Question

Among the following compounds of Xenon, identify those having exactly one lone pair on Xenon:
$XeF_2$, $XeO_2F_2$, $XeO_3$, $XeO_3F_2$, $XeOF_2$.

Hint:

For Xenon compounds, count total $\sigma$ bonds first. Number of lone pairs = 4 − (number of $\sigma$ bonds). Then verify using VSEPR notation AX$_m$E$_n$.

Answer 1

Step 1: Understanding the Topic
This question requires applying VSEPR theory to several compounds of Xenon to determine the number of non-bonding electron pairs (lone pairs) on the central Xenon atom. As a noble gas, Xenon has 8 valence electrons available for bonding and for forming lone pairs.
Step 2: Key Approach - Electron Counting 
For each molecule, we'll determine how many of Xenon's 8 valence electrons are used in forming bonds. 

Each fluorine (F) atom forms a single bond, using 1 electron from Xe. 
Each oxygen (O) atom forms a double bond, using 2 electrons from Xe. 
The number of lone pairs is the number of remaining electrons divided by two.
Step 3: Detailed Analysis of Each Compound 
(i) $XeF_2$: 

Electrons used for bonding = 2 (for 2 F atoms) $\times$ 1 = 2. 
Electrons remaining on Xe = $8 - 2 = 6$. 
Number of lone pairs = $6 / 2 = \mathbf{3}$. 
(ii) $XeO_2F_2$: 

Electrons used for bonding = (2 for 2 O atoms $\times$ 2) + (2 for 2 F atoms $\times$ 1) = $4 + 2 = 6$. 
Electrons remaining on Xe = $8 - 6 = 2$. 
Number of lone pairs = $2 / 2 = \mathbf{1}$. 
(iii) $XeO_3$: 

Electrons used for bonding = 3 (for 3 O atoms) $\times$ 2 = 6. 
Electrons remaining on Xe = $8 - 6 = 2$. 
Number of lone pairs = $2 / 2 = \mathbf{1}$. 
(iv) $XeOF_2$: 

Electrons used for bonding = (1 for 1 O atom $\times$ 2) + (2 for 2 F atoms $\times$ 1) = $2 + 2 = 4$. 
Electrons remaining on Xe = $8 - 4 = 4$. 
Number of lone pairs = $4 / 2 = \mathbf{2}$. (Note: The original solution also seems to have an error here. Let's re-verify. VSEPR: Central Xe, steric number = 1(O) + 2(F) + lone pairs. Bonding electrons = 4. Remaining on Xe = 4 -> 2 lone pairs. Total domains = 3 bonds + 2 lone pairs = 5. VSEPR type is AX$_3$E$_2$, which is T-shaped. So, 2 lone pairs.) 
*Correction based on re-analysis:* Following standard electron counting rules, the compounds with exactly one lone pair are $XeO_2F_2$ and $XeO_3$. The provided solution key appears to contain errors for $XeO_3F_2$ and $XeOF_2$. I will proceed by giving the answer based on correct chemical principles.

 Step 4: Final Answer 
Based on a systematic VSEPR analysis, the compounds having exactly one lone pair on the central Xe atom are: \[ \boxed{ XeO_2F_2 \text{ and } XeO_3 } \]