Question

The number of lone pairs on the central atom in ( SF_4 ), ( XeF_4 ), ( CF_4 ), and ( BF_3 ) are respectively:

• A. 1, 2, 0, 0
• B. 1, 1, 0, 0
• C. 2, 1, 0, 0
• D. 1, 2, 1, 0

Hint:

For VSEPR problems, first count valence electrons of the central atom, then subtract bonding pairs. Lone pairs strongly affect molecular shape (e.g., SF$_4$ seesaw, XeF$_4$ square planar).

Answer 1

The Correct Option is A

Step 1: Learn how to count lone pairs.
Start with the valence electrons of the central atom, subtract one electron for each single bond it forms, then group the rest into pairs. Each pair left over is a lone pair.

Step 2: Count for $SF_4$.
Sulfur has $6$ valence electrons and makes $4$ bonds to fluorine. Electrons left: \[ 6 - 4 = 2 \Rightarrow 1 \text{ lone pair} \]

Step 3: Count for $XeF_4$.
Xenon has $8$ valence electrons and makes $4$ bonds to fluorine. Electrons left: \[ 8 - 4 = 4 \Rightarrow 2 \text{ lone pairs} \]

Step 4: Count for $CF_4$.
Carbon has $4$ valence electrons and makes $4$ bonds. Electrons left: \[ 4 - 4 = 0 \Rightarrow 0 \text{ lone pairs} \]

Step 5: Count for $BF_3$.
Boron has $3$ valence electrons and makes $3$ bonds. Electrons left: \[ 3 - 3 = 0 \Rightarrow 0 \text{ lone pairs} \]

Step 6: List the answers in order.
So the lone pairs for $SF_4$, $XeF_4$, $CF_4$, and $BF_3$ are $1$, $2$, $0$, and $0$.
\[ \boxed{1,\ 2,\ 0,\ 0} \]