Question

Match List-I with the List-II

List-I (Molecule / Species)	List-II (Property / Shape)
(A) \(SO_2Cl_2\)	(I) Paramagnetic
(B) NO	(II) Diamagnetic
(C) \(NO^{-}_{2}\)	(III) Tetrahedral
(D) \(I^{-}_{3}\)	(IV) Linear

Choose the correct answer from the options given below:

• A. A-IV, B-I, C-III, D-II
• B. A-III, B-I, C-II, D-IV
• C. A-II, B-III, C-I, D-IV
• D. A-III, B-IV, C-II, D-I

Answer 1

The Correct Option is B

(A) SO$_2$Cl$_2$: Hybridization is sp$^3$, Tetrahedral shape.
(B) NO: Unpaired electron, hence Paramagnetic.
(C) NO$_2^-$: Paired electrons, hence Diamagnetic.
(D) I$_3^-$: Linear shape with sp$^3$d hybridization.

Answer 2

To solve the problem of matching molecules/species with their properties or shapes, let's systematically analyze each molecule listed under List-I and match them with the appropriate category from List-II.

1. (A) \(SO_2Cl_2\): Sulfuryl chloride (\(SO_2Cl_2\)) has a tetrahedral molecular geometry. This is because the central sulfur atom forms four bonds—two with oxygen and two with chlorine. Hence, this molecule should be matched with "Tetrahedral," corresponding to option III.
2. (B) NO: Nitric oxide (NO) is a molecule that possesses one unpaired electron, making it a radical. The presence of an unpaired electron gives NO paramagnetic properties. Therefore, this molecule should be matched with "Paramagnetic," which is option I.
3. (C) \(NO^-_2\): The nitrite ion (\(NO^-_2\)) has no unpaired electrons; its electronic configuration leads to all electrons being paired. Consequently, this ion is diamagnetic. Therefore, it should be matched with "Diamagnetic," corresponding to option II.
4. (D) \(I^-_3\): The triiodide ion (\(I^-_3\)) has a linear geometry. It consists of three iodine atoms with the central iodine atom forming bonds with the two outer iodine atoms, forming a linear structure. Hence, this species should be matched with "Linear," which corresponds to option IV.

After analyzing each molecule/species, the correct matches are established as follows:

• \(SO_2Cl_2\) - Tetrahedral (III)
• NO - Paramagnetic (I)
• \(NO^-_2\) - Diamagnetic (II)
• \(I^-_3\) - Linear (IV)

Therefore, the correct match-answer is: A-III, B-I, C-II, D-IV.