1. Goal: Count how many given xenon species have exactly one lone pair on the central Xe atom.

2. Method: Use the steric number / electron pair counting on Xe.

   For xenon compounds:

   • Each bonded atom contributes one sigma bond around Xe.
   • A double bond to O still counts as one sigma bond domain for VSEPR.
   • Lone pairs on Xe are found from the total valence electron count / known geometry.

3. Check each species one by one

(i) $\mathrm{XeF_5^+}$

• Valence electrons: $8 + 5\times 7 - 1 = 42$
• Five Xe–F bonds use 10 electrons, fluorines take 30 electrons as lone pairs.
• Left on Xe: $42 - 10 - 30 = 2$
• So Xe has 1 lone pair.

✅ Counts

(ii) $\mathrm{XeO_3}$

• Xenon trioxide has structure with three Xe–O bonds and one lone pair on Xe.
• VSEPR type: $AX_3E$.

✅ Counts

(iii) $\mathrm{XeO_2F_2}$

• Around Xe: 2 O and 2 F = 4 sigma bonds.
• Known geometry is seesaw, corresponding to $AX_4E$.
• Hence Xe has 1 lone pair.

✅ Counts

(iv) $\mathrm{XeF_5^-}$

• Valence electrons: $8 + 5\times 7 + 1 = 44$
• Five Xe–F bonds use 10 electrons, fluorines take 30 electrons.
• Left on Xe: $44 - 10 - 30 = 4$
• So Xe has 2 lone pairs.

❌ Does not count

(v) $\mathrm{XeO_3F_2}$

• Around Xe: 3 O and 2 F = 5 sigma bonds.
• This gives trigonal bipyramidal electron arrangement with no lone pair on Xe.
• VSEPR type: $AX_5$.

❌ Does not count

(vi) $\mathrm{XeOF_4}$

• Around Xe: 1 O and 4 F = 5 sigma bonds.
• Known geometry is square pyramidal, corresponding to $AX_5E$.
• Hence Xe has 1 lone pair.

✅ Counts

(vii) $\mathrm{XeF_4}$

• Valence electrons: $8 + 4\times 7 = 36$
• Four Xe–F bonds use 8 electrons, fluorines take 24 electrons.
• Left on Xe: $36 - 8 - 24 = 4$
• So Xe has 2 lone pairs.

❌ Does not count

4. Total species with exactly one lone pair on Xe

They are: $\mathrm{XeF_5^+},\ \mathrm{XeO_3},\ \mathrm{XeO_2F_2},\ \mathrm{XeOF_4}$

So the number is $\boxed{4}$

5. Comparison with stored answer

• Derived answer: $4$
• Stored correct answer: $4$
• They agree.