1. Identify compound $A$ from the first reaction

A compound that gives chromyl chloride, $\mathrm{CrO_2Cl_2}$, on heating with a little $\mathrm{NaCl}$ and concentrated $\mathrm{H_2SO_4}$ is a chromate/dichromate compound. This is the well-known chromyl chloride test for chromium.

So effectively, chromium in $A$ is converted to chromyl chloride: $\mathrm{CrO_4^{2-}/Cr_2O_7^{2-} \xrightarrow[H_2SO_4]{NaCl} CrO_2Cl_2}$

2. Reaction of chromyl chloride with NaOH

Chromyl chloride on hydrolysis/alkaline treatment gives chromate: $\mathrm{CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O}$

Hence, $B = \mathrm{Na_2CrO_4}$

3. Acidification of chromate

On adding acid, chromate converts to dichromate: $2\mathrm{CrO_4^{2-}} + 2\mathrm{H^+} \rightleftharpoons \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O}$

Thus, $C = \mathrm{Cr_2O_7^{2-}}$

4. Count the terminal oxygen atoms in $C$

Structure of dichromate ion, $\mathrm{Cr_2O_7^{2-}}$:

• It contains one bridging oxygen between the two chromium atoms.
• Total oxygen atoms = $7$
• Therefore, terminal oxygens = $7 - 1 = 6$

So, the number of terminal oxygen atoms in $C$ is: $\boxed{6}$