Question

In ${\left[ {{\text{Co}}{{\text{F}}_{\text{6}}}} \right]^{ - 3}}$, ${\text{C}}{{\text{o}}^{ + 3}}$ uses outer d orbits (4d) in hybridization. The number of unpaired $s{p^3}{d^2}$electrons present in complex ions is.
A) 0
B) 4
C) 2
D) 3

Answer 1

Cobalt is the transition element placed in the 4th period and 9th group of the periodic table.The atomic number of the cobalt is 27 indicates there are 27 electrons as well as the 27 protons are there in the neutral cobalt atom.In the given question the oxidation state of the cobalt given is +3 indicates three electrons are removed from the neutral cobalt atom.Therefore, 24 electrons are present in the ${\text{C}}{{\text{o}}^{ + 3}}$ ion.

Complete answer:
Here, the compound given is ${\left[ {{\text{Co}}{{\text{F}}_{\text{6}}}} \right]^{ - 3}}$ indicates cobalt is present in a +3 oxidation state.
The atomic number of the cobalt is 27 and its electronic configuration is $\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^{\text{7}}}{\text{4}}{{\text{s}}^{\text{2}}}$.
The ${\text{C}}{{\text{o}}^{ + 3}}$ ion is formed from the cobalt atom by the removal of the three valence electrons. Therefore, two electrons are removed from the 4s orbital and one electron from the 3d orbital. Therefore, the electronic configuration of the ${\text{C}}{{\text{o}}^{ + 3}}$ ion is as follows:
$\left[ {{\text{Ar}}} \right]{\text{3}}{{\text{d}}^6}{\text{4}}{{\text{s}}^0}$
It indicates there are six electrons in the ${\text{C}}{{\text{o}}^{ + 3}}$ ion. Five d orbitals can occupy a maximum of ten electrons.
Here, the complex ligand ${{\text{F}}^{ - \,}}$ is present which is a weak field ligand hence, does not cause the pairing of the electrons.
Here, we can see that there are four unpaired electrons and two are paired electrons.
Now, option(A)0 is the incorrect answer to the given question.
Here, option(C)2 is incorrect.
Now, option(D)3 is also an incorrect answer to the question.
Now, option(B)4 is the correct answer to the given question.

Note: Here, it is important to know that the strength and nature of the ligand affects the paring of the electrons present in the metal orbitals.
The strong field ligands like ${\text{CO,NO,C}}{{\text{N}}^ - }\,{\text{etc}}$ causes the pairing of metal electron while the weak field ligand like ${{\text{H}}_{\text{2}}}{\text{O, N}}{{\text{H}}_{\text{3}}}{\text{, C}}{{\text{l}}^ - }{\text{, }}{{\text{I}}^ - }{\text{, B}}{{\text{r}}^ - }{\text{, }}{{\text{F}}^{ - \,}}\,{\text{etc}}$does not causes the pairing of the metal electron.