Question

Which of the following has the maximum number of unpaired electrons
A.$$M{g}^{2+}$$
B.$$T{i}^{3+}$$
C.$$V^{3+}$$
D.$$F{e}^{2+}$$

Answer 1

Hint:To answer this question, you should recall the concept of d-block elements. First write the electronic configuration of ionized states and use configuration to figure out the unpaired electrons.

Complete step by step answer:
The d block contains 3 complete series which are:
$$3d-Sc,Ti,V,Cr,Mn,Fe,Co,Ni,Cu,Zn$$;
$$4d-Y,Zr,Nb,Mo,Tc,Ru,Rh,Pd,Ag,Cd$$;
$$5d-La,Hf,Ta,W,Re,Os,Ir,Pt,Au,Hg$$ and
$$6d-\text{ incomplete}\text{.}$$
There are 10 elements filling up the ‘d’ orbital in each series.
The laws which dictate the filling of electrons in atomic orbitals are Aufbau principle, Hund's Rule of Maximum Multiplicity and Pauli exclusion principle. This generates a specific filling order of electrons which is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s and so on. Let us compare the electronic configuration of the elements given in the question:
$$M{g}^{2+}:1s^{2}2s^{2}2p^6:$$ no unpaired electron.
$$T{i}^{3+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^1:$$ one unpaired electron.
$$V^{3+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^2:$$ two unpaired electrons.
$$F{e}^{2+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6:$$ four unpaired electrons
Thus, we see that $$F{e}^{2+}$$ has the most unpaired electrons.

Hence, we conclude that the correct answer for this question is option D.

Note:
The electronic configuration of elements is based on majorly 3 rules:
-According to the Pauli exclusion principle in an atom, no two electrons will have an identical set or the same quantum numbers. There salient rules of Pauli Exclusion Principle are that only two electrons can occupy the same orbital and the two electrons that are present in the same orbital should be having opposite spins.-According to Hund’s Rule of Maximum Multiplicity rule for a given electronic configuration of an atom, the electron with maximum multiplicity falls lowest in energy.
According to the Aufbau principle, the electrons will start occupying the orbitals with lower energies before occupying higher energy orbitals.