Question

What are the number of lone pairs of electrons in \[S{F_4}\], \[C{F_4}\] and \[Xe{F_4}\] ?

Answer 1

Hint: The number of lone pair and bond atoms present in between the atoms determine the shape of the molecule. The number of lone pairs present in the central atom and the number of electrons involved in bonding determine the shape.

Complete answer:
Lone pair refers to a pair of valence electrons that are not shared with another atom in a covalent bond and is sometimes called an unshared pair or non-bonding pair. Lone pairs are found in the outermost electron shell of atoms. The VSEPR theory is used to predict the geometry of individual molecules from the number of electron pairs surrounding their central atoms.
Let’s solve the given question one by one
 \[S{F_4}\] has one lone pair and four sigma bonds of \[\;fluorine\] . Here the central atom is \[Sulphur\] . Thus it is bonded by four \[\;fluorine\] atoms by using eight of the valence electrons of \[Sulphur\] . Apart from that \[\;fluorine\] have three lone pairs of electrons in its octet. During \[S{F_4}\] formation the bonding occurs between \[Sulphur\] and \[\;fluorine\] by forming a single bond between these two atoms.
In \[C{F_4}\] the central atom has six electrons and out of them four are valence electrons. \[\;fluorine\] atom forms bond with these four valence electrons of the \[carbon\] and thus there are no lone pair of electrons present in the \[C{F_4}\] molecule and all the \[\;fluorine\] atoms are bonded by central atom through single bond.
In \[Xe{F_4}\] there are eight valence electrons around the central atom \[xenon\] out of which four are involved in bond formation with the \[\;fluorine\] atom. Thus four valence electrons left in \[xenon\] form two lone pairs. Thus \[Xe{F_4}\] has two lone pairs.

Note:
According to VSEPR theory the valence electron pair surrounding an atom tends to repel each other. The order of repulsion between electron pairs is as follows: Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.