Find the total number of cations for which $I.P.$ of cation is lower than corresponding atom.${\mathrm{K}}^{+}, {\mathrm{O}}^{+}, {\mathrm{Ne}}^{+}, {\mathrm{P}}^{+}, {\mathrm{Be}}^{+}, {\mathrm{Na}}^{2+}, {\mathrm{Fe}}^{+}, {\mathrm{Sn}}^{2+}$

According to $I. C.$ Slater effective nuclear charge, $Z*,$ due to screening, is not exactly equal to the actual nuclear charge $Z$ of the nucleus of the atom.$Z*$ depends on the type of orbital In which the electron is housed, and on the ability of other electrons
in more penetrating orbitals to screen the electron in question from the nucleus.
The relative extent to which the various orbitals penetrate the electron clouds of other orbitals is$s>p>d>f.$ The effective nuclear charge $Z*$ due to screening is given by $Z^{*}=Z-\sigma$ where $Z$ is the atomic number and $\sigma$ is the slater screening constant values:$0<\sigma <Z$

Screening effect of one electron in the outermost orbital, is not considered in calculation of $\sigma .$

What is $Z^{*}$ for $\mathrm{Cu}\left(29\right)?$

The general electronic configuration of the outermost and penultimate shell of an element is $(\mathrm{n}-1){\mathrm{s}}^2(\mathrm{n}-1){\mathrm{p}}^6(\mathrm{n}-1){\mathrm{d}}^{\mathrm{x}}{\mathrm{ns}}^1.$

If $\mathrm{n}=5$ and $\mathrm{x}=5,$ then the number of protons in the nucleus of the element will be: