Question

Total number of lone pairs and $\sigma$ bond pairs formed by central atom (Xe) in $XeO_6^{4-}$ is:

Hint:

For hypervalent molecules like xenon compounds, Xe can expand its octet. Count electrons carefully and subtract bonding electrons to find lone pairs.

Answer 1

Correct Answer: 6

Step 1: Understanding the Question:
We need to find the count of \(\sigma\) bonds and lone pairs on the central Xenon atom in the perxenate ion, \(XeO_6^{4-}\).
Step 2: Key Formula or Approach:
Use VSEPR theory or the steric number formula to determine hybridization and geometry.
Number of valence electrons of Xenon (\(Xe\)) = 8.
Step 3: Detailed Explanation:
In \(XeO_6^{4-}\) (perxenate ion):
The oxidation state of \(Xe\) is \(+8\), which means all its 8 valence electrons are involved in bonding.
The structure is octahedral, where \(Xe\) is bonded to 6 oxygen atoms.
Number of \(\sigma\) bond pairs = 6 (one with each oxygen atom).
Since all 8 valence electrons are used to form 6 \(\sigma\) bonds and potentially 2 \(\pi\) bonds (to minimize formal charge), there are no leftover electrons on \(Xe\).
Number of lone pairs on \(Xe\) = 0.
Total number of (\(\sigma\) bonds + lone pairs) = \(6 + 0 = 6\).
Step 4: Final Answer:
The total number of lone pairs and \(\sigma\) bond pairs on \(Xe\) is 6.