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Arrange the hydrides of the p-Block Group 15 elements in order of bond angle:

$\text{NH}_3 \gt \text{PH}_3 \gt \text{AsH}_3 \gt \text{SbH}_3 \gt \text{BiH}_3 $ $\text{NH}_3 \gt \text{PH}_3 \lt \text{AsH}_3 \lt \text{SbH}_3 \gt \text{BiH}_3 $ $\text{NH}_3 \gt \text{PH}_3 \lt \text{AsH}_3 \lt \text{SbH}_3 \lt \text{BiH}_3 $ $\text{NH}_3 \lt \text{PH}_3 \lt \text{AsH}_3 \lt \text{SbH}_3 \lt \text{BiH}_3 $
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Answer: $\text{NH}_3 \gt \text{PH}_3 \gt \text{AsH}_3 \gt \text{SbH}_3 \gt \text{BiH}_3$
In all the hydrides of Group-15 elements, the central atom is surrounded by four electron pairs, three bond pairs and one lone pair.
As we move down the group, the size of the central atom increases while its electronegativity decreases.
As a result, the force of repulsion between bond pairs decreases and therefore the bond angle also decreases.
answered Jul 14, 2014 by balaji.thirumalai
edited Jul 14, 2014 by balaji.thirumalai
 

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