Concept: The bond angle between two hybrid orbitals is related to the percentage s-character and p-character of the hybrid orbitals. For hybrid orbitals, the cosine of the bond angle ($\theta$) is related to the s-character ($s$) and p-character ($p$) by the formula: $\cos \theta =\frac{s}{s-1}$ or $\cos \theta =-\frac{s}{p}$ where $s+p=1$.

Why (D) is correct:

We are given the bond angle $\theta =105^{\circ }$. We can use the formula relating the bond angle to the s-character:

$\cos \theta =\frac{s}{s-1}$

Substitute the given angle:

$\cos 105^{\circ }=\frac{s}{s-1}$

We know that $\cos 105^{\circ }\approx -0.2588$.

So, $-0.2588=\frac{s}{s-1}$

$-0.2588(s-1)=s$

$-0.2588s+0.2588=s$

$0.2588=s+0.2588s$

$0.2588=1.2588s$

$s=\frac{0.2588}{1.2588}\approx 0.2056$

The percentage s-character is $0.2056\times 100$. This value is closest to the range $22-23$ among the given options. The slight discrepancy might be due to rounding or the specific model used for the relationship, but option D is the closest.

Option Analysis:

• A) between $30-35$: This range would correspond to a larger bond angle (closer to $109.5^{\circ }$ for $s{p}^3$ with 25% s-character).
• B) between $50-60$: This range would correspond to a much larger bond angle (e.g., $sp$ hybridization has 50% s-character and a $180^{\circ }$ bond angle).
• C) between $25-30$: This range is close to $s{p}^3$ hybridization (25% s-character, $109.5^{\circ }$ bond angle). Our calculated value is lower.
• D) between $22-23$: Our calculated s-character of approximately $20.56$ falls closest to this range, indicating a bond angle slightly less than $109.5^{\circ }$.

Correct Answer: (D)