Which of the following sets are correctly matched? (i) $\mathrm{B}_2 \mathrm{H}_6$ - electron deficient hydride (ii) $\mathrm{NH}_3$ - electron precise hydride (iii) $\mathrm{H}_2 \mathrm{O}$ - election rich hydride

Diborane have not enough valency electrons to form conventional two bonds between all of the adjacent pairs of atoms and so there compounds are termed electron deficient. Ammonia have lone pair of electron so it is electron rich hydride. Water have two lone pair of electron so it is electron rich hydride. So (i) and (iii) are correct.