Question

How many 3d electrons does the manganese (II) ion, $M{n}^{2+}$, have?

Answer 1

Hint: Manganese is the element of group 7 of the d-block and its atomic number is 25. So, there are 25 electrons in the shell of the manganese and according to the increasing energy of the orbitals. In $M{n}^{2+}$ ion, there will be a reduction of two electrons, so remove two electrons from the last orbital.

Complete answer:
Manganese is the element of group 7 of the d-block and its atomic number is 25. So, there are 25 electrons in the shell of the manganese and according to the increasing energy of the orbitals.
There are many orbitals that are placed according to their increasing order of energies. The order is $\text{1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d 7p}$. And we know that the s-orbital can accommodate only 2 electrons, the p-orbital can accommodate 6 electrons, the d-orbital can accommodate 10 electrons, and the f-orbital can accommodate 14 electrons. So, when the 25 electrons are filled in these orbitals, the electronic configuration will be:
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^5$
In $M{n}^{2+}$ion, there will be a reduction of two electrons, so remove two electrons from the last orbital. So, the number of electrons will be 23, and it seems that the electrons will be removed from 3d orbital but the electrons will be removed from the 4s orbital because it has higher energy, so the configuration will be:
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{0}3d^5$
So, the total number of electrons in the 3d-orbital is 5 electrons. Therefore, the answer is 5.

Note:
In all the transition metals, if the ion has +2 charge then first the electrons will be removed from the s-shell having the highest number, then the removal from d-orbital will take place.