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Fifth group elements form hydrides to ty...

Fifth group elements form hydrides to type `AH_(3)`. The hydrides have a lone pair of electrons. The hydries are reducing in nature and the reducing power is related to the stability of A-H bonds. The hydrides are covalent and low boiling. Their boiling points depends on their ability to from hydrogen bond and their molecular size which decide the intermolcular forces in the hydrides .
Reducing power of V-group hydrides are in order :

A

`NH_3 gt PH_3 gt AsH_3 gt SbH_3 gt BiH_3`

B

`BiH_3 gt SbH_3 gt AsH_3 gt PH_3 gt NH_3 `

C

` PH_3 gt NH_3 gt AsH_3 gt SbH_3 gt BiH_3`

D

`BiH_3 gt SbH_3 gt AsH_3 gt NH_3 gt PH_3 `

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Verified by Experts

The correct Answer is:
B

A-H bond energy decreases with increase in size of A.Hence Bi-H bond readily breaks to make it the best reducer and the reducing power decrease of size of A.
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Fifth group elements form hydrides to type AH_(3) . The hydrides have a lone pair of electrons. The hydries are reducing in nature and the reducing power is related to the stability of A-H bonds. The hydrides are covalent and low boiling. Their boiling points depends on their ability to from hydrogen bond and their molecular size which decide the intermolcular forces in the hydrides . The H-M-H bond angle of V group hydrides decrease from 107^(circ) to 90^(circ) for NH_(3) to SbH_(3) , this is due to:

Fifth group elements form hydrides to type AH_(3) . The hydrides have a lone pair of electrons. The hydries are reducing in nature and the reducing power is related to the stability of A-H bonds. The hydrides are covalent and low boiling. Their boiling points depends on their ability to from hydrogen bond and their molecular size which decide the intermolcular forces in the hydrides . The boiling points of the hydrides of V-group elements are in the order :

Which of the following is the correct order of reducing power of hydrides ?

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