The correct order of bond angle in the given | vedclass

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(A) $1$. $SiO_4^{4-}$ has $sp^3$ hybridization with a tetrahedral geometry,so the bond angle is $109^{\circ} 28'$.$2$. In $NCl_3$ and $PCl_3$,both hav

Vedclass · Accepted Answer

$SiO_4^{4-} > NCl_3 > PCl_3 > SbH_3 > H_2Te$

Vedclass · Answer

(A) $1$. $SiO_4^{4-}$ has $sp^3$ hybridization with a tetrahedral geometry,so the bond angle is $109^{\circ} 28'$.$2$. In $NCl_3$ and $PCl_3$,both have a lone pair. The bond angle in $NCl_3$ $(107^{\circ})$ is greater than in $PCl_3$ $(100^{\circ})$ because the electronegativity of $N$ is higher,pulling bonding electrons closer to the central atom,and the $s$-character in the hybrid orbitals decreases down the group.$3$. In $SbH_3$,the bond angle is close to $90^{\circ}$ because the bonding involves almost pure $p$-orbitals.$4$. In $H_2Te$,the bond angle is approximately $90^{\circ}$ (slightly higher due to lone pair-lone pair repulsions). Comparing $SbH_3$ and $H_2Te$,$H_2Te$ has a slightly larger bond angle due to the presence of two lone pairs on $Te$ compared to one on $Sb$.$5$. Thus,the correct order is $SiO_4^{4-} > NCl_3 > PCl_3 > SbH_3 > H_2Te$.

The correct order of bond angle in the given species is:

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