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In permanganate ion $Mn{O_4}^ - $, manganese has an oxidation number of $ + 7$. Therefore it is:
A. $s{p^3}{d^3}$ hybridised
B. $s{p^3}$ hybridised
C. $ds{p^2}$hybridised
D. ${d^3}s{p^3}$ hybridised

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Last updated date: 20th Jun 2024
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Answer
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Hint: Hybridisation: It is defined as the process of combining two or more atomic orbitals from the same atom to form a new orbital having different components.
Oxidation number: It is defined as the oxidation state which is defined as the total number of electrons that an atom either loses or gains to form a chemical bond.

Complete step by step answer:
First of all we will read about transition elements.
Transition elements: Those elements which are in groups from three to eleven. They are called a transition because they are in between the s- block elements and p-block elements. For example: Scandium, iron, zinc, etc. They have fully or at least one electron in their d-orbits. For example: scandium has one d-electron, zinc has ten d-electrons. Due to the different number of electrons present in their d-shells they show different valencies. For example: Scandium having one electron can show three valencies as one, two and three. Because the atomic number of scandium is $21$and its electronic configuration is $Ar3{d^1}4{s^2}$. By losing only one s-electron it will attain one valency and by losing both the s-electrons it will attain valency two. And by losing two s-electrons and one d-electron it will achieve three valencies which is a stable state of scandium. Because after losing three electrons it will attain electronic configuration of noble gas argon.
Hybridisation: It is defined as the process of combining two or more atomic orbitals from the same atom to form a new orbital having different components.
In $s{p^3}$ hybridisation the mixing of one $2s$ orbital with three $2p$ orbitals takes place to form four hybrid orbitals.
In $s{p^3}{d^3}$ hybridisation the mixing of one $s$ orbital with three $p$ orbitals and three $d$ orbitals takes place to form seven hybrid orbitals.
In ${d^3}s{p^3}$ hybridisation the mixing of three $d$ orbitals with one $s$ orbital and three $p$ orbitals takes place to form seven hybrid orbitals.
In $ds{p^2}$ hybridisation the mixing of one $d$ orbital with one $s$ orbital and two $p$ orbitals takes place to form four hybrid orbitals.
In the question we are given the compound of manganese. The atomic number of manganese is $25$. So its electronic configuration will be $Ar3{d^5}4{s^2}$. So manganese can show the valencies equal to seven i.e. five due to d-electrons and two due to s-electrons. When manganese shows oxidation number equal to seven i.e. manganese loses all the five d-electrons as well as both the s-electrons. So its hybridisation will be $s{p^3}$. Because orbitals of oxygen form $s{p^3}$ hybridisation.

So, the correct answer is Option B .

Note:
The number of hybrid orbitals after hybridisation is equal to the number of orbitals which takes part in the hybridisation. For example: in $s{p^3}$ the orbitals which take part in the hybridisation is four. So the number of hybrid orbitals formed will be four.