\(\Delta_{\mathrm{r}} G^{\ominus}=-n F \mathrm{E}_{\text {(cell) }}^{\ominus}\) 
n in the above equation is 2, F = 96487 C mol ^–1 and \(\mathrm{E}_{\text {(cell })}^{\ominus}\) = 1.1 V
Therefore, \(\Delta_{\mathrm{r}} G^{\ominus}\) = – 2 × 1.1V × 96487 C mol ^–1
= – 21227 J mol^–1
= – 212.27 kJ mol^–1

Z = 27
\(\Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^{7} 4 \mathrm{~s}^{2}\) 
\(\therefore \mathrm{M}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{7}\) 

i.e., 3 unpaired electrons
\(\therefore n=3\) 
\(\Rightarrow \sqrt{n(n+2)}=\mu\) 
\(\Rightarrow \sqrt{3(3+2)}=\mu\) 
\(\Rightarrow \sqrt{15}=\mu\) 
\(\mu \approx 4 \mathrm{BM}\)

The cell will be represented as:
Zn (s) I Zn²⁺(aq) II Ag²⁺ (aq) I Ag (s)
(i) Anode, i.e., zinc electrode will be negatively charged.
(ii) The current will flow from silver to copper in the external circuit.
(iii) At Anode: Zn (s)⇾ Zn2+(aq) + 2 e^-
At Cathode: Ag⁺ (aq) + e ⇾ Ag

Hinsberg's test is used for the identification of primary, secondary, and tertiary amines.
Hinsberg's reagent is benzenesulphonyl chloride (C_6​H₅​SO_2​Cl).
It reacts differently with primary, secondary, and tertiary amines.
(i) Hinsberg's reagent reacts with primary amines to form N− alkylbenzenesulphonyl amide which is acidic in nature and soluble in alkali.
Note: N− alkylbenzenesulphonyl amide contains a strong electron-withdrawing sulphonyl group. Due to this, the H− atom attached to nitrogen can be removed easily. Hence, it is acidic.
(ii) Hinsberg's reagent reacts with secondary amines to form a sulphonamide which is insoluble in alkali.
Note: As there is no hydrogen atom attached to the N atom in the sulphonamide, it is not acidic and insoluble in alkali.
(iii) Hinsberg's reagent does not react with tertiary amines.

The IUPAC names of various compounds are given below in order:
(i) 2-Bromo-3-chlorobutane
(ii) 1-Bromo-1-chloro-1,22-trifluoroethane
(iii) 1-Bromo-4-chlorobut-2-yne
(iv) 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane
(v) 2-Bromo-3,3-bis (4-chlororphenyl) butane
(vi) 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law. Thus:
\(x(\text { Nitrogen })=\frac{p \text { (nitrogen) }}{K_{\mathrm{H}}}=\frac{0.987 \mathrm{bar}}{76,480 \mathrm{bar}}=1.29 \times 10^{-5}\)
As 1 litre of water contains 55.5 mol of it, therefore if n represents number of moles of N₂ in solution,
\(x(\text { Nitrogen })=\frac{n \text { mol }}{n \text { mol }+55.5 \text { mol }}=\frac{n}{55.5}=1.29 \times 10^{-5}\)
(n in denominator is neglected as it is < < 55.5)
Thus n = 1.29 × 10^–5 \(\times\) 55.5 mol = 7.16\(\times\) 10^–4 mol
\(=\frac{7.16 \times 10^{-4} \mathrm{~mol} \times 1000 \mathrm{ \ mmol}}{1 \mathrm{~mol}}=0.716 \mathrm{ \ mmol}\)

(i) 

(ii)
 
(iii)
 
(iv)
 
(v)

(vi)

(a) The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.
(b) Most of the compounds of transition elements contain unpaired electrons in their (n-1) d-subshells. Therefore, they are paramagnetic in nature and are attracted by the magnetic field. The magnetic character is expressed in terms of magnetic moment. The larger the number of unpaired electrons in a substance, the greater is the paramagnetic character and larger is the magnetic moment. The magnetic moment is expressed in Bohr magneton abbreviated as B.M. For example, Ti²⁺ has 2 unpaired electrons and has less magnetic moment than Mn²⁺which has 3 unpaired electrons. Mn²⁺has 5 unpaired electrons and has maximum magnetic moment among the divalent transition metal ions because d-subshell can have a maximum of 5 unpaired electrons.
(c) Most of the transition metal ions are colored both in the solid state and in aqueous solutions.The color of these ions is attributed to the presence of incomplete (n - 1) d subshell. The electrons in these metal ions can be easily promoted from one energy level to another in the same d-subshell. The amount of energy required to excite the electrons to higher energy states within the same d-subshell corresponds to the energy of certain colors of visible light. Therefore, when white light falls on a transition metal compound, some of its energy corresponding to a certain color, is absorbed causing promotion of d-electrons. This is known as d-d transitions. The remaining colors of white light are transmitted and the compound appears colored. For example, hydrated cupric compounds absorb radiations corresponding to red light and the transmitted color is greenish blue (which is complementary color to red color). Thus, cupric compounds have greenish-blue color
(d) Some transition metals and their compounds act too good catalysts for various reactions. This is due to their ability to show multiple oxidation states. The common examples are Fe, Co, Ni, V, Cr, Mn, Pt, etc.The transition metals form reaction intermediates with the substrate by using empty d-orbitals. These intermediates give reaction paths of lower activation energy and therefore, increase the rate of reaction. 

