Answer
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Hint: We must have to know that in a coordination complex, there are metal, coordination complexes and ligands attached to it. The metal, ligands and counter ion together forms a coordination complex. Bidentate ligands are the one in which two atoms coordinate directly to the central atom forming a coordination complex.
Complete answer:
Before starting this question we need to know what coordination number and oxidation state is?
We must remember that the coordination number is derived for the central atom present in any complex, that central atom is bonded to many ligands around it. So the number of atoms, molecules or ions which are bonded to central atoms defines the coordination number of any complex.
For example: A complex \[[CuC{l_2}]\]- has a coordination number two since it has two chlorine atoms.
Now, what is the oxidation state? Oxidation state can be defined on the basis of gaining and losing an electron. When an atom loses or gains an electron so as to form a chemical bond with another atom in a molecule, it defines oxidation state.
For example: A complex \[Mn{O_2}\] has \[ + 2\] oxidation state.
Now let’s solve for the given complex
Oxalate and en are both bidentate ligands, thus in this molecule there are two \[en\] ligands thus four coordinate bonds while oxalate will give two coordinate bonds, thus in total \[6\] will be the coordination number for the given complex.
To find the oxidation state of M, let the oxidation state of M be \[x\]
\[x + \left( {2 \times 0} \right) + \left( { - 2} \right) + \left( { - 1} \right) = 0\]
\[ \Rightarrow x = + 3\]
\[en\] is a neutral ligand thus the charge on it will be zero. Oxalate has \[ + 2\] charge on it and \[N{O_2}\] carries \[ - 1\] charge, thus the oxidation state of the metal comes to be \[ + 3\].
Thus we can conclude it by saying the coordination number on this complex is 6 and oxidation state is\[ + 3\].
Note:
We have to know that \[N{O_2}\] in this complex is a nitrate ion which has a molecular name nitrate ion. When an atom loses or gains an electron so as to form a chemical bond with another atom in a molecule, it defines oxidation state.
Complete answer:
Before starting this question we need to know what coordination number and oxidation state is?
We must remember that the coordination number is derived for the central atom present in any complex, that central atom is bonded to many ligands around it. So the number of atoms, molecules or ions which are bonded to central atoms defines the coordination number of any complex.
For example: A complex \[[CuC{l_2}]\]- has a coordination number two since it has two chlorine atoms.
Now, what is the oxidation state? Oxidation state can be defined on the basis of gaining and losing an electron. When an atom loses or gains an electron so as to form a chemical bond with another atom in a molecule, it defines oxidation state.
For example: A complex \[Mn{O_2}\] has \[ + 2\] oxidation state.
Now let’s solve for the given complex
Oxalate and en are both bidentate ligands, thus in this molecule there are two \[en\] ligands thus four coordinate bonds while oxalate will give two coordinate bonds, thus in total \[6\] will be the coordination number for the given complex.
To find the oxidation state of M, let the oxidation state of M be \[x\]
\[x + \left( {2 \times 0} \right) + \left( { - 2} \right) + \left( { - 1} \right) = 0\]
\[ \Rightarrow x = + 3\]
\[en\] is a neutral ligand thus the charge on it will be zero. Oxalate has \[ + 2\] charge on it and \[N{O_2}\] carries \[ - 1\] charge, thus the oxidation state of the metal comes to be \[ + 3\].
Thus we can conclude it by saying the coordination number on this complex is 6 and oxidation state is\[ + 3\].
Note:
We have to know that \[N{O_2}\] in this complex is a nitrate ion which has a molecular name nitrate ion. When an atom loses or gains an electron so as to form a chemical bond with another atom in a molecule, it defines oxidation state.
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