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(D) The relationship between the bond angle $(	heta)$ and the percentage of $s-$ character $(s\%)$ in hybrid orbitals is given by the formula: $\cos

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$23 - 25\%$

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(D) The relationship between the bond angle $(	heta)$ and the percentage of $s-$ character $(s\%)$ in hybrid orbitals is given by the formula: $\cos 	heta = \frac{s}{s-1}$.For a bond angle of $105^o$,$\cos(105^o) \approx -0.2588$.Substituting this into the formula: $-0.2588 = \frac{s}{s-1}$.Solving for $s$: $-0.2588(s-1) = s \implies -0.2588s + 0.2588 = s \implies 1.2588s = 0.2588 \implies s \approx 0.2056$.This corresponds to approximately $20.56\%$ $s-$ character.Among the given options,the range $23 - 25\%$ is the closest to the calculated value,as bond angles in molecules like $H_2O$ $(104.5^o)$ involve $sp^3$ hybridization with slightly less than $25\%$ $s-$ character due to Bent's rule.

In the formation of a certain compound,the central atom uses its hybrid orbitals and has a bond angle around it of $105^o$. The percentage of $s-$ character in the hybrid orbitals is likely to be

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