Question

Which one of the following is an inner orbital complex as well as diamagnetic in behaviour?
[Atomic number of Zn=30, Cr=24, Co=27, Ni=28]
(A) ${\left[ {Zn{{(N{H_3})}_6}} \right]^{ + 2}}$
(B) ${\left[ {Cr{{(N{H_3})}_6}} \right]^{ + 3}}$
(C) ${\left[ {Co{{(N{H_3})}_6}} \right]^{ + 3}}$
(D) ${\left[ {Ni{{(N{H_3})}_6}} \right]^{ + 2}}$

Answer 1

The inner orbital complexes are those which use the inner d-orbitals to form complexes. Also the compounds which have unpaired electrons are paramagnetic and those with paired electrons are diamagnetic. We should check for strong field ligands in the option to identify inner orbital complexes because strong ligands cause pairing of electrons.

Complete step by step answer:
-First of all we will talk about magnetism of these coordination compounds.
We already know that the metallic complexes have unpaired electrons. Due to the presence of these unpaired electrons these metals become magnetic in nature. Based on their magnetism the ions may be diamagnetic or paramagnetic in nature. The diamagnetic materials are usually repelled by the magnetic field and such materials have no atomic dipoles due to the pairing between the electrons. When we apply an external magnetic field, dipoles are induced in the diamagnetic materials in such a way that induced dipoles oppose the external magnetic field according to Lenz’s law.
-Let us also take a look at what are inner orbital complexes.
If any complex is formed using the inner d-orbitals they will be known as inner orbital complexes or low spin complexes. While those complexes formed using outer d-orbitals are known as outer orbital complexes or high spin complexes.
-Hence to find out the complex which is inner orbital complex as well as diamagnetic, we need to check their hybridisation and number of unpaired electrons.

-For (A) ${\left[ {Zn{{(N{H_3})}_6}} \right]^{ + 2}}$: The ligand $N{H_3}$ is a neutral ligand and so its oxidation state is 0. We will now find out the oxidation state of zinc in the complex and write down its electronic configuration.
Atomic number of Zn is 30 and its electronic configuration is: $\left[ {Ar} \right]3{d^{10}}4{s^2}$
Let the O.S. of Zn be x.
O.S. of complex = O.S. of Zn + 6 ( O.S. of $N{H_3}$)
+2 = x + 6 (0)
x = +2
So, the electronic configuration of $Z{n^{ + 2}}$ will be: $\left[ {Ar} \right]3{d^{10}}$
Since the inner d-orbitals are completely filled the 6 ligands of $N{H_3}$ will attach to outer d orbitals. The hybridisation of the complex ${\left[ {Zn{{(N{H_3})}_6}} \right]^{ + 2}}$ would be $s{p^3}{d^2}$ and hence it will an outer orbital complex. So, this option is not correct.

-For (B) ${\left[ {Cr{{(N{H_3})}_6}} \right]^{ + 3}}$: We will now find out the oxidation state of zinc in the complex and write down its electronic configuration.
Atomic number of Cr is 24 and its electronic configuration is: $\left[ {Ar} \right]3{d^5}4{s^1}$
Let the O.S. of Cr be x.
O.S. of complex = O.S. of Cr + 6 ( O.S. of $N{H_3}$)
+3 = x + 6 (0)
x = +3
So, the electronic configuration of $C{r^{ + 3}}$ will be: $\left[ {Ar} \right]3{d^3}$
Since the 2 of the inner d-orbitals are available for bonding the 6 ligands of $N{H_3}$ will begin to attach from the inner d orbitals. The hybridisation of the complex ${\left[ {Cr{{(N{H_3})}_6}} \right]^{ + 3}}$ would be ${d^2}s{p^3}$ and hence it will be an inner orbital complex. But in the 3d orbital the 3 electrons present are unpaired and so this complex is paramagnetic. So, this option is also not correct.

-For (C) ${\left[ {Co{{(N{H_3})}_6}} \right]^{ + 3}}$: We will now find out the oxidation state of Cobalt in the complex and write down its electronic configuration.
Atomic number of Co is 27 and its electronic configuration is: $\left[ {Ar} \right]3{d^7}4{s^2}$
Let the O.S. of Co be x.
O.S. of complex = O.S. of Co + 6 ( O.S. of $N{H_3}$)
+3 = x + 6 (0)
x = +3
So, the electronic configuration of $C{o^{ + 3}}$ will be: $\left[ {Ar} \right]3{d^6}$
Since there are 6 electrons in the inner 3d orbitals, the pair with each other ($N{H_3}$ is a strong field ligand) and then the 2 remaining orbitals of the 3d-orbitals are available for bonding. So the ligand ($N{H_3}$) will begin to attach from the inner d orbitals. The hybridisation of the complex ${\left[ {Co{{(N{H_3})}_6}} \right]^{ + 3}}$ would be ${d^2}s{p^3}$ and hence it will be an inner orbital complex. Also the 3d orbital has 6 electrons in it which are paired with each other and there are no unpaired electrons present. So, this complex would be diamagnetic in nature.
Since the complex ${\left[ {Co{{(N{H_3})}_6}} \right]^{ + 3}}$ is both an inner orbital complex and diamagnetic, this option is correct.

-For (D) ${\left[ {Ni{{(N{H_3})}_6}} \right]^{ + 2}}$: We will now find out the oxidation state of nickel in the complex and write down its electronic configuration.
Atomic number of Ni is 28 and its electronic configuration is: $\left[ {Ar} \right]3{d^8}4{s^2}$
Let the O.S. of Ni be x.
O.S. of complex = O.S. of Ni + 6 ( O.S. of $N{H_3}$)
+2 = x + 6 (0)
x = +2
So, the electronic configuration of $N{i^{ + 2}}$ will be: $\left[ {Ar} \right]3{d^8}$
There are 8 electrons present in the inner 3d orbital, 2 of which are unpaired. The bonding of the ligand ($N{H_3}$) will begin from the s orbital. The hybridisation of the complex ${\left[ {Ni{{(N{H_3})}_6}} \right]^{ + 2}}$ would be $s{p^3}{d^2}$ and hence it will be an outer orbital complex. Also there are 2 unpaired electrons present in the 3d orbital and hence this complex is paramagnetic. So, this option is also not correct.
So, the correct answer is “Option C”.

Note: In the diamagnetic materials there are no atomic dipoles present and are repelled by magnets. Also the resultant magnetic moment of each atom of diamagnetic materials is zero due to paired electrons.
Also ${\left[ {Co{{(N{H_3})}_6}} \right]^{ + 3}}$ is a diamagnetic and low spin $3{d^6}$ octahedral complex which is used in various biological study methods like X-ray crystallography or nuclear magnetic resonance to study structures of nucleic acids like RNA and DNA.