BF₃ → electron deficient →  Lewis acid
PF₃ → electron rich →  Lewis base
CF₄ → neutral →  neither Lewis acid nor base
SiF₄ → neutral →  neither Lewis acid nor base

H₂O + H₂O ⇌  H₃O⁺ + OH^-
acid 1  base 1  acid 2   base 2
HF + H₂O ⇌  H₃O⁺ + F^-
acid 1  base 1  acid 2   base 2
∴ Conjugate bases are OH^- and F^- respectively.

Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-
Given that pH = 9
pOH = 14 - 9 = 5
[pOH = - log₁₀ [OH]]
[OH^-] = 10 ^[pOH]
[OH] = 10^-5 M
Ksp = [Ca²⁺] [OH^-]
= 10^-5/2 x (10^-5)²  = 0.5 x 10^-15

Cr ⇒ [Ar]3d⁵4s¹

Aldehydes can be prepared by the hydrogenation of acid chloride, in the presence of Pd supported by barium sulphate

(i) Aqueous solution of carboxylic acids turn blue litmus into red colour.
(ii) Carboxylic acids give brisk effervescence with sodium bicarbonate due to the I evolution of CO₂,

\(\underset { acetylhloride }{ { CH }_{ 3 }COCl } \overset { { H }_{ 2 }/Pd/Ba{ SO }_{ 4 } }{ \longrightarrow } \underset { Acetaldehude(A) }{ { CH }_{ 3 }CHO } \) 
This reaction is Rosenmund reduction.
\({ CH }_{ 3 }CHO\overset { { NH }_{ 2 }OH }{ \underset { -{ { H }_{ 2 }O } }{ \longrightarrow } } \underset { Acetaldoxime }{ { CH }_{ 3 }CH=NOH } \) 
The above reaction is nucleophilic addition reaction followed by elimination of water.

(i) Lewis acid: It is a species that accepts an electron pair. Eg: \(\mathrm{Ag}^{+} ; \mathrm{BF}_{3} ; \mathrm{A} / \mathrm{Cl}_{3}\)
(ii) Lewis base: It is a species that donates an electron pair. Eg: \( \mathrm{Cl}^{-} ; \mathrm{NH}_{3} ; \mathrm{H}_{2} \mathrm{O}\)

Electronic configuration of Ce⁴⁺ = [Xe] 4f⁰5d⁰6s⁰
Electronic configuration of Co²⁺ = [Ar]3d⁷

An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. They are usually non-stoichiometric compounds. Transition metals form a number of interstitial compounds such as TiC, ZrH_1.92, Mn₄N etc.
Properties of interstitial compound
(i) They are hard and show electrical and thermal conductivity.
(ii) They have high melting points higher than those of pure metals.
(iii) Transition metal hydrides are used as powerful reducing agents 
(iv) Metallic carbides are chemically inert.

Benzaldehyde condenses with tertiary aromatic amines like N, N - dimethyl aniline in the presence of strong acids to from triphenyl methane dye or malachite green dye.

\(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightleftharpoons 3 \mathrm{Ca}^{2+}_{(aq)}+2 \mathrm{PO}_{4_{(aq)}}^{3-}\\\)
(s)                           (3s)             (2s)
\(K_{sp}=[Ca^{2+}]^{3}[PO_{4}^{3-}]^{2}\)
\(K_{sp}=(3s)^{3}(2s)^{2}\)
\(K_{sp}=27s^{3}.4s^{2}\)
\(K_{sp}=108s^{5}\)
(or)
\(K_{s p} =m^{m} \cdot n^{n} \cdot(s)^{m+n} \)
\(K_{\text {sp }} =3^{3} \cdot 2^{2} \cdot(s)^{3+2} \)
\(K_{s p} =27 \times 4 \times(s)^{5} \)
\(=108(s)^{5}=108 s^{5}\)

Considering Ba(OH)₂ to be a strong base:
\(\text { Normality } =\text { Molarity } \times \text { Acidity }  \)
\(=1.5 \times 10^{-3} \times 2\)
\({\left[\mathrm{OH}^{-}\right] } =3 \times 10^{-3}  \)
\(\mathrm{pOH} =-\log _{10}[\mathrm{OH}^-]  \)
\(=-\log _{10}\left(3 \times 10^{-3}\right)  \)
\(=-\left[\log _{10} 3+3 \log 10\right]  \)
= 3-log 3 
= 3-0.4771 
= 2.5229 
\(\mathrm{pH} =14-\mathrm{pOH} \)
= 14-2.5229 
pH = 11.4771 = 11.48

(i) Fe³⁺ - electronic configuration - [Ar] 3d⁵
(ii) It has exactly half-filled stable electronic configuration.
(iii) Fe²⁺ - electronic configuration -[Ar]3d⁶
(iv) It has only partially filled d-orbitals.
Hence Fe³⁺ is more stable than Fe²⁺.

