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Chemical Bonding and Molecular Structure: Chemistry | JEE Main

Amongst the following, the number of species having the linear shape is

\underline{\quad\quad\quad}

\mathrm{XeF}_2, \mathrm{I}_3^+, \mathrm{C}_3\mathrm{O}_2, \mathrm{I}_3^-, \mathrm{CO}_2, \mathrm{SO}_2, \mathrm{BeCl}_2 \text{ and } \mathrm{BCl}_2^\ominus

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Correct answer

Among the 8 given species, exactly 5 are linear: $\mathrm{XeF}_2$ , $\mathrm{C}_3\mathrm{O}_2$ , $\mathrm{I}_3^-$ , $\mathrm{CO}_2$ , and $\mathrm{BeCl}_2$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Given chemical species: $\mathrm{XeF}_2, \mathrm{I}_3^+, \mathrm{C}_3\mathrm{O}_2, \mathrm{I}_3^-, \mathrm{CO}_2, \mathrm{SO}_2, \mathrm{BeCl}_2, \mathrm{BCl}_2^-$ .

02find

Find the total number of species with a linear molecular geometry.

03strategise

Use VSEPR theory and hybridization: calculate steric number $\mathrm{SN} = \sigma\text{-bonds} + \text{lone pairs}$ on central atoms. A species is linear if $\mathrm{SN} = 2$ with 0 lone pairs, or $\mathrm{SN} = 5$ with 3 lone pairs in equatorial positions (such as $AX_2E_3$ ). For cumulenes like $\mathrm{C}_3\mathrm{O}_2$ , each carbon has $\mathrm{SN} = 2$ ( $sp$ -hybridized), giving an overall linear skeleton.

04execute

Evaluate each species systematically: 1. $\mathrm{XeF}_2$ : Central $\mathrm{Xe}$ has 8 valence electrons, forms 2 single bonds with F, leaving 3 lone pairs. $\mathrm{SN} = 2 + 3 = 5$ ( $sp^3d$ ). Lone pairs occupy equatorial sites; shape is linear (linear species #1). 2. $\mathrm{I}_3^+$ : Central $\mathrm{I}$ has 7 valence electrons, minus 1 gives 6. Forms 2 single bonds, leaving 2 lone pairs. $\mathrm{SN} = 2 + 2 = 4$ ( $sp^3$ ). Shape is bent/angular. 3. $\mathrm{C}_3\mathrm{O}_2$ (carbon suboxide): Structure is $\mathrm{O=C=C=C=O}$ . All carbon atoms are $sp$ -hybridized with 0 lone pairs; shape is linear (linear species #2). 4. $\mathrm{I}_3^-$ : Central $\mathrm{I}$ has 7 valence electrons plus 1 gives 8. Forms 2 single bonds, leaving 3 lone pairs. $\mathrm{SN} = 2 + 3 = 5$ ( $sp^3d$ ). Shape is linear (linear species #3). 5. $\mathrm{CO}_2$ : Central $\mathrm{C}$ forms two double bonds $\mathrm{O=C=O}$ , $\mathrm{SN} = 2 + 0 = 2$ ( $sp$ ). Shape is linear (linear species #4). 6. $\mathrm{SO}_2$ : Central $\mathrm{S}$ forms 2 $\sigma$ -bonds and has 1 lone pair. $\mathrm{SN} = 2 + 1 = 3$ ( $sp^2$ ). Shape is bent/V-shaped. 7. $\mathrm{BeCl}_2$ (monomer in vapour): Central $\mathrm{Be}$ forms two $\sigma$ -bonds with no lone pairs. $\mathrm{SN} = 2 + 0 = 2$ ( $sp$ ). Shape is linear (linear species #5). 8. $\mathrm{BCl}_2^-$ : Central $\mathrm{B}$ has 3 valence electrons plus 1 gives 4. Forms 2 single bonds with Cl, leaving 1 lone pair. $\mathrm{SN} = 2 + 1 = 3$ ( $sp^2$ ). Shape is bent/angular. Total linear species: $1 + 0 + 1 + 1 + 1 + 0 + 1 + 0 = 5$ .

✓verify

The species determined to be linear are $\mathrm{XeF}_2$ , $\mathrm{C}_3\mathrm{O}_2$ , $\mathrm{I}_3^-$ , $\mathrm{CO}_2$ , and $\mathrm{BeCl}_2$ . Exactly 5 species are linear, matching all standard VSEPR rules.

Hints that build this answer step by step

What criterion determines whether a polyatomic species has a linear molecular geometry according to VSEPR theory?
The bonded atoms lie in a straight line ( $180^\circ$ bond angle), typically resulting from $sp$ hybridization (2 bond pairs, 0 lone pairs) or $sp^3d$ hybridization with 3 equatorial lone pairs and 2 axial bond pairs.
What are the steric numbers and lone pair counts for $\mathrm{XeF}_2$ , $\mathrm{I}_3^-$ , and $\mathrm{I}_3^+$ ?
$\mathrm{XeF}_2$ : 2 bonds + 3 lone pairs; $\mathrm{I}_3^-$ : 2 bonds + 3 lone pairs; $\mathrm{I}_3^+$ : 2 bonds + 2 lone pairs.
Which species among $\mathrm{C}_3\mathrm{O}_2$ , $\mathrm{CO}_2$ , $\mathrm{SO}_2$ , $\mathrm{BeCl}_2$ , and $\mathrm{BCl}_2^-$ have linear geometries?
$\mathrm{C}_3\mathrm{O}_2$ , $\mathrm{CO}_2$ , and $\mathrm{BeCl}_2$ are linear; $\mathrm{SO}_2$ and $\mathrm{BCl}_2^-$ are bent.
Summing all linear species from the set $\{\mathrm{XeF}_2, \mathrm{I}_3^+, \mathrm{C}_3\mathrm{O}_2, \mathrm{I}_3^-, \mathrm{CO}_2, \mathrm{SO}_2, \mathrm{BeCl}_2, \mathrm{BCl}_2^-\}$ , what is the total count?
5 (namely $\mathrm{XeF}_2$ , $\mathrm{C}_3\mathrm{O}_2$ , $\mathrm{I}_3^-$ , $\mathrm{CO}_2$ , and $\mathrm{BeCl}_2$ )

✓ 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: 9 September 2026

Quick checks

Students also ask

Is C3O2 truly linear?

