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FIRST OF all use the following formula for finding out the hybridising

by H= S+(1/2)[E - V +/- C]

HERE H is the hybridisation

S is the no of surrounding atoms to the central atom

E is the no of outer shell es in the central atom

V is the no of valence electrones of the central atom

and C is the charge on the molecule

C= +ve for -ve charge and -ve for the + ve charge.

if H= 2,3,4,5,6,7 then the hybridization will be sp sp2 sp3 sp3d sp3d2 and sp3d3 respectively and if

sp the structure is linear

sp2 then trigonal planer

v shape for the one lone pair

sp3 then tetrahedral

pyramidal (1lone pair)

v shape (2lone pair)

and linear (3lone pair)

sp3d trigonal bipyramidal

distorted bipyramidal (1 lone pair)

t shape (2 lone pair)

linear (3 lone pair)

sp3d2 octahedral

square planer(1 lone pair)

planer (2 lone pair)

sp3d3 pentagonal bipyramidal

distorted pentagonal bipyramidal (1 lone pair)

thus after knowing the hybridization we can get the structure closer to some. now for knowing the no of lone pairs we follows following path

example we take NH3

HERE WE find first the hybridisation

by the above mentioned formula

here central atom is N and S= 3 atoms

and E= 5 (outer shell es) and C=0 and V= 3 thus

H= 3 + 1/2*[ 5-3+0]= 3+1=4 means sp3 hybridization.now

find out the no of lone pair

we know the no of outer es =5

use of es by H atom = 3 then remaining atoms= 5-3=2/2=1 lone pair

so the structure is pyramidal.

SOLVED BY

DEVENDRA KUMAR VERMA

(RESEARCH SCHOLAR),

DELHI TECHNOLOGICAL UNIVERSITY, DELHI.

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12y ago
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Q: How do you find the lone pairs in a molecule?
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