Question

The geometry and the number of unpaired electron (s) of \[{\left[ {{\rm{MnB}}{{\rm{r}}_{\rm{4}}}} \right]^{2 - }}\] respectively, are
A. Tetrahedral and 1
B. Square planar and 1
C. Tetrahedral and 5
D. Square planar and 5

Answer 1

Hint: To find the geometry and number of unpaired electrons, we have to first calculate the oxidation state of the central metal atom of the complex, that is, Manganese (Mn). After that, we have to find the geometry.

Complete Step by Step Answer:
Let's first calculate the oxidation state of Mn. We know that bromine possesses an oxidation state of -1 and the charge of the complex is -2. We take x as the oxidation state of Mn.
\[x + \left( { - 4} \right) = - 2\]
\[x = 4 - 2 = 2\]

Therefore, Mn has an oxidation state of Mn is +2.
So, the electronic configuration of Mn is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^5}\] .
The electronic configuration of \[{\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}\] is \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}\].

So, the number of electrons in the outermost shell is five. As the ligand Bromine is a weak field ligand, the paring of the d- electrons is not possible, therefore, hybridization of the complex is \[s{p^3}\] and the geometry of the complex is tetrahedral.

Therefore, the count of unpaired electrons in the given complex is 5 and the geometry is tetrahedral.
Hence, the option C is right.

Note: There are some ligands that are called strong ligands such as cyanide (CN), carbon monoxide (CO), EDTA, etc. They are called strong ligands because they can pair electrons. They also can split the d-orbitals. If the coordination number is four and the ligands are strong ligands, then the geometry of the molecule is square planar. And if the coordination number is four and ligands are weak, then the complex has tetrahedral geometry.