Match List I with List II : | List I (Molecule/Species) | List II (Property/Shape) | |---|---| | A. $\mathrm{SO}_{2} \mathrm{Cl}_{2}$ | I. Paramagnetic | | B. NO | II. Diamagnetic | | C. $\mathrm{NO}_{2}^{-}$ | III. Tetrahedral | | D. $\mathrm{I}_{3}^{-}$ | IV. Linear | Choose the correct answer from the options given below :

: S is , so the shape is tetrahedral (III). NO has an odd number of electrons (one unpaired), so it is paramagnetic (I). has all electrons paired, so it is diamagnetic (II). has a central iodine, giving a linear ion (IV). So A-III, B-I, C-II, D-IV.

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Chemical Bonding and Molecular Structure

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Match List I with List II :

List I (Molecule/Species)	List II (Property/Shape)
A. $SO_{2} Cl_{2}$	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 :

Answer: A → III, B → I, C → II, D → IV

Solution

S O_{2} C l_{2}

: S is

s p^{3}

, so the shape is tetrahedral

\to

(III).
NO has an odd number of electrons (one unpaired), so it is paramagnetic

\to

(I).

N O_{2}^{-}

has all electrons paired, so it is diamagnetic

\to

(II).

I_{3}^{-}

has a central

s p^{3} d

iodine, giving a linear ion

\to

(IV).
So A-III, B-I, C-II, D-IV.

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Match the columns

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