In the nitrogen family,the H-M-H bond angle i | vedclass

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(B) As we move from $NH_3$ to $SbH_3$,the electronegativity of the central atom $M$ decreases. Due to the decrease in electronegativity,the bond pair

Vedclass · Accepted Answer

Almost pure $p$-orbitals are used for $M-H$ bonding

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(B) As we move from $NH_3$ to $SbH_3$,the electronegativity of the central atom $M$ decreases. Due to the decrease in electronegativity,the bond pair of electrons shifts away from the central atom towards the hydrogen atom. According to Drago's rule,as the electronegativity of the central atom decreases,the hybridization of the central atom becomes less significant,and the $M-H$ bonds are formed using almost pure $p$-orbitals of the central atom. Since $p$-orbitals are oriented at $90^o$ to each other,the bond angle approaches $90^o$.

In the nitrogen family,the $H-M-H$ bond angle in the hydrides $MH_3$ gradually becomes closer to $90^o$ on going from $N$ to $Sb$. This shows that gradually:

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