Question

Which of the following has the maximum number of unpaired d-electrons?
(A) $Z{n^{2 + }}$
(B) $F{e^{2 + }}$
(C) $N{i^{3 + }}$
(D) $C{u^ + }$

Answer 1

To calculate the number of unpaired electrons of the given ions, we have to write and check the electronic configuration of these atoms and then compare to find out which element has the most number of unpair

Complete step by step answer:
Elements from $Sc(scandium)$ to $Zn(zinc)$ come in the $3d -$series of the D block elements. These elements can be found from the ${3^{rd}}$ group to the ${12^{th}}$ group of the modern periodic table.
Now we will write the electronic configuration of each element and then compare the number of unpaired electrons present in their d orbital.
The atomic number of zinc is $29$. Its electronic configuration is $[Ar]3{d^{10}},4{s^2}$. When zinc will lose its two electrons and form $Z{n^{2 + }}$ions, the electronic configuration is changed. The E.C. of $Z{n^{2 + }}$will be $[Ar]3{d^{10}}$. As we can see in the d orbital all the electrons are paired, hence it does not have any unpaired electrons.
The atomic number of iron is $26$. Its electronic configuration is $[Ar]3{d^6},4{s^2}$. When iron will lose its two electrons and form $F{e^{2 + }}$ ions, the electronic configuration is changed. The E.C. of $F{e^{2 + }}$will be $[Ar]3{d^6}$. As we can see in the d orbital all the electrons are not paired, hence it has $4$ unpaired electrons.
The atomic number of Nickel is $28$. Its electronic configuration is $[Ar]3{d^8},4{s^2}$. When nickel will lose its three electrons and form $N{i^{3 + }}$ ions, the electronic configuration is changed. The E.C. of $N{i^{3 + }}$ will be $[Ar]3{d^7}$. As we can see in the d orbital all the electrons are not paired, hence it has $3$ unpaired electrons.
The atomic number of copper is $29$. Its electronic configuration is $[Ar]3{d^{10}},4{s^1}$. When copper will lose its electrons and form $C{u^ + }$ ions, the electronic configuration is changed. The E.C. of $C{u^ + }$ $[Ar]3{d^{10}}$ will be. As we can see in the d orbital all the electrons are paired, hence it has zero unpaired electrons.
Hence, out of the given options $F{e^{2 + }}$ has the most number of unpaired electrons i.e. $4$.

Therefore, Option (B) is correct.

Note:
Elements of the $3d -$series shows great properties of metals like ductility, malleability, and lustre. They also show high melting and boiling points. Due to their small size, they have high ionization enthalpies.