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: Count bond pairs and lone pairs for each option using VSEPR.
XeF\(_2\): Xe has 8 valence e\(^-\); 2 bonds use 4 e\(^-\), leaving 4 e\(^-\) = 3 lone pairs on Xe. Bond pairs = 2, lone pairs = 3. Lone pairs > bond pairs.

Step 2: Check others quickly.
ClF\(_3\): 3 bonds, 2 lone pairs -- equal or fewer lone pairs. XeF\(_4\): 4 bonds, 2 lone pairs -- fewer. SF\(_4\): 4 bonds, 1 lone pair -- fewer. Only XeF\(_2\) satisfies the condition. \[ oxed{XeF_2} \]