The brown ring complex compound is formulated as $\left[ {{\text{Fe(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{5}}}{\text{NO}}} \right]{\text{S}}{{\text{O}}_{\text{4}}}$. The oxidation state of iron is:
A. $ + 1$
B. $ + 2$
C. $ + 3$
D. $ + 6$
Answer
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Hint:To determine the oxidation number of the central metal we will assume that the oxidation number of the metal is x. Then take the sum of oxidation numbers of all ligands and metal. Put them equal to the charge of the complex or equal to the oxidation state of the counterion and determine the value of x. The value of x is the oxidation number of the metal.
Complete answer:
Metal combines with ligands to form a complex. During the formation of the complex metal gets oxidized so metal gets a positive charge. The charge of the metal is known as the oxidation number of the metal. The positive charge of the metal cancels out by anionic ligand. The remaining charge (may be positive or negative) is represented as the charge of the complex which is satisfied by counter ions.
Assume that the oxidation state of iron metal is x.
The oxidation state of the sulphate ion is $ - 2$, oxidation state of water is zero and the oxidation state of nitroso ${\text{NO}}$ is $ + 1$.
$x\, + \,0 + \left( { + 1} \right)\, = \, - 2$
$x = + 1$
So, The oxidation state of iron in brown ring complex compound formulated as $\left[ {{\text{Fe(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{5}}}{\text{NO}}} \right]{\text{S}}{{\text{O}}_{\text{4}}}$ is $ + 1$.
Therefore, the correct answer is option (A) i.e, $ + 1$ is correct.
Note:The sum of the oxidation number of the metal and ligands is equal to the charge of the complex which is equal to the charge of the counter ion. If no charge is given on the complex then put the sum of the oxidation number of the metal and ligands, equal to the zero. The metal has a positive charge whereas the ligands can be neutral, anionic or neutral. Only the anionic ligands cancel the positive charge of the metal. The anionic ligand can be identified by their names. The name of the anionic ligand has the suffix ‘O’, ‘ato’, or ‘ito’.
Complete answer:
Metal combines with ligands to form a complex. During the formation of the complex metal gets oxidized so metal gets a positive charge. The charge of the metal is known as the oxidation number of the metal. The positive charge of the metal cancels out by anionic ligand. The remaining charge (may be positive or negative) is represented as the charge of the complex which is satisfied by counter ions.
Assume that the oxidation state of iron metal is x.
The oxidation state of the sulphate ion is $ - 2$, oxidation state of water is zero and the oxidation state of nitroso ${\text{NO}}$ is $ + 1$.
$x\, + \,0 + \left( { + 1} \right)\, = \, - 2$
$x = + 1$
So, The oxidation state of iron in brown ring complex compound formulated as $\left[ {{\text{Fe(}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{)}}_{\text{5}}}{\text{NO}}} \right]{\text{S}}{{\text{O}}_{\text{4}}}$ is $ + 1$.
Therefore, the correct answer is option (A) i.e, $ + 1$ is correct.
Note:The sum of the oxidation number of the metal and ligands is equal to the charge of the complex which is equal to the charge of the counter ion. If no charge is given on the complex then put the sum of the oxidation number of the metal and ligands, equal to the zero. The metal has a positive charge whereas the ligands can be neutral, anionic or neutral. Only the anionic ligands cancel the positive charge of the metal. The anionic ligand can be identified by their names. The name of the anionic ligand has the suffix ‘O’, ‘ato’, or ‘ito’.
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