) and (b) are wrong
For salt MY, MY ⇌ M⁺ + Y^-
K_sp = (s) (s)
6.2 × 10^-13 = s²
\(\therefore\) s = \(\sqrt{6.2 \times 10^{-13}} = 10^{-7}\)
For salt NY₃,
NY₃ ⇌ N³⁺ + 3Y^-
K_sp = (s) (3s)³
K_sp = 27s⁴
\(s = (\frac{6.2 \times 10^{-13}}{27})^{1/4}\)
s = 10^-4
The molar solubility of MY in water is less than of NY₃

A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

Mn²⁺ ⇒ 3d⁵ contains 5 unpaired electrons
n = 5,
\( \sqrt{n(n+ 2)} \) BM
\(= \sqrt{5(5+ 2)} = \sqrt{35} = 5.92 BM\)

The reciprocal of the resistance \((\frac{l}{R})\) gives the conductance of an electrolytic solution. The SI unit of conductance is Siemen (S).

(ii) benzoic acid

(iii) α - hydroxyphenylaceticacid.

(i) Neutralisation of chargers of ion will taken place and hence precipitation will take place (ie) Fe³⁺ and S^2- ion changes are neutralized. No new compounds are formed.
(ii) Fe(OH)₃ is a positive Sol
(iii) As₂S₃ is a negative Sol

[H₃O⁺] = 4 \(\times\) 10^-5M
pH = - log₁₀[H₃O⁺]
pH=-log₁₀[4 \(\times\) 10^-5]
pH = -log₁₀[4] - log₁₀[10^-5]        log₁₀ 10 = 1
pH = -log 4 + 5log₁₀10
= 5 - log 4
= 5 - 0.6021
=4.3979
Since pH is less than 7, the solution is acidic.

Overall order of this reaction = 1+ 2 = 3

(i) The electron gain enthalpy for oxygen is less negative because of its small size due to which the electron repulsions in the relatively small 2p-subshell are comparatively large.
(ii) Hence the Incoming electrons are not accepted with the same ease as in case of sulphur as it has relatively large size.

(i) An anionic complex compound contains a complex anion and simple cation.

(ii) A cationic complex contains complex cation and simple anion

Acidic nature of nitro alkanes:
(i) The α -H atom of 1⁰ & 2⁰ nitroalkanes show acidic character because of the eIectron with drawing effect of NO₂ group.
(ii) These are more acidic than aldehydes, ketones, ester and cyanides.
(iii) Nitroalkanes dissolve in NaOH solution to form a salt.
(iv) Aci - nitro derivatives are more acidic than nitro form.
(v) When the number of alkyl group attached to α carbon increases, acidity decreases, due to +1 effect of alkyl groups

(i) Sucrose: D - glucose and D - fructose linked by α. - 1, 2 glycosidic bond.
(ii) Maltose: Two molecules of α. - D - glucose linked by α. - 1, 4 glycosidic bond.
(iii) Lactose: f3 - D - glucose and f3 - D galactose linked by β - 1, 4 glycosidic bond

Glycerol can give rise to a variety of oxidation products depending on the nature of the oxidising agent used for oxidation.
(i) Oxidation of glycerol with dil. HNO₃ gives glyceric acid and tartronic acid.
(ii) Oxidation of glycerol with Cone. HNO₃ gives mainly glyceric acid.
(iii) Oxidation of glycerol with bismuth nitrate gives as meso oxalic acid.
(iv) Oxidation of glycerol with Br/H₂O (or) NaOBr (or) Fenton reagent (FeSO₄ + H₂O₂) gives a mixture of glyceraldehyde I and dihydroxy acetone (This mixture is named as glycerose).

(i) When it is placed on the right-hand side of the zinc electrode, the hydrogen electrode reaction is,
2H⁺+ 2e^- ➝ H²
The electrons flow to the SHE and it acts as the cathode.
(ii) When the SHE is placed on the left hand side, the electrode reaction is
H₂ ➝ 2H⁺+ 2e^- 
The electrons flow to the copper electrode and the hydrogen electrode acts as the anode.

