Question

The number of lone pair of electrons and the hybridization of Xenon (Xe) in XeOF$_2$ are

• A. 1, sp$^3$
• B. 1, dsp$^2$
• C. 3, dsp$^3$
• D. 2, sp$^3$d

Hint:

Using a simple frame or just bolding for the box
Key Points:
Central atom: Xe (8 valence e$^-$).
Bonded atoms: 1 O (double bond), 2 F (single bonds).
Electrons used in bonds: 2 (for =O) + 2$\times$1 (for F) = 4.
Non-bonding electrons: 8 - 4 = 4.
Lone pairs: 4 / 2 = 2.
Steric Number (SN) = (Bonded atoms) + (Lone pairs) = 3 + 2 = 5.
Hybridization for SN=5 is sp$^3$d.

Answer 1

The Correct Option is D

The VSEPR (Valence Shell Electron Pair Repulsion) theory is employed to determine the lone pairs and hybridization of the central Xenon (Xe) atom in XeOF₂.
(A) Valence Electrons of Central Atom (Xe): Xenon, a Group 18 noble gas, has 8 valence electrons.
(B) Bonding Electrons: Xe uses a total of 4 electrons in bonding: 2 (for O) + 2 (for F). 
(E) Non-bonding Electrons on Xe: Non-bonding electrons = (Total valence electrons of Xe) - (Electrons used in bonding) = 8 - 4 = 4 electrons. 
(F) Lone Pairs on Xe: 
Lone pairs = (Non-bonding electrons) / 2 
Lone pairs = 4 / 2 = 2 lone pairs. 
(G) Steric Number (SN): 
SN = (Atoms bonded to Xe) + (Lone pairs on Xe) 
SN = (1 O + 2 F) + 2 = 3 + 2 = 5. 
(H) Hybridization: A steric number of 5 corresponds to sp³d hybridization. 
Therefore, Xenon in XeOF₂ has 2 lone pairs and sp³d hybridization. This matches option (D). The electron geometry is trigonal bipyramidal, and the molecular geometry is T-shaped.