1. Identify the element
   An ion with 55 protons has atomic number $Z=55$, which is cesium (Cs).

2. Write the electronic configuration of neutral cesium
   Cesium has 55 electrons in the neutral state: $\mathrm{Cs}: 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^{10}\,5p^6\,6s^1$ Or briefly, $\mathrm{Cs} = [Xe]6s^1$

3. Form the unipositive ion
   Unipositive state means $\mathrm{Cs}^+$, so one electron is removed from the outermost orbital $6s$: $\mathrm{Cs}^+ = [Xe]$ Thus the configuration becomes: $1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^{10}\,4p^6\,5s^2\,4d^{10}\,5p^6$

4. Count the total number of $s$-electrons
   The occupied $s$-subshells are:

• $1s^2$ $\to 2$
• $2s^2$ $\to 2$
• $3s^2$ $\to 2$
• $4s^2$ $\to 2$
• $5s^2$ $\to 2$
• $6s^0$ $\to 0$ (removed in $\mathrm{Cs}^+$)

Total number of $s$-electrons: $2+2+2+2+2=10$

5. Evaluate options

• A: $10$ ✅
• B: $9$ ❌
• C: $12$ ❌
• D: $8$ ❌

Therefore, the correct answer is A.