NaCl reacts with conc. $$H_2SO_4$$ and $$K_2Cr_2O_7$$ to give reddish fumes (B), which react with NaOH to give yellow solution (C). (B) and (C) respectively are :

This is the chromyl chloride test. When NaCl reacts with concentrated $$H_2SO_4$$ and $$K_2Cr_2O_7$$, reddish-brown fumes of chromyl chloride are produced:

$$4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2\uparrow + 4NaHSO_4 + K_2SO_4 + 3H_2O$$

So compound [B] = $$CrO_2Cl_2$$ (reddish fumes).

When chromyl chloride reacts with NaOH, it produces the yellow solution of sodium chromate:

$$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$$

So compound [C] = $$Na_2CrO_4$$ (yellow solution).

The answer is Option A: $$CrO_2Cl_2, \; Na_2CrO_4$$.