The number of lone pairs on the central atom in ( SF_4 ), ( XeF_4 ), ( CF_4 ), and ( BF_3 ) are respectively:
Show 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).
Concept:
The number of lone pairs on the central atom is determined using the valence electrons of the central atom and the number of sigma bonds formed in the molecule. In VSEPR theory, lone pairs are calculated after distributing bonding pairs.
Step 1: SF$_4$
Sulfur (Group 16) has 6 valence electrons. It forms 4 sigma bonds with fluorine atoms.
Remaining electrons on S:
\[
6 - 4 = 2 \text{ electrons} \Rightarrow 1 \text{ lone pair}
\]
Thus, SF$_4$ has 1 lone pair.
Step 2: XeF$_4$
Xenon (Group 18) has 8 valence electrons. It forms 4 sigma bonds.
Remaining electrons:
\[
8 - 4 = 4 \text{ electrons} \Rightarrow 2 \text{ lone pairs}
\]
Thus, XeF$_4$ has 2 lone pairs.
Step 3: CF$_4$
Carbon (Group 14) has 4 valence electrons. It forms 4 sigma bonds.
Remaining electrons:
\[
4 - 4 = 0 \Rightarrow 0 \text{ lone pairs}
\]
Thus, CF$_4$ has 0 lone pairs.
Step 4: BF$_3$
Boron (Group 13) has 3 valence electrons. It forms 3 sigma bonds.
Remaining electrons:
\[
3 - 3 = 0 \Rightarrow 0 \text{ lone pairs}
\]
Thus, BF$_3$ has 0 lone pairs.
