MCQ Detailed View

Xenon oxytetrafluoride (XeOF₄) is a well-known compound of the noble gas xenon. To determine the number of lone pairs of electrons on xenon in XeOF₄, we apply the Valence Shell Electron Pair Repulsion (VSEPR) theory.

Xenon (Xe) belongs to Group 18 of the periodic table and has 8 valence electrons. In XeOF₄, xenon forms one bond with oxygen and four bonds with fluorine atoms. Each bond involves two electrons. Thus, xenon uses 5 pairs (10 electrons) in bonding with oxygen and fluorine.

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  Total valence electrons of Xe = 8

  
  


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  Bonding pairs used with O and 4 F = 5 pairs = 10 electrons

  
  


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  Electrons used by xenon in bonds = 10

  
  


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  Remaining electrons on xenon = 8 – 5 = 3 pairs (6 electrons)

  
  

But here’s the important point: out of these, one pair is counted as a lone pair on xenon, while the others are distributed in bonding. Oxygen is double-bonded to xenon (Xe=O), while each fluorine forms a single bond. This arrangement leaves one lone pair on xenon.

According to VSEPR theory, this results in a square pyramidal geometry. The central xenon atom is surrounded by five bonded atoms and one lone pair, giving a total steric number of 6. The geometry closely resembles an octahedral structure, but one position is occupied by a lone pair instead of a bonded atom.

This lone pair strongly influences the molecular shape by repelling the bonded pairs, slightly distorting bond angles. The presence of a lone pair also explains the high polarity and reactivity of XeOF₄ compared to other xenon compounds.

Therefore, the number of lone pairs of electrons on the central xenon atom in XeOF₄ is 1.