Q.Draw the shape of the following molecules:

Concept: VSEPR theory – molecular geometry is determined by minimizing electron-pair repulsions around the central atom, with lone pairs occupying equatorial positions in trigonal bipyramidal and octahedral arrangements.

(a) $XeOF_4$

Xenon has 8 valence electrons. With one double-bonded oxygen and four fluorines, the steric number is 6 (5 bonds + 1 lone pair). The electron geometry is octahedral. The lone pair occupies one position, leaving five atoms arranged in a square pyramidal shape: oxygen and the four fluorines form the base square and apex. Oxygen typically takes an axial position due to its stronger bonding.

(b) $BrF_3$ …

Why VSEPR theory determines molecular shape

The shape of a molecule is governed by the repulsion between electron pairs around the central atom. Both bonding pairs (shared electrons) and lone pairs (unshared electrons) occupy space, but lone pairs repel more strongly because they're held closer to the nucleus. The geometry minimizes these repulsions.

The strategy is always the same: count the total electron pairs (bonding + lone), determine the electron-pair geometry, then describe the molecular shape by looking only at where the atoms sit.

(a) $\mathrm{XeOF_4}$

Step 1: Count valence electrons and determine the steric number

Xenon has 8 valence electrons. It forms one double bond with oxygen (using 2 pairs) and four single bonds with fluorine atoms (using 4 pairs). That accounts for 6 bonding pairs.

Total electron pairs around Xe = bonding pairs + lone pairs. Xenon started with 8 electrons; the bonds use $2 + 4 = 6$ electrons, leaving $8 - 6 = 2$ electrons as 1 lone pair.

Steric number = $5$ bonding regions (one O, four F) + $1$ lone pair = $6$.

Step 2: Determine electron-pair geometry

Six electron pairs arrange themselves in an octahedral geometry to minimize repulsion.

Step 3: Place the lone pair

In an octahedral arrangement, all six positions are equivalent. The lone pair occupies one position.

Step 4: Place the atoms

The oxygen and four fluorines occupy the remaining five positions. Because oxygen forms a double bond (stronger, shorter), it typically occupies an axial position. The four fluorines then sit in the equatorial square plane plus one axial position opposite to the lone pair.

The resulting molecular shape is square pyramidal: four F atoms form a square base, the O atom sits at the apex, and the lone pair is on the opposite side.

         O
         |
    F----Xe----F
        /|\
       F   F
    (lone pair below)

(b) $\mathrm{BrF_3}$

Step 1: Count valence electrons and determine the steric number

Bromine has 7 valence electrons. It forms three single bonds with fluorine atoms, using 3 electrons. That leaves $7 - 3 = 4$ electrons, which form 2 lone pairs.

Steric number = $3$ bonding pairs + $2$ lone pairs = $5$. …