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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) $C{{u}^{+}}$

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Last updated date: 04th Mar 2024
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IVSAT 2024
Answer
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Hint: As electronic configuration increases from $\left[ Ar \right]3{{d}^{1}}$ to $\left[ Ar \right]3{{d}^{5}}$,number of unpaired d electrons increases from 1 to 5. As electronic configuration moves from $\left[ Ar \right]3{{d}^{6}}$ to $\left[ Ar \right]3{{d}^{10}}$, number of unpaired d electrons decreases from 4 to 0.

Complete solution step by step:
- Atomic number of zinc is 30. So, electronic configuration of zinc is \[\left[ Ar \right]3{{d}^{10}}4{{s}^{2}}\].
- In divalent ion $Z{{n}^{+2}}$, Zinc loses two electrons from 4s and electron configuration leads to\[\left[ Ar \right]3{{d}^{10}}\]. As all electrons are paired in d orbital, number of unpaired d electrons are zero.
- Atomic number of Iron is 26. Electronic configuration of Iron is $\left[ Ar \right]3{{d}^{6}}4{{s}^{2}}$.
- In divalent ion $F{{e}^{+2}}$,iron loses two electrons from 4s and electron configuration becomes $\left[ Ar \right]3{{d}^{6}}$,As only two electrons are paired in 3d orbital, number of unpaired d electrons are four.
- Atomic number of Nickel is 28. Electronic configuration if nickel is\[\left[ Ar \right]3{{d}^{8}}4{{s}^{2}}\].
- In trivalent ions, nickel loses three electrons, two electrons from 4s and one electron from 3d. Electronic configuration becomes $\left[ Ar \right]3{{d}^{7}}$. Number of unpaired d electrons are three.
- Atomic number of Copper is 29. Expected electronic configuration of copper is \[\left[ Ar \right]3{{d}^{9}}4{{s}^{2}}\]and observed electronic configuration of copper is \[\left[ Ar \right]3{{d}^{10}}4{{s}^{1}}\]. To gain extra stability, one electron from 4s is transferred to 3d.
In Monovalent ion, Copper loses one electron from 4s and electronic configuration of copper becomes $\left[ Ar \right]3{{d}^{10}}$.As all electrons are paired in 3d orbital, number of unpaired d electron is zero.
As, $F{{e}^{+2}}$ has four unpaired electrons, option (B) $F{{e}^{+2}}$ has a maximum number of unpaired d-electrons.

Note: Chromium and copper have an exception of half-filled and completely filled d orbitals respectively. Zinc has completely filled d orbital and zero unpaired d electrons. To gain extra stability, one 4s electron is transferred to 3d so copper has completely filled d orbitals and zero unpaired d electrons. Chromium has half-filled d orbitals and the number of unpaired d electrons is five.