Question #15909

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Oct 25, 2017

The hybridisation of #P# IN #PCl_5# is #sp^3d# and the geometry of the compound is Trigonal bi-pyramidal

Explanation:

Hybridisation#=# Number of #sigma# bonds #+# Number of lone pairs.

In the case of #PCl_5# , #P # has #5# #sigma# bonds and #0# lone pairs.

So, the hybridisation is #5#.

Therefore it is #sp^3d# hybridized.

Now compounds with #sp^3d# hybridisation have Trigonal bi-pyramidal geometry.

There are #3# #Cl# bonds in the equitorial plane at an angle of #120# degrees from each other and #2# #Cl# bonds in the axial plane at an angle of #90# degrees from the equitorial plane.

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