Question

Which of the following has maximum unpaired d electrons?
(A) \[Z{n^{2 + }}\]
(B) \[F{e^{2 + }}\]
(C) \[N{i^{2 + }}\]
(D) \[C{u^ + }\]

Answer 1

Hint: In the above question, the total number of electrons in the subshell of d orbital need to be known. The number of electrons and the electronic configuration will be needed to answer the given question. The transition metal will have electrons in the d and s orbitals. The conversion of positive will be achieved by atom by losing electrons from s orbital in transition metal atoms. General electronic configuration of transition metal is \[\left( {n - 1} \right){d^{1-10}}{\rm{ }}n{s^{1-2}}\] ,here \[n\] is \[4\] and \[\left( {n - 1} \right)\] is \[3\] for first transition series.

Step by step answer: As we can see all the atoms in the options belongs to the d-orbital of the first transition series the number of electrons will lie from\[21{\rm{ }}to{\rm{ }}30\] for scandium to Zinc. The valence orbitals will be d-orbitals in which the maximum number can be occupied will be ten, for s-orbital it will be two. The electronic configuration of the element is the distribution of the electron in the subshell (s, p, d and f) as per Aufbau, Pauli and Hund’s rule. The electrons will remain unpaired after filling the orbitals as per these rules hence the atom with unpaired electron will have paramagnetic property. The electronic configuration of zinc, iron, nickel and copper are as follows:
\[Zn= \left[ {Ar} \right]{\rm{ }}3{d^{10}}4{s^2}\]
\[Fe = \left[ {Ar} \right]3{d^6}4{s^2}\]
\[Ni= \left[ {Ar} \right]{\rm{ }}3{d^8}4{s^2}\]
\[Cu=\left[ {Ar} \right]{\rm{ }}3{d^{10}}4{s^1}\]
The electron are removed as per the charge on the ions, here there is \[ + 2\] charge on the atom the electrons will get two electron will be removed from the s orbitals and then from the d orbitals if needed, hence the electronic configuration after removal of two electrons will be as follows:
\[Z{n^{ + 2}} = \left[ {Ar} \right]{\rm{ }}3{d^{10}}\] , the number of unpaired electrons are zero.
\[F{e^{ + 2}}{\rm{ = }}\left[ {Ar} \right]{\rm{ }}3{d^6}\], the number of unpaired electrons are four.
\[N{i^{ + 2}} = \left[ {Ar} \right]{\rm{ }}3{d^8}\], the number of unpaired electrons are two.
\[C{u^{ + 1}} = \left[ {Ar} \right]{\rm{ }}3{d^{10}}\], the number of unpaired electrons are zero.

Hence, the maximum number of unpaired electrons are four, so the correct option is: (B) \[F{e^{ + 2}}\].

Note: Here the electronic configuration of the atom should be known as we can see that the copper has different way of filling the orbital, as all the d orbitals are filled first in copper then electron will be filled in the s-orbital as fully filled d-orbital is more stable than that of the partially filled.