The molecules having square pyramidal geometry are

Core Logic

Let's find the number of bond pairs (σ$\sigma$-bonds or regions) and lone pairs on the central atom of each species:

• ICl₂^-$\mathrm{ICl}_2^-$:
• Central atom Iodine has 7$7$ valence electrons + 1$+ 1$ negative charge = 8$= 8$ electrons.
• Forms 2$2$ single bonds (bond pairs = 2$= 2$).
• Remaining 6$6$ electrons form 3$3$ lone pairs.
• Ratio is 2 : 3$2 : 3$ (Matches LIST-II, III).



• H₂O$\mathrm{H}_2\mathrm{O}$:
• Oxygen has 6$6$ valence electrons.
• Forms 2$2$ bond pairs with Hydrogens.
• Remaining 4$4$ electrons form 2$2$ lone pairs.
• Ratio is 2 : 2$2 : 2$ (Matches LIST-II, IV).



• SO₂$\mathrm{SO}_2$:
• Sulfur has 6$6$ valence electrons.
• Forms 2$2$ double bonds (which are counted as 4$4$ bonding pairs of electrons/bond pairs in typical VSEPR representations here).
• Remaining 2$2$ electrons form 1$1$ lone pair.
• Ratio is 4 : 1$4 : 1$ (Matches LIST-II, II).



• XeF₄$\mathrm{XeF}_4$:
• Xenon has 8$8$ valence electrons.
• Forms 4$4$ bond pairs with Fluorines.
• Remaining 4$4$ electrons form 2$2$ lone pairs.
• Ratio is 4 : 2$4 : 2$ (Matches LIST-II, I).



Thus, the correct mapping is: A-III, B-IV, C-II, D-I.

Pattern Recognition

For match-the-column with VSEPR structures:

• Always find steric number: Steric Number = (1)/(2)(V + M - C + A)$\text{Steric Number} = \frac{1}{2}(V + M - C + A)$.
• Water is 2$2$ bond pairs, 2$2$ lone pairs (sp³$sp^3$) arrow$\rightarrow$ B-IV. This alone helps eliminate multiple incorrect options immediately.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Core Logic

Let's compute the total valence electrons for the nitrite ion (NO₂^-$NO_2^-$):

In the valid Lewis structural representation:

• The central nitrogen atom forms one single bond and one double bond with the terminal oxygens, consuming 2 + 4 = 6$2 + 4 = 6$ bonding electrons.
• Remaining non-bonded valence electrons = 18 - 6 = 12$18 - 6 = 12$ electrons.

Step 1: Account for Lone Pairs

Distribution of non-bonded electrons across the individual atoms:

• Central Nitrogen atom has 1$1$ lone pair (2$2$ electrons).
• Single-bonded Oxygen atom has 3$3$ lone pairs (6$6$ electrons).
• Double-bonded Oxygen atom has 2$2$ lone pairs (4$4$ electrons).

Pattern Recognition

Non-bonded electrons can always be obtained directly by subtracting total bonding electrons from total valence electrons.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Step 1: Determine 'n'

ClF₃$\mathrm{ClF}_3$ has a central Chlorine atom with 7 valence electrons, bound to 3 Fluorine atoms. This leaves 2 lone pairs. The geometry is trigonal bipyramidal (T-shaped molecule). In its most stable geometry, both lone pairs lie in the equatorial plane to minimize lone pair-bonding pair repulsions. Thus, n = 2$n = 2$.

Step 2: Find ions with 2 unpaired electrons

Let us compute the number of unpaired electrons for each configuration:

• A. V³⁺$\mathrm{V}^{3+}$: [Ar] 3d² arrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons.
• B. Ti³⁺$\mathrm{Ti}^{3+}$: [Ar] 3d¹ arrow 1$[\mathrm{Ar}] 3d^1 \rightarrow 1$ unpaired electron.
• C. Cu²⁺$\mathrm{Cu}^{2+}$: [Ar] 3d⁹ arrow 1$[\mathrm{Ar}] 3d^9 \rightarrow 1$ unpaired electron.
• D. Ni²⁺$\mathrm{Ni}^{2+}$: [Ar] 3d⁸ arrow 2$[\mathrm{Ar}] 3d^8 \rightarrow 2$ unpaired electrons.
• E. Ti²⁺$\mathrm{Ti}^{2+}$: [Ar] 3d² arrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons.

Thus, A, D, and E have exactly n=2$n=2$ unpaired electrons.

Pattern Recognition

Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember ClF₃$\mathrm{ClF}_3$ is T-shaped with 2 equatorial lone pairs. Look for d²$d^2$ or d⁸$d^8$ configurations among the transition metal ions.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements

Core Logic

Let us evaluate each central atom configuration systematically:

• A. PF₅$\text{PF}_5$: Phosphorus has 5 valence electrons, forming 5σ$5\sigma$ bonds with zero lone pairs. Steric number = 5 sp³d$= 5 \implies sp^3d$ hybridisation.
• B. SF₆$\text{SF}_6$: Sulfur has 6 valence electrons, forming 6σ$6\sigma$ bonds with zero lone pairs. Steric number = 6 sp³d²$= 6 \implies sp^3d^2$ hybridisation.
• C. Ni(CO)₄$\text{Ni}(\text{CO})_4$: Nickel is in a 0 oxidation state (3d⁸ 4s²$3d^8 4s^2$). Carbon monoxide is a strong field ligand, forcing rearrangement into a filled 3d¹⁰$3d^{10}$ state. The vacant 4s$4s$ and three 4p$4p$ orbitals hybridise to give an sp³$sp^3$ configuration.

  

• D. [PtCl₄]²⁻$[\text{PtCl}_4]^{2-}$: Platinum is in the +2$+2$ oxidation state (5d⁸$5d^8$). Since it belongs to the 5d$5d$ transition series, all ligands behave as strong field elements, leading to interior spin-pairing and an inner orbital square-planar dsp²$dsp^2$ hybridisation state.

  

Pattern Recognition

Shortcut: Match main-group species first: PF₅ arrow sp³d (II)$\text{PF}_5 \rightarrow sp^3d\text{ (II)}$, SF₆ arrow sp³d² (III)$\text{SF}_6 \rightarrow sp^3d^2\text{ (III)}$. This immediately isolates Option (A) without needing to evaluate coordination fields.

Evaluation Rubric / Model Answer

Option (A)

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds