Question

s-character in hybrid orbitals of $\text{C}{{\text{H}}_{\text{4}}}$ and $\text{BeC}{{\text{l}}_{\text{2}}}$ are respectively:
A.\[0.75\], \[0.5\]
B.\[0.25\], \[0.5\]
C.\[0.75\], $0.334$
D.\[0.25\], $0.334$

Answer 1

Hint: When two or more atomic orbitals in an atom combine together to form a set of orbitals that have the character of all the atomic orbitals involved then that is known as hybridization.

Complete Answer:
In the formation of the hybrid orbitals, there are two or more than two orbitals involved which are different from each other. The “s-character” refers to the percentage of s-atomic orbitals that were involved in the formation of the hybrid orbitals. Similarly, the “p-character” is the percentage of the p-orbitals that are involved in the hybridization.
In methane,$\text{C}{{\text{H}}_{\text{4}}}$, the hybridization of the central element, i.e., carbon is $\text{s}{{\text{p}}^{\text{3}}}$. Hence, there it can be seen that there are one s and three p orbitals that make up the four hybrid orbitals. So the s-character in hybrid orbitals of $\text{C}{{\text{H}}_{\text{4}}}$ is \[0.25\].
In beryllium chloride,$\text{BeC}{{\text{l}}_{\text{2}}}$, the hybridization of the central element, i.e., beryllium is $\text{s}{{\text{p}}^{2}}$. Hence, there it can be seen that one s and two p orbitals make up the three hybrid orbitals. So the s-character in hybrid orbitals of $\text{C}{{\text{H}}_{\text{4}}}$ is , \[0.334\].

So, the correct answer is option D.

Notes: For the transition metals in the periodic table, there are mainly two types of hybridization, one is the ${{\text{d}}^{\text{2}}}\text{s}{{\text{p}}^{\text{3}}}$, and the other is $\text{s}{{\text{p}}^{\text{3}}}{{\text{d}}^{\text{2}}}$. In the former one the 3d, 4s and 4p orbitals are involved in hybridization while the later one 4s, 4p, and 4d orbitals are involved in the hybridization.