Question

how to hybridisation P4O10?

Answer 1

hi,

The hybridisation of 'P' in P2O5 is 'sp3'

P4O10 is more stable than P2O5

According to electronic configuration of P, (2,8,5)

Ground state 1s2 2s2 2p6 3s2 3p3

Excited state 1s2 2s2 2p6 3s1 3p3 3d1 (low-lying 3d orbital)

P-containing compound is stable only when it is sp3d hybridized which is trigonal bipyrimidal

Hence P tends to form 5 bonds.

In P4O10 case,

O

/ \

O = P - O - P = O

/ \

O O

\ /

O = P - O - P = O

\ /

O

In P2O5 case,

O O

ll ll

P - O - P

ll ll

O O

If P4O10 is formed, it is sp3 hybridized and it is a form of tetrahedral.

(if sd3 hybridized, square planar)

If P2O5 is formed, it is sp2 hybridized, it is a form of trigonal planar.

Actually both p and d orbital can form sigma bond (head-to-head overlap) and sigma bond (sideway overlap), but s orbital only can form sigma bond.

Since P is odd species (2,8,5) whereas O is even species (2,6). It is not possible for P to form 5 sigma bond. To minimize the change in geometry (originally it is trigonal bipyrimidal (angle between axial-equatorial = 90 and that between equatorial-equatorial = 120), tetrahedral is adopted (all angles are about 109.5)

hope this helps you...

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