Which of the following has sp³d hybridization?

Concept:

Hybridization:

• The atomic orbitals s, p, d,and  f of the central atom having compatible energies mix up.

• This leads to the formation of new hybrid orbitals which have now different energies, shapes, and orientations than the component atomic orbitals.
• The number of hybrid orbitals is equal to atomic orbitals mixing.
•  After the formation of hybrid orbitals, they overlap with atomic orbitals of suitable side atoms and lead to the formation of covalent chemical bonds.
• The formula for calculating hybridization - H = ½ ( V + M -C + A )
• Where V = number of valence electrons on the central atom. M = No. Of monovalent atoms C = cationic charge A = anionic charge.

Few types of hybridization, their modes of mixing, and the geometry of molecules are-

Explanation:

SF4

• Contribution of electrons of sulfur toward the bonding = 6
• Contribution from Fluorine = 4
• The total number of electron pairs is 5, which corresponds to sp³d hybridization.
• However, out of these 5 pairs, 4 are bond pairs and one is lone pair.
• The geometry is trigonal bipyramidal but the shape is a see-saw .

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Hence, SF4 has sp³d hybridization.

Additional Information 

XeF4

• Contribution of electrons of Xe toward the bonding = 8
• Contribution from Fluorine = 4
• The total number of electron pairs is 6, which corresponds to sp³d² hybridization.
• Out of 6 pairs, 4 are bond pairs and two are lone pairs.

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XeOF4

• Contribution of electrons of Xe toward the bonding = 8
• Contribution from Fluorine = 4 , Oxygen = 2
• The total number of electron pairs is 7 out of which one is a pi bond, which corresponds to sp³d² hybridization.
• Out of 6 pairs, 5 are bond pairs, and one is a lone pair and the shape is square pyramidal.

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NH4+

• Contribution of electrons toward bonding by Ammonia = 5 , H = 3
• A total number of electron pairs = 4; which corresponds to sp³ hybridization.
• There are three bond pairs and one lone pair in ammonia.
• The lone pair is donated to an H+ ion to form a co-ordinate bond. It occupies the fourth position of a tetrahedron,

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