QJEEAdv05Chfa29: JEE Advanced 2005 Chemistry - Class 11 JEEAdv05Chfa29 Solution: Which species has the maximum number of lone pair of electrons…

To determine which species has the maximum number of lone pairs of electrons on the central atom, we'll examine the electron configuration of each molecule's central atom:

Option A: $[ClO_3]^-$

The central atom is chlorine (Cl).

Chlorine has 7 valence electrons. In the $[ClO_3]^-$ ion, chlorine forms three bonds with oxygen atoms and there is an additional negative charge (one extra electron). Thus, chlorine has:

$7 + 1 - 3 = 5$ valence electrons left, which form 2.5 lone pairs. However, due to the requirement to minimize formal charges, typically chlorine in $[ClO_3]^-$ is involved in resonance structures reducing the average lone pairs per resonance structure to 1.

So, $[ClO_3]^-$ has 1 lone pair on the central atom.

Option B: $XeF_4$

The central atom is xenon (Xe).

Xenon has 8 valence electrons. In $XeF_4$ , xenon forms four bonds with four fluorine atoms. Thus, xenon has:

$8 - 4 = 4$ valence electrons left, which form 2 lone pairs.

So, $XeF_4$ has 2 lone pairs on the central atom.

Option C: $SF_4$

The central atom is sulfur (S).

Sulfur has 6 valence electrons. In $SF_4$ , sulfur forms four bonds with four fluorine atoms. Thus, sulfur has:

$6 - 4 = 2$ valence electrons left, which form 1 lone pair.

So, $SF_4$ has 1 lone pair on the central atom.

Option D: $[I_3]^-$

The central atom is iodine (I).

Iodine has 7 valence electrons. In the $[I_3]^-$ ion, iodines form a linear structure where the central iodine forms two bonds with two other iodine atoms and there is an additional negative charge (one extra electron). Thus, the central iodine has:

$7 + 1 - 2 = 6$ valence electrons left, which form 3 lone pairs.

So, $[I_3]^-$ has 3 lone pairs on the central atom.

Hence, the species with the maximum number of lone pairs of electrons on the central atom is:

Option D: $[I_3]^-$