Question

According to slater’s rule, the order of effective nuclear charge $ \left( {{Z_{eff}}} \right) $ for the last electron in case of $ {{\text{O}}^{2 - }},\;M{g^{2 + }},\;A{l^{3 + }} $ is
(A) $ {{\text{O}}^{2 - }} > \;M{g^{2 + }} > \;A{l^{3 + }} $
(B) $ A{l^{3 + }} > \;M{g^{2 + }} > \;{{\text{O}}^{2 - }} $
(C) $ {{\text{O}}^{2 - }} = \;M{g^{2 + }} = \;A{l^{3 + }} $
(D) $ M{g^{2 + }} > \;A{l^{3 + }} > {{\text{O}}^{2 - }} $

Answer 1

Hint: We are talking about effective nuclear charge $ \left( {{Z_{eff}}} \right) $ in this question. Effective Nuclear charge is defined as the total charge experienced by an atom containing more than one electron. It is calculated by $ {{\text{Z}}_{eff}} = {\text{Z}} - {\text{S}} $ where Z is the atomic number or number of protons in the atom and S is the shielding effect of electrons in that atom.

Complete answer:
First of all, let’s see what the slater rule says:
State the electronic configuration of the atom.
Electronic configuration of Oxygen Anion,
 $ {{\text{O}}^{2 - }} = 1{{\text{s}}^2}\;2{{\text{s}}^2}\;2{{\text{p}}^6} $
Electronic configuration of Magnesium Cation,
 $ M{g^{2 + }} = 1{{\text{s}}^2}\;2{{\text{s}}^2}\;2{{\text{p}}^6} $
Electronic configuration of Aluminium Cation,
 $ A{l^{3 + }} = 1{{\text{s}}^2}\;2{{\text{s}}^2}\;2{{\text{p}}^6} $
Calculate Shielding effect of electrons in the atom. It is calculated for each orbital separately if n is the total number of orbitals than number of electron in $ \left( {{\text{n}} - 3} \right) $ orbital multiplied by 1, number of electron in $ \left( {{\text{n}} - 2} \right) $ orbital multiplied by $ 1 $ , number of electron in $ \left( {{\text{n}} - 1} \right) $ multiplied by $ 0.85 $ , and number of electron in nth orbital multiplied by $ 0.35 $ .
So, value of Shielding effect for Oxygen
 $ {{\text{O}}^{2 - }} = \left( {8 \times 0.35} \right) + \left( {2 \times 0.85} \right) = 4.5 $
Value of Shielding effect for Magnesium
 $ M{g^{2 + }} = \left( {8 \times 0.35} \right) + \left( {2 \times 0.85} \right) = 4.5 $
Value of Shielding effect for Aluminium
 $ A{l^{3 + }} = \left( {8 \times 0.35} \right) + \left( {2 \times 0.85} \right) = 4.5 $
Since the number of electrons is the same for the Oxygen, Magnesium and Aluminium their electron configuration is same they have equal shielding effect.
Now, Calculating Effective Nuclear Charge
for Oxygen
 $ {{\text{Z}}_{eff}}\left( {{{\text{O}}^{2 - }}} \right) = 8 - 4.5 = 3.5 $
for Magnesium
 $ {{\text{Z}}_{eff}}\left( {M{g^{2 + }}} \right) = 12 - 4.5 = 7.5 $
for Aluminium
 $ {{\text{Z}}_{eff}}\left( {A{l^{3 + }}} \right) = 13 - 4.5 = 8.5 $
Effective is highest for Aluminium Cation, then Magnesium Cation, Oxygen anion has the least effective nuclear charge among them.
So, option (B) is the correct answer.

Note:
Carefully doing the electronic configuration is necessary for calculating shielding effect as it requires the number of electrons in each orbital to be known. remember which orbital electrons are to be multiplied by which number.