how to hybridisation P4O10?
Answers
Answered by
12
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...
if need more information visit the link below
https://www.discuss.com.hk/archiver/?tid-8801777.html
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...
if need more information visit the link below
https://www.discuss.com.hk/archiver/?tid-8801777.html
Anonymous:
please mark as brainliest
Oye.. net se chaap diya.. xD
didi ke book se chap hai
haha.. I was scared ki tu itni hosiyar kabse Ho gai.. haha
Similar questions