MCQ
Strong reducing behaviour of $H_3PO_2$ is due to
- AHigh oxidation state of phosphorus
- BPresence of two $-OH$ groups and one $P-H$ bond
- ✓Presence of one $-OH$ group and two $P-H$ bonds
- DHigh electron gain enthalpy of phosphorus
All oxy-acid of phosphorus which contain $\mathrm{P}-\mathrm{H}$ bond act as reductant.
presence of one - $\mathrm{OH}$ group and two $\mathrm{P}-\mathrm{H}$ bonds
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${N_2}(g)\, + 3{H_2}(g)\, \rightleftharpoons \,2N{H_3}(g)$
The equilibrium constant of the above reaction is $K_3$. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that $P_{NH_3}<\,< P_{total}$ at equilibrium)

