Chemical Bonding and Molecular Structure: Chemistry | JEE Main

The maximum number of lone pairs of electrons on the central atom from the following species is

\ldots\dots

\mathrm{ClO}_3^{-}, \mathrm{XeF}_4, \mathrm{SF}_4

and

\mathrm{I}_3^{-}

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Correct answer

The maximum number of lone pairs on the central atom among the given species is 3, found in the triiodide ion ( $\mathrm{I}_3^{-}$ ).

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Given chemical species: $\mathrm{ClO}_3^{-}$ , $\mathrm{XeF}_4$ , $\mathrm{SF}_4$ , and $\mathrm{I}_3^{-}$ .

02find

The maximum number of lone pairs of electrons on the central atom among the given species.

03strategise

Use the formula for lone pairs on the central atom: $\text{Lone pairs} = \frac{V - B}{2}$ or Steric Number $SN = \frac{1}{2}(V + M - C + A)$ , where $V$ is valence electrons of central atom, $B$ is the number of bonding electrons used (or $M$ is monovalent atoms bonded, $C$ is cationic charge, $A$ is anionic charge). Find the lone pairs for each species and take the maximum.

04execute

1. For $\mathrm{ClO}_3^{-}$ : Central atom $\mathrm{Cl}$ has 7 valence electrons $+ 1$ negative charge $= 8$ valence electrons. Three oxygen atoms form bonds using $3 \times 2 = 6$ electrons. Remaining non-bonding electrons $= 8 - 6 = 2$ , which gives $\frac{2}{2} = 1$ lone pair. 2. For $\mathrm{XeF}_4$ : Central atom $\mathrm{Xe}$ has 8 valence electrons. Four fluorine atoms form 4 single bonds using 4 electrons. Remaining non-bonding electrons $= 8 - 4 = 4$ , which gives $\frac{4}{2} = 2$ lone pairs. 3. For $\mathrm{SF}_4$ : Central atom $\mathrm{S}$ has 6 valence electrons. Four fluorine atoms form 4 single bonds using 4 electrons. Remaining non-bonding electrons $= 6 - 4 = 2$ , which gives $\frac{2}{2} = 1$ lone pair. 4. For $\mathrm{I}_3^{-}$ : Central atom $\mathrm{I}$ has 7 valence electrons $+ 1$ negative charge $= 8$ valence electrons. Two peripheral iodine atoms form 2 single bonds using 2 electrons. Remaining non-bonding electrons $= 8 - 2 = 6$ , which gives $\frac{6}{2} = 3$ lone pairs. The maximum number of lone pairs is $\max(1, 2, 1, 3) = 3$ .

✓verify

Check central atom of $\mathrm{I}_3^-$ : Steric number $= \frac{7 + 2 + 1}{2} = 5$ ( $sp^3d$ hybridization with 2 bond pairs and 3 equatorial lone pairs). The value 3 is an integer within valid range.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why does I3- have 3 lone pairs on the central iodine?

In the triiodide ion [I-I-I]-, the central iodine has 7 valence electrons. It gains 1 extra electron from the single negative charge. This gives a total of 8 valence electrons. It forms 2 single covalent bonds with two terminal iodine atoms using 2 electrons. This leaves 6 non-bonding electrons, which equal 3 lone pairs. These 3 lone pairs sit in equatorial positions, giving a linear geometry.

Answer

The maximum number of lone pairs on the central atom among the given species is 3, found in the triiodide ion ( $\mathrm{I}_3^{-}$ ).

Why each option works or fails

Step-by-step solution

given: Given chemical species: $\mathrm{ClO}_3^{-}$ , $\mathrm{XeF}_4$ , $\mathrm{SF}_4$ , and $\mathrm{I}_3^{-}$ .
find: The maximum number of lone pairs of electrons on the central atom among the given species.
strategise: Use the formula for lone pairs on the central atom: $\text{Lone pairs} = \frac{V - B}{2}$ or Steric Number $SN = \frac{1}{2}(V + M - C + A)$ , where $V$ is valence electrons of central atom, $B$ is the number of bonding electrons used (or $M$ is monovalent atoms bonded, $C$ is cationic charge, $A$ is anionic charge). Find the lone pairs for each species and take the maximum.
execute: 1. For $\mathrm{ClO}_3^{-}$ : Central atom $\mathrm{Cl}$ has 7 valence electrons $+ 1$ negative charge $= 8$ valence electrons. Three oxygen atoms form bonds using $3 \times 2 = 6$ electrons. Remaining non-bonding electrons $= 8 - 6 = 2$ , which gives $\frac{2}{2} = 1$ lone pair. 2. For $\mathrm{XeF}_4$ : Central atom $\mathrm{Xe}$ has 8 valence electrons. Four fluorine atoms form 4 single bonds using 4 electrons. Remaining non-bonding electrons $= 8 - 4 = 4$ , which gives $\frac{4}{2} = 2$ lone pairs. 3. For $\mathrm{SF}_4$ : Central atom $\mathrm{S}$ has 6 valence electrons. Four fluorine atoms form 4 single bonds using 4 electrons. Remaining non-bonding electrons $= 6 - 4 = 2$ , which gives $\frac{2}{2} = 1$ lone pair. 4. For $\mathrm{I}_3^{-}$ : Central atom $\mathrm{I}$ has 7 valence electrons $+ 1$ negative charge $= 8$ valence electrons. Two peripheral iodine atoms form 2 single bonds using 2 electrons. Remaining non-bonding electrons $= 8 - 2 = 6$ , which gives $\frac{6}{2} = 3$ lone pairs. The maximum number of lone pairs is $\max(1, 2, 1, 3) = 3$ .
verify: Check central atom of $\mathrm{I}_3^-$ : Steric number $= \frac{7 + 2 + 1}{2} = 5$ ( $sp^3d$ hybridization with 2 bond pairs and 3 equatorial lone pairs). The value 3 is an integer within valid range.

Chemical Bonding and Molecular Structure: Chemistry | JEE Main

The maximum number of lone pairs on the central atom among the given species is 3, found in the triiodide ion (I3−\mathrm{I}_3^{-}I3−​).

01given

02find

03strategise

04execute

✓verify

Students also ask

The maximum number of lone pairs on the central atom among the given species is 3, found in the triiodide ion ( $\mathrm{I}_3^{-}$ ).