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No. of covalent bond present in the Lewis structure of \[IC{l_4}^ - \] ​are:
A. \[4\]
B.\[3\]
C.\[5\]
D.\[2\]

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Last updated date: 20th Jul 2024
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Answer
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Hint: A covalent bond is formed when the electrons from both participating atoms are shared equally. Molecular bonds are another name for covalent bonds.

Complete answer:
We checked each of the options given and hence we found that the correct option is Option A- .
Actually, we know the definition of Hybridization as the process of combining two atomic orbitals with the same energy levels to produce a degenerated new class of orbitals. Quantum mechanics is used to explain this intermixing. And actually, only atomic orbitals with the same energy level can participate in hybridization, and both complete and half-filled orbitals can participate in this process if their energies are similar.
Actually, the atomic orbitals of similar energy are mixed together during the hybridization process, such as mixing two ‘s' orbitals or two ‘p' orbitals, or mixing a ‘s' orbital with a ‘p' orbital or a ‘s' orbital with a ‘d' orbital.
Here \[IC{l_4}^ - \] is \[s{p^3}{d^2}\] hybridized.
The \[1s,\,3p\]and \[2d\] orbitals of \[s{p^3}{d^2}\] hybridization are intermixed to form six identical $sp_3d_2$ hybrid orbitals.
These six orbitals are aimed at the octahedron's corners. They are angled at a \[90\]-degree angle to one another.
Iodine forms four covalent double bonds with four chlorine atoms because the electronegativity difference between Cl and I is so small.
As a result, the proper answer is A.

Note:
Hybridization occurs when the energies of atomic orbitals are equal. So we should show that the number of hybrid orbitals produced is equal to the number of mixed atomic orbitals. Hybridization does not require the participation of all half-filled orbitals. Also orbitals that are fully filled but have slightly different energies may participate.