Question

Match the molecules in Group I with the type of bonds present in them, in Group II

Group I	Group II
P) NaCl	1) Coordination bond
Q) $H_2$	2) Polar covalent bond
R) $Pd-P$ bond in $Pd(PPh_3)_4	3) Covalent bond
S) $C-Cl$ bond in $CH_3Cl $	4) Ionic bond

• A. P-4, Q-1, R-3, S-2
• B. P-2, Q-3, R-1, S-4
• C. P-4, Q-3, R-1, S-2
• D. P-4, Q-3, R-2, S-1

Answer 1

The Correct Option is C

To solve this question, we need to identify the type of bond present in each molecule listed in Group I and match it with the correct bond type in Group II. 

1. NaCl: Sodium chloride is a classic example of an ionic compound. It forms an ionic bond because it consists of positively charged sodium ions (\(\text{Na}^+\)) and negatively charged chloride ions (\(\text{Cl}^−\)). Hence, it matches with 4) Ionic bond.
2. \( H_2 \): Molecular hydrogen consists of two hydrogen atoms sharing a pair of electrons, forming a covalent bond. This matches with 3) Covalent bond.
3. \( \text{Pd-P} \) bond in \( \text{Pd}(\text{PPh}_3)_4 \): The palladium-phosphorus bond in this compound is a result of a coordination bond, where the phosphorus atom donates a lone pair of electrons to the palladium. This matches with 1) Coordination bond.
4. \( C-Cl \) bond in \( \text{CH}_3\text{Cl} \): The bond between carbon and chlorine in chloromethane (methyl chloride) is a polar covalent bond. This is due to the significant electronegativity difference between carbon and chlorine, which causes a partial charge separation. This matches with 2) Polar covalent bond.

Based on the above analysis, the correct matches are:

• P-4: NaCl is an ionic compound.
• Q-3: \( H_2 \) has a covalent bond.
• R-1: \( \text{Pd-P} \) bond in \( \text{Pd}(\text{PPh}_3)_4 \) is a coordination bond.
• S-2: \( C-Cl \) bond in \( \text{CH}_3\text{Cl} \) is a polar covalent bond.

Therefore, the correct answer is: P-4, Q-3, R-1, S-2.