for n ≠ 1 t_1/2 = \(\frac{2^{n-1} -1}{(n- 1) k[A_{0}]^{-1}}\)
for n = 0; t_1/2 = \(\frac{1}{2 k[A_{0}]^{-1}}\)
t_1/2 = \(\frac{[A_{0}]}{2 k}\)
t_1/2 α [A₀] ...(1)
Given [A₀] = 0.002 M; t_1/2 = 10 min
[A₀] = 0.04M; t_1/2 = ?
Substitute in (1)
10 min α 0.02 M...(2)
t_1/2 α 0.04M ....(3)
(3)(2)
⇒ t_1/2 / 10 min
= 0.04 M/0.02 M
t_1/2 = 2 x 10 min = 20 min

(i) The electrolytic cell consists of two iron electrodes dipped in molten sodium chloride and they are connected to an external DC power supply via a key as shown in the figure. The electrode which is attached to the negative end of the power supply is called the cathode, and the one which attached to the positive end is called the anode. Once the key is closed, the external DC power supply drives the electrons to the cathode and at the same time pull the electrons from the anode.

Cell reactions:
Na⁺ ions are attracted towards cathode, where they combines with the electrons and reduced to liquid sodium.

Cathode (reduction)
\(N a_{(l)}^{+}+e^{-} \rightarrow N a_{(l)} \quad ; \quad E^{0}=-2.71 V\)
Similarly, Cl^- ions are attracted towards anode where they lose their electrons and oxidised to chlorine gas.
Anode (oxidation)
2CI^-_(l) ⟶ CI_2(g) + 2e^- E⁰ = -1.36V
The overall reaction is
2Na⁺_(l) + 2Cl^-_(l)➝ 2Na_(l) + Cl_2(g) ; E° = - 4.07V
(ii) The negative E° value shows that the above reaction is a non-spontaneous one.
(iii) Hence, we have to supply a voltage greater than 4.07V to cause the electrolysis of molten NaCI.
(iv) In electrolytic cell, oxidation occurs at the anode and reduction occur at the cathode as in a galvanic cell.
(v) But the sign of the electrodes is the reverse i.e., in the electrolytic cell cathode is -ve and anode is +ve.

(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⁺.

(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) Gangue: The ores are associated with nonmetallic impurities, rocky materials and siliceous matter which are collectively known as gangue.
Eg: SiO₂ is the gangue present in the iron ore (Fe₂O₃)
(ii) Slag: In the smelting process, a flux combines with Silica gangue forming slag.
CaO_(s) + SiO_2(s) → CaSiO_3(s)
Flux     + gangue → Slag

(ii) 2,4-dichlorobenzaldehyde.

(iii) ethanal

(i)  According to Lewis theory a species that donates a pair of electron is called Lewis base.
(ii) Nitrogen more in NH₃ is less electro negative than oxygen in water. So the non - bonded electron pair on nitrogen is more available for sharing than a non - bonded electron pair on oxygen atom. So NH₃ is a stronger lewis base than H₂O.

(i) Let A = 100M, [A₀] - [A] = 20M,
For the zero order reaction
\(k=\left( \frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \right) \)
(i) 20% completion  \(k=\left( \frac { 20M }{ 20min } \right) \) = 1 mol L^-1 min^-1 
(ii) 80% completion
\(\mathrm{K}=1 \mathrm{~mol} \mathrm{~L}{ }^{-1} \mathrm{~s}^{-1} ;\left[\mathrm{A}_{0}\right]=100 \mathrm{M} ;\left[\mathrm{A}_{0}\right]-[\mathrm{A}]=80 \mathrm{M} ; \mathrm{t}=?\)
\(\therefore t=\left(\frac{\left[A_{O}\right]-[A]}{K}\right)=\frac{80}{1}=80 \mathrm{mins}\)

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

In this method, the metal is treated with a suitable reagent which can form a volatile compound with the metal.
Then the volatile compound is decomposed to give the pure metal.

(i) [Co(en₃)][Cr(CN)₆] - Exhibits coordination isomerism
(a) [Co(en₃)][Cr(CN)₆]

b) [Cr(en₃)][Co(CN)₆]

(ii) [Co(NH₃)₅(NO₂)]²⁺ - Exhibits linkage isomerism
a) [Co(NH₃)₅(NO₂)]²⁺ - N attached

b) [Co(NH₃)₅(ONO₂)]²⁺ - O attached

(iii) [Pt(NH₃)₃(NO₂)]Cl - Exhibits ionisation isomerism
a) [Pt(NH₃)₃(NO₂)]Cl

(i) Acetic anhydride from acetic acid:

Carboxylic acid on heating in the presence of a strong dehydrating agent such as P₂O₅ forms acid anhydride.
(ii) Ethylacetate from methyl acetate:
1. In presence of a little acid, methyl acetate is cleaved by ethyl alcohol to form ethyl acetate.
\(\underset{Methyl \ acetate }{\mathrm{CH}_{3} \mathrm{COOCH}_{3}}+ \underset{Propyl \ alcohol} {\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}} \stackrel{\mathrm{H}^{+}}{\rightleftharpoons} \underset { Ethyl \ acetate}{\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}}+\mathrm{CH}_{3} \mathrm{OH} \)
2. This is called 'trans esterification'.
Acetamide from methyl cyanide:
Partial hydrolysis of alkyl cyanides with cold con HCl gives amides.

(iv) Lactic acid from ethanal

Acetophenone frorr acetylchloride:
Friedel Crafts acetylation of, benzene with CH₃COCI/AICI₃

Ethane from sodium acetate:
\(\underset{Sodium \ acetate}{2 \mathrm{CH}_{3}} \mathrm{COONa}+2 \mathrm{H}_{2} \mathrm{O} \stackrel{\text { Electrolysis }}{\longrightarrow} \underset{Ethane}{\mathrm{CH}_{3}}-\mathrm{CH}_{3}+2 \mathrm{CO}_{2}+\mathrm{H}_{2}+2 \mathrm{NaOH} \)
This is Kolbe's electrolysis.
Benzoic acid from toluene:
Toluene is oxidised by acidified KMnO₄
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3} \frac{\mathrm{H}^{+} / \mathrm{KMnO}_{4}}{(\mathrm{O})} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\)
Malachite green from benzaldehyde:
Benzaldehyde condenses with tertiary aromatic amines like N, N - dimethyl aniline in the presence of strong acids to form triphenyl methane dye.

Cinnamic acid from benzaldehyde:

Acedaldehyde from ethyne:

(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 %