Question

The correct order of decreasing second ionization enthalpy of is:
$(Ti=22,V=23,Cr=24,Mn=25)$
(A) $Mn>Cr>Ti>V$
(B) $Ti>V>Cr>Mn$
(C) $Cr>Mn>V>Ti$
(D) $V>Mn>Cr>Ti$

Answer 1

Hint:In order to solve these types of questions we should just do the electronic configuration and have the look that which of the elements need less or more energy to be just in its stable state. And the factors which are affecting should be analyzed carefully.

Complete answer:So first of all for solving this question we will have to do the electronic configuration according to aufbau rule one by one so let's do this :
For Mn since it is given that the atomic number of the Mn is 25 so its electronic configuration will be
Mn $$\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^2$$
For Cr since it is given that the atomic number of the Cr is 24 so its electronic configuration will be
Cr $\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{5}4s^1$
For Ti since it is given that the atomic number of the Ti is 22 so its electronic configuration will be
Ti $\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{2}4s^2$
For V since it is given that the atomic number of the V is 23 so its electronic configuration will be
V $\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^2$
Now we all the configurations so we can now make the analysis according to the factors affecting the ionization enthalpy;Generally we have studied that the ionization energy ( talking about 1st and 2nd ) increases as we move from right to left across the period because effective nuclear charge increases. So by the above basis the order should be;Mn > Cr > V > Ti.But this is incorrect there are some more facts which changes the order of this;Here we can see that in chromium after it loses one electron in its d shell there will be 5 electrons the number 5 is exactly half of 10 and we know that in d shell there can be 10 electrons so this will be half-filled due to this its stability increases. So here it is an exception.By knowing this now we will write the correct order of it which will be Cr > Mn > V > Ti.So this is the correct decreasing order of second ionization enthalpy of above elements.

Hence option (C) is correct.

Note:This effect is also valid in some other elements also.One thing should be kept in mind that first it was noticed practically then in order to satisfy this it was made these approaches. This is also seen on the f orbitals also.