Step 1: Understanding the Concept:
Molecular geometry is determined by the Valence Shell Electron Pair Repulsion (VSEPR) theory.
VSEPR theory states that electron pairs (both bonding and lone pairs) surrounding a central atom arrange themselves in space to minimize electrostatic repulsion.
The primary factor is the "Steric Number" (SN), which is the sum of the number of atoms bonded to the central atom and the number of lone pairs on the central atom.
Linear geometry occurs specifically when the Steric Number is 2 (sp hybridization) or when the Steric Number is 5 with 3 lone pairs (sp3d hybridization, with lone pairs in equatorial positions).
Step 2: Key Formula or Approach:
The Steric Number (SN) for a central atom can be calculated as:
\[ SN = \frac{1}{2} [V + M - C + A] \]
Where $V$ is valence electrons, $M$ is monovalent atoms, $C$ is cationic charge, and $A$ is anionic charge.
- $SN = 2$: Linear molecular shape.
- $SN = 3$: Trigonal planar (if 0 LP), Bent (if 1 LP).
- $SN = 4$: Tetrahedral (if 0 LP), Pyramidal (if 1 LP), Bent (if 2 LP).
- $SN = 5$: Trigonal bipyramidal (if 0 LP), Seesaw (1 LP), T-shape (2 LP), Linear (3 LP).
Step 3: Detailed Explanation:
1. $N_2O$ (Nitrous Oxide):
The structure is $N \equiv N \rightarrow O$. The central Nitrogen is bonded to another Nitrogen and an Oxygen.
The central N atom has no lone pairs and 2 sigma bonding regions (treating multiple bonds as one region for VSEPR).
$SN = 2 + 0 = 2$. Hybridization is $sp$. The geometry is strictly linear.
2. $ClF_2^-$:
Central atom Cl has 7 valence electrons. Add 1 for the negative charge = 8.
It forms 2 bonds with F atoms (monovalent). $M = 2$.
$SN = \frac{1}{2} [7 + 2 - 0 + 1] = 5$.
Hybridization is $sp^3d$. With 2 bond pairs and 3 lone pairs, the lone pairs occupy the equatorial positions of a trigonal bipyramid to minimize repulsion.
The two F atoms occupy the axial positions, resulting in a linear molecular shape.
3. $SO_2$:
Central atom S (6 valence electrons) forms 2 double bonds with O.
It uses 4 electrons for bonding and has 1 lone pair remaining.
$SN = 2$ atoms $+ 1$ lone pair $= 3$. Hybridization is $sp^2$.
The geometry is bent or V-shaped due to the lone pair-bond pair repulsion.
4. $I_3^+$:
Central Iodine (7 valence electrons). Subtract 1 for positive charge = 6.
It bonds with 2 other Iodine atoms. $SN = \frac{1}{2} [7 + 2 - 1 + 0] = 4$.
Hybridization is $sp^3$. With 2 bond pairs and 2 lone pairs, the shape is bent (similar to water).
Step 4: Final Answer:
Comparing the results, $N_2O$ and $ClF_2^-$ are the only species with linear structures. This corresponds to Option (A).