
Give the oxidation number and coordination number of the central metal atom in the complex compound ${K_3}[Fe{({C_2}{O_4})_3}].$
Answer
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Hint: Oxidation number is defined as the amount of charge present on the atom in combined form whereas the coordination number is defined as the number of nearest neighbouring atoms.
Complete answer: or Complete step by step answer:
Let us consider the above coordination compound ${K_3}[Fe{({C_2}{O_4})_3}]$. Here iron is the central metal atom. Let the oxidation state of the central metal atom is x. Since Potassium is the alkali metal and we know the oxidation state of alkali metal is 1. ${C_2}{O_4}$ , i.e. oxalate ions act as a ligand. It is an anionic ligand and its denticity is 2 so the oxidation state of oxalate ion is 2.
As we know the oxidation state of the whole compound is equal to the sum of the oxidation state of all the atoms present in a compound.
So the oxidation state $Fe$ is calculated as shown below:
$
+ 3 + x + 3 \times ( - 2) = 0 \\
\Rightarrow 3 + x - 6 = 0 \\
\Rightarrow x - 3 = 0 \\
\Rightarrow x = 3 \\
$
Hence the oxidation state of iron (Fe) is +3.
In the coordination compound, the coordination number is also defined as the number of ligands that are bonded to the central metal atom by the coordinate bond. Here three oxalate ion is coordinated with central metal atom i.e. $Fe$. Hence the coordination number is 3.
Note:
Alternatively we can also find the oxidation number as shown below:
The given coordination compound is ${K_3}[Fe{({C_2}{O_4})_3}]$. Since potassium carries a +1 charge and as the complex is neutral so the charge on complexion should be -3. Hence the complex ion will be written as ${[Fe{({C_2}{O_4})_3}]^{ - 3}}$. Let us take the oxidation state $Fe$ is x. Oxalate ion is bidentate ligand hence, carries the charge -2. On adding the charge of all the atom we get the overall oxidation state as shown below:
$
x + 3 \times ( - 2) = - 3 \\
\Rightarrow x - 6 = - 3 \\
\Rightarrow x = - 3 + 6 \\
\Rightarrow x = 3 \\
$
Complete answer: or Complete step by step answer:
Let us consider the above coordination compound ${K_3}[Fe{({C_2}{O_4})_3}]$. Here iron is the central metal atom. Let the oxidation state of the central metal atom is x. Since Potassium is the alkali metal and we know the oxidation state of alkali metal is 1. ${C_2}{O_4}$ , i.e. oxalate ions act as a ligand. It is an anionic ligand and its denticity is 2 so the oxidation state of oxalate ion is 2.
As we know the oxidation state of the whole compound is equal to the sum of the oxidation state of all the atoms present in a compound.
So the oxidation state $Fe$ is calculated as shown below:
$
+ 3 + x + 3 \times ( - 2) = 0 \\
\Rightarrow 3 + x - 6 = 0 \\
\Rightarrow x - 3 = 0 \\
\Rightarrow x = 3 \\
$
Hence the oxidation state of iron (Fe) is +3.
In the coordination compound, the coordination number is also defined as the number of ligands that are bonded to the central metal atom by the coordinate bond. Here three oxalate ion is coordinated with central metal atom i.e. $Fe$. Hence the coordination number is 3.
Note:
Alternatively we can also find the oxidation number as shown below:
The given coordination compound is ${K_3}[Fe{({C_2}{O_4})_3}]$. Since potassium carries a +1 charge and as the complex is neutral so the charge on complexion should be -3. Hence the complex ion will be written as ${[Fe{({C_2}{O_4})_3}]^{ - 3}}$. Let us take the oxidation state $Fe$ is x. Oxalate ion is bidentate ligand hence, carries the charge -2. On adding the charge of all the atom we get the overall oxidation state as shown below:
$
x + 3 \times ( - 2) = - 3 \\
\Rightarrow x - 6 = - 3 \\
\Rightarrow x = - 3 + 6 \\
\Rightarrow x = 3 \\
$
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