In the formation of hydrides of $V^{t h}$ and $V I^{t h}$ group elements, only in the hydride of $1^{s t}$ member the central atom involves in $s p^{3}$ hybridization but in other atoms the central atom undergo only partial hybridization or only pure orbitals participate in bonding. This is because

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In $N$ and $O$ the hybrid orbitals are small in size and can overlap effectively compared to pure orbitals with s-orbital of hydrogen atoms

In the heavier members of $N$ and $O$ families the hybrid orbitals are larger and cannot overlap effectively compared to pure orbitals with s-orbital of hydrogen atoms

Compared to the pure p-orbital, the electron distribution in s-orbitals is much more diffuse than in $s p^{n}$ hybrids or p-orbitals alone

Hybridization of orbitals of an atom takes place only in the atoms of $2^{n d}$ period elements

answer is A, B, C.

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Detailed Solution

Due to hybridization the size of orbitals change Electron distribution in s-orbitals is much more diffuse than in $s p^{n}$ hybrids or p-orbitals alone. The hybridization of 3s and 3p orbitals on sulphur or phosphorous would produce more favorable bond angles with regard to repulsion, but the overlap between those orbitals and the hydrogen 1s orbitals is less effective. The hydrogen orbitals can overlap better with a smaller unhybridized p-orbital on sulphur or phosphorous . The result is that the orbitals used by the central atom have a very slight degree of hybridization but closely resemble pure p-orbitals

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