Bent's rule of hybridization:

Conceptual facts: The more electronegative atom prefer to stay in that hybrid orbital having less S-character or more p- character and more electropositive atom prefer 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                                  

(1) APICOPHIOLICITY IN TBP GEOMETRY:

(1) Trigonal bipiramidal geometry more electronegative atom prefers to stay in low electronegative P_Zdz² orbital of sp³d hybridization. You can also say that more electronegative atom prefer 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 atom.  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 straingth 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 Difloromethan (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:

What is Bent’s rule of hybridization?

Which of the following compound have longest (S=O)bond length , O=SF2, O=SCl2, O=SBr2.

Why Bond length of O-O is greater in H2O2 than O2F2?

How to arrange increasing (C-H) bond length in increasing order and H-C-F bond angle in the given compounds, CH4, CH3F, CH2F2 and CHF3 ?

Dipole moment of PCl3F2 molecule is zero while dipole moment of PCl2F3 molecule is non zero why?

Dipole moment of P(CH3)2(CF3)2 molecule is zero while dipole moment of P(CH3)2(CF)3 molecule is non zero why?