Interstitial compounds are chemically inert.

$$\eqalign{ & M{n^{ + + }} - 5\,\,{\text{unpaired}}\,{\text{electrons}} \cr & {\text{F}}{{\text{e}}^{ + + }} - 4\,{\text{unpaired}}\,{\text{electrons}} \cr & T{i^{ + + }} - 2\,{\text{unpaired}}\,{\text{electrons}} \cr & C{r^{ + + }} - 4\,{\text{unpaired}}\,{\text{electrons}} \cr} $$
hence maximum no. of unpaired electron is present in $$M{n^{ + + }}.$$
NOTE : Magnetic moment $$ \propto $$ number of unpaired electrons

When $$KI$$  is added to mercuric iodide it disssolve in it and form complex.
\[\begin{align} & \underset{\begin{smallmatrix} \text{red, solid} \\ \left( \text{insoluble} \right) \end{smallmatrix}}{\mathop{Hg{{I}_{2}}}}\,+KI\to \underset{\left( \text{soluble} \right)}{\mathop{{{K}_{2}}\left[ Hg{{I}_{4}} \right]}}\, \\ & \text{On heating }Hg{{I}_{2}}\text{, decomposes as } \\ & Hg{{I}_{2}}\rightleftharpoons Hg+\underset{\left( \text{violet}\,\text{vapours} \right)}{\mathop{{{I}_{2}}}}\, \\ \end{align}\]

In a period on moving from left to right ionic radii decreases. So, the order of cationic radii is
$$C{r^{2 + }} > M{n^{2 + }} > F{e^{2 + }} > N{i^{2 + }}$$
In $$Sc > Ti > Cr > Mn$$     (correct order of atomic radii)
\[\begin{align} & \text{In}\,\,\underset{\overset{\downarrow }{\mathop{\begin{smallmatrix} \text{Five} \\ \text{unpaired} \\ \text{electrons} \end{smallmatrix}}}\,}{\mathop{M{{n}^{2+}}}}\,>\underset{\overset{\downarrow }{\mathop{\begin{smallmatrix} \text{Two} \\ \text{unpaired} \\ \text{electrons} \end{smallmatrix}}}\,}{\mathop{N{{i}^{2+}}}}\,<\underset{\overset{\downarrow }{\mathop{\begin{smallmatrix} \text{Three} \\ \text{unpaired} \\ \text{electrons} \end{smallmatrix}}}\,}{\mathop{C{{o}^{2+}}}}\,<\underset{\overset{\downarrow }{\mathop{\begin{smallmatrix} \text{Four} \\ \text{unpaired} \\ \text{electrons} \end{smallmatrix}}}\,}{\mathop{F{{e}^{2+}}}}\, \\ & \underset{\overset{\downarrow }{\mathop{4}}\,}{\mathop{\text{In}\,\,F{{e}^{2+}}}}\,\,\,\,\,\,>\,\,\,\,\,\underset{\overset{\downarrow }{\mathop{3}}\,}{\mathop{C{{o}^{2+}}}}\,\,\,\,>\,\,\,\underset{\overset{\downarrow }{\mathop{2}}\,}{\mathop{N{{i}^{2+}}}}\,\,\,\,\,\,\,>\,\,\,\,\,\,\underset{\overset{\downarrow }{\mathop{1}}\,}{\mathop{C{{u}^{2+}}}}\, \\ \end{align}\]
Number of unpaired electrons

$$2Fe + 3C{l_2}\,\left( {{\text{dry}}} \right) \to 2FeC{l_3}\left( {{\text{anhydrous}}} \right)$$

In $$TiF_6^{2 - }$$  titanium is in + 4 oxidation state. In $$C{u_2}C{l_2},$$  the copper is in + 1 oxidation state. Thus in both cases, transition from one $$d$$ - orbital to other is not possible.
$$\eqalign{ & Ti - 3{d^2}4{s^2} \to T{i^{4 + }} - 3{d^0}4{s^0} \cr & Cu - 3{d^{10}}4{s^1} \to C{u^{1 + }} - 3{d^{10}}4{s^0} \cr} $$

Let the oxidation state of $$Cr$$  is $$x$$
$$\eqalign{ & {K_2}C{r_2}{O_7} \cr & \therefore \,\,2\left( { + 1} \right) + 2x + 7\left( { - 2} \right) = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 + 2x - 14 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x - 12 = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x = 12 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{{12}}{2} = + 6 \cr} $$

Most of the $$L{n^{3 + }}$$ compounds except $$L{a^{3 + }}$$  and $$L{u^{3 + }}$$ are coloured due to the presence of $$f$$ -electrons.

Magnetic moment, $$\mu = \sqrt {n\left( {n + 2} \right)} BM$$     where,
$$n=$$ number of unpaired electrons
$$\eqalign{ & \mu = 2.84\left( {{\text{given}}} \right) \cr & \therefore 2.84 = \sqrt {n\left( {n + 2} \right)} B.M \cr & {\left( {2.84} \right)^2} = n\left( {n + 2} \right) \cr & 8 = {n^2} + 2n \cr & {n^2} + 2n - 8 = 0 \cr & {n^2} + 4n - 2n - 8 = 0 \cr & n\left( {n + 4} \right) - 2\left( {n + 4} \right) = 0 \cr & n = 2 \cr} $$
$$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}$$     ( two unpaired electrons )
$$T{i^{3 + }} = \left[ {Ar} \right]3{d^1}4{s^0}$$     ( one unpaired electrons )
$$C{r^{3 + }} = \left[ {Ar} \right]3{d^3},$$     ( three unpaired electrons )
$$C{o^{2 + }} = \left[ {Ar} \right],3{d^7},4{s^0}$$     ( three unpaired electrons )
So, only $$N{i^{2 + }}$$ has 2 unpaired electrons.

The reaction of aqueous $$KMn{O_4}$$  with $${H_2}{O_2}$$  in acidic medium is
$$3{H_2}S{O_4} + 2KMn{O_4} + 5{H_2}{O_2} \to $$       $$5{O_2} + 2MnS{O_4} + 8{H_2}O + {K_2}S{O_4}$$
In the above reaction, $$KMn{O_4}$$  oxidises $${H_2}{O_2}$$  to $${O_2}$$  and itself i.e. $$\left[ {MnO_4^ - } \right]$$  gets reduced to $$M{n^{2 + }}$$  $$ion$$  as $$MnS{O_4}.$$   Hence, aqueous solution of $$KMn{O_4}$$  with $${H_2}{O_2}$$  yields $$M{n^{2 + }}$$  and $${O_2}$$  in acidic conditions.