Match Column-I with Column-II and choose the correct answer from the code given below:

Column-I

Column-II

A.

Linear shape

I.

SO₂

B.

sp³d-hybridisation

II.

XeF₄

C.

sp²

III.

BeCl₂

D.

sp³d²

IV.

I₃^–

Question

Match Column-I with Column-II and choose the correct answer from the code given below:

	Column-I		Column-II
A.	Linear shape	I.	SO₂
B.	sp³d-hybridisation	II.	XeF₄
C.	sp²	III.	BeCl₂
D.	sp³d²	IV.	I₃^–

Accepted Answer

Let's analyze each option in Column-I and match it with the corresponding molecule in Column-II based on its shape or hybridization.

Concept: VSEPR theory and hybridization are used to predict the geometry and shape of molecules. Hybridization determines the number of hybrid orbitals, while VSEPR theory considers both bonding and lone pairs to predict the molecular shape.

A. Linear shape:

BeCl $_{2}$ : Beryllium in BeCl $_{2}$ undergoes sp hybridization, forming two sigma bonds with chlorine atoms. There are no lone pairs on the central beryllium atom. This results in a linear geometry.
I $_{3}^{-}$ : The central iodine atom in I $_{3}^{-}$ has 3 bond pairs (with two other iodine atoms) and 2 lone pairs. Its hybridization is sp $_{3}$ d, and the electron geometry is trigonal bipyramidal. However, due to the presence of lone pairs in the equatorial positions, the molecular shape is linear.

Therefore, both BeCl $_{2}$ and I $_{3}^{-}$ can have a linear shape. Looking at the options, I $_{3}^{-}$ is paired with A-IV in option D, which is correct.

B. sp $_{3}$ d-hybridisation:

I $_{3}^{-}$ : As discussed above, the central iodine atom in I $_{3}^{-}$ has 3 bond pairs (two I-I bonds and one lone pair effectively acting as a bond pair for hybridization calculation) and 2 lone pairs. The steric number is 5 (3 bond pairs + 2 lone pairs), which corresponds to sp $_{3}$ d hybridization.

So, B matches with IV (I $_{3}^{-}$ ).

C. sp $_{2}$ :

SO $_{2}$ : The central sulfur atom in SO $_{2}$ forms two sigma bonds with oxygen atoms and has one lone pair. The steric number is 3 (2 bond pairs + 1 lone pair), which corresponds to sp $_{2}$ hybridization.

So, C matches with I (SO $_{2}$ ).

D. sp $_{3}$ d $_{2}$ :

XeF $_{4}$ : The central xenon atom in XeF $_{4}$ forms four sigma bonds with fluorine atoms and has two lone pairs. The steric number is 6 (4 bond pairs + 2 lone pairs), which corresponds to sp $_{3}$ d $_{2}$ hybridization.

So, D matches with II (XeF $_{4}$ ).

Combining these matches:

A - IV (I $_{3}^{-}$ has a linear shape)
B - IV (I $_{3}^{-}$ has sp $_{3}$ d hybridization)
C - I (SO $_{2}$ has sp $_{2}$ hybridization)
D - II (XeF $_{4}$ has sp $_{3}$ d $_{2}$ hybridization)

Let's re-evaluate the options provided in the question and the correct answer. The question asks to match Column-I with Column-II. The provided correct answer is D.

Let's check option D: A-II, B-IV, C-III, D-I

A. Linear shape: Matches with II (XeF $_{4}$ ). This is incorrect. XeF $_{4}$ has a square planar shape.
B. sp $_{3}$ d-hybridisation: Matches with IV (I $_{3}^{-}$ ). This is correct.
C. sp $_{2}$ : Matches with III (BeCl $_{2}$ ). This is incorrect. BeCl $_{2}$ has sp hybridization.
D. sp $_{3}$ d $_{2}$ : Matches with I (SO $_{2}$ ). This is incorrect. SO $_{2}$ has sp $_{2}$ hybridization.

There seems to be a discrepancy between the question's options/correct answer and standard chemical principles. Let's re-evaluate the matches based on the most common understanding and then see if any option fits.

Re-evaluation:

A. Linear shape: BeCl $_{2}$ (sp hybridization, 0 lone pairs) and I $_{3}^{-}$ (sp $_{3}$ d hybridization, 2 lone pairs) both have linear shapes.
B. sp $_{3}$ d-hybridisation: I $_{3}^{-}$ (steric number 5: 2 bond pairs + 3 lone pairs, or 3 bond pairs + 2 lone pairs if considering the central atom's bonding to two other I atoms and 3 lone pairs on the central I atom for hybridization calculation, leading to 5 electron domains). Let's be precise: Central I has 7 valence electrons. It forms single bonds with two other I atoms. So, 2 bond pairs. Remaining electrons = 7 - 2 = 5. These form 3 lone pairs. Total electron domains = 2 (bond pairs) + 3 (lone pairs) = 5. Thus, sp $_{3}$ d hybridization.
C. sp $_{2}$ : SO $_{2}$ (2 bond pairs + 1 lone pair = 3 electron domains) has sp $_{2}$ hybridization.
D. sp $_{3}$ d $_{2}$ : XeF $_{4}$ (4 bond pairs + 2 lone pairs = 6 electron domains) has sp $_{3}$ d $_{2}$ hybridization.

Based on this, the correct matches should be:

A. Linear shape $\to$ III (BeCl $_{2}$ ) or IV (I $_{3}^{-}$ )
B. sp $_{3}$ d-hybridisation $\to$ IV (I $_{3}^{-}$ )
C. sp $_{2}$ $\to$ I (SO $_{2}$ )
D. sp $_{3}$ d $_{2}$ $\to$ II (XeF $_{4}$ )

Let's check the given options again with these correct matches:

A) A-IV, B-III, C-I, D-II: A-IV (Linear - I $_{3}^{-}$ ) is correct. B-III (sp $_{3}$ d - BeCl $_{2}$ ) is incorrect (BeCl\(_

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