MCQ
Select best reagent to carry out above conversion


- A$LiAlH_4$
- ✓$NaBH_4$
- C$NH_2-NH_2+H_2O_2$
- D$H_2\, (excess) / Ni, \Delta $

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This is called:$\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{CO}_2, \mathrm{H}_2 \mathrm{O}, \mathrm{CH}_4, \mathrm{BF}_3$
$MnO_4^ - + 5F{e^{2 + }} + 8{H^ + } \to M{n^{2 + }} + 5F{e^{3 + }} + 4{H_2}O$,
here $10\, ml$ of $0.1\, M$ $KMn{O_4}$ is equivalent to