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: Calculate the total number of valence electrons.
Xenon (Xe) has 8 valence electrons.
Each oxygen atom contributes 6 electrons, so for 6 oxygen atoms: \(6 \times 6 = 36\).
The charge on the ion is \(4-\), so add 4 extra electrons.
\[ \text{Total electrons} = 8 + 36 + 4 = 48 \] Step 2: Determine the bonding around Xe.
In \(XeO_6^{4-}\), xenon forms bonds with 6 oxygen atoms.
Each Xe–O bond is a $\sigma$ bond.
Thus, number of $\sigma$ bonds formed by Xe = 6.
Step 3: Determine lone pairs on Xe.
Xenon uses its valence electrons to form 6 bonds.
Each bond involves one electron from Xe, so 6 electrons are used in bonding.
Remaining electrons on Xe: \[ 8 - 6 = 2 \text{ electrons} \] These 2 electrons form 1 lone pair.
Step 4: Final count.
Lone pairs on Xe = 1
$\sigma$ bond pairs formed by Xe = 6
Final Answer:
\[ \boxed{\text{Lone pairs = 1,\ \sigma \text{ bonds} = 6} } \]