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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Note: py $=$ pyridine
Given: Atomic numbers of $Fe , Co , Ni$ and $Cu$ are $26,27,28$ and $29$ , respectively)
$(A)$ $\left[ FeCl _4\right]^{-}$and $\left[ Fe ( CO )_4\right]^{2-}$
$(B)$ $\left[ Co ( CO )_4\right]^{-}$and $\left[ CoCl _4\right]^{2-}$
$(C)$ $\left[ Ni ( CO )_4\right]$ and $\left[ Ni ( CN )_4\right]^{2-}$
$(D)$ $\left[ Cu ( py )_4\right]^{+}$and $\left[ Cu ( CN )_4\right]^{3-}$
