Cathodic protection:
In this technique, unlike galvanising the entire surface of the metal to be protected need not be covered with a protecting metal. Instead, metals such as Mg or zinc which is corroded more easily than iron can be used as a sacrificial anode and the iron material acts as a cathode. So iron is protected, but Mg or Zn is corroded. This known as sacrificial protection. (or) Cathodic protection.

(i) A Colloid depends on the size of the particle. A Colloid is formed when the size of the particle lies between 1 nm and 100 nm. For example soap dissolves in water to form colloidal soap solution whereas it dissolves in alcohol to form a true solution. Thus change of state takes place. A colloidal state maybe an intermediate between a true solution and a suspension.
(ii) Also some crystalloids under certain conditions can be colloids. NaCl is a crystalloid in aqueous medium; but when mixed with benzene it acts as a colloid.

K_sp = 1.8 \(\times\)10^-10, [AgNO₃]= 1 M
\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \)
                    s            s
\(\mathrm{AgNO}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{Ag}_{(\text {aq })}^{+}+\mathrm{NO}_{3_{(\text {aq })}}^{-}\\ 1 \mathrm{M} \quad \quad \quad \quad \quad 1 \mathrm{M} \quad \quad 1 \mathrm{M} \)
\(\left[\mathrm{Ag}^{+}\right]=(\mathrm{s}+1) \approx 1 \quad(\therefore \mathrm{s}<<1) \)
\(\left[\mathrm{Cl}^{-}\right]=\mathrm{s} \)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \)
\(1.8 \times 10^{-10}=(1)(s) \)
\(\therefore \mathrm{s}=1.8 \times 10^{-10} \mathrm{M}\)

α-D (+) glucophyranose

K = 1.54 \(\times\) 10^-3s^-1; t_1/2 = ?
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 0.693 }{ k } \)
\(=\frac { 0.693 }{ 1.54\times { 10 }^{ -3 } } \)
= 450 s

Chromium (24), the electronic configuration is 3d⁵4s¹. It ready to lose its outer most electron (4s¹) to get exactly half-filled stable electronic configuration. The electronic configuration of Zinc is 3d¹⁰4s².ie., It has completely filled stable configuration. From this configuration, the removal of 1e^- from 4s orbital is very difficult & it required more ionisation enthalpy. Due to this reason Zn has higher first Ionisation enthalpy (1.E₁) than that of chromium.

The CFSE is defined as the energy of the electronic configuration in the ligand field minus the energy of the electronic configuration in the isotropic field.
CFSE (\(\Delta\)E_o) = {E_Lf}-{E_iso}
={[n_t2g(-0.4) + n_eg(0.6)]\(\Delta\)_o + n_pP} - {n'_pP}
Here, n_t2g is the number of electrons in t_2g orbitals;
n_eg is number of electrons in e_g orbitals;
n_p is number of electron pairs in the ligand field; &
n'_p is the number of electron pairs in the isotropic field (barycenter).
P - pairing energy

Oxidation number of F = -1
Oxidation number of H = +1
Oxidation number of O in HOF =x
(+1) + x + (-1) = 0
x = 0
Oxidation number of O in HOF = 0

(i) The extraction of metals from their oxides can be carried out by using different reducing agents.
(ii) Consider the following reaction
\(\frac{2}{\mathrm{y}} \mathrm{M}_{\mathrm{x}} \mathrm{O}_{\mathrm{y}(\mathrm{s})} \rightarrow \frac{2 \mathrm{x}}{\mathrm{y}} \mathrm{M}_{(s)}+\mathrm{O}_{ 2(\mathrm{~g})}\)            (1)
(iii) The above reduction may be carried out with carbon. In this case the reducing agent carbon may be oxidized to either CO or CO₂
\(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}  \)                  (2)
\(2 \mathrm{C}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{(\mathrm{g})} \)                (3)
(iv) If CO is used as a reducing agent
\(2 \mathrm{CO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}\)           (4)
(v) A suitable reducing agent is selected based on the thermodynamics considerations.
(vi) We know that for a spontaneous reaction, the change in free energy (\(\triangle\)G) should be negative.
(vii) Therefore, thermodynamically, the reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is negative.
(viii) Hence, the reducing agent is selected in such a way that it provides a large negative \(\triangle\)G value for the coupled reaction.

Cell reactions:
Oxidation at anode: \(Cd_{(s)}\rightarrow Cd^{2+}_{(aq)}+2e^{-}\);    (E⁰_ox)_cd|cd²⁺ = 0.40V ; (E⁰)_cd|cd²⁺ = -0.40V
Reduction at cathode: \(Cu^{2+}_{(aq)}+2e^{-}\rightarrow Cu_{(s)} \);  (E⁰_red)_cu²⁺|Cu = +0.34V
Adding: \(Cd_{(s)}+Cu^{2+}_{(aq)}\rightarrow Cd^{2+}_{(aq)}+Cu_{(s)}\)
E⁰_cell=(E⁰_ox)+(E⁰_red)
=(-0.4) + 0.34V
= 0.74V.
Emf is +ve, so \(\Delta G\) is -ve, the cell reaction is feasible.

(i) A polymer containing two or more different kinds of monomer units is called a copolymer.
(ii) For example, SBR rubber(Buna-S) contains styrene and butadiene monomer units.
(iii) Copolymers have properties quite different from the homopolymers.
(iv) Mixture of styrene and 1,3 butadiene to form a copolymer (Buna -S)
Preparation of Buna-S:
It is a co-polymer. It is obtained by the polymerisation of buta -1,3 - diene and styrene in the ratio 3: 1 in the presence of sodium.

According to Arrhenius equation
\(\log\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right) \)
T₂ = 400K ; k₂ = 0.04 s^-1
T₁ = 200K ; k₁ = 0.02 s^-1
\(\log\left( \frac { 0.04}{ 0.02} \right) =\frac { { E }_{ a } }{ 2.303\times 8.314} =\left( \frac { 400-200 }{ 200\times 400 } \right) \)
\(\log(2)=\frac { { E }_{ a } }{ 2.303\times 8.314} =\left( \frac { 1 }{ 400 } \right) \)
E_a = log(2) \(\times\) 2.303 \(\times\) 8.314 \(\times\) 400
= 0.3010 \(\times\) 2.303 \(\times\) 8.314 \(\times\) 400
E_a = 2305 J mol^-1 = 2.305 kJ mol^-1

Zirconium is a 4d element and Hafnium is a 5d element. There is an unexpected observation in the atomic radius of 5d elements which have nearly the same atomic radius as that of corresponding 4d elements. This is due to lanthanoid contraction. Hence they exhibit similar properties.

For cubic closed packed structure
\(r =\frac{a \sqrt{2}}{4}  \)
\(\therefore a =\frac{4 r}{\sqrt{2}}  \)
\(=\frac{4 \times 1.25 \times 10^{-8}}{1.414}=3.53 \times 10^{-8} \mathrm{~cm} \)
= 353 pm

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:

Compound	Formula	Name
A	LiH	Lithium hydride
B	B2H6	Diborane
C	LiBH4	Lithium borohydride

(i) Compound A giving violet colour with neutral ferric chloride is phenol (C₆H₅OH).
(ii) (B) and (C) isomers are formed when alkaline solution of (A) is refluxed with CCI₄.

(iii) (A) when treated with C₆H₅N₂CI, forms a red orange dye (D).

Compound	Compound Name	Formula
A	Phenol	C6H5OH
B	o-Hydroxy benzoic acid
C	p-Hydroxy benzoic acid
D	p-Hydroxy azo benzene

Compound A,B,C and D