Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Among the triatomic molecules / ions, ${\text{BeCl}}_2$, ${\text{N}}_3^{-}$, ${\text{N}}_2\text{O}$, ${\text{NO}}_2^{+}$, ${\text{O}}_3$, ${\text{SCl}}_2$, ${\text{ICl}}_2^{-}$, ${\text{I}}_3^{-}$ and ${\text{XeF}}_2$ the total number of linear molecules(s) / ion(s) where the hybridization of the central atoms does not have contribution from the d-orbital(s) is : (Atomic number : S = 16, Cl = 17, I = 53 and Xe = 54)

When tackling complex molecular geometry problems in JEE, the secret lies in breaking down the constraints. In this problem, we are given a list of nine triatomic species and asked to find the ones that satisfy two strict conditions simultaneously:

1. The molecule or ion must be geometrically linear. 2. The hybridization of its central atom must not involve any d-orbitals.

To solve this, we must calculate the steric number for each central atom. The steric number is the sum of the number of sigma bonds and the number of lone pairs on the central atom. Let's evaluate our candidates group by group.

Perfect Linearity

Let's examine the first group of species: ${\text{BeCl}}_2$, ${\text{N}}_3^{-}$, ${\text{N}}_2\text{O}$, and ${\text{NO}}_2^{+}$.

In beryllium chloride (${\text{BeCl}}_2$), beryllium has 2 valence electrons and forms two single bonds with chlorine, leaving zero lone pairs. The steric number is $2+0=2$. This corresponds to sp hybridization, which naturally results in a linear geometry with a bond angle of $180^{\circ }$.

Similarly, the azide ion (${\text{N}}_3^{-}$) features a central nitrogen double-bonded to two other nitrogens, with no lone pairs. Nitrous oxide (${\text{N}}_2\text{O}$) and the nitronium ion (${\text{NO}}_2^{+}$) follow the exact same pattern: two sigma bonds and zero lone pairs. All four of these species are sp hybridized, perfectly linear, and completely free of d-orbital involvement. They are our prime candidates!

Next, we look at ozone (${\text{O}}_3$) and sulfur dichloride (${\text{SCl}}_2$).

In ozone, the central oxygen forms one double bond and one single coordinate bond, and it retains one lone pair. This gives a steric number of $2+1=3$, meaning it is sp${}^2$ hybridized. The lone pair repels the bond pairs, bending the molecule into a V-shape.

Sulfur dichloride has a central sulfur atom with two single bonds and two lone pairs. Its steric number is $2+2=4$, leading to sp${}^3$ hybridization. Like water, it is heavily bent. Both of these species fail the linearity test immediately.

Linear but Disqualified

Finally, we encounter the trickiest group: ${\text{ICl}}_2^{-}$, ${\text{I}}_3^{-}$, and ${\text{XeF}}_2$.

Let's take Xenon difluoride (${\text{XeF}}_2$) as an example. Xenon has 8 valence electrons. It uses two to bond with fluorine, leaving 6 electrons, or 3 lone pairs. The steric number is $2+3=5$, which corresponds to sp${}^3$d hybridization.

According to VSEPR theory, to minimize repulsion, the three lone pairs occupy the equatorial positions of a trigonal bipyramid, while the two fluorine atoms are pushed to the axial positions. This makes the molecule perfectly linear! However, because their hybridization is sp${}^3$d, they utilize a d-orbital. This directly violates the second condition of our problem, so we must ruthlessly disqualify them.

After filtering out the bent molecules and the d-orbital deceivers, we are left with exactly four species that are both linear and lack d-orbital contribution: ${\text{BeCl}}_2$, ${\text{N}}_3^{-}$, ${\text{N}}_2\text{O}$, and ${\text{NO}}_2^{+}$. The final answer is 4.