In the hydrides of the nitrogen group,the H-M | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In the hydrides of group $15$ $(NH_3, PH_3, AsH_3, SbH_3, BiH_3)$,the central atom undergoes $sp^3$ hybridization in $NH_3$,leading to a bond angl

Vedclass · Accepted Answer

Almost pure $p$-orbitals are used in $M-H$ bonds.

Vedclass · Answer

(B) In the hydrides of group $15$ $(NH_3, PH_3, AsH_3, SbH_3, BiH_3)$,the central atom undergoes $sp^3$ hybridization in $NH_3$,leading to a bond angle of $107^o$. As we move down the group,the electronegativity of the central atom decreases. Consequently,the bond pairs of electrons move further away from the central atom and closer to the hydrogen atoms. This results in the central atom using almost pure $p$-orbitals for bonding,which have an inter-orbital angle of $90^o$. Therefore,the bond angle decreases from $107^o$ in $NH_3$ to nearly $90^o$ in $SbH_3$ and $BiH_3$.

In the hydrides of the nitrogen group,the $H-M-H$ bond angle gradually approaches $90^o$ as we move from $N$ to $Sb$. This indicates that gradually.....

Similar Questions

The reaction that does not produce nitrogen is

Which one of the following can be used as an anaesthetic?

Which statement is incorrect for nitric oxide $(NO)$?

Which of the following molecules has the highest number of $P-H$ bonds?

Vedclass Test Series

Exam Paper Generator

Online Exam Module