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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Trigonal pyramidal
Bent, like water.
Consider: Number of bonding domains on the central atom Number of non-bonding electron pairs (lone pairs) on the central atom
Lone pairs typically have the greater repulsion because lone pairs want to be as far apart from one another as possible, even more so than bonding pairs. This is because the lone pairs consist of free-moving electrons.
No lone pairs