Question

Which of the following complexes is an example of strongest reducing agent?
A.${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{4 - }}$
B.${\left[ {Co{{\left( {CN} \right)}_6}} \right]^{3 - }}$
C.${\left[ {Ag{{\left( {CN} \right)}_2}} \right]^{1 - }}$
D.${\left[ {Cu{{\left( {CN} \right)}_4}} \right]^{2 - }}$

Answer 1

First, we will find the oxidation state of the metal. Here in all the given complexes the ligand is an anionic ligand (cyano). So we will assume the oxidation state of metal to be x. Then add the oxidation state of the metal and ligand and equate it to the charge on the complex. Then we will determine which metal can easily give its electron because that compound is the strongest reducing agent.

Complete step by step answer:
We know that Cyano ligand is anionic ligand so it will carry negative charge.
In compound A, We have to find the oxidation state of Cobalt .Let the oxidation number of Cobalt be x then-
$\Rightarrow x + 6\left( { - 1} \right) = - 4$
On solving, we get-
$\Rightarrow x = 6 - 4 = + 2$
Now we know that the atomic number of Cobalt is $27$ and its electronic configuration is $\left[ {Ar} \right]3{d^7}4{s^2}$
Here since the oxidation state of cobalt is $+ 2$ so the electronic configuration becomes-
$\Rightarrow C{o^{2 + }} = \left[ {Ar} \right]3{d^7}$
Here to complete its octet it will either accept $3$ electrons or donate $2$ electrons. So it is not a strong reducing agent.
In compound B, the oxidation state of Cobalt is –
$\Rightarrow x + 6\left( { - 1} \right) = - 3$
On solving, we get-
$\Rightarrow x = 6 - 3 = + 3$
Now here we can see that the oxidation state of cobalt is $+ 3$ so the electronic configuration becomes-
$\Rightarrow C{o^{3 + }} = \left[ {Ar} \right]3{d^6}$
Now to complete its octet cobalt will have to donate its $1$ electron so it can easily give one electron and hence acts as a strong reducing agent.
In compound C, the oxidation state of Silver is-
$\Rightarrow x + 2\left( { - 1} \right) = - 1$
On solving, we get-
$\Rightarrow x = 2 - 1 = + 1$
Now we know that the atomic number of Silver is $47$ and its electronic configuration is $\left[ {Kr} \right]4{d^{10}}5{s^1}$
Here since the oxidation state of silver is $+ 1$ so the electronic configuration becomes-
$\Rightarrow A{g^{1 + }} = \left[ {Kr} \right]4{d^{10}}$
Since the d-orbital is fulfilled so it will be most stable. Hence it will not donate electrons easily. So it is not a reducing agent.
In compound D, the oxidation state of copper is-
$\Rightarrow x + 4\left( { - 1} \right) = - 2$
On solving, we get-
$\Rightarrow x = 4 - 2 = + 2$
Now we know that the atomic number of Copper is $29$ and its electronic configuration is $\Rightarrow C{u^{2 + }} = \left[ {Ar} \right]3{d^{10}}4{s^1}$
Here since the oxidation state of silver is $+ 2$ so the electronic configuration becomes-
$\Rightarrow C{u^{2 + }} = \left[ {Ar} \right]3{d^9}$
Since the d-orbital needs only one electron to become stable, hence it will accept an electron as it is not easy to donate four electrons. So it is not a reducing agent.

Hence the correct answer is B.

Note:
Here we have simply calculated the oxidation state as we generally do. Many students may make the mistake of equating the addition of oxidation state of metal and ligand to zero because we do that for neutral molecules. Here all the complexes carry charge so we will equate the sum to the charge present on the complex.