Question

What is true about ${[MnC{l_6}]^{3 - }}$, ${[Fe{F_6}]^{3 - }}$ and ${[Co{F_6}]^{3 - }}$?
A.Each of these are outer orbital complexes.
B.Each of these have $s{p^3}{d^2}$ hybridisation
C.Each of these are paramagnetic
D.${[MnC{l_6}]^{3 - }}$, ${[Fe{F_6}]^{3 - }}$ and ${[Co{F_6}]^{3 - }}$ have four, five and four unpaired electrons respectively.

Answer 1

Hint: In the above given question we are first required to see the valency of the d block elements in each element and then see the number of electrons in their outermost shell by writing their configuration. The pairing and unpairing would talk about their paramagnetic or diamagnetic nature.

Complete answer: Before starting to answer the above given questions we need to first understand what is the meaning of paramagnetic. Paramagnetic means when the electrons present in the orbitals are unpaired while diamagnetism means the element has all paired electrons, the total spin is zero.
Now let’s see each and every compound individually and define them.
First of all, let’s see ${[MnC{l_6}]^{3 - }}$. Now in this compound we have manganese with chlorine. The chlorine has an oxidation state of negative one, using that lets find out the oxidation state of manganese keeping in mind the total charge on the compound is $ - 3$, let the oxidation state of manganese be $x$
\[x + ( - 1) \times 6 = - 3\]
 \[ \Rightarrow x = 3\]
Now knowing that manganese has the oxidation state of $ + 3$ and the atomic number of manganese is twenty-five let’s write the electronic configuration of manganese in $ + 3$oxidation state
$M{n^{ + 3}} = [Ar]3{d^4}$
Now from the configuration we can say that the compound has four electrons in the outermost orbital and all of them are unpaired which makes it paramagnetic. The hybridisation is defined as the fusing of atomic orbitals in Presence of any ligand, then the position which the anion occupies helps us define the hybridisation. Here the ligand chlorine is a weak ligand which is not able to pair all the electrons and therefore would be occupying the later orbitals and making its hybridisation $s{p^3}{d^2}$.
Now let’s see the compound ${[Fe{F_6}]^{3 - }}$, here iron is bonded with fluorine which is a very weak ligand and electronegative and needs only electron to reach its noble gas configuration, valency of fluorine would be $ - 1$ and the total charge on the compound is $ - 3$, therefore valency of iron would be let’s say $y$
$y + ( - 1) \times 6 = - 3$
$ \Rightarrow y = + 3$
Now knowing that iron has the oxidation state of $ + 3$ and the atomic number of irons is twenty-six let’s write the electronic configuration of iron in $ + 3$ oxidation state
$F{e^{ + 3}} = [Ar]3{d^5}$
It is in its one of the most stable state therefore irrespective of fluorine being a weak ligand, the electrons wont pair and the fluorine would assume the spaces of later orbitals thus making the hybridisation $s{p^3}{d^2}$. Now all the electrons are unpaired therefore the compound is paramagnetic.
Now let’s see the compound ${[Co{F_6}]^{3 - }}$, here cobalt is bonded with fluorine which is a very weak ligand and electronegative and needs only electron to reach its noble gas configuration, valency of fluorine would be $ - 1$ and the total charge on the compound is $ - 3$, therefore valency of cobalt would be let’s say $z$
$z + ( - 1) \times 6 = - 3$
$ \Rightarrow z = + 3$
Now knowing that cobalt has the oxidation state of $ + 3$ and the atomic number of cobalt is twenty-seven let’s write the electronic configuration of cobalt in $ + 3$ oxidation state
$C{o^{ + 3}} = [Ar]3{d^6}$
It has six electrons in its outermost shell, which means it would have four unpaired and two paired electrons in the outermost shell and fluorine being a weak ligand, the electrons wont pair and the fluorine would assume the spaces of later orbitals thus making the hybridisation $s{p^3}{d^2}$. Now there are four unpaired electrons in the outermost orbital of cobalt therefore it is paramagnetic.
All the statements are true about ${[MnC{l_6}]^{3 - }}$, ${[Fe{F_6}]^{3 - }}$ and ${[Co{F_6}]^{3 - }}$.

Note:
We need to remember that the most stable configuration of an atom would be when either all the electrons are paired or all the electrons are unpaired. The ligands need very high energy to pair these unpaired electrons and for that hybridisation happens and the shapes of orbitals also change.