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: Basic idea.
Xenon has 8 valence electrons = 4 electron pairs.
Number of lone pairs on Xe = 4 − (number of bonds formed by Xe).
(Each $\sigma$ bond counts as one bonding pair.)
Step 2: Analyze each compound.
(i) $XeF_2$
Number of $\sigma$ bonds = 2.
Lone pairs on Xe = $4 - 2 = 2$ pairs left? Actually total electron domains = 5 (AX$_2$E$_3$).
Thus, 3 lone pairs are present.
(ii) $XeO_2F_2$
Total $\sigma$ bonds = 4 (2 Xe=O and 2 Xe–F; each contributes one $\sigma$ bond).
Thus bonding pairs = 4.
Remaining lone pairs = $4 - 4 = 0$?
But electron domain counting gives AX$_4$E$_1$.
Hence, 1 lone pair is present.
(iii) $XeO_3$
Three Xe=O bonds → 3 $\sigma$ bonds.
Thus bonding pairs = 3.
Remaining lone pairs = $4 - 3 = 1$.
Hence, 1 lone pair is present.
(iv) $XeO_3F_2$
Total $\sigma$ bonds = 5.
Bonding pairs = 5.
Remaining lone pairs = $4 - 5$ not possible → actually AX$_5$E$_1$.
Hence, 1 lone pair present.
(v) $XeOF_2$
Total $\sigma$ bonds = 3.
Thus bonding pairs = 3.
Remaining lone pairs = $4 - 3 = 1$.
Hence, 1 lone pair is present.
Step 3: Final Answer.
Compounds having exactly one lone pair on Xe are:
\[ \boxed{ XeO_2F_2,\; XeO_3,\; XeO_3F_2,\; XeOF_2 }. \]