Question

The number of lone pairs of electrons on the central atom of $\text{XeO}_3, \text{XeOF}_4$ and $\text{XeF}_6$ respectively is

• A. 1, 1, 1
• B. 3, 2, 1
• C. 2, 1, 0
• D. 1, 2, 1

Hint:

A general formula for the number of lone pairs (LP) on the central atom is: $LP = \frac{1}{2} [\text{Valence } e^- \text{ on central atom} + (\text{number of attached atoms} \times \text{valency}) - 2 \times \text{charge}]$. A simpler way is: $LP = \frac{1}{2} [\text{Valence } e^- - \text{Bonding } e^-]$. For $\text{O}$, count 2 bonding $e^-$.

Answer 1

The Correct Option is A

Step 1: Calculate the steric number and lone pairs for $\text{XeO_3$.}
Central atom: Xenon ($\text{Xe}$) from Group 18, so 8 valence electrons. Surrounding atoms: 3 Oxygen ($\text{O}$) atoms. $\text{O}$ forms double bonds ($\sigma + \pi$), but only the $\sigma$ bonds and lone pairs contribute to the steric number. The number of $\sigma$ bonds is 3. Bonding electrons (effective): $3 \times 2 = 6$ (since $\text{O}$ is double bonded, but the lone pair calculation treats $\text{O}$ as taking 2 electrons). A simpler method is $\text{Steric Number} = (\text{Valence } e^- + \text{Monovalent atoms}) / 2$. This doesn't work here. Using the total number of valence electrons around $\text{Xe}$: Valence $e^-$ on $\text{Xe}$: 8. Electrons used in $\sigma$ bonds: 3 (one $\sigma$ bond to each $\text{O}$). Electrons used in $\pi$ bonds: 3 ($\text{O}$ forms a total of 6 bonds, 3 $\sigma$, 3 $\pi$). Number of $\sigma$ bonds = 3. Lone pairs $\text{LP} = \frac{1}{2} (\text{Valence } e^- - 2 \times \text{number of } \sigma \text{ bonds})$. This is not right either. The correct, simplified approach is: $\text{LP} = \frac{1}{2} (\text{Valence } e^- - \text{number of bonds})$. This works only for single bonds. The formal method: $\text{Steric Number (SN)} = \text{number of } \sigma \text{ bonds} + \text{number of lone pairs}$. For $\text{XeO}_3$: $\text{O}$ forms 2 bonds. Total bonds on $\text{Xe}$ are $3 \times 2 = 6$. $\text{Xe}$ has 8 valence $e^-$. $6$ are used in bonding (3 $\sigma$ and 3 $\pi$). Remaining $e^- = 8-6=2$. Lone pairs = $2/2 = 1$. ($\text{SN}=3+1=4$, shape is trigonal pyramidal).

Step 2: Calculate the steric number and lone pairs for $\text{XeOF_4$.}
Central atom: $\text{Xe}$, 8 valence $e^-$. Surrounding atoms: 1 $\text{O}$ (double bond), 4 $\text{F}$ (single bonds). Number of $\sigma$ bonds = $1 (\text{O}) + 4 (\text{F}) = 5$. Electrons used in $\sigma$ and $\pi$ bonds: $4(\text{F}) + 2(\text{O}) = 6$. Remaining $e^- = 8-6=2$. Lone pairs = $2/2 = 1$. ($\text{SN}=5+1=6$, shape is square pyramidal).

Step 3: Calculate the steric number and lone pairs for $\text{XeF_6$.}
Central atom: $\text{Xe}$, 8 valence $e^-$. Surrounding atoms: 6 $\text{F}$ (single bonds). Electrons used in $\sigma$ bonds: 6. Remaining $e^- = 8-6=2$. Lone pairs = $2/2 = 1$. ($\text{SN}=6+1=7$, shape is distorted octahedral).

Step 4: Conclude the final answer.
The number of lone pairs on the central atom of $\text{XeO}_3, \text{XeOF}_4$ and $\text{XeF}_6$ are 1, 1, and 1 respectively.