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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(Given atomic number: $\mathrm{Sc}=21, \mathrm{Ti}=22, \mathrm{~V}=23$, $\mathrm{Cr}=24, \mathrm{Mn}=25$ and $\mathrm{Zn}=30)$
Among the following, the option representing change the freezing point is
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