In potassium hexacyanoferrate(II) the number of unpaired electron in the metal complex is:
Answer
262.5k+ views
Hint: Transition metals show variable oxidation state. The presence of unfilled d-orbitals helps them to show variable oxidation state. They form complexes for stabilizing the positive charge generated on them.
Complete step by step solution:
The above question is asking about the number of unpaired electrons present in metal complexes.
First, let us see what the formula of the metal complex given in the question is.
The formula of the metal complex can be written as:
\[{{K}_{4}}\left[ Fe{{(CN)}_{6}} \right]\]
\[\Rightarrow {{\left[ Fe{{(CN)}_{6}} \right]}^{-4}}\]
Now, we will be finding the oxidation state of Fe
Atomic number of Fe = 26
Let’s assume \[x\] be the oxidation state of Fe in the given metal complex.
Now, charge on \[C{{N}^{-}}\] is -1
So, \[x-6=-4\]
\[\Rightarrow x=+2\]
Hence, the oxidation state of Fe is +2
Now, we know that the configuration of Fe = \[\left[ Ar \right]3{{d}^{6}}4{{s}^{2}}\]
Now, Fe will form \[F{{e}^{+2}}\]in the metal complex
Hence, we will write the configuration of \[F{{e}^{+2}}\] = \[\left[ Ar \right]3{{d}^{6}}\]
So, the d-orbital configuration of \[F{{e}^{+2}}\] is \[{{d}^{6}}\]in the metal complex
Now, we will show the filling of electrons in the d-orbital
d-orbital configuration of \[F{{e}^{+2}}\] = \[{{d}^{6}}={{\underline{\uparrow \downarrow }}_{{}}}{{\underline{\uparrow }}_{{}}}{{\underline{\uparrow }}_{{}}}{{\underline{\uparrow }}_{{}}}\underline{ \uparrow }\]
Since, \[C{{N}^{-}}\] is a strong field ligand, thus pairing of electrons will take place
Hence, the new configuration will be
\[{{d}^{6}}={{\underline{\uparrow \downarrow }}_{{}}}{{\underline{\uparrow \downarrow }}_{{}}}{{\underline{\uparrow \downarrow }}_{{}}}{{\underline{{}}}_{{}}}\underline{{}}\]
Now, we can see that there are no unpaired electrons in the metal complex central atom.
Hence, number of unpaired electrons = 0
Note: The nature of ligands should not be confused. Weak field ligands do not cause pairing of electrons. Potassium hexacyanoferrate(II) is used in commercial blood glucose meters used by diabetics.
Complete step by step solution:
The above question is asking about the number of unpaired electrons present in metal complexes.
First, let us see what the formula of the metal complex given in the question is.
The formula of the metal complex can be written as:
\[{{K}_{4}}\left[ Fe{{(CN)}_{6}} \right]\]
\[\Rightarrow {{\left[ Fe{{(CN)}_{6}} \right]}^{-4}}\]
Now, we will be finding the oxidation state of Fe
Atomic number of Fe = 26
Let’s assume \[x\] be the oxidation state of Fe in the given metal complex.
Now, charge on \[C{{N}^{-}}\] is -1
So, \[x-6=-4\]
\[\Rightarrow x=+2\]
Hence, the oxidation state of Fe is +2
Now, we know that the configuration of Fe = \[\left[ Ar \right]3{{d}^{6}}4{{s}^{2}}\]
Now, Fe will form \[F{{e}^{+2}}\]in the metal complex
Hence, we will write the configuration of \[F{{e}^{+2}}\] = \[\left[ Ar \right]3{{d}^{6}}\]
So, the d-orbital configuration of \[F{{e}^{+2}}\] is \[{{d}^{6}}\]in the metal complex
Now, we will show the filling of electrons in the d-orbital
d-orbital configuration of \[F{{e}^{+2}}\] = \[{{d}^{6}}={{\underline{\uparrow \downarrow }}_{{}}}{{\underline{\uparrow }}_{{}}}{{\underline{\uparrow }}_{{}}}{{\underline{\uparrow }}_{{}}}\underline{ \uparrow }\]
Since, \[C{{N}^{-}}\] is a strong field ligand, thus pairing of electrons will take place
Hence, the new configuration will be
\[{{d}^{6}}={{\underline{\uparrow \downarrow }}_{{}}}{{\underline{\uparrow \downarrow }}_{{}}}{{\underline{\uparrow \downarrow }}_{{}}}{{\underline{{}}}_{{}}}\underline{{}}\]
Now, we can see that there are no unpaired electrons in the metal complex central atom.
Hence, number of unpaired electrons = 0
Note: The nature of ligands should not be confused. Weak field ligands do not cause pairing of electrons. Potassium hexacyanoferrate(II) is used in commercial blood glucose meters used by diabetics.
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