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Question: The atomic number of Vanadium, Chromium, manganese and iron are respectively \(23\), \(24\), \(25\),...

The atomic number of Vanadium, Chromium, manganese and iron are respectively 2323, 2424, 2525, 2626 . Which of these may be expected to have the highest second ionization energy?
A.V{\text{V}}
B.Cr{\text{Cr}}
C.Mn{\text{Mn}}
D.Fe{\text{Fe}}

Explanation

Solution

Ionization energy is the energy required to remove the electron from the outermost shell of the orbital of an element. The more the attraction power is between the nucleus and electrons; the more energy is required to remove the electron.

Complete step by step answer:
Given the atomic number of vanadium is 2323 and its electronic configuration is given as- (Ar)3d34s2\left( {{\text{Ar}}} \right)3{{\text{d}}^{\text{3}}}4{{\text{s}}^{\text{2}}}
The atomic number of chromium is 2424 and its electronic configuration is (Ar)3d54s1\left( {{\text{Ar}}} \right)3{{\text{d}}^{\text{5}}}{\text{4}}{{\text{s}}^{\text{1}}}
The atomic number of manganese is 2525 and its electronic configuration is (Ar)3d54s2\left( {{\text{Ar}}} \right)3{{\text{d}}^{\text{5}}}{\text{4}}{{\text{s}}^2}
And the atomic number of Iron is 2626 and its electronic configuration is (Ar)3d64s2\left( {{\text{Ar}}} \right)3{{\text{d}}^6}{\text{4}}{{\text{s}}^2}
On observing the electronic configuration we can say that the ionization energy for removal of the first electron and second electron in the vanadium, manganese and iron will be almost the same as in all of them the electrons are being released from s- orbital.
But in chromium the first electron is removed from s- orbital as it has only one electron. After removal of the first electron the next electron will have to be taken from the d-sub shell which is in half filled state. So, more energy will be required to remove the second electron due to the second ionization energy.
Hence chromium is expected to have second highest ionization energy.

The correct answer is B.

Note:
Ionization energy depends upon the atomic radius. As the atomic radius increases the ionization energy also increases. The trend of ionization energy is-
Across the period the ionization energy increases due to increase in nuclear charge.
Down the period the ionization energy decreases because of increase in atomic size.