Question
Predict the shape, hybridisation, and bond angle of using VSEPR theory. Then explain why it differs from even though both have four fluorine atoms attached.
Solution — Step by Step
Xe has 8 valence electrons. Four are used in bonds with F atoms (one each). The remaining 4 electrons form two lone pairs.
Total electron pairs around Xe: 4 bonding + 2 lone = 6 pairs.
Six electron pairs hybridisation.
With 6 electron pairs and 2 lone pairs, the lone pairs occupy positions opposite to each other (axial of the octahedron) to minimise repulsion. The four bonding pairs lie in the equatorial plane.
Shape: square planar. Bond angle: .
Carbon has 4 valence electrons, all used in bonding with F. No lone pairs. So is hybridised, tetrahedral, bond angle .
Xenon (period 5) has empty orbitals available for hybridisation, allowing it to expand its octet to 12 electrons. Carbon (period 2) has no orbitals in its valence shell, so it cannot exceed 8 electrons. The lone pairs on Xe force the geometry to flatten into a square plane.
: , square planar, . : , tetrahedral, .
Why This Works
VSEPR theory says electron pairs (bonding and lone) repel each other and arrange to minimise this repulsion. The basic geometry is set by the total number of pairs; lone pairs distort the shape by pushing harder than bonding pairs.
Period 3+ elements can use orbitals to host extra electrons. This is why , , exist while their period-2 counterparts (e.g. ) do not.
Alternative Method
Use the steric number formula: . For : . For : . Skips the electron-pair counting. Faster for MCQs.
Common Mistake
Predicting tetrahedral geometry for because it has 4 F atoms. The bonded atoms count alone does not determine geometry — lone pairs count too. Always include lone pairs when computing the steric number.