MCQ
${P_4} + 3NaOH + 3{H_2}O \to A + 3Na{H_2}P{O_2}$ here, $'A'$ is
- A$N{H_3}$
- ✓$P{H_3}$
- C${H_3}P{O_4}$
- D${H_3}P{O_3}$
${P_4} + 3NaOH + 3{H_2}O \to P{H_3} + 3Na{H_2}P{O_2}$
In this reaction, phosphorus disproportionate into phosphine and sodium hydrogen phosphite.
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$\mathrm{H}_3 \mathrm{PO}_4 \ \ \ \ \mathrm{H}_2 \mathrm{SO}_4 \ \ \ \ \mathrm{H}_3 \mathrm{PO}_3 \ \ \ \ \mathrm{H}_2 \mathrm{CO}_3 \ \ \ \ \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 $
$ \mathrm{H}_3 \mathrm{BO}_3 \ \ \ \ \mathrm{H}_3 \mathrm{PO}_2 \ \ \ \ \mathrm{H}_2 \mathrm{CrO}_4 \ \ \ \ \mathrm{H}_2 \mathrm{SO}_3 $