Potassium permanganate is prepared from pyrolusite (MnO₂).
(i) Conversion of (MnO₂) to potassium manganate
Powdered ore is fused with KOH in the presence of air, green coloured potassium manganate is formed.
\(2 \mathrm{MnO}_{2}+4 \mathrm{KOH}+\mathrm{O}_{2} \rightarrow 2 \mathrm{~K}_{2} \mathrm{MnO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad Potassium \ manganate \ (green)\)
(ii) Oxidation of potassium manganate to (O₃) potassium permanganate
(a) Chemical oxidation
In this method, potassium manganate is treated with ozone (O₃) or chlorine to get potassium permanganate.
\(2 \mathrm{MnO}_{4}^{2-}+\mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{MnO}_{4}^{-}+2 \mathrm{OH}^{-}+\mathrm{O}_{2}  \)
\(2 \mathrm{MnO}_{4}^{2-}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{MnO}_{4}^{-}+2 \mathrm{Cl}^{-}\)
(b) Electrolytic oxidation
In this method, aqueous solution of potassium manganate is electrolysed in the presence of alkali.
\(\mathrm{K}_{2} \mathrm{MnO}_{4} \rightleftharpoons 2 \mathrm{~K}^{+}+\mathrm{MnO}_{+}^{2-} \)
\(\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}\)
Manganate ions are converted into permanganate ions at anode.
\(2 \mathrm{MnO}_{4}^{2-} \rightleftharpoons 2 \mathrm{MnO}_{4}^{-}+2 \mathrm{e}^{-}\\ Green \quad \quad Purple\)
H₂ is liberated at the cathode.
\(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \uparrow\)
The purple coloured solution is concentrated by evaporation and forms crystals of potassium permanganate on cooling.

This reaction is catalysed by base. The carbanion generated is nucleophilic in nature. Hence it can bring about nucleophilic attack on carbonyl group
Step 1: The carbanion is formed as the a-hydrogen atom is removed as a proton by the base

Step 2: The carbanion attacks the carbonyl carbon of another unionised aldehyde molecule

Step 3: The alkoxide ion formed is protonated by water to give 'aldol'.

(i) Ostwald's dilution law relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c).
where  \(\alpha=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
(ii) The dissociation of acetic acid can be represented as
\(\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\)
The dissociation constant of acetic acid is,
\({ K }_{ a }=\frac { \left[ { H }^{ + } \right] \left[ { CH }_{ 3 }COO^{ - } \right] }{ \left[ { CH }_{ 3 }COOH \right] } \)  ........(1) 

Substituting the equilibrium concentration in equation (1)
\({ K }_{ a }=\cfrac { \left( \alpha C \right) \left( \alpha C \right) }{ \left( 1-\alpha \right) C } \) 
\({ K }_{ a }=\cfrac { { \alpha }^{ 2 }C }{ 1-\alpha } \) .......(2) 
(iii) We know that weak acid dissociates only to a very small extent compared to one, a is so small and hence in the denominator (1 - α) ⋍1. The above expression (2) now becomes,
k_a =a²C \(\Rightarrow { \alpha }^{ 2 }=\cfrac { { k }_{ a } }{ C } \) ; \(\alpha =\sqrt { \cfrac { { K }_{ a } }{ C } } \) 
(iv) When dilution increases, the degree of dissociation of weak electrolyte also increases. This is called Ostwald's dilution law
Also \(;\left[\mathrm{H}^{+}\right]=\alpha \mathrm{C}\) and \(\left[\mathrm{H}^{+}\right]=\left(\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}\right) \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{a}} \mathrm{C}^{2}}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}\)
Similarly for a weak base
\(\begin{aligned} & \mathrm{K}_{\mathrm{b}}=\alpha^2 \mathrm{C} ; \quad \therefore \alpha=\sqrt{\frac{\mathrm{k}_{\mathrm{b}}}{\mathrm{C}}}, \\ \end{aligned}\)
\(\begin{aligned} & {\left[\mathrm{OH}^{-}\right] \alpha \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}}} \times \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}} \mathrm{C}^2}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}} \end{aligned}\)