(1)

In vanadium, there is gradual change of basic character from the basic ${\mathrm{V}}_2{\mathrm{O}}_3$ to less basic ${\mathrm{V}}_2{\mathrm{O}}_4$ and to amphoteric ${\mathrm{V}}_2{\mathrm{O}}_5$. ${\mathrm{V}}_2{\mathrm{O}}_4$ dissolves in acids to give ${\mathrm{VO}}^{+2}$ salts.

(1)

Reaction of ${\mathrm{MnO}}_4^{-}$ with ${\mathrm{I}}^{-}$ in neutral or faintly alkaline solution:

$\begin{array}{r}\stackrel{+7}{2{\mathrm{MnO}}_4^{-}}+{\mathrm{H}}_2\mathrm{O}+\stackrel{-1}{{\mathrm{I}}^{-}} ⟶ \stackrel{+4}{2{\mathrm{MnO}}_2}+2{\mathrm{OH}}^{-}+\stackrel{+5}{{\mathrm{IO}}_3^{-}}\end{array}$

(2)

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In manganate and permanganate ions, $\mathrm{\pi }$-bonding takes place by overlap of p-orbitals of oxygen with d-orbitals of manganese.

(3)

In neutral or faintly alkaline solutions:

$2{\mathrm{MnO}}_4^{-}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}^{-}⟶2{\mathrm{MnO}}_2+2{\mathrm{OH}}^{-}+\underset{\text{'}X\text{'}}{{\mathrm{IO}}_3^{-} }$

(4)

In ${\mathrm{CrO}}_4^{2-}$, ${\mathrm{Cr}}^{6+}(n=0)$ diamagnetic
In ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$, ${\mathrm{Cr}}^{6+}(n=0)$ diamagnetic
In ${\mathrm{MnO}}_4^{-}$, ${\mathrm{Mn}}^{7+}(n=0)$ diamagnetic 
In ${\mathrm{MnO}}_4^{2-}$, ${\mathrm{Mn}}^{6+}(n=1)$ paramagnetic 
In ${\mathrm{MnO}}_4^{2-}$, one unpaired electron ($n$) is present in $d$-orbital so, $d-d$ transition is possible.

(1)

$S{O}_2$ readily decolourises pink violet colour of acidified ${\mathrm{KMnO}}_4$ solution.

$\underset{(\mathrm{Pink} \mathrm{violet})}{2{\text{KMnO}}_4}+5{\text{SO}}_2+2{\text{H}}_2\text{O}⟶K_2{\text{SO}}_4+\underset{\left(\mathrm{Colourless}\right)}{2{\text{MnSO}}_4}+2{\text{H}}_2{\text{SO}}_4$

(2)

${\text{HgCl}}_{2 \left(aq\right)}+4{\text{I}}_{\left(aq\right)}^{-} ⟶ {\text{HgI}}_{4\left(aq\right)}^{2-}+2{\text{Cl}}_{\left(aq\right)}^{-}$

${\text{I}}_{2 \left(s\right)}+{\text{I}}_{\left(aq\right)}^{-} ⟶ {\text{I}}_{3\left(aq\right)}^{-}$

(2)

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{SO}}_4+3{\mathrm{SO}}_2 ⟶ {\mathrm{K}}_2{\mathrm{SO}}_4+\underset{\left(\mathrm{Green}\right)}{{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3}+{\mathrm{H}}_2\mathrm{O}$

(4)

${\mathrm{KMnO}}_4\left({\mathrm{Mn}}^{7+}\right)$ changes to ${\mathrm{Mn}}^{2+}$ i.e., number of electrons involved per mole of ${\mathrm{KMnO}}_4$ is 5.  

(a) For ${\mathrm{FeSO}}_3$,  

${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}$ (No. of $e^{-}s$ involved = 1)

${\mathrm{SO}}_3^{2-}\to {\text{SO}}_4^{2-}$ (No. of $e^{-}s$ involved = 2)

Total number of $e^{-}s$ involved = 1 + 2 = 3

(b) For ${\mathrm{FeC}}_2{\mathrm{O}}_4$,  

${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}$ (No. of $e^{-}s$ involved = 1)

$C_2{\text{O}}_4^{2-}\text{→}2{\text{CO}}_2$ (No. of $e^{-}s$ involved = 2)

Total number of $e^{-}s$ involved = 1 + 2 = 3

(c) For $\mathrm{Fe}{\left({\mathrm{NO}}_2\right)}_2$,  

${\mathrm{Fe}}^{2+}\to {\mathrm{Fe}}^{3+}$ (No. of $e^{-}s$ involved = 1)

$2{\text{NO}}_2^{-}\text{→}2{\text{NO}}_3^{-}$ (No. of $e^{-}s$ involved = 4)

Total number of $e^{-}s$ involved = 1 + 4 = 5

(d) For ${\mathrm{FeSO}}_4$,  

${\mathrm{Fe}}^{2+}{\to \mathrm{Fe}}^{3+}$ (No. of $e^{-}s$ involved = 1)

Total number of $e^{-}s$ involved = 1  

As ${\mathrm{FeSO}}_4$ requires the least number of electrons, thus, it will require the least amount of ${\mathrm{KMnO}}_4$.

(2)

Hydrogen peroxide is oxidised to $H_{2}O$ and $O_2$.

$2{\text{KMnO}}_4+3{\text{H}}_2{\text{SO}}_4+5{\text{H}}_2{\text{O}}_2 ⟶ K_2{\text{SO}}_4+2{\text{MnSO}}_4+8{\text{H}}_2\text{O}+5{\text{O}}_2$

or,    $2{\text{MnO}}_4^{-}+5{\text{H}}_2{\text{O}}_2+6{\text{H}}^{+} ⟶ 2{\text{Mn}}^{2+}+8{\text{H}}_2\text{O}+5{\text{O}}_2$