Chemistry Of Noble Gases Pseudohalogens And Interhalogen Compounds

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 bond counts as one bonding pair.)
Step 2: Analyze each compound.
(i)
Number of bonds = 2.
Lone pairs on Xe = pairs left? Actually total electron domains = 5 (AX E ).
Thus, 3 lone pairs are present.
(ii)
Total bonds = 4 (2 Xe=O and 2 Xe–F; each contributes one bond).
Thus bonding pairs = 4.
Remaining lone pairs = ?
But electron domain counting gives AX E .
Hence, 1 lone pair is present.
(iii)
Three Xe=O bonds → 3 bonds.
Thus bonding pairs = 3.
Remaining lone pairs = .
Hence, 1 lone pair is present.
(iv)
Total bonds = 5.
Bonding pairs = 5.
Remaining lone pairs = not possible → actually AX E .
Hence, 1 lone pair present.
(v)
Total bonds = 3.
Thus bonding pairs = 3.
Remaining lone pairs = .
Hence, 1 lone pair is present.
Step 3: Final Answer.
Compounds having exactly one lone pair on Xe are:

Step 1: Understand valence electrons of Xenon.
Xenon (Xe) has 8 valence electrons.
While forming compounds, some electrons are used in bonding and remaining electrons form lone pairs.
Step 2: Check common Xenon compounds.
(i)
Electron pairs around Xe = 5 (2 bonding + 3 lone pairs).
Hence, 3 lone pairs are present.
(ii)
Electron pairs around Xe = 6 (4 bonding + 2 lone pairs).
Hence, 2 lone pairs are present.
(iii)
Electron pairs around Xe = 7 (6 bonding + 1 lone pair).
Hence, 1 lone pair is present.
(iv)
Total electron domains = 4 (3 bonding + 1 lone pair).
Hence, 1 lone pair is present.
(v)
Total electron domains = 6 (5 bonding + 1 lone pair).
Hence, 1 lone pair is present.
Step 3: Final Answer.
Compounds of Xenon having one lone pair are: