Question

The molecule in which hybrid $\mathrm{AO}$s involve only one $\mathrm{d}$- orbital of the central atom is:

• A. ${\left[{\mathrm{CrF}}_6\right]}^{3-}$
• B. ${\mathrm{XeF}}_4$
• C. ${\mathrm{BrF}}_5$
• D. ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$

Answer 1

The correct option is D

${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$

Explanation for correct option:

D. ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$

$\mathrm{d}$ - orbital of the central atom:

1. This function can be used to determine the likelihood of discovering any atom's electron in any particular area surrounding the nucleus.
2. Particularly with transition metal ions, the $ds{p}^2$form of hybridization is observed.

${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$:

1. This form of hybridization involves the orbitals $d_{x^2-y^2},$ one $\mathrm{s}$ and two $\mathrm{p}$
2. The square planar shape is adopted by the four $ds{p}^2$ hybrid orbitals.

$ds{p}^2$ Hybridization is involved with the complex ion ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$

Explanation for incorrect options:

A. ${\left[{\mathrm{CrF}}_6\right]}^{3-}$

1. In order to create $d^{2}s{p}^3$ hybrid orbitals, s and p atomic orbitals from the same electron shell are mixed with d orbitals from a different electron shell.
2. Six hybrid orbitals are the product of this hybridization. The geometry of these hybrid orbitals is octahedral.

B. ${\mathrm{XeF}}_4$

1. In this one s orbital, three p orbitals and one d orbital are mixed together.
2. In this case, three orbitals are perpendicular to the plane of the triangle, and the other two point towards the corner of an equilateral triangle.
3. Option b is incorrect

C. ${\mathrm{BrF}}_5$

1. In this structure, the three p orbitals, one s orbital, and two d orbitals adopt octahedral hybridization
2. Option c is incorrect

The molecule in which hybrid $\mathrm{AO}$s involve only one $\mathrm{d}$ - orbital of the central atom is ${\left[\mathrm{Ni}{\left(\mathrm{CN}\right)}_4\right]}^{2-}$ .

Hence, Option D is correct.