(i) [CoCl₄]^2-
IUPAC name Tetrachloridocobalt (II) ion In [CoCl₄]^2-. Co is in +2 state and has an Outer Outer electronic configuration of 3d⁷ Cl^- is a weak field ligand so pairing of electrons will not occur

Four pairs of electrons one from each Cl^- ion occupy the four Sp³ hybrid orbitals. Therefore. the complex has tetrahedral geometry and paramagnetic nature as it contains three unpaired electrons. Thus,
Hybridisation - Sp³
Shape - Tetrahedral
Magnetic nature - Paramagnetic
(ii) [Ni(CN)₄]^2-
IUPAC name Tetracyanonickelate (II) ion
Hybridisation - dsp²
Shape Square planar
Magnetic nature Diamagnetic [For detailed structure refer to solution 16 of Topic 2.]
(iii) [Cr(H₂O)₂(C₂O₄)₂]^-
IUPAC name
Diaquadioxalatochroma te (iii ) ion
Hybridisation - d²Sp³
Shape Octahedral
Magnetic behaviour Paramagnetic
[For detailed structure, refer to solution 18 (i) subpart of Topic 2 ]

(i) \({ CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -O-{ CH }_{ 3 }+HI\longrightarrow { CH }_{ 3 }-\overset { \underset { | }{ { CH }_{ 3 } } }{ \underset { \overset { | }{ { CH }_{ 3 } } }{ C } } -I+\underset { Methanol }{ { CH }_{ 3 }OH } \)
(ii) \({ CH }_{ 3 }-CH={ CH }_{ 2 }\overset { (i){ B }_{ 2 }{ H }_{ 6 } }{ \underset { (ii)3{ H }_{ 2 }{ O }_{ 2 }/{ H }^{ - } }{ \longrightarrow } } { CH }_{ 3 }{ CH }_{ 2 }{ CH }_{ 2 }OH\)
 

(i) Vitamin C. 
(ii) Peplide linkange. 
(iii) n-hexane or its structure.

Step-I : Calculation for concentration of the solution.
According to Van't Hoff's equation
p = CRT or C = \(\frac { \pi }{ RT } \)
p = 8.8 atm, T = 40\(°\)C = 313 K
R = 0.0821 L atm K^-1 mol^-1
\(C=\frac { 8.8 }{ 0.0821\times 313 } \)=0.34 molL^-1(M)
Step-II: Calculation of freezing point of the solution depression in freezing point of solution
\(\Delta\)T_f = K_f \(\times\) m
m = 0.34 mol kg^-1
(same as molarity as given)
K_f = 1.86 K kg mol^-1
\(\Delta\)T_f = 1.86 \(\times\) 0.84 (\(\because\) \(\Delta\) T_f = K_f m)
 = 0.63 K = 0.63\(°\)C
Freezing point of the solution = 0 - 0.63 \(°\)C
= - 0.63\(°\) C

During SNIreaction, the ionization gives 2° carbocation which rearranges by a hydride shift to a more stable 3"-carbocation to form 3°-alcohol as:

(a) Chemotherapy: Chemotherapy is the science in which chemicals are used in treatment of diseases due to bacterial invasion. These chemicals destroy the microorganisms without affecting, to any material extent, the tissues (of the host).
(b) The antacids prevents the interaction of histamine with receptors present in stomach wall which results in lesser amount of acid. Antacids and antiallergic drugs have different receptors, therefore, they do not interfere with the function of each other.

ΔT_b = 100.0832 - 100.0 = 0.0832°C,
K_b = 0.52 K kg mol^-1,
W_A = 250 g, W_B = 3.12 g,
M_B = 137 + 71 = 208 9 mol^-1
Now \(ΔT_b=iK_b\times{W_B\over M_B}\times{1000\over W_A}\)
⇒ \(0.0832 = i \times 0.52 \times{3.12\over4 6.489}\times{1000\over 250}\)
\(⇒ i={0.0832\times208\over 4\times3.12\times0.52}={17.30\over 6.489}=2.66\)
\(\alpha={i-1\over n-1}={2.66-1\over 3-1}={1.66\over 2}=0.83\)
Degree of dissociation (α) = 83%

\(d=\frac { Z\times M }{ { a }^{ 3 }\times { N }_{ A } } \)
\(\\ M=\frac { d\times { a }^{ 3 }\times { N }_{ A } }{ Z }\)
\( Z=\frac { 8.0g{ cm }^{ -3 }\times { \left( 250 \right) }^{ 3 }\times { 10 }^{ -30 }{ cm }^{ 3 }\times 6.023\times { 10 }^{ 23 } }{ 2 }\)
\(Z=\frac { 8\times 15.625\times 6.023 }{ 20 } \)
\(=\frac { 752.875 }{ 20 } \)
\(\\ 37.64 \ g{ \ mol }^{ -1 }\)
\(For \ bcc, \ 4r=\sqrt { 3 } a\)
\(\ r=\frac { \sqrt { 3 } a }{ 4 } =\frac { 1.732\times 250 }{ 4 } =108.25 \ pm\)

Higher the melting point more will be the stability of crystals. Methane has a lower melting point because it is non-polar and has least intermolecular forces of attraction. Diethyl ether is polar, therefore, its melting point is higher than that of methane due to greater intermolecular forces of attraction. Ethyl alcohol has higher melting point than CH4 as well as diethyl ether because molecules of ethyl alcohol are associated with intermolecular hydrogen bonding. Water has the highest melting point because water molecules are associated with' intermolecular hydrogen bonding more than ethyl alcohol.

p-Nitroaniline < Aniline < p- Toluidine

'A' is carboxylic acid, 'B' is alcohol, 'C' is easter.

\(\pi =CRT\)
\(\therefore\) In the first case, \(4.98=\frac { 36 }{ 180 } \times R\times 300=60R\)
In the second case,\(1.52=C\times R\times 300\)
Dividing (ii) by (i), we get C = 0.061 M.