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  4. (NH4)2Cr2O7 on heating liberates a gas. The same gas will be obtained by

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Q. $(NH_4)_2Cr_2O_7$ on heating liberates a gas. The same gas will be obtained by

The d-and f-Block Elements

Solution:

$(NH_4)_2Cr_2O_7 \xrightarrow{\text{heat}} Cr_2O_3 + N_2 + 4H_2O $
$ NH_4NO_2 \rightarrow N_2 + 2H_2O $
$NH_4NO_3 \rightarrow N_2O + 2H_2O $
$Mg_3N_2 + H_2O \rightarrow 3Mg(OH)_2 + 2NH_3 $
$2NaNO_2 + H_2O_2 \rightarrow Na_2O_2 + 2HNO_2 $
$N_2$ gas is liberated by heating $(NH_4)_2Cr_2O_7$ and the same gas i.e., $N_2$ is liberated by heating $NH_4NO_2$. In all other cases $N_2$ is not a product.