Which is the correct statement?As the s-chara | vedclass

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(A) The $s$-character in hybrid orbitals is related to the hybridization as follows: $sp$ $(50\% \ s)$,$sp^2$ $(33.3\% \ s)$,and $sp^3$ $(25\% \ s)$.$

Vedclass · Accepted Answer

$(I)$,$(III)$ and $(IV)$

Vedclass · Answer

(A) The $s$-character in hybrid orbitals is related to the hybridization as follows: $sp$ $(50\% \ s)$,$sp^2$ $(33.3\% \ s)$,and $sp^3$ $(25\% \ s)$.$(I)$ As $s$-character decreases,the bond angle decreases (e.g.,$180^{\circ}$ $\rightarrow 120^{\circ}$ $\rightarrow 109.5^{\circ}$),so $(I)$ is correct.$(II)$ Bond strength is directly proportional to $s$-character. As $s$-character decreases,bond strength decreases,so $(II)$ is incorrect.$(III)$ Bond length is inversely proportional to $s$-character. As $s$-character decreases,bond length increases,so $(III)$ is correct.$(IV)$ Orbitals with higher $p$-character (lower $s$-character) are larger in size because $p$-orbitals are more diffuse than $s$-orbitals. Thus,as $s$-character decreases,the size of the orbital increases,so $(IV)$ is correct.Therefore,statements $(I)$,$(III)$,and $(IV)$ are correct.

Which is the correct statement?
As the $s$-character of a hybrid orbital decreases:
$(I)$ The bond angle decreases
$(II)$ The bond strength increases
$(III)$ The bond length increases
$(IV)$ Size of orbitals increases

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Which is the correct statement?As the $s$-character of a hybrid orbital decreases:$(I)$ The bond angle decreases$(II)$ The bond strength increases$(III)$ The bond length increases$(IV)$ Size of orbitals increases