The number of s-electrons present in an ion with $55$ protons in its unipositive state is

The atom with 55 protons is caesium(Cs). The electronic configuration of Cs is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^{10}4{\mathrm{p}}^{6}5{\mathrm{s}}^{2}4{\mathrm{d}}^{10}5{\mathrm{p}}^{6}6{\mathrm{s}}^1$.

The mono positive ion means one electron is removed from outermost orbital. Hence, the sum of electrons in $1\mathrm{s}, 2\mathrm{s}, 3\mathrm{s}, 4\mathrm{s}, 5\mathrm{s}$ orbitals are$10$.