# $H_2$ gas is liberated when $H_2O_2$ reacts with

$\begin{array}{1 1}(a)\;\text{Acidified }KMNO_4\\(b)\;O_3\\(c)\;\text{Acidified }K_2Cr_2O_7\\(d)\;HCHO\end{array}$

$2HCHO+H_2O_2\rightarrow 2HCOOH+H_2(g)$
Hence (d) is the correct answer.