Question

Hybrid orbital with least s-character is:

• A. ${\mathrm{sp}}^3\mathrm{d}$
• B. ${\mathrm{sp}}^2$
• C. ${\mathrm{sp}}^3$
• D. $\mathrm{sp}$

Answer 1

The correct option is A

${\mathrm{sp}}^3\mathrm{d}$

Explanation for Correct Option:

Option (A): ${\mathrm{sp}}^3\mathrm{d}$

The percentage of any orbit can be calculated by diving that number of particular orbitals by the total number of orbitals present.

For ${\mathrm{sp}}^3\mathrm{d}$, the percentage of s-character is

$$\begin{gathered} \mathrm{s}=\frac{1}{5}\times 100 \\ \mathrm{s}=.2\times 100 \\ \mathrm{s}=20\% \end{gathered}$$

Hence, the percentage of s-character in ${\mathrm{sp}}^3\mathrm{d}$ hybrid orbital is 20%.

Explanation for Incorrect Option:-

Option (B): ${\mathrm{sp}}^2$

For ${\mathrm{sp}}^2$, the percentage of s-character is

$$\begin{gathered} \mathrm{s}=\frac{1}{3}\times 100 \\ \mathrm{s}=.33\times 100 \\ \mathrm{s}=33\% \end{gathered}$$

Hence, the percentage of s-character in ${\mathrm{sp}}^2$ hybrid orbital is 33%.

Option (C): ${\mathrm{sp}}^3$

For ${\mathrm{sp}}^3$, the percentage of s-character is

$$\begin{gathered} \mathrm{s}=\frac{1}{4}\times 100 \\ \mathrm{s}=.25\times 100 \\ \mathrm{s}=25\% \end{gathered}$$

Hence, the percentage of s-character in ${\mathrm{sp}}^3$ hybrid orbital is 25%.

Option (D): $\mathrm{sp}$

For $\mathrm{sp}$, the percentage of s-character is

$$\begin{gathered} \mathrm{s}=\frac{1}{2}\times 100 \\ \mathrm{s}=.5\times 100 \\ \mathrm{s}=50\% \end{gathered}$$

Hence, the percentage of s-character in sp hybrid orbital is 50%.

Therefore, the correct option is (A).