JEE Chemistry 2025 Question with Solution

Given: The species are $\mathrm{SO_2}$, $\mathrm{BeCl_2}$, $\mathrm{CO_2}$, $\mathrm{N_3^-}$, $\mathrm{NO_2}$, $\mathrm{F_2O}$, $\mathrm{XeF_2}$, $\mathrm{NO_2^+}$, $\mathrm{I_3^-}$, and $\mathrm{O_3}$.

Find: The number of molecules/ions that have linear geometry.

Using VSEPR theory, analyze each species:

• $\mathrm{SO_2}$: Bent geometry due to one lone pair on the sulfur atom.
• $\mathrm{BeCl_2}$: Linear geometry. Beryllium has no lone pairs and is bonded to two chlorine atoms.
• $\mathrm{CO_2}$: Linear geometry. Carbon has no lone pairs and is double bonded to each oxygen atom.
• $\mathrm{N_3^-}$: Linear geometry as the resonance structures contribute to a linear arrangement.
• $\mathrm{NO_2}$: Bent geometry due to the presence of a lone electron.
• $\mathrm{F_2O}$: Bent geometry due to two lone pairs on the oxygen atom.
• $\mathrm{XeF_2}$: Linear geometry, due to three lone pairs on xenon that arrange themselves to minimize repulsion.
• $\mathrm{NO_2^+}$: Linear geometry. Nitronium ion has no lone pairs on nitrogen.
• $\mathrm{I_3^-}$: Linear geometry, with iodine atoms bonded symmetrically around a central iodine atom, compensated by three lone pairs.
• $\mathrm{O_3}$: Bent geometry due to a lone pair on the central oxygen atom.

Therefore, the linear species are $\mathrm{BeCl_2}$, $\mathrm{CO_2}$, $\mathrm{N_3^-}$, $\mathrm{XeF_2}$, $\mathrm{NO_2^+}$, and $\mathrm{I_3^-}$.

So, the total number of linear molecules/ions is $6$.

Given: A list of ten molecules/ions is provided.

Find: How many among them are linear.

The key idea is that linear geometry commonly arises for $sp$-hybridized central atoms or for trigonal bipyramidal electron geometry where three lone pairs occupy equatorial positions, as in $\mathrm{XeF_2}$ and $\mathrm{I_3^-}$.

Classify the species one by one:

1. $\mathrm{SO_2}$ → bent
2. $\mathrm{BeCl_2}$ → linear
3. $\mathrm{CO_2}$ → linear
4. $\mathrm{N_3^-}$ → linear
5. $\mathrm{NO_2}$ → bent
6. $\mathrm{F_2O}$ → bent
7. $\mathrm{XeF_2}$ → linear
8. $\mathrm{NO_2^+}$ → linear
9. $\mathrm{I_3^-}$ → linear
10. $\mathrm{O_3}$ → bent

Count the linear ones:

Hence, the required numerical answer is $6$.