Question

The molecule which has more number of lone pair of electrons than the bond pair of electrons in its central atom is:

• A. \(XeF_2\)
• B. \(ClF_3\)
• C. \(XeF_4\)
• D. \(SF_4\)

Hint:

Remember the VSEPR configurations: \[ XeF_2 : AX_2E_3 \] \[ ClF_3 : AX_3E_2 \] \[ XeF_4 : AX_4E_2 \] \[ SF_4 : AX_4E \] Only \(XeF_2\) has more lone pairs (\(3\)) than bond pairs (\(2\)).

Answer 1

The Correct Option is A

Step 1: Determine bond pairs and lone pairs in \(XeF_2\).
Xenon has \(8\) valence electrons.
In \(XeF_2\), xenon forms two \(Xe-F\) bonds. Therefore, \[ \text{Bond pairs}=2 \] The remaining electrons on xenon form \[ \text{Lone pairs}=3 \] Thus, \[ \text{Lone pairs} \gt \text{Bond pairs} \] since \[ 3\gt 2 \]

Step 2: Check \(ClF_3\).
For \(ClF_3\), \[ \text{Bond pairs}=3 \] \[ \text{Lone pairs}=2 \] Therefore, \[ 2\lt 3 \] Hence, it does not satisfy the condition.

Step 3: Check \(XeF_4\).
For \(XeF_4\), \[ \text{Bond pairs}=4 \] \[ \text{Lone pairs}=2 \] Therefore, \[ 2\lt 4 \] Hence, it does not satisfy the condition.

Step 4: Check \(SF_4\).
For \(SF_4\), \[ \text{Bond pairs}=4 \] \[ \text{Lone pairs}=1 \] Therefore, \[ 1\lt 4 \] Hence, it does not satisfy the condition.

Step 5: Final conclusion.
Only \(XeF_2\) has more lone pairs than bond pairs on the central atom. \[ \boxed{XeF_2} \] Therefore, the correct option is \[ \boxed{(1)} \]