A and B formed in the following reactions are: {CrO2Cl2 + 4NaOH - A + 2NaCl + 2H2O} {A + 2HCl + 2H2O2 - B + 3H2O}

Step 1: Analyze the First Reaction

When chromyl chloride vapours ($\ce{CrO2Cl2}$) are passed into a basic solution like sodium hydroxide ($\ce{NaOH}$), a yellow solution of sodium chromate is formed.

$\ce{CrO2Cl2 + 4NaOH -> Na2CrO4 + 2NaCl + 2H2O}$

Here, chromyl chloride acts as if it were the acid chloride of chromic acid. The neutralisation generates the chromate salt. Therefore, compound A is $\ce{Na2CrO4}$ (Sodium chromate).

Step 2: Analyze the Second Reaction

The yellow solution of sodium chromate (compound A) is acidified with $\ce{HCl}$ and treated with hydrogen peroxide ($\ce{H2O2}$). The acidified chromate ion (which exists mostly in equilibrium with dichromate, $\ce{Cr2O7^{2-}}$) reacts with $\ce{H2O2}$ to form a deep blue compound called chromium pentoxide ($\ce{CrO5}$).

$\ce{Na2CrO4 + 2HCl + 2H2O2 -> CrO5 + 2NaCl + 3H2O}$ (Note: A more accurate depiction often traces through dichromate: $\ce{Cr2O7^{2-} + 4H2O2 + 2H+ -> 2CrO5 + 5H2O}$, but chemically B remains identical).

Therefore, compound B is $\ce{CrO5}$ (Chromium pentoxide).

Step 3: Conclusion

Compound A is $\ce{Na2CrO4}$ and Compound B is $\ce{CrO5}$.

Key Points to Remember:

• $\ce{CrO2Cl2}$ (red vapours) + base $\rightarrow$ yellow chromate ($\ce{CrO4^{2-}}$).
• Acidified chromate / dichromate + $\ce{H2O2}$ $\rightarrow$ deep blue chromium pentoxide ($\ce{CrO5}$).
• $\ce{CrO5}$ has a butterfly structure with an oxidation state of $+6$ for Chromium.