Question

What is the oxidation state of \[Mn\] in \[{\left[ {Mn{{\left( {CN} \right)}_5}} \right]^{2 - }}\]?
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: Oxidation state means a number which is assigned to an element to represent the number of electrons lost from that element or the number of electrons get consumed by the element. Oxidation states can be positive or negative. Transition elements show more than one oxidation state. This is mainly because they have filled d orbitals because of which no unpaired electron is available.

Complete step by step answer:
To calculate the oxidation number of the Mn in \[{\left[ {Mn{{\left( {CN} \right)}_5}} \right]^{2 - }}\] you have to sum all the oxidation number of the ions present in the compound to find out the total charge of this ion which is -2.
Let assume the oxidation number of metal Mn is x. Each cyanide shows a -1 oxidation state.
Therefore, the sum of this oxidation charge is \[x + ( - 1 \times 5) = - 2\]
Therefore, the oxidation state of Mn is,
\[
  x + ( - 1 \times 5) = - 2 \\
  x + ( - 5) = - 2 \\
  x = - 2 + 5 \\
  x = 3 \\
 \]

So, the correct option is C.

Note:
The elements which lie in the middle of the group II-A and group II B elements in the periodic table are transition elements. They are known as d block elements. The transition elements mean the elements that lie between the metals and non-metals of the periodic table. These metals usually have high melting and boiling points. Because of the unavailability of unpaired d-orbital electrons, these metals do not undergo covalent bonding.