Question

What is ${{Z}_{eff}}$ of 4s electrons of Cr?
 (A) 3.04
 (B) 2.59
 (C) 3.60
 (D) 2.95

Answer 1

Hint: In order to solve this question we need to understand the term ${{Z}_{eff}}$ or effective nuclear charge. The term ${{Z}_{eff}}$ refers to the number of protective electrons which surround the nucleus by the outermost or valence electrons of multi-electron atoms.

Complete step by step answer:
 - The effective nuclear charge can be found by using Slater rule. The principle behind Slater's rule is that the actual charge which is felt by an electron will be equal to what we will expect the charge to be from a certain number of protons and subtracted by a certain amount of charge from other electrons.
- This rule allows us to calculate the effective nuclear charge ${{Z}_{eff}}$ from the real number of protons in the nucleus and the effective shielding of electrons in each orbital and the rule can be represented as follows
 \[{{Z}_{eff}}=Z-S\]
Where, Z is the number of protons in the nucleus and S is the average electron density between electron and the nucleus and as we already know ${{Z}_{eff}}$ is the effective nuclear charge.
- The atomic number of Chromium is 24 and thus there will be 24 electrons in its orbitals. The electronic configuration of Chromium can be written as follows
\[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}4{{s}^{1}}\]

 By using the Slater’s rule the ${{Z}_{eff}}$ of 4S electron can be written as follows
\[{{Z}_{eff}}=Z-\left( \left( 4s-1 \right)\times 0.35 \right)-\left( pre\text{ }penultimate\text{ }shell\text{ }{{e}^{-}}\times 0.85 \right)-\left( remaining\text{ }{{e}^{-}}\times 1 \right)\]

The above equation means that the electrons in the 4s−1 orbital will contribute 0.35, electrons in the pre-penultimate shell will contribute 0.85 and finally the remaining electrons need to be subtracted. From the electronic configuration, we can see that there is only one electron in the 4s orbital and thus electrons in the 4s−1 orbital will be zero.
\[{{Z}_{eff}}=24-\left( 0\times 0.35 \right)-\left( 13\times 0.85 \right)-\left( 10\times 1 \right)\]
\[{{Z}_{eff}}=2.95\]
 Thus the value of ${{Z}_{eff}}$ of 4s electrons of Cr is 2.95.
So, the correct answer is “Option D”.

Note: It should be noted that if an atom has a very less number of electrons in its orbitals such as one or two electrons, then the ${{Z}_{eff}}$ (effective nuclear charge) experienced by the electrons will be high since the distance between the electron and nucleus along with the shielding by other electrons is less.