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(A) $B_2H_6$ (Diborane) is an electron-deficient hydride because it has fewer electrons than required for conventional bonding.$NH_3$ (Ammonia) has on

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$i, iii$ only

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(A) $B_2H_6$ (Diborane) is an electron-deficient hydride because it has fewer electrons than required for conventional bonding.$NH_3$ (Ammonia) has one lone pair of electrons on the nitrogen atom,making it an electron-rich hydride,not electron-precise.$H_2O$ (Water) has two lone pairs of electrons on the oxygen atom,making it an electron-rich hydride.Therefore,only statements $(i)$ and $(iii)$ are correct.

Which of the following sets are correctly matched?
$(i)$ $B_2H_6$ - electron deficient hydride
$(ii)$ $NH_3$ - electron precise hydride
$(iii)$ $H_2O$ - electron rich hydride

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Which of the following sets are correctly matched?$(i)$ $B_2H_6$ - electron deficient hydride$(ii)$ $NH_3$ - electron precise hydride$(iii)$ $H_2O$ - electron rich hydride