If true enter 1, else enter 0.
The $H - N - H$ bond angle in ${N H}_{3}$ is greater than $H - A s - H$ bond angle in $A s H_{3}$ .
1

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Solution

The correct option is A 1
Hydride
$N H_{3}$
$P H_{3}$
$A s H_{3}$
$S b H_{3}$
H-M-H angle
107
92
91
90
This variation in bond angle can be explained on the basis of electronegativity of the central atom.
Since nitrogen atom is highly electronegative. So in NH3 there is high electron density around the N atom which causes greater repulsions between the electron pairs around the N atom resulting in maximum bond angle.
Since electronegativity decreases down the group, the electron density also decreases and consequently the repulsive interactions between the electron pairs also decrease thereby decreasing the bond angle H-M-H.

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A s H_{3} < P H_{3} < N H_{3}

Consider the molecules $C H_{4}$ , $N H_{3}$ and $H_{2} O$ . Which of the given statements is false?

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Hydride	$N H_{3}$	$P H_{3}$	$A s H_{3}$	$S b H_{3}$
H-M-H angle	107	92	91	90