Question

Which of the following is incorrectly matched?

COMPLEX	OXIDATION NUMBER	ELECTRONIC CONFIGURATION
(A) ${{K}_{3}}\left[ Co{{\left( {{C}_{2}}{{O}_{4}} \right)}_{3}} \right]$	+3	${{t}_{2g}}^{6}{{e}_{g}}^{0}$
(B) ${{\left( N{{H}_{4}} \right)}_{2}}\left[ Co{{F}_{4}} \right]$	+2	${{t}_{2g}}^{5}{{e}_{g}}^{2}$
(C) $Cis-\left[ Cr{{\left( en \right)}_{2}}C{{l}_{2}} \right]Cl$	+3	${{t}_{2g}}^{3}{{e}_{g}}^{0}$
(D) $\left[ Mn{{\left( {{H}_{2}}O \right)}_{6}} \right]S{{O}_{4}}$	+2	${{t}_{2g}}^{3}{{e}_{g}}^{2}$

Answer 1

To get the proper and correct match, all the sections in the above given table should be consistent simultaneously i.e. the given complex must have corresponding oxidation number and electronic configuration necessarily.
The complex which won’t satisfy the criteria will be incorrectly matched, hence the answer.

Complete step by step solution:
Let us see what we mean by coordinate complex; it’s oxidation number, geometry and electronic configuration one by one to reach our answer easily.
Coordinate complex- The coordination complex is the molecule which contains a central metal atom which is surrounded by the nonmetal atoms or groups of atoms i.e. ligands.
These complexes can be either neutral or have charge on them. Those who are charged, they are stabilised by the attaching counter ions.
Types of ligands- Ligands donate the electrons to the central metal atom to form a coordinate bond. So, depending upon the number of donor atoms, the ligands can be classified as;
1. Monodentate ligands – one donor atom. For ex- ${{F}^{-}},C{{l}^{-}}.$ (Negative ions); ${{H}_{2}}O,CO.$ (Neutral molecules); $N{{O}^{+}},N{{H}_{4}}^{+}.$ (Positive ligands) and some organic free radicals.
2. Bidentate ligands – two donor atoms. For ex- Oxalate ion and ethylenediamine (en).
3. Tridentate ligands – three donor atoms. For ex- Diethylene triamine (dien) and triaminopropane.
4. Polydentate ligands – more than three donor atoms. For ex- Triethylenetetramine (trien) and ethylene diamine tetraacetic acid ion (EDTA ion).
Nature of bonding in coordination compounds- Werner’s theory and Sidwick’s Electronic theory tried to explain the nature of bonding but Valence Bond Theory explains it more efficiently. VBT explained the type of hybridisation and geometry of complexes according to the coordination number.
We came across some limitations of VBT so Crystal Field Theory (CFT) was introduced. This theory explains the crystal field splitting of d-orbital for the types of complexes easily.
Types of complexes (as described by VBT and CFT):
Octahedral complex- This has a coordination number of 6 and would have a general formula as $M{{L}_{6}}$ (M-metal atom and L- ligands). The geometry of such complexes will be in the form of octahedrons.
Tetrahedral complex- This has a coordination number of 4 and would have a general formula as $M{{L}_{4}}$ (M-metal atom and L- ligands). The geometry of such complexes will be in the form of tetrahedral.
Square planar complex- This has a coordination number of 4 and would have a general formula as $M{{L}_{4}}$ (M-metal atom and L- ligands). The geometry of such complexes will be square planar in nature. The electronic configuration of these types of complexes will depend upon the geometry they show and the extent of splitting of d-orbitals within them.
Crystal field splitting- The separation of the five-degenerate d-orbitals into two groups under the influence of ligands is known as crystal field splitting.
For octahedral complexes- The five-degenerate d-orbitals splits and form two groups as,
1. ${{e}_{g}}$ - consisting of two orbitals i.e. ${{d}_{{{x}^{2}}-{{y}^{2}}}}$ and ${{d}_{{{z}^{2}}}}$ having higher energy.
2. ${{t}_{2g}}$ - consisting of three orbitals i.e. ${{d}_{xy}},{{d}_{yz}}$ and ${{d}_{zx}}$ having lower energy.
For tetrahedral complexes- The five-degenerate d-orbitals splits and form two groups as,
1. $e$ - consisting of two orbitals i.e. ${{d}_{{{x}^{2}}-{{y}^{2}}}}$ and ${{d}_{{{z}^{2}}}}$ having lower energy.
2. ${{t}_{2}}$ - consisting of three orbitals i.e. ${{d}_{xy}},{{d}_{yz}}$ and ${{d}_{zx}}$ having higher energy.
Oxidation number- The oxidation number of a central metal atom is determined the same as we determine for any molecule considering the charges on the contributing atoms. Here, the oxidation number of the metal atom will give us the number of electrons in d-orbital which will then describe the specific electronic configuration. Now, in accordance with the theory explained let us move towards the illustration given;
For (A)- Complex- ${{K}_{3}}\left[ Co{{\left( {{C}_{2}}{{O}_{4}} \right)}_{3}} \right]$
Coordination number – 6 i.e. octahedral geometry.
Oxidation number-
$\begin{aligned} & x+3\left( -2 \right)=-3 \\\ & x-6=-3 \\\ & x=+3 \\\ \end{aligned}$
Thus, Electronic configuration- ${{t}_{2g}}^{6}{{e}_{g}}^{0}$
For (B)- Complex- ${{\left( N{{H}_{4}} \right)}_{2}}\left[ Co{{F}_{4}} \right]$
Coordination number – 4 i.e. tetrahedral geometry.
Oxidation number-
$\begin{aligned} & x+\left( -4 \right)=-2 \\\ & x=+2 \\\ \end{aligned}$
Thus, Electronic configuration- ${{e}^{4}}{{t}_{2}}^{3}$
For (C)- Complex- $Cis-\left[ Cr{{\left( en \right)}_{2}}C{{l}_{2}} \right]Cl$
Coordination number – 6 i.e. octahedral geometry.
Oxidation number-
$\begin{aligned} & x+2\left( 0 \right)+2\left( -1 \right