Question

Match List - I with List - II.

List - I	List - II
(A) ICl	(IV) Linear
(B) ICl3	(I) T-Shape
(C) ClF5	(II) Square pyramidal
(D) IF7	(III) Pentagonal bipyramidal

Choose the correct answer from the options given below:

• A. (A)–(I), (B)–(IV), C–(III), D–(II)
• B. (A)–(I), (B)–(III), C–(II), D–(IV)
• C. (A)–(IV), (B)–(I), C–(II), D–(III)
• D. (A)–(IV), (B)–(III), C–(II), D–(I)

Answer 1

The Correct Option is C

To determine the correct matches, we analyze the molecular geometry of each compound based on the number of bond pairs and lone pairs of electrons around the central atom.
Step 1: Molecular geometry of \(\text{ICl}\)
\(\text{ICl}\) consists of only two atoms (iodine and chlorine), forming a diatomic molecule. Its molecular geometry is linear. Thus:
\[(A) \, \text{ICl} \rightarrow \text{(IV) Linear}.\]
Step 2: Molecular geometry of \(\text{ICl}_3\)
\(\text{ICl}_3\) has 3 bond pairs and 2 lone pairs around the iodine atom. According to the VSEPR theory, this results in a T-shaped molecular geometry. Thus:
\[(B) \, \text{ICl}_3 \rightarrow \text{(I) T-Shape}.\]
Step 3: Molecular geometry of \(\text{ClF}_5\)
\(\text{ClF}_5\) has 5 bond pairs and 1 lone pair around the central chlorine atom. According to the VSEPR theory, this configuration forms a square pyramidal geometry. Thus:
\[(C) \, \text{ClF}_5 \rightarrow \text{(II) Square pyramidal}.\]
Step 4: Molecular geometry of \(\text{IF}_7\)  
\(\text{IF}_7\) has 7 bond pairs and no lone pairs around the iodine atom. This configuration corresponds to a pentagonal bipyramidal geometry.
Thus:
\[(D) \, \text{IF}_7 \rightarrow \text{(III) Pentagonal bipyramidal}.\]
Final Matches:
\[(A) \rightarrow \text{(IV)}, \, (B) \rightarrow \text{(I)}, \, (C) \rightarrow \text{(II)}, \, (D) \rightarrow \text{(III)}.\]
Correct Answer: (3).

Answer 2

Step 1: Understanding the question.
We are asked to match the given interhalogen compounds (List-I) with their molecular geometries (List-II) based on the VSEPR theory.

Step 2: Analyze each compound.
(A) ICl → It consists of two atoms, hence the molecule is linear.
So, (A) → (IV) Linear.

(B) ICl₃ → Iodine has 7 valence electrons, and 3 are used in bonding with chlorine atoms, leaving 2 lone pairs. Thus, it has 5 electron pairs (AX₃E₂ type), giving a T-shaped structure.
So, (B) → (I) T-Shape.

(C) ClF₅ → Chlorine has 7 valence electrons and forms 5 bonds with fluorine, leaving 1 lone pair (AX₅E type). According to VSEPR theory, this results in a square pyramidal structure.
So, (C) → (II) Square pyramidal.

(D) IF₇ → Iodine forms 7 bonds with fluorine and has no lone pairs (AX₇ type). The geometry is pentagonal bipyramidal.
So, (D) → (III) Pentagonal bipyramidal.

Step 3: Final Matching.
(A)–(IV), (B)–(I), (C)–(II), (D)–(III)

Final Answer: (A)–(IV), (B)–(I), (C)–(II), (D)–(III)