Hybrididation and VSEPR theory - Free MCQ Practice Test with solutions,

(Trigonal planar geometry); Bond angle

In these, order of bond angle:
BCl₃ > PCl₃ > AsCl₃ > BiCl₃

SF₆ is octahedral and F − S − F bond angle is 90^∘

s -character ∝ bond angle

For 25%s character (as in sp³ hybrid orbital), bond angle is 109.5^∘, for 33.3%s character (as in sp² hybrid orbital), bond angle is 120∘ and for 50%s character (as in sp hybrid orbital), bond angle is 180^∘. Similarly, when the bond angle decreases below 109.5∘, the s -character will decrease accordingly. Decrease in angle = 120^∘ − 109.5^∘ = 10.5^∘

Decrease in s -character = 33.3 − 25 = 8.3
Actual decrease in bond angle = 109.5^∘ − 105^∘ = 4.5^∘
Expected decrease in s -character

Thus, the s -character should decrease by about 3.56%
i.e.,s -character = 25 − 3.56 = 21.44%

and orbitals do not have proper orientation to overlap and hence no bond is formed.

In the tetrahedral model the top X atom is labeled as A, the bottom three are labeled as B, C and D. The atom C is above the plane and the atom D is inside the plane.
Thus the bond angles in MX4 are written as
∠AMB
∠AMC
∠AMD
∠BMC
∠BMD
∠CMD

Thus here bond angles between

It has two nodal planes. It is

Ethyne
Number of -bonds at -atom
Number of lone pair of electrons at -atom
∴ Total
∴ Hybridisation is .
is linear
Methane
Number of bonds at -atom
Number of-lone pair of electrons at C-atom
∴ Total
∴ Hybridisation is .
is tetrahedral.