The bond length of the S=O bond is maximum in | vedclass

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(B) According to $Bent's$ rule,more electronegative atoms prefer to occupy hybrid orbitals having less $s$-character,while less electronegative atoms

Vedclass · Accepted Answer

$SOBr_2$

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(B) According to $Bent's$ rule,more electronegative atoms prefer to occupy hybrid orbitals having less $s$-character,while less electronegative atoms prefer orbitals with more $s$-character.In $SOX_2$ molecules,the $S=O$ bond has more $s$-character as the electronegativity of the halogen atom $(X)$ increases.As the electronegativity of $X$ decreases $(F > Cl > Br)$,the $s$-character in the $S-X$ bonds increases,which means the $s$-character in the $S=O$ bond decreases.Since bond length is inversely proportional to $s$-character,the $S=O$ bond length increases as the $s$-character in the $S=O$ bond decreases.Therefore,the $S=O$ bond length is maximum in $SOBr_2$ because $Br$ is the least electronegative among $F, Cl,$ and $Br$.

The bond length of the $S=O$ bond is maximum in which of the following compounds?

Similar Questions

Which of the following pairs does not contain isoelectronic species?

Which of the following statements is correct?

The formal charges of $N_{(1)}$,$N_{(2)}$ and $O$ atoms in the structure shown below are respectively:
$:N_{(1)}=N_{(2)}=\ddot{O}:$

In which of the following pairs of species is the bond length as well as the bond energy of the $I^{st}$ species less than the $II^{nd}$ species?

Arrange the bonds in order of increasing ionic character in the molecules: $LiF, K_2O, N_2, SO_2$ and $ClF_3$.

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