Question

Find the number of unpaired electrons in $C{r}^{3+}$ ion.

Answer 1

Hint:To calculate the number of unpaired electrons, you should know the total no. of electrons in that atom, then express the electronic configuration of the atom using nearest noble gas configuration and the valence shell configuration. Then look if electrons are present in pairs in orbitals.

Complete step by step answer:Unpaired electrons are the electrons that occupy the orbit of an atom singly without being part of an electron pair.
The expected electronic configuration of chromium Cr is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^4$
But is actual, the electronic configuration of Cr is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^5$
Basically, the one electron of $4s$ orbital will shift to $3d$ orbital so that the $3d$ and $4s$ orbital will become half filled orbitals.
As half filled and fully filled orbitals provide extra stability so that’s why the second electronic configuration for chromium is correct.
Now, the electronic configuration for $C{r}^{3+}$ ion is
$Cr\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^5$
$C{r}^{3+}\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$
The two electrons will be removed from $3d$ subshell and one electron will be removed from $4s$ subshell. According, to the rule, the electrons are first removed from the highest subshell.
As, $C{r}^{3+}\to 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^3$, the number of unpaired electrons are $3$.

So, the number of unpaired electrons are $3$ in case of $C{r}^{3+}$.

Note:The configuration of Cr is somewhat different from the rest of the elements and does not follow the general trend of filling of electrons like other atoms, which is due to the possibility of a more stable electronic configuration.