According to Henderson – Hasselbalch equation,
\(pH=pK_{a}+\log\frac{[salt]}{[acid]}\)
\(p{K_{a}}=-\log K_{a}=-\log(1.8\times10^{-5})=4.74\)
[Salt]=\(\frac{\text {Number of moles of sodium acetate}}{\text {Volume of the solution (litre)}}\)
Number of moles of sodium acetate =\(\frac{\text {mass of sodium acetate}}{\text {molar mass of sodium acetate}}\)
\(=\frac{8.2}{82}=0.1\)
\(\therefore [Salt]=\frac{0.1\ mole}{1/2 \ Litre}=0.2M\)
\([acid]=\frac{(\frac{mass \ of \ CH_{3}COOH}{molar \ mass \ of \ CH_{3}COOH})}{\text{Volume of solution in litre}}\)
=\(\frac{(\frac{6}{60})}{\frac{1}{2}}\)=0.2 M
\(\therefore pH=4.74+log\frac{(0.2)}{(0.2)}\)
pH = 4.74 + log1
pH = 4.74 + 0 = 4.74

(i) The colour of the transition elements is due to the d-d transition.
(ii) Since the oxidation state and polarising power of Mn in Mn₂O₇ is higher, it is acidic in nature
(iii) μ = \(\sqrt{n(n+2)}=\sqrt{4(4+2)}\)
= 4.90 BM

(i) Vapour phase method, the metal is treated with a suitable reagent which can form a volatile compound with the metal.
(ii) Then the volatile compound is decomposed to give the pure metal.
(iii) Vapour phase method is used for refining nickel.

(i) Ionic solids have high melting points.
(ii) These solids do not conduct electricity, because the ions are fixed in their lattice positions.
(iii) They are hard so strong external force can change the relative positions of ions.

A hydride of 2^nd period alkali metal (A) is lithium hydride (LiH).
Lithium hydride (A) reacts with diborane (B) to give lithium borohydride (C) which is acts as a reducing agent.
B₂H₆ + 2 LiH \(\xrightarrow[]{ether}\) 2 LiBH₄
[Diborane (B)]   [Lithium hydride (A)]  [Lithium borohydride (C)]
Result:

(i) NH₃ with HCHO:
Formaldehyde forms hexamethylene tetramine with NH₃.
6HCHO + 4NH₃ ➝ (CH₂)₆N₄ + 6Hp
(ii) NH₃ with CH₃CHO:
Acetaldehyde reacts with NH₃ to form aldimine.
\(CH_{ 3 }CHO+{ NH }_{ 3 }\longrightarrow { H }_{ 3 }C-\overset { \underset { | }{ { NH }_{ 2 } } }{ \underset { \overset { | }{ H } }{ C } } -OH\overset { -{ H }_{ 2 }O }{ \longrightarrow } \underset { aldimine }{ CH_{ 3 }-CH=NH } \) 
(iii) NH₃ with C₆H₅CHO:
Benzaldehyde undergoes condensation with ammonia to form hydrobenzamide

(iv) NH₃ with CH₃CO - CH₃:
Acetone with ammonia forms acetone ammonia initially at room temperature. On heating, it forms diacetone amine 
\({ H }_{ 3 }C-\overset { \underset { | }{ { CH }_{ 3 } } }{ C } =O+{ HNH }_{ 2 }\longrightarrow \underset { acetone \ ammonia }{ CH_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { NH }_{ 2 } } }{ C } } -{ OH }_{ 2 } } \) 
\({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { NH }_{ 2 } } }{ C } } -OH+{ CH }_{ 3 }-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 } \overset { high \ temp }{ \underset { -{ H }_{ 2 }O }{ \longrightarrow } } \underset {Diacetone \ amine}{{ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } } { \underset { \overset { | }{ { NH }_{ 3 } } }{ C } } -{ CH }_{ 2 }}-\underset { \overset { || }{ O } }{ C } -{ CH }_{ 3 }\)

(i) An Organic compound (A) with molecular formula CH₂O is formaldehyde (HCHO).
(ii) Formaldehyde reacts with CH₃MgI to give compound (B) ethanol.