Yes, carbon suboxide is an allene-like cumulene O=C=C=C=O where all carbon atoms are sp hybridized, making the molecular framework linear.

Why is I3- linear when it has 5 electron domains?

In a trigonal bipyramidal electron geometry (steric number 5), lone pairs occupy equatorial positions with 120° angles to each other. This arrangement minimizes electron-electron repulsion. As a result, the two axial bonding pairs are forced into a 180° straight line.

Answer

Among the 8 given species, exactly 5 are linear: $\mathrm{XeF}_2$ , $\mathrm{C}_3\mathrm{O}_2$ , $\mathrm{I}_3^-$ , $\mathrm{CO}_2$ , and $\mathrm{XeF}_2, \mathrm{I}_3^+, \mathrm{C}_3\mathrm{O}_2, \mathrm{I}_3^-, \mathrm{CO}_2, \mathrm{SO}_2, \mathrm{BeCl}_2 \text{ and } \mathrm{BCl}_2^\ominus$ .

Why each option works or fails

Step-by-step solution

given: Given chemical species: $\mathrm{XeF}_2, \mathrm{I}_3^+, \mathrm{C}_3\mathrm{O}_2, \mathrm{I}_3^-, \mathrm{CO}_2, \mathrm{SO}_2, \mathrm{BeCl}_2, \mathrm{BCl}_2^-$ .
find: Find the total number of species with a linear molecular geometry.
strategise: Use VSEPR theory and hybridization: calculate steric number $\mathrm{SN} = \sigma\text{-bonds} + \text{lone pairs}$ on central atoms. A species is linear if $\mathrm{SN} = 2$ with 0 lone pairs, or $\mathrm{SN} = 5$ with 3 lone pairs in equatorial positions (such as $AX_2E_3$ ). For cumulenes like $\mathrm{C}_3\mathrm{O}_2$ , each carbon has $\mathrm{SN} = 2$ ( $sp$ -hybridized), giving an overall linear skeleton.
execute: Evaluate each species systematically: 1. $\mathrm{XeF}_2$ : Central $\mathrm{Xe}$ has 8 valence electrons, forms 2 single bonds with F, leaving 3 lone pairs. $\mathrm{SN} = 2 + 3 = 5$ ( $sp^3d$ ). Lone pairs occupy equatorial sites; shape is linear (linear species #1). 2. $\mathrm{I}_3^+$ : Central $\mathrm{I}$ has 7 valence electrons, minus 1 gives 6. Forms 2 single bonds, leaving 2 lone pairs. $\mathrm{SN} = 2 + 2 = 4$ ( $sp^3$ ). Shape is bent/angular. 3. $\mathrm{C}_3\mathrm{O}_2$ (carbon suboxide): Structure is $\mathrm{O=C=C=C=O}$ . All carbon atoms are $sp$ -hybridized with 0 lone pairs; shape is linear (linear species #2). 4. $\mathrm{I}_3^-$ : Central $\mathrm{I}$ has 7 valence electrons plus 1 gives 8. Forms 2 single bonds, leaving 3 lone pairs. $\mathrm{SN} = 2 + 3 = 5$ ( $sp^3d$ ). Shape is linear (linear species #3). 5. $\mathrm{CO}_2$ : Central $\mathrm{C}$ forms two double bonds $\mathrm{O=C=O}$ , $\mathrm{SN} = 2 + 0 = 2$ ( $sp$ ). Shape is linear (linear species #4). 6. $\mathrm{SO}_2$ : Central $\mathrm{S}$ forms 2 $\sigma$ -bonds and has 1 lone pair. $\mathrm{SN} = 2 + 1 = 3$ ( $sp^2$ ). Shape is bent/V-shaped. 7. $\mathrm{BeCl}_2$ (monomer in vapour): Central $\mathrm{Be}$ forms two $\sigma$ -bonds with no lone pairs. $\mathrm{SN} = 2 + 0 = 2$ ( $sp$ ). Shape is linear (linear species #5). 8. $\mathrm{BCl}_2^-$ : Central $\mathrm{B}$ has 3 valence electrons plus 1 gives 4. Forms 2 single bonds with Cl, leaving 1 lone pair. $\mathrm{SN} = 2 + 1 = 3$ ( $sp^2$ ). Shape is bent/angular. Total linear species: $1 + 0 + 1 + 1 + 1 + 0 + 1 + 0 = 5$ .
verify: The species determined to be linear are $\mathrm{XeF}_2$ , $\mathrm{C}_3\mathrm{O}_2$ , $\mathrm{I}_3^-$ , $\mathrm{CO}_2$ , and $\mathrm{BeCl}_2$ . Exactly 5 species are linear, matching all standard VSEPR rules.

Chemical Bonding and Molecular Structure: Chemistry | JEE Main

What feels right?

Among the 8 given species, exactly 5 are linear: $\mathrm{XeF}_2$ , $\mathrm{C}_3\mathrm{O}_2$ , $\mathrm{I}_3^-$ , $\mathrm{CO}_2$ , and $\mathrm{BeCl}_2$ .

Why each option works or fails

Solution

01given

02find

03strategise

04execute

✓verify

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