Question

What is the oxidation number of metals (i) \[{[Fe{\left( {CN} \right)_6}]^{4 - }}\](ii) $Mn{O_4}^ - $

Answer 1

Hint: The charge that would exist on an atom if the bonding were fully ionic is known as its oxidation number. The charge on the ion is equal to the atom's oxidation number in simple ions. The oxidation number of a molecule or compound is the sum of the oxidation numbers of its component atoms. Atoms containing electrons in the d-shell can have a variety of oxidation values.

Complete answer:
Atoms containing electrons in the d-shell can have a variety of oxidation values. The oxidation numbers of these atoms in complex ions or molecules may be determined if the oxidation numbers of the other atoms in the species are fixed. Roman numerals are used to indicate the oxidation numbers of metals with several oxidation states. The charge that would exist on an atom if the bonding were fully ionic is known as the oxidation number, or oxidation state. The degree of oxidation (loss of electrons) of an atom in a chemical compound is indicated by this oxidation number. The oxidation state, which can be positive, negative, or zero, is the hypothetical charge that an atom would have if all of its connections to other atoms were entirely ionic, with no covalent component. Small integers are commonly used to indicate oxidation numbers. Many atoms can have several oxidation numbers, including most atoms having d subshells. The oxidation numbers of these atoms in complex ions or molecules may be determined if the oxidation numbers of the other atoms in the species are fixed.
(i) Fe in ${\left[ {{\text{Fe}}{{({\text{CN}})}_6}} \right]^{4 - }}$let the oxidation number be
Fe in${\left[ {{\text{Fe}}{{({\text{CN}})}_6}} \right]^{4 - }}$be x and CN be -1
Hence
${\text{x }} - 1$
${\text{Fe = x}}$
CN = -1
So the Sum of the oxidation numbers of all the atoms in${\left[ {{\text{Fe}}{{({\text{CN}})}_6}} \right]^{4 - }} = {\text{x}} + 6( - {\text{I}}) = - 4 = {\text{x}} - 6 = - 4;{\text{x}} = + 2$
Thus, the oxidation number of Fe in ${\left[ {{\text{Fe}}{{({\text{CN}})}_6}} \right]^{4 + }}$is 2
(ii) Mn is${\text{MnO}}_4^ - $
Now let the oxidation number of Mn in${\text{MnO}}_4^ - $be x.
Writing the oxidation number of each atom above in its symbol, we get.
Mn = x
O = -2
${\text{i}}.{\text{e}} \cdot {\text{x}} + 4( - 2) = - 1$
X =7

Note:
In coordination chemistry, the oxidation number has a somewhat different meaning. Transition metals are the elements in Groups 3 through 12 of the periodic table's d block. Because unpaired d electrons have a low reactivity, these metals can create many oxidation states and hence have multiple oxidation numbers. Unfortunately, there is no straightforward rule for identifying the oxidation states of transition metals, thus memorising the common states of each element is the best option.