Answer
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Hint: The first nitrosyl complex was discovered by J Priestley in 1790, $\mathop K\nolimits_4 \left[ {\mathop {Fe(CN)}\nolimits_6 } \right]$ . Here, we first write the oxidation state of each molecule individually leaving only Fe.
Complete step by step answer:
The oxidation state is the number of electrons that are lost, gained, or shared by a certain atom with other atoms in the same molecule
Here, we will focus on all the fixed components and then work out the oxidation state of $Fe$.
The net charge on the molecule is 0. Therefore, all the oxidation states will add up to 0 after they are multiplied by the number of molecules.
\[(4 \times {\text{oxidation state of }}K) + ({\text{oxidation state of Fe}}) + (6 \times {\text{oxidation state of CN}}) = 0\]
Let the oxidation state of $Fe$ be $x$
We know that almost all elements from group 1 have an oxidation state of $ + 1$. This rule never changes. Thus, we will assume that the oxidation state of $K$ is $ + 1$. And the oxidation state of $CN$ is $ - 1$.
Now plugging in the numbers in the equation given above, we get,
\[[4 \times ( + 1)] + [x] + [6 \times ( - 1)] = 0\]
$ \Rightarrow + 4 + x - 6 = 0$
$ \Rightarrow x - 2 = 0$
$\therefore x = + 2$
So, the oxidation state of $Fe$ is $ + 2$..
Thus, the oxidation state of $Fe$ is +2 when the oxidation states of $K$ and $CN$ are $ + 1$ and $ - 1$respectively.
Note:
Always check what the net charge on the given ion or molecule is, this will change the number present on the R.H.S. of the equation.
For example: If we consider \[\mathop {SO}\nolimits_4^{2 - } \] then to calculate the oxidation state of $S$, we need to consider the equation.
\[({\text{oxidation state of S}}) + (4 \times {\text{oxidation state of O}}) = - 2\]
Let the oxidation state of $S$ be $x$
Then, $[x] + [4 \times ( - 2)] = - 2$
$ \Rightarrow x + ( - 8) = - 2$
$ \Rightarrow x = + 6$
Here the oxidation state of S will be +6. If we do not pay attention to the net charge, the answer will turn out to be +8, which is not possible.
Complete step by step answer:
The oxidation state is the number of electrons that are lost, gained, or shared by a certain atom with other atoms in the same molecule
Here, we will focus on all the fixed components and then work out the oxidation state of $Fe$.
The net charge on the molecule is 0. Therefore, all the oxidation states will add up to 0 after they are multiplied by the number of molecules.
\[(4 \times {\text{oxidation state of }}K) + ({\text{oxidation state of Fe}}) + (6 \times {\text{oxidation state of CN}}) = 0\]
Let the oxidation state of $Fe$ be $x$
We know that almost all elements from group 1 have an oxidation state of $ + 1$. This rule never changes. Thus, we will assume that the oxidation state of $K$ is $ + 1$. And the oxidation state of $CN$ is $ - 1$.
Now plugging in the numbers in the equation given above, we get,
\[[4 \times ( + 1)] + [x] + [6 \times ( - 1)] = 0\]
$ \Rightarrow + 4 + x - 6 = 0$
$ \Rightarrow x - 2 = 0$
$\therefore x = + 2$
So, the oxidation state of $Fe$ is $ + 2$..
Thus, the oxidation state of $Fe$ is +2 when the oxidation states of $K$ and $CN$ are $ + 1$ and $ - 1$respectively.
Note:
Always check what the net charge on the given ion or molecule is, this will change the number present on the R.H.S. of the equation.
For example: If we consider \[\mathop {SO}\nolimits_4^{2 - } \] then to calculate the oxidation state of $S$, we need to consider the equation.
\[({\text{oxidation state of S}}) + (4 \times {\text{oxidation state of O}}) = - 2\]
Let the oxidation state of $S$ be $x$
Then, $[x] + [4 \times ( - 2)] = - 2$
$ \Rightarrow x + ( - 8) = - 2$
$ \Rightarrow x = + 6$
Here the oxidation state of S will be +6. If we do not pay attention to the net charge, the answer will turn out to be +8, which is not possible.
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