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:$0<\sigma <Z$

Electrons in orbitals	$\sigma$ per electron in orbital
	$n$	$\left(n-1\right)$	$(n-2), (n-3),$ etc.
$s$ or $p$ orbital	$0.35$	$0.85$	$1.00$
$d$ or $f$orbital	$0.35$	$1.00$	$1.00$

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

What is the value of $Z*$ for $\mathrm{Rb} \left(37\right)?$

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 $\mathrm{Cu}\left(29\right)?$

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 $\sigma .$

In which of the following cases the concept of $Z*$ is applicable?

The first $\left({\mathrm{IE}}_1\right)$ and the second $\left({\mathrm{IE}}_2\right)$ ionisation enthalpies $(\mathrm{kJ} {\mathrm{mol}}^{-1})$ of a few elements designated by Roman numerals are shown below:

Which of the above elements is likely to be:
A reactive metal:

The first $\left({\mathrm{IE}}_1\right)$ and the second $\left({\mathrm{IE}}_2\right)$ ionisation enthalpies $(\mathrm{kJ} {\mathrm{mol}}^{-1})$ of a few elements designated by Roman numerals are shown below:

Which of the above elements is likely to be:
A reactive non-metal:

The first $\left({\mathrm{IE}}_1\right)$ and the second $\left({\mathrm{IE}}_2\right)$ ionisation enthalpies $(\mathrm{kJ} {\mathrm{mol}}^{-1})$ of a few elements designated by Roman numerals are shown below:

Which of the above elements is likely to be:
A noble gas.

The first $\left({\mathrm{IE}}_1\right)$ and the second $\left({\mathrm{IE}}_2\right)$ ionisation enthalpies $(\mathrm{kJ} {\mathrm{mol}}^{-1})$ of a few elements designated by Roman numerals are shown below:

Which of the above elements is likely to be:
A metal that forms a stable binary halide of the formulae ${\mathrm{AX}}_2$(${\mathrm{X}}_2=$halogen).