Question

Match the molecule/ion (List-I) with their number of bond pair (BP) and lone pair (LP) on the central metal atom (List-II). \[\begin{array}{|l|l|} \hline \textbf{List-I} & \textbf{List-II} \\ \hline \textbf{Molecule/ion} & \textbf{BP/LP on central metal atom} \\ \hline \text{(A) SO$_2$} & \text{(I) BP = 4 and LP = 2} \\ \hline \text{(B) ClF$_3$} & \text{(II) BP = 5 and LP = 1} \\ \hline \text{(C) BrF$_5$} & \text{(III) BP = 2 and LP = 1} \\ \hline \text{(D) XeF$_4$} & \text{(IV) BP = 3 and LP = 2} \\ \hline \end{array}\]

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

Hint:

The number of bond pairs and lone pairs can be predicted based on the structure of the molecule using the VSEPR theory.

Answer 1

The Correct Option is A

Step 1: SO\(_2\) (Sulfur Dioxide).
SO\(_2\) has 2 bond pairs and 1 lone pair on the sulfur atom, thus corresponding to option (III) with BP = 2 and LP = 1.

Step 2: ClF\(_3\) (Chlorine trifluoride).
ClF\(_3\) has 3 bond pairs and 2 lone pairs on the chlorine atom, corresponding to option (IV) with BP = 3 and LP = 2.

Step 3: BrF\(_5\) (Bromine pentafluoride).
BrF\(_5\) has 5 bond pairs and 1 lone pair on the bromine atom, corresponding to option (II) with BP = 5 and LP = 1.

Step 4: XeF\(_4\) (Xenon tetrafluoride).
XeF\(_4\) has 4 bond pairs and 2 lone pairs on the xenon atom, corresponding to option (I) with BP = 4 and LP = 2.

Step 5: Conclusion.
Thus, the correct matching is: (A) - (III), (B) - (II), (C) - (I), (D) - (IV). The correct answer is option (1).