wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

It was told that dueidu the formation of SO2 the oxidation state of S is +4 that is it is in its first exited state.it forms sp2 hybridisation and the other p and d orbitals from pπ-pπ and pπ-dπ bonds.then where did the lonepair come from?
​​​

Open in App
Solution

Yes, you are right. One electron from 3p orbital is unpaired and excited to a 3d orbital. The hybridization of sulfur atom is sp2 hence a lone pair and two bond pairs(due to sigma bonding) reside in these hybrid orbitals.

The unpaired electrons are 3p and 3d hybridized orbitals are used in pi bonding with oxygen's unhybridized 2p orbitals. Hence two pi bonds are formed which are p-p pi and d-p pi bonds. Because of delocalisation these bonds are identical and have properties intermediate to the two different pi bonds.

To understand how the electron is able to make a jump with high energy difference, you will have to understand that d orbitals usually participate in bonding only when central atom is joined to many highly electronegative atoms. Because of polarization, there is a slight positive charge on the sulfur atom due to which orbital contract. This contraction is very very significant in d orbitals. Because of this electrons can be excited to the less energetic and tightly held d orbital. Also, the bonds formed will have a greater extent of overlap because d orbitals are not that diffused anymore. This is the reason that PF5 exists but PH5 does not.

Also, note that we cannot use dz^2 orbital for pi bonds as its lobes have the same phase and hence net overlap during pi bond will become zero


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction to Chemical Bonding
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon