Question

The shape of $Mn{O}_4^{-}$ ion and the hybridisation of $Mn$ in $Mn{O}_4^{-}$ is:
A.Tetrahedral, $s{p}^3$
B.Tetrahedral, $d^{3}s$
C.Square planar, $ds{p}^2$
D.Square planar, $s{p}^3$

Answer 1

Hint: To answer this question, we must find the hybridisation of the central atom and the arrangement of the bond pairs and lone pairs on the central atom. To do that, you must recall the VSEPR (Valence shell electron pair repulsion) theory. It suggests that all valence shell electron pairs surrounding the central atom arrange themselves in such a manner so as to be as far away from each other as possible.

Complete step by step solution:
Atomic number of manganese is 25. So we can write the electronic configuration as:
$Mn:\left[Ar\right]4s^{2}3d^5$
It has seven electrons in total in its 4s and 3d orbital. In the permanganate ion, the manganese is bonded to four oxygen atoms. Oxygen has six valence electrons in its valence shell, two short of a complete octet or a noble gas like configuration. Thus, each oxygen has a valency of 2 and tends to get two more electrons either by sharing or by accepting an electron from a metal. Manganese forms a double bond with three oxygen atoms and a single bond with the fourth. The fourth oxygen satisfies its valency by accepting an electron which is represented by the overall charge on the ion.
Thus, four single bonds are formed and three pi bonds are formed. So, four hybrid orbitals are required. The configuration in the excited state is given as:
$Mn:\left[Ar\right]3d^{3}4s^1$
So the hybridisation of manganese is $d^{3}s$ and it is tetrahedral in shape

Hence, the correct option is option (B).

Note: During bond formation, the atomic orbitals of an atom are mixed in such a manner as to produce equivalent orbitals. This mixing of orbitals is known as hybridisation. The arrangement of these hybrid orbitals according to the VSEPR theory gives us the shape of the molecule.