Question

Oxidation number of \[Mn\] in $Mn{O_2},MnO_4^{2 - },MnO_4^ - $ are $4, + 6$ and $ + 7$ respectively. If true, enter 1, else enter 0.
A. 1
B. 0

Answer 1

Hint:Oxidation state indicates the degree of oxidation for an atom in a chemical compound. It is the imaginary charge that an atom would have if all bonds to atoms of different elements were completely ionic. Oxidation states are typically represented by integers, which can be positive, negative, or zero.

Complete step by step answer:
Manganese is a d- block element and the d- block elements have a special characteristic of having the most number of variable oxidation states in the periodic table. It has an atomic number equal to 25 and has five electrons in the outermost $3d$ subshell. Let us determine the oxidation number of manganese in each of the compounds or ions one by one.
(i) $Mn{O_2}$ : Let the oxidation number of manganese in manganese dioxide be $x$ .
As we know, the oxidation state of oxygen atom is = $ - 2$
Thus, $x + 2( - 2) = 0$
$ \Rightarrow x = + 4$
(ii) $MnO_4^{2 - }$ : Let the oxidation number of manganese in the manganate ion be $y$ .
As we know, the oxidation state of oxygen atom is = $ - 2$
Thus, $y + 4( - 2) = - 2$
$ \Rightarrow y - 8 = - 2$
Hence, $y = + 6$
(iii) $MnO_4^ - $ : Let the oxidation number of manganese in the permanganate ion be $z$ .
As we know, the oxidation state of oxygen atom is = $ - 2$
Thus, $z + 4( - 2) = - 1$
Hence, $z = + 7$
Thus, the oxidation numbers of \[Mn\] in $Mn{O_2},MnO_4^{2 - },MnO_4^ - $ are $4, + 6$ and $ + 7$ respectively.
The correct option is A. 1.


Note:Manganese has the highest number of oxidation states in the periodic table. It has five electrons in the $3d$ sub-shell and two electrons in the $4s$ subshell. Due to the presence of five unpaired electrons in the $3d$ sub-shell it can show different oxidation states. Along with that, the energy level of $3d$ sub-shell and $4s$ subshell is the same, so the transition of electrons takes place easily from the outermost valence shells.