(iii) Ethanol on dehydration with cone. H₂SO₄ at 410K gives compound (C)

(i) For a chemical reaction, catalyst is needed in very small quantity.
(ii) There may be some physical changes, but the catalyst remains unchanged in mass and chemical composition in a chemical reaction.
(iii) A catalyst itself cannot initiate a reaction. 
(iv) A solid catalyst will be more effective if it is taken in a finely divided form.
(v) A catalyst are specific in nature.
(vi) In an equilibrium reaction, presence of catalyst reduces the time for attainment of equilibrium and hence it does not affect the position of equilibrium and the value of equilibrium constant.
(vii) A catalyst is highly effective at a particular temperature called as optimum temperature.
(viii) Presence of a catalyst generally does not change the nature of products

On the basis of magnetic properties, materials can be broadly classified as
(a) paramagnetic materials
(b) diamagnetic materials, besides these there are ferromagnetic and antiferromagnetic materials
(i) Materials with no elementary magnetic dipoles are diamagnetic, in other words a species with all paired electrons exhibits diamagnetism.
(ii) This kind of materials are repelled by the magnetic field because the presence of external magnetic field, a magnetic induction is introduced to the material which generates weak magnetic field that oppose the applied field
(iii) Paramagnetic solids having unpaired electrons possess magnetic dipoles which are isolated from one another.
(iv) In the absence of external magnetic field, the dipoles are arranged at random and hence the solid shows no net magnetism.
(v) But in the presence of magnetic field, the dipoles are aligned parallel to the direction of the applied field and therefore, they are attracted by an external magnetic field.
(vi) Ferromagnetic materials have domain structure and in each domain the magnetic dipoles are arranged.
(vii) But the spin dipoles of the adjacent domains are randomly oriented.
(viii) Some transition elements or ions with unpaired d electrons show ferromagnetism.

Density of the element, d = 2.7 x 10³ kg m^-3
Molar mass, M = 2.7 x 10^-2 kg mol^-1
Edge length, a = 405 pm
= 405 x 10^-12 m
= 4.05 x 10^-10 m
Avogadro's number, N_A= 6.022 x 10²³ mol^-1
\(\therefore d=\frac { Z\times M }{ { a }^{ 3 }\times { N }_{ A } } \)
\(\Rightarrow Z=\frac { d\times { a }^{ 3 }{ N }_{ A } }{ M } \)
\(=\frac { 2.7\times { 10 }^{ 3 }kg\quad { m }^{ -3 }{ (4.05\times { 10 }^{ -10 }m) }^{ 3 }\times 6.022\times { 10 }^{ 23 }{ mol }^{ -1 } }{ 2.7\times { 10 }^{ -2 }kg\quad { mol }^{ -1 } } \)
= 4.004 = 4.
This implies that four atoms of the element are present per unit cell. Hence the unit cell is face centred cubic.

(i) [NiCI₄]^2-
IUPAC name - Tetrachloridonickelate (II) ion
(ii) Ni²⁺ = 3d⁸,4s⁰

Cl^- being a weak field ligand cannot pair up the unpaired electron. So, it is sp³ hybridised, and it has tetrahedral geometry.

(i) Order of the reaction =1; \(\mathrm{t}_{1 / 2}=60 \mathrm{~s} ; \mathrm{k}=?\)
\(\mathrm{k}=\frac{0.6932}{\mathrm{t}_{\frac{1}{2}}} \)
\(=\frac{0.6932}{60}  \)
\(k =1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
(ii) \(\left[\mathrm{A}_{0}\right]=100 \% ; \mathrm{t}=180 \mathrm{~s} ;[\mathrm{A}]=? ; \mathrm{k}=1.155 \times 10^{-2} \mathrm{~s}^{-1}\)
For first order reaction
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]} \)
\(1.155 \times 10^{-2} =\frac{2.303}{180} \log \left(\frac{100}{[A]}\right)  \)
\(\frac{0.01155 \times 180}{2.303} =\log \left(\frac{100}{[A]}\right)  \)
\(0.9027 =\log 100-\log [\mathrm{A}]  \)
\(\log [\mathrm{A}] =\log 100-0.9027 \)
\(\log [A]=2-0.9027 \)
\(\log [A]=1.0972 \)
[A] = antilog  of (1.0972)
[A] =12.51 %

B) if both assertion and reason are true but reason is not the correct explanation of assertion.