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Why does the basicity of Group 15 hydrides ($\ce{NH_3,~ PH_3,~ etc}$) decreases down the group ?

I know that's because of its Lewis Base nature due to the lone pair but wouldn't the more EN element be less reluctant to give the lone pair ?
And nitrogen is most EN in that group .

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It can be explained by the very helpful Bent's rule. As nitrogen is more electronegative than phosphorus, there is a bigger s character of the sp3 hybrid orbitals that are used to form the bond to hydrogen than the ones of phosphorus. As a result, the sp3 hybrid orbital that holds the electron pair, has a bigger p character for nitrogen than for phosphorus, meaning that the pair is more distant from the nucleus and more susceptible to donation.

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  • $\begingroup$ If s character is more for N , p character should be more for P right ? $\endgroup$ – Rajesh Feb 22 '15 at 3:50
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    $\begingroup$ In the sp3 orbitals used to form the bond with hydrogen the s character will increase for both nitrogen and phosphorus (they are both more electronegative than hydrogen), but much more for nitrogen. The s character for a standard sp3 orbital is 25%, for phosphorus it should be little higher than that and for nitrogen much higher. On the other hand, the sp3 orbital containing the electron pair will have a much higher p character for nitrogen than for phosphorus. $\endgroup$ – RBW Feb 22 '15 at 9:31

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