Question

Which is the correct ionization order?
 $\begin {align}
(A) & Z {{n} ^ {+}}>C {{a} ^ {+}}>C {{r} ^ {+}} \\
(B) & C {{o} ^ {2+}}>M {{n} ^ {2+}}>C {{r} ^ {2+}} \\
(C) & M {{n} ^ {2+}}>N {{i} ^ {2+}}>C {{a} ^ {2+}} \\
(D) & Z {{n} ^ {+}}>C {{u} ^ {+}}>N {{i} ^ {+}} \\
\end{align}$

Answer 1

Hint Write down the electronic configuration for all the cations and then check for their stability, s-subshell contains 1 orbital, p-subshell contains 3 orbitals and d-subshell contains 5 orbitals. There are maximum 2 electrons in each orbital according to the Pauli Exclusion Principle. The spin of both the electrons should be opposite.

Complete Step By Step Solution:
The electronic configuration for $Z{{n}^{{}}}\,is\,1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}$
When one electron is lost then it becomes $Z{{n}^{+}}$, the configuration for it becomes$[Ar]\,3{{d}^{10}}4{{s}^{1}}$. The electron is removed from the s-subshell and now it becomes half filled. Ionization energy is the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom. So, the ionization energy for $Z{{n}^{+}}$ is quite high as it attains half- filled stable configuration due to which it is difficult to remove its outer electron and thus requires high ionization energy.
For $C{{a}^{+}}-\,[Ar]\,4{{s}^{1}}$, but the 4s electrons are closer to the zinc than in Calcium because of its greater atomic no. due to which the pull of protons is more on electrons and so the ionization energy is more for $Z{{n}^{+}}$ as compared to $C{{a}^{+}}$. In case of $C {{r} ^ {+}} $ we have$[Ar]\, 3{{d} ^ {5}} $, this is half filled d-subshell which is most stable among the other two. So, it will have the highest ionization energy. So, option (A) is incorrect.
Checking the second series- Configuration for $M {{n} ^ {2+\,}}\, is\,[Ar]3{{d}^{5}}$which is again half – filled stable configuration. The configuration of $C{{o}^{2+}}\,is\,[Ar]3{{d}^{7}}$, it contains 3 unpaired electrons so is less stable than manganese cation. So, its ionization energy will be less than $M{{n}^{2+}}$i.e. $M{{n}^{2+}}>C{{o}^{2+}}$. So, option (B) is also incorrect.
Now checking the third series- Configuration for $ M {{n} ^ {2+\,}}\, is\,[Ar]3{{d}^{5}}$, it contains half- filled d-orbitals so it is stable.
Now for $N{{i}^{2+}}\,we\,have\,[Ar]3{{d}^{8}}$, it contains 2 unpaired electrons so it is less stable than manganese ion, hence requires less ionization energy for the removal of electrons.
For $C{{a}^{2+}}\,we\,have\,1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$, it has fully filled p orbitals so it should be the highest stable among the others but due to fully filled orbitals the electrons create repulsion amongst one another which leads to the decrease in its ionization energy even lower than both manganese and nickel ion. So, the order given in the third series is correct.

Hence the correct choice is (C).

In option (D), the ionization energy of nickel ion is more than copper ion, so it is also incorrect.

Note: Study the periodic table well to know the atomic no. of various elements to write their electronic configuration correct. The 4s orbitals are filled earlier than 3d orbitals because their energy is lower than 3d orbitals. Take care for the configuration of $C {{a} ^ {2+}} $ because fully filled orbitals result in repulsion which decreases their ionization energy.