Consider the following reactions
$$A + NaCl + H_2SO_4 \to CrO_2Cl_2 + \text{Side Products}$$ (Little amount)
$$CrO_2Cl_{2(Vapour)} + NaOH \to B + NaCl + H_2O$$
$$B + H^+ \to C + H_2O$$

The number of terminal 'O' present in the compound 'C' is _________.

First write the usual laboratory method for making chromyl chloride.
When a dichromate such as $$K_2Cr_2O_7$$ is heated with solid $$NaCl$$ and concentrated $$H_2SO_4$$, deep-red vapours of chromyl chloride are obtained.

Balanced form: $$K_2Cr_2O_7 + 4\,NaCl + 3\,H_2SO_4 \rightarrow 2\,CrO_2Cl_2 + 2\,KHSO_4 + 2\,NaHSO_4 + 3\,H_2O$$

Hence compound $$A$$ is $$K_2Cr_2O_7$$, and the first product of interest is $$CrO_2Cl_2$$ (chromyl chloride).

Next, chromyl chloride undergoes alkaline hydrolysis:
$$CrO_2Cl_2 + 4\,NaOH \rightarrow Na_2CrO_4 + 2\,NaCl + 2\,H_2O$$

Thus compound $$B$$ is $$Na_2CrO_4$$, which contains the chromate ion $$CrO_4^{2-}$$.

On acidifying chromate we get dichromate:
$$2\,CrO_4^{2-} + 2\,H^+ \rightarrow Cr_2O_7^{2-} + H_2O$$

Therefore compound $$C$$ is the dichromate ion $$Cr_2O_7^{2-}$$.

Structure of $$Cr_2O_7^{2-}$$: it consists of two $$CrO_4$$ tetrahedra sharing one oxygen atom (bridging O).
Total O atoms = 7.
Bridging O atoms = 1.
Terminal O atoms = $$7 - 1 = 6$$.

Hence the number of terminal oxygen atoms present in compound $$C$$ is $$6$$.