 \({ t }_{ { 1 }/{ 2 } }=5730 \ years\)
\({ t }_{ { 1 }/{ 2 } }=\frac { 0.693 }{ k } \)
\(k=\frac { 0.693 }{ 5730 } { years }^{ -1 }\)
\(k=\frac { 2.303 }{ t } \log { \frac { { \left[ N \right] }_{ 0 } }{ \left[ N \right] } }\)
\(t=\frac { 2.303 }{ k } \log { \frac { 100 }{ 80 } }\)
\(=\frac { 2.303 }{ 0.693 } \times 5730\left[ \log { 5 } -\log { 4 } \right] \)
\(=\frac { 5730 }{ 0.3010 } \times \left[ 0.6990-0.6021 \right] \)
\(=\frac { 5730 }{ 0.3010 } \times 0.0969=\frac { 55.237 }{ 0.3010 } \)
\(=1844.64 \ years\approx 1845 \ years\)

(i) Clemmensen reduction It involves reduction of carbonyl group of an aldehyde or ketone to methylene group to form a hydrocarbon. Clemmensen reduction is carried out in presence of Zn amalgam and conc. HCl.

It is widely used for the reduction of aldehydes and ketones which are sensitive to alkalies.
(ii) Cannizzaro reaction Aldehydes which do not have ∝-H atoms undergo Cannizzaro reaction on treatment with cone. alkali. In this reaction, one molecule of aldehyde is reduced to alcohol while another molecule is oxidised to salt of carboxylic acid.

(i) Pentaamminenitrito-O-cobalt(III) chloride
(ii) Potassium hexacyanochromate(III)

3

Benzenediazonium chloride is unstable. Therefore, it cannot be stored. Instead it is used immediately after its preparation.

The temperature at which the trend of solubility changes (e.g., first increases and then decreases) is called the transition temperature.

This is because it reaches a state of equilibrium where rate of evaporation = rate of condensation.

Amines form benzenesulphonamides which are soluble in alkali, i.e., option (a) is correct.

B) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Explanation:
Cysteine can cross link peptide chains through disulphide bridge. Cross linking by disulphide bridge can occur either between the distant, properly oriented parts of the same polypeptide chain (as in oxytocin or vasopressin) or between different polypeptide chains.

(c): Mn²⁺ = [Ar]3d⁵,      Mn³⁺ = [Ar]3d⁴
      Fe²⁺= [Ar]3d⁶,         Fe³⁺= [Ar]3d⁵
Thus, Mn²⁺ has more stable configuration than Mn³⁺ while Fe³⁺has more stable configuration than Fe²⁺. Hence, reduction potential for Mn³⁺/Mn²⁺ couple is more positive than Fe³⁺/ Fe²⁺. As we move across the period, ionisation potential increases, thus, ionisation potential of Fe is more than that of Mn.

(c) Mercury cell gives the constant voltage throughout its life because the overall reaction does not involve any ion in the solution, concentration of which changes during its whole life.Thus, (A) is correct but (R) is incorrect.

(i) (d): Glycinate ion is an example of bidentate ligand. It contains Nand O as donor atoms.
(ii) (a)
(iii) (c) : H₂N - NH₂ does not act as chelating ligand.
The coordination by hydrazine leads to a three member highly unstable strained ring and thus it does not act as chelating agent.
(iv) (d): In Zeises salt, coordination no. of Pt is 4. Ethylene is a mono dentate ligand.

(i) (b) : \(\begin{array}{l} M \longrightarrow M^{+}+e^{-} \\ (1 \cdot M)(0.05 M) \end{array}\) 
For concentration cell, \(E_{\text {cell }}=-\frac{0.059}{1} \log \frac{0.05}{1}\) 
\(E_{\text {cell }}=-\frac{0.059}{1} \log \left(5 \times 10^{-2}\right)\) 
\(E_{\text {cell }}=-\frac{0.059}{1}[(-2)+\log 5]-0.059(-2+0.698)\) 
= -0.059(-1.302) = 0.0768
\(\Delta G=-n F E_{\text {cell }}\) 
If E_cell is positive, \(\Delta G\) is negative.
(ii) (c) : \(K=\operatorname{antilog}\left(\frac{n E^{\circ}}{0.0591}\right)\) 
For feasible cell, E^o is positive, hence from the above equation K > 1 for a feasible cell reaction.
(iii) (b)
(iv) (a)