(i) [CoNH₃CI(en)₂]²⁺
amminechloridobis(ethane-1,2-diamine)cobalt (Ill) ion

Co³⁺ : 4s⁰3d⁶ 
(ii) [Ni(H₂O)₂(OX)₂]^2-
diaquadioxalatonickelate (II) ion
Ni²⁺ : 4s⁰3d⁸

It is because 'N' is smaller in size and N-H bond is most polar and sp³ hybridised whereas in others there is s-p bonding.  Secondly, polarity is higher in NH₃ than in other hydrides.

Resonating structures of \({ NO }_{ 2 }\) are :
 
Resonating structures of \({ N }_{ 2 }{ O }_{ 5 }\)  are : 

(i) it is due to increase in size of central atom as a result of which bond length increases and bp-bp repulsion decreases and decrease repulsion decreases and decrease in polarity due to less difference in electronegativity.
(ii) SiF₆^2- is known because 'F₂' is best oxidising agent, smaller in size than chloride and can supply energy for excitation of electron whereas CI₂ is not, therefore, SiCl₆^2- does not exist.

(i) \(Xe +\underset {excess} {2F_{2}} \xrightarrow [7 bar] {873 k} XeF_{4}\) 
(ii) \(6Li + N_{2} \rightarrow 2Li_{3}N\)

\(\)

(i) It is due to increases in size of group 15 elements, the bond angle decreases as bond length increases and bond pair-bond pair repulsion decreases.
(ii) It is because it is inert.

Mn³⁺ = 3d⁴ = 4 unpaired electron, Cr³⁺ = 3d³ = 3 unpaired electrons, V³⁺ = 3d² = 2 unpaired electrons, Ti³⁺= 3d ¹ = 1 unpaired electron. Cr³⁺ is most stable out of these in aqueous solution because it has half filled t_2g level (i.e., t³_2g) .

(i)

(ii)

Preparation of sulphuric acid:
By Contact Process: Burning of sulphur or sulphide ores in presence of oxygen to produce SO₂. Catalytic oxidation of SO₂ with O₂to give SO₃ in the presence of V₂O₅.
\(2 S O_{2}(g)+O_{2}(g) \stackrel{v, 0,}{\longrightarrow} 2 S O_{3}(g)\) 
Then SO₃ made to react with sulphuric acid of suitable normality to obtain a thick oily liquid called oleum.
\(\mathrm{SO}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(l)\) 
Then oleum is diluted to obtain sulphuric acid of desired concentration
\(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{4}(l)\) 
The sulphuric acid obtained by contact process is 96-98% pure.

(i) High pressure (2 bar),
(ii) 720 K temperature,
(iii) V₂O₅ as catalyst.

(i) (a) Due to the comparatively smaller size of the metal ions, their high ionic charges and the availability of vacant d-orbitals for bond formation, transition metals form a  large number of complex compounds.
(b) As oxidation number (or oxidation state) of an element increases ionic character decreases. In general, the oxides in lower oxidation states of metals are basic and in their higher oxidation state, the oxides are amphoteric.
In lower oxidation state of the metal, some of the valence electrons of the metal atom are not involved in bonding. Hence, it can donate electrons and behave as a base. In higher oxidation state, valence electrons are involved in bonding and hence, electrons are not available for donation. Instead, their effective nuclear charge is high and hence they behave as acids.
(c) Mn³⁺(3d⁴) is less stable than Mn²⁺(3d⁵) because Mn²⁺ has stable half-filled configuration. Cr³⁺ has stable 3d³(t³_2g) configuration, therefore, Cr³⁺ cannot be reduced to Cr²⁺. That's why, EO value for the Mn³⁺ / Mn²⁺ couple is much more positive than Cr³⁺ /Cr²⁺. In other words, Mn³⁺ is a strong oxidising agent.
(ii) Similarity Both lanthanoids and actinoids exhibit +3 oxidation state predominantly.Difference Lanthanoids have less tendency towards complex formation while actinoids have greater tendency towards complex formation.

(a) (i) \(6NaOH+3C1_{ 2 }\longrightarrow 5NaC1+NaC1O_{ 3 }+3H_{ 2 }O\)
(ii) \(XeF_{ 4 }+O_{ 2 }F_{ 2 }\overset { 143k }{ \longrightarrow } XeF_{ 6 }+O_{ 2 }\)
(b) (i) H₃PO₂:

(ii) H₂S₂O₇:

(iii) XeOF₄:

(i) The transition elements exhibit variable oxidation states. The variable oxidation states of transition metals are due to the participation of ns and (n - 1) d-electrons. This is because of the very small difference between the energies of (n - 1) d and ns orbitals. For the first five elements, the minimum oxidation state is equal to the number of electrons ---in the 4s orbitals and the other oxidation states are equal to the sum of 4s and some of the 3d-electrons. The highest oxidation state is equal to the sum of 4s and 3d electrons. For the remaining elements, the minimum oxidation state is equal to electrons in 4s-orbitals and the maximum oxidation state is not equal to the sum of 4s and 3d electrons. In general, the oxidation state increases up to the middle and then decreases.

(a) (i) Due to small size of F atom, there are strong repulsions between the non-bonding electrons of F atoms in the small sized \({ F }_{ 2 }\)  molecule. Therefore, bond enthalpy of \({ F }_{ 2 }\)  is lower than relatively \({ CI }_{ 2 }\) larger  molecule in which repulsions between non bonding electrons are less.
(ii) Ammonia exists as associated molecules due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike \({ NH }_{ 3 }\) , phosphine (\({ PH }_{ 3 }\)) molecules are not associated through hydrogen bonding in liquid state. This is because of low electronegativity of P than N. As a result, the boiling point of \({ PH }_{ 3 }\)  is lower than that of \({ NH }_{ 3 }\) .
(b) (i)
​​​​​

(a) (i)

(ii)

(b)
(i) The size of S is more than that of 0. Therefore, the distance between S and H i.e., S-H bond length is more than O-H bond length. As a result, the bond dissociation enthalpy of S-H will be less and it will be easier to break the bond in  than O-H bond in water. Therefore,  will be more acidic than .
(ii) Fluorine is the most electronegative element and therefore, it shows oxidation state of -1 only. It does not show any positive oxidation states.
(iii) Helium does not form compounds because it has very high ionisation enthalpy and smallest size. Its electron gain enthalpy is also almost zero.

\(2Pb{ { { { (NO }_{ 3 }) } } }_{ 2 }\xrightarrow { heat } 2PbO(s)+{ NO }_{ 2 }+{ O }_{ 2 }\)
                                            Brown (A)
\({ 2NO }_{ 2 }(g)\overset { cooling }{ \rightleftharpoons } { N }_{ 2 }{ O }_{ 4 }(s)\) 
                         (B) Colourless
\({ N }_{ 2 }{ O }_{ 4 }+2NO\overset { heat }{ \underset { 250k }{ \rightleftharpoons } } { 2N }_{ 2 }{ O }_{ 3 }(s)\) 
                             (C) Blue solid

Resonating Structures of  \({ N }_{ 2 }{ O }_{ 4 }\)

(a) (i) Sodium chlorate and sodium chloride are formed.
\({ 3CI }_{ 2 }+6NaOH\longrightarrow 5NaCI+{ NaCIO }_{ 3 }+{ 3H }_{ 2 }O\)
(ii)_Fe (III) salt is reduced to Fe (II) salt.
\({ 2Fe }^{ 3+ }+{ SO }_{ 2 }+{ 2H }_{ 2 }O\longrightarrow { 2Fe }^{ 2+ }+{ SO }_{ 4 }^{ 2- }+{ 4H }^{ + }\)
(b) (i) \({ H }_{ 3 }{ PO }_{ 3 } \)  is dibasic and has basicity of two because it has two P-OH bonds which are ionisable. The third H atom is linked to P and is non-ionisable.

(ii) Fluorine does not play the role of a central atom in interhalogen compounds because it is highly electronegative. Moreover, it has only one electron less than the octet and does not have vacant d-orbitals in its valence shell, Therefore, it can form only one bond with other halogen atoms and cannot act as central atom in interhalogen compounds.
(iii) Noble gases have very low boiling points because only weak van der Waals' forces are present between the atoms of the noble gases in the liquid state.

(b) \((i)\quad { 3HgCI }_{ 2 }+{ 2PH }_{ 3 }\longrightarrow { Hg }_{ 3 }{ p }_{ 2 }+6HCI\)
\(\\ (ii)\quad { SO }_{ 2 }+{ H }_{ 2 }{ SO }_{ 4 }\longrightarrow { H }_{ 2 }{ S }_{ 2 }{ O }_{ 7 }\)
\(\\ (iii)\quad { 6XeF }_{ 4 }+{ 12H }_{ 2 }O\longrightarrow 4Xe+{ 2XeO }_{ 3 }+24HF+{ 3O }_{ 2 }\)