CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Describe the hybridisation in case of $$PCl_5$$. Why are the axial bonds longer as compared to.


Solution


The ground state and the excited state outer electronic configuration of phosphorus $$(Z=15)$$ are $$1s^2 2s^2 2p^6 3s^2 3p^3$$ and $$1s^2 2s^2 2p^6 3s^1 3p^3 3d^1$$ respectively.
Now the five orbitals are available for hybridisation to yield a set of five $$sp^3d$$ hybrid orbitals which are directed towards the five corners of a trigonal bipyramidal. It should be noted that all the bond angles in trigonal bipyramidal geometry are not equivalent . In $$PCl_5$$, the five $$sp^3 d$$ orbitals of phosphorus overlap with the singly occupied $$p$$ orbitals of chlorine atoms to from five $$P-Cl$$ sigma bond. Three $$P-Cl$$ bond lie in one plane and make an angle of $$120^o$$ with each other, these bond are termed as equatorial bonds. The remaining two $$P-Cl$$ bonds one lying about and the other lying below the equatorial plane, make an angle of $$90^o$$ with the plane. These bonds are called axial bonds. s axial bonds suffer more repulsion than equatorial bond, these are found to be slightly longer and hence slightly weaker than equatorial bonds.
1741414_1829609_ans_a5c06299772940bfa6483366e500e723.png

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image