Oxygen rich compounds like chlorate, perchlorate, $${K_2}C{r_2}{O_7},$$   etc. when heated gives oxygen but ammonium dichromate gives nitrogen gas when heated.
\[\begin{align} & 2KCl{{O}_{3}}\xrightarrow{\Delta }2KCl+3{{O}_{2}}\uparrow \\ & 2Zn{{\left( Cl{{O}_{3}} \right)}_{2}}\xrightarrow{\Delta }ZnC{{l}_{2}}+3{{O}_{2}} \\ \end{align}\]
\[\underset{\begin{smallmatrix} \text{Potassium} \\ \text{dichromate} \end{smallmatrix}}{\mathop{4{{K}_{2}}C{{r}_{2}}{{O}_{7}}}}\,\xrightarrow{\Delta }\underset{\begin{smallmatrix} \text{Potassium} \\ \text{chromate} \end{smallmatrix}}{\mathop{4{{K}_{2}}Cr{{O}_{4}}}}\,\]       \[+\underset{\begin{smallmatrix} \text{Chromic} \\ \text{oxide} \end{smallmatrix}}{\mathop{2C{{r}_{2}}{{O}_{3}}}}\,+3{{O}_{2}}\uparrow \]
\[{{\left( N{{H}_{4}} \right)}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\Delta }\]     \[{{N}_{2}}\uparrow C{{r}_{2}}{{O}_{3}}+4{{H}_{2}}O\]

The electronic configuration of $${X^{3 + }}$$  is $$\left[ {Ar} \right]3{d^5}.$$
∴ Atomic no. of $$X = 18 + 5 + 3 = 26$$

$$N{a_2}C{r_2}{O_7}$$   is hygroscopic.

$$2C{u^{2 + }} + 4{I^ - } \to 2Cu{I_2}$$
The $$Cu{I_2}$$  immediately decomposes to liberate $${I_2}$$  and insoluble copper$$\left( {\text{I}} \right)$$  iodide.
$$2Cu{I_2} \to 2CuI + {I_2}$$

Only $$Na$$  gives golden colour to bunsen flame.
So, $$A$$  is $$Na$$
$$\mathop {2Na}\limits_A + 2{H_2}O \to \mathop {2NaOH}\limits_C + \mathop {{H_2}}\limits_B \uparrow $$
$$\mathop {Zn}\limits_D + \mathop {2NaOH}\limits_C \to N{a_2}Zn{O_2} + \mathop {{H_2}}\limits_B \uparrow $$
$$\mathop {Zn}\limits_D + dil.{H_2}S{O_4} \to ZnS{O_4} + \mathop {{H_2}}\limits_B \uparrow $$
$$Na,$$  produces golden yellow colour with smokeless flame of Bunsen burner

$$2C{u_2}O + C{u_2}S \to 6Cu + S{O_2}$$
self reduction.

$${K_2}C{r_2}{O_7} + 4{H_2}S{O_4} + 3{H_2}S \to {K_2}S{O_4} + C{r_2}{\left( {S{O_4}} \right)_3} + 7{H_2}O + 3S$$

The colour exhibited by transition metal ions is due to the presence of unpaired electrons in $$d$$ - orbitals which permits the $$d - d$$  excitation of electrons.
$${\text{In}}{\left[ {Ti{F_6}} \right]^{2 - }}\, - Ti\,\,{\text{is}}\,{\text{in }} + 4\,O.S.\,\,{\text{;}}$$       $$3{d^0} - {\text{colourless}}$$
$${\text{In}}{\left[ {Co{F_6}} \right]^{3 - }}\, - Co\,\,{\text{is}}\,{\text{in }} + 3\,O.S\,\,{\text{;}}$$       $$3{d^5} - {\text{coloured}}$$
$${\text{In}}\,C{u_2}C{l_2} - Cu\,\,{\text{is}}\,{\text{in }} + 1\,O.S.\,\,{\text{;}}$$       $$3{d^{10}} - {\text{colourless}}$$
$${\text{In}}\,{\left[ {NiC{l_4}} \right]^{2 - }} - Ni\,\,{\text{is}}\,{\text{in }} + 2\,O.S\,\,{\text{;}}$$       $$3{d^8} - {\text{coloured}}$$

$$Zn$$  and $$Hg$$  have $$d$$ - shells completely filled hence they do not exhibit variable valency.

When $${H_2}{O_2}$$  is added to an acidified solution of a dichromate $$C{r_2}O_7^{2 - },$$  a deep blue coloured complex, chromic peroxide $$Cr{O_5}\left[ {{\text{or}}\,\,CrO{{\left( {{O_2}} \right)}_2}} \right]$$    is formed.
$$C{r_2}O_7^{2 - } + 2{H^ + } + 4{H_2}O \to $$     $$\underbrace {2CrO{{\left( {{O_2}} \right)}_2}}_{{\text{Chromic}}\,\,{\text{peroxide}}}$$    $$\left[ {{\text{blue}}\,\,{\text{coloured}}\,\,{\text{complex}}} \right]$$     $$ + 5{H_2}O$$
This deep blue coloured complex has the following structure

Oxidation state of $$Cr$$  in $$Cr{O_5}$$  is + 6 due to the presence of two peroxide linkages which can be calculated as
$$\eqalign{ & \mathop X\limits_{\left[ {{\text{For}}\,\,Cr} \right]} + \mathop {\left( { - 1} \right) \times 4}\limits_{\left[ {{\text{For}}\,\,0 - 0} \right]} + \mathop {1 \times \left( { - 2} \right)}\limits_{\left[ {{\text{For}}\,\,0} \right]} \cr & x - 6 = 0 \cr & \,\,\,\,\,\,\,\,\,x = + 6 \cr} $$