Question

The tetrahedral shape is.
A-$s{p^2}$
B-$s{p^3}$
C-$ds{p^2}$
D- None of the above.

Answer 1

We know that if a molecule has four bonds and no lone pairs around the central atom then the molecule has tetrahedral geometry and the bond angle is ${109.5^ \circ }$. Ammonium ion and the methane molecule are some of the examples of the tetrahedral molecule..

Complete step by step answer:
First, we discuss hybridization.
Hybridization means the mixing of two atomic orbitals of the same energy to give a new hybrid molecular orbital of the same energy. Based on the type of orbitals involved, hybridization is classified as $sp$,$s{p^2}$,$s{p^3}$,$ds{p^2}$,$ds{p^3}$ etc.
Let us discuss the shape and bond angle of given hybridization with examples.
A-$s{p^2}$ Hybridization.
If one $s$ and two $p$ orbitals having the same energy mixed to form three equivalent hybrid orbitals then the newly formed orbital is known as $s{p^2}$ orbital. An example of $s{p^2}$ hybridized molecule is Boron trifluoride. The bond angle of $s{p^2}$ orbital is ${120^ \circ }$. The shape of $s{p^2}$ hybridized molecule is a trigonal planar. Thus option (A) is incorrect.
The structure of $B{F_3}$ is,

B-$s{p^3}$ Hybridization.
If one $s$ and three $p$ orbitals having the same energy mixed to form four equivalent hybrid orbitals then the newly formed orbital is known as $s{p^3}$ orbital. An example of $s{p^3}$ hybridized molecule is methane. The bond angle of $s{p^3}$orbital is ${109.5^ \circ }$ . The shape of $s{p^3}$ hybridized molecule is a tetrahedral. Thus option (B) is correct.
The structure of methane is,

C-$ds{p^2}$ Hybridization.
If one $d$ , one $s$ , and two $p$ orbitals having the same energy mixed to form four equivalent hybrids orbital then the newly formed orbital is known as $ds{p^2}$ orbital. An example of $ds{p^2}$ hybridized molecule is nickel complex. The bond angle of $ds{p^2}$orbital is ${90^ \circ }$. The shape of $ds{p^2}$ hybridized molecule is a square planar. Thus option (C) is incorrect.
The structure of ${\left[ {Ni{{\left( {CN} \right)}_4}} \right]^{2 - }}$ is,

Option B is the correct answer for this question.

Note:
We must know that the shape of a molecule can be identified by calculating the steric number. The total number of bonded and lone pairs of electrons in a molecule is called a steric number. It can be calculated using the formula,
$Steric\,number = \dfrac{{Valence\,electrons\,of\,the\,central\,atom\, + number\,of\,Bonded\,atom}}{2}$
Let us calculate the steric number of methane. Using the formula,
$Steric\,number = \dfrac{{4\, + 4}}{2} = 4$
The steric number of methane is four. Thus it has four bonded atoms and no lone pair around the central atom. Thus it is $s{p^3}$ hybridized tetrahedral molecule.