The H-N-H angle vlaue is higher than H-P-H, H-AsH and H-SbH angles. Why?

Open in App

Solution

The central atom E (Where, E =N,P, As, Sb, Bi) in all given hydrides is \(sp-3) hybridized. However, its electronegativity decreases and atomic size increases on moving down the group. Therefore, there is a gradual moving down the group. Therefore, there is a gradual decrease in the force of repulsion between the shared electron pairs around the central atom. Thus, bond angle decreases as we move down the group.

$\begin{matrix} M o l e c u l e & N H_{3} & P H_{3} & A s H_{3} & S b H_{3} & B i H_{3} B o n d a n g l e & {107.8}^{\circ} & {93.6}^{\circ} & {91.8}^{\circ} & {91.3}^{\circ} & 90^{\circ} \end{matrix}$

Suggest Corrections

Similar questions

The correct statement with regard to $H_{2}^{+}$ & $H_{2}^{-}$ is

H_{2} S, N H_{3}, B F_{3}

S i H_{4}

H_{2}

H_{2} {S O}_{4}

H_{2}

H_{2} {S O}_{4}

Z

H_{2}

C O_{2}

C O_{2}

H_{2}

H_{2} S, N H_{3}, B F_{3} a n d S i H_{4}

Join BYJU'S Learning Program