(1) The more electronegative atom prefers
to stay in that hybrid orbital having less S-character or more p- character.
(2) And more electropositive atom prefers to that hybrid orbital
which have more S-character or less p-character.
Explaination:
The more electropositive atom or group will withdraw the bond pair more from central atom.
It is easy when hybrid orbital is having less S-character and more P-character.
S-orbital
is closer to the nucleus so it electronegativity is more than p-orbital.
Illustrative example (1):
PCl3F2 (sp3d
Hybridization) and TBP (Trigonal bi pyramidal)
Where angle
< F-P-F =180 and <Cl-P-Cl =120 two P-Cl bond is axial while three P-Cl
bond is equatorial
(B) Apicophiolicity
in TBP geometry:
(1) Trigonal bipyramidal geometry more
electronegative atom prefers to stay in low electronegative PZdz2 orbital
of sp3d hybridization. You can also say that more electronegative
atom prefers that hybrid orbital having low s-character or no s-character.
(2) If electronegativity difference
between central atom and surrounding atom is large then due to polarity some
ionic character is developed and covalent character decreased.
(3) The poor covalence is not only due to
electronegativity difference between bonding atoms. it is also generated due to
poor overlapping, steric hindrance or mismatch of the overlapping orbitals.
(4) Position of lone pair: Bent rule is very important in predicting position of lone pair. lone pair is attracted by only one nucleus while bond pair is attracted by two nucleuses. Central atom hold lone pair cloud tightly if central atom is having more S-Character.
illustrative
example (2):
SF4 (Sp3d) TBP and SF2Cl2 (Sp3d)
illustrative
example (3): XeF2 and
XeO3F2
(2) Orbital
analysis: (calculation of (% s ) and (
%p) character:
S%
character is equatorial orbital at1 200c :
S%
character for orbital at 900 C:
(3)
Effect of the strength of covalency: (Alternate Statement of bent):
RULE: The more electronegative atom
not only prefer to stay in that orbital which having less % S character
(more p-character) but it also decreases % S-character and increases %
P-character in its attached orbital from the central atom depending on circumstance.
On
increasing % s character in hybrid orbital, the bond length decreases
while bond angle increases.
illustrative
example (3): Explain
C-H bond length of CH4 is longer than C-H bond length of Difluoromethane (CH2F2)
?
EXCEPTIONS OF BENT’S RULE:
(1): Bent’s rule is applicable in those molecules where central atoms are same and they are also in same Hybridization. For example N-N bond length cannot be compared in N2H4 and N2O4 using Bent rule.
illustrative
example (4): Arrange
PF3 ASH3, PH3, NH3, H2Se
in decreasing order by Bent’s rule here we used DRAGO’S RUEE”
solution: NH3> NF3>PF3>
PH3>AsH3>H2Se
(2): Bent’s rule violets in those molecules where steric factor’s plays dominating rule.
illustrative
example (5): Compared
Bond angle among H2O OF2, OMe2, OCl2
SOLUTION:
illustrative
example (6): IN
CH2SF4
Which of the
following option is correct regarding?
(1) 1800> 120
(2) 1800> >120
(3) 1200> >900
(4) 900 > >00
Solution: S -character present in equatorial so
more decrease in Bond angle in equatorial orbital than in axial. Because no
S-character in present in axial orbital.
So 1200 > >
90 and 1800 > > 120
Related
Questions:
(1) What
is Bent’s rule of hybridization?
(2) Which
of the following compound have longest (S=O)bond length , O=SF2, O=SCl2,
O=SBr2.
(3) Why
Bond length of O-O is greater in H2O2 than O2F2?
(5) Dipole
moment of PCl3F2 molecule is zero while dipole moment of PCl2F3 molecule is non
zero why?
This is very useful. Good content.
ReplyDeleteGood explanation
DeleteAmazing article !! Excellent explanation of Bent's rule.
ReplyDelete