Question

What is an effective nuclear charge?

Answer 1

Hint: The effective nuclear charge is denoted as ${{Z}^{*}}\text{ or }{{\text{Z}}_{eff}}$ . It is a result of the shielding effect provided by the inner orbital electrons in an atom. The outermost orbital electrons experience only a portion of the nuclear charge.

Complete step by step solution:
We know that the nuclear charge of an atom is the electrical charge on the nucleus and is equal to the product of the total number of protons present in the nucleus and the elementary charge on the atom.
However, we can describe an effective nuclear charge as the total charge faced by an electron in a poly-electronic atom. We use the symbol ${{Z}^{*}}\text{ or }{{\text{Z}}_{eff}}$ to denote the effective nuclear charge on an electron. We know that electrons are arranged in different energy levels around the nucleus. The inner electrons face higher attraction by the nucleus compared to the electrons in the outer orbitals. This causes a shielding effect provided by the inner electrons to the outer electrons.
We use the term effective here because of the shielding effect. Due to this effect, the inner negatively charged electrons prevent the outer orbital electrons from experiencing the full nuclear charge of the nucleus. The outer orbital electrons experience only a portion of the actual nuclear charge. This effective nuclear charge is also known as the core charge.
The effective nuclear charge is very important while studying the periodic table because we can explain many physical and chemical properties of the elements based on it.
In poly-electronic atoms, we can calculate the effective nuclear charge using Slater’s rule as-${{Z}_{eff}}=Z-\sigma $ where Z is the atomic number of the element and sigma ($\sigma $ ) is the shielding constant.
To calculate the shielding constant, all the electrons should be written in groups like-$\left( 1s \right),\left( 2s,2p \right),\left( 3s,3p \right),\left( 3d \right),\left( 4s,4p \right),\left( 4d \right),\left( 4f \right)...$
The contribution for the electrons present on the right of the one under consideration will be zero.

Same group	(n-1)	(n-2)	(n-3)
0.30	-			1s
0.35	0.85	1	1	(ns,np)
0.35	1	1	1	(nd) or (nf)

The effective nuclear charge is the highest for the innermost electrons and the least for the outermost electrons.

Note: If we want to calculate the effective nuclear charge for 3p-orbital electrons for iron, firstly we have to write the electronic configuration which will be $Fe\to {{\left( 1s \right)}^{2}}{{\left( 2s,2p \right)}^{8}}{{\left( 3s,3p \right)}^{8}}{{\left( 3d \right)}^{6}}{{\left( 4s,4p \right)}^{2}}$
For 3p, $\sigma =\left( 7\times 0.35 \right)+\left( 8\times 0.85 \right)+\left( 2\times 1 \right)=11.25$
While calculation the shielding constant, one less electron of the group we are considering because we are calculating for that, one electron. Therefore, here we have considered 7 for (3s,3p) instead of 8.
Therefore, ${{Z}_{eff}}=26-11.25=14.75$ .