Consider the following species

Then calculate total number of species which have back bond or back bond.

Question

Consider the following species
$O(Me)_2, N(SiH_3)_3, CO, O(SiH_3)_2, CCl_2( ext{Singlet}), CCl_3^-, H_4SiO_4, OCl_2, MeNCS$
Then calculate total number of species which have $(p_\pi - p_\pi)$ back bond or $(p_\pi - d_\pi)$ back bond.

Prashant Kotian · Accepted Answer

Correct answer: 7

**SOLUTION:** IN $O(ME)_2$ AND $MENCS$ BACK BONDING IS ABSENT.. Step 1: Apply Bent's rule: more electronegative substituents (like F) occupy orbitals with greater p-character, leaving more s-character for the S-O bond. Step 2: Greater s-character in the S-O bond leads to a shorter and stronger S-O bond. Thus, SOF2 has the shortest S-O bond. Step 3: Since Bromine is the least electronegative substituent among F, Cl, and Br, it leaves the least s-character for the S-O bond, making the S-O bond length maximum in SOBr2, corresponding to option (b).

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