Question

Highest oxidation state of Manganese in fluorides is +4 $\left( Mn{{F}_{4}} \right)$ but highest oxidation state in oxides is +7 $\left( M{{n}_{2}}{{O}_{7}} \right)$ because
(A) Fluorine is most electronegative than oxygen.
(B) Fluorine doesn’t possess d orbitals
(C) Fluorine stabilises lower oxidation states
(D) In covalent compounds, fluorine can form a single bond only while oxygen forms a double bond.

Answer 1

Hint:
Firstly , let’s see the definition of oxidation state and about oxidation state of Manganese element-

-Oxidation State- It is a number which is assigned to an element in a chemical combination that represents the number of electrons lost, or we can say the number of electrons gained, (if the number is negative), by an atom of that element in the compound. It is also sometimes termed as oxidation number.
-Oxidation state of Manganese- Manganese belongs to d- block elements that are also called as transition elements. Transition metals are well known for having more than one stable oxidation state. Hence, as we know that Manganese is a transition element and that’s why it has more than any one oxidation state . Here the point also should be noted that, in its compounds manganese exhibits oxidation states from +2 to +7. The various common oxidation states are +2, +4, and +7, but the less common are +3, +5, and +6 states that can be easily prepared.
     \[\]

Complete step by step solution:
Let’s see how to find out the oxidation state of Mn in above given compounds -
1-Oxidation state of Mn in fluoride that is $\left( Mn{{F}_{4}} \right)$ -
Let’s consider Mn as x,
\[\begin{align}
  & x+4\left( -1 \right)=0 \\
 & x-4=0 \\
 & x=+4 \\
\end{align}\]
hence, oxidation state in Mn is = +4.

2-Oxidation state of Mn in oxide that is $\left( M{{n}_{2}}{{O}_{7}} \right)$is-
     \[\begin{align}
  & 2x+7\left( -2 \right)=0 \\
 & 2x+\left( -14 \right)=0 \\
 & 2x-14=0 \\
 & x=\dfrac{14}{2} \\
 & x=7 \\
\end{align}\]
     \[\]

Hence, the oxidation state of Mn in fluorides is +7.

-Manganese has highest oxidation state in oxides that is +7 $\left( M{{n}_{2}}{{O}_{7}} \right)$ because-
Mn forms $p\pi -d\pi $ multiple bonds with oxygen by using 2p orbitals of oxygen and 3d orbitals of Mn. Let’s understand the meaning of $p\pi -d\pi $ bond , it is a bond which was formed between two atoms, where one atom will have one lone pair of electrons and other atom will have vacant orbital . Here in this case multiple bonds are formed between oxygen and manganese. Hence, we can say that oxygen has a greater ability than fluorine to stabilise the higher oxidation states of transition metal ions.
-Highest oxidation state of Manganese in fluorides is +4 $\left( Mn{{F}_{4}} \right)$ because-
 Fluorine in the various covalent compounds can form a single bond. Here also, we can see that it forms a single bond with Mn and also doesn't show multiple bonding.
Hence, we can conclude that the correct option is (d).
Additional information:
-Manganese has an electronic configuration of- $\left[ Ar \right]3{{d}^{5}}4{{s}^{2}}$ It is having atomic number 25. Mn is not found as a free element in nature, and is often found in minerals in combination with iron.
-Harmful health effects of Mn – Nowadays ,it is found that higher levels of exposure to manganese in drinking water are associated with reduced intelligence quotients and intellectual impairment. And also leads to neurological disorders.
Note:
-The $p\pi -d\pi $ and bond are formed depending on the orbital to which electron pair is donated and from which electron pair is donated. And as we have discussed, to form this bond it is necessary to have a vacant orbital in an atom, and a lone pair of electrons should be present in another atom. In $\left( M{{n}_{2}}{{O}_{7}} \right)$ , Mn is having a vacant d orbital, and oxygen is having two unpaired electrons or we can say a lone pair of electron.

-The next point to be noted is that, Manganese element that was denoted by the symbol Mn should not be confused with the Magnesium element denoted by Mg .