We match each species in List I with the correct property/shape in List II.

1. $\mathrm{SO_2Cl_2}$

This is sulfuryl chloride.

• Central atom: S
• Around S, there are four sigma bonds: two $\mathrm{S=O}$ and two $\mathrm{S-Cl}$.
• Hence, electron pair geometry and molecular shape are tetrahedral.

So, $A \to \text{III (Tetrahedral)}$

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2. $\mathrm{NO}$

Nitric oxide has an odd number of electrons.

Total valence electrons: $5 + 6 = 11$

Since it has one unpaired electron, it is paramagnetic.

So, $B \to \text{I (Paramagnetic)}$

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3. $\mathrm{NO_2^-}$

This is nitrite ion.

Total valence electrons: $5 + 2(6) + 1 = 18$

• Central atom: N
• It has two bond pairs and one lone pair.
• Electron pair geometry: trigonal planar
• Molecular shape: bent/angular
• All electrons are paired, so it is diamagnetic.

So, $C \to \text{II (Diamagnetic)}$

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4. $\mathrm{I_3^-}$

Triiodide ion:

• Central atom: I
• It has 5 electron pairs around central iodine: 2 bond pairs + 3 lone pairs.
• Trigonal bipyramidal electron arrangement with three lone pairs occupying equatorial positions.
• Therefore molecular shape is linear.

So, $D \to \text{IV (Linear)}$

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5. Final matching

Thus the correct matching is: $A-\text{III},\; B-\text{I},\; C-\text{II},\; D-\text{IV}$

This corresponds to Option B.

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6. Comparison with stored answer

Stored correct answer: B

My derived answer: B

Hence, the answer agrees with the stored correct answer.