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Chemistry Question 71 – JEE-MAIN 2026

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According to Lewis theory, the total number of σ bond-pairs and lone pair of electrons around the central atom of XeO64− ion is _______.

Calculate the total number of valence electrons in the XeO64− ion, considering the charge.

Step 1: Calculate Total Valence Electrons✦ Active

The central atom is Xenon (Xe), which is in Group 18 and has 8 valence electrons. Each Oxygen (O) atom is in Group 16 and has 6 valence electrons. The ion has a 4- charge, adding 4 electrons. Total valence electrons in XeO64− are:

8(from Xe)+6×6(from 6 O atoms)+4(for 4- charge)=8+36+4=48 electrons
Step 2: Determine Sigma Bond Pairs○ Expand

Xenon is bonded to 6 oxygen atoms. Each bond between the central atom and a terminal atom is considered a σ bond. Therefore, there are 6 σ bond pairs around the central Xe atom. These 6 bonds utilize 6×2=12 electrons.

Step 3: Determine Lone Pairs on Central Atom and Total Count○ Expand

Remaining electrons to be distributed = 48−12=36 electrons. Each of the 6 oxygen atoms needs 6 more electrons to complete its octet (since it already has 2 electrons from the Xe-O bond). Thus, 6×6=36 electrons are used to complete the octets of the terminal oxygen atoms. Since all remaining electrons are used by the terminal atoms, there are no lone pairs on the central Xe atom. The total number of σ bond-pairs and lone pairs around the central atom is the sum of these two values:

6(σ bond pairs)+0(lone pairs)=6
💡 Teacher's Secret Hint

Remember that double or triple bonds count as one electron domain for VSEPR, but only one of them is a sigma bond. Here, assuming single bonds for initial electron distribution is a good starting point.

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