Question

molecular shape for xef₄ (hint: start with a lewis structure) square pyramidal t - shaped square planar tetrahedral octahedral

Explanation:

To determine the molecular shape of \( \text{XeF}_4 \):

1. Lewis Structure: Xenon (Xe) is the central atom. Xe has 8 valence electrons, and each F contributes 7, so total valence electrons: \( 8 + 4\times7 = 36 \). Xe forms 4 single bonds with F (using 8 electrons), leaving \( 36 - 8 = 28 \) electrons. These are distributed as 3 lone pairs on Xe (since \( \frac{28}{2} = 14 \), but after bonding, Xe has \( 8 - 4 = 4 \) non - bonding electrons? Wait, no: correct calculation - Xe valence electrons: 8, 4 F atoms: \( 4\times7 = 28 \), total: 36. In \( \text{XeF}_4 \), Xe is bonded to 4 F atoms (4 single bonds, 8 electrons used), so remaining electrons: \( 36 - 8 = 28 \). But Xe can have an expanded octet. The electron - pair geometry (considering bonding and lone pairs) around Xe: number of bonding pairs (BP) = 4, number of lone pairs (LP) = \( \frac{8 - 4\times1}{2}= 2 \) (since each bond uses 1 electron from Xe, Xe has 8 valence electrons, 4 used in bonding, so 4 left, forming 2 lone pairs). So electron - pair geometry is octahedral (BP + LP = 4+2 = 6, octahedral arrangement).
2. Molecular Shape: The molecular shape is determined by the arrangement of the bonding pairs, ignoring the lone pairs. In an octahedral electron - pair geometry with 2 lone pairs, the lone pairs are placed opposite each other (to minimize repulsion). The 4 bonding pairs (Xe - F bonds) are in a plane, forming a square planar shape.

Answer:

square planar