Question

In which of the following pairs, the central atoms have the same number of lone pairs of electrons?

• A. $PC{l}_5,Br{F}_5$
• B. $Xe{F}_2,ICl$
• C. $Xe{F}_4,{Cl{O}_4}^{-}$
• D. $SC{l}_4,C{H}_4$

Answer 1

The correct option is B $Xe{F}_2,ICl$

Explanation:

A 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

Option A: $PC{l}_5$ has $0$ lone pair, as all-electron groups are bonding pairs, so the structure is designated as $A{X}_5$

whereas in $Br{F}_5$ there are $7$ electrons in $Br$ valence shell from which $5$ electrons went to make bond with $F$, while the remaining two makes a lone pair. and thus only $1$ lone pair exist in $Br{F}_5$

Option B: $Xe{F}_2$ there are two pairs of bonded electrons and $3$ lone pairs of electrons whereas in $ICl$ number of lone pairs are $3$ and one pair of bonded electrons

Option C: $Xe{F}_4$ there are 3 lone pairs present on each $F$ atom and two lone pairs present on $Xe$. So, the total number of lone pairs are: 4 x 3 + 2 = 14

whereas in $Cl{O}_4^{-}$ there are nine lone pairs on atoms. There are three oxygen atoms that are connected through double bonds to the chlorine atom. Each of those oxygen atoms has two lone pairs and 3 in one.

Option D: In $SC{l}_4$ $1$ lone pair is present. $S$ has 10 valence electrons in $SC{l}_4$ from 4 bonds and 1 lone pair.

whereas in $C{H}_4$ all are bond pair $0$ lone pair. If these are all bond pairs the molecular geometry is tetrahedral

Hence option B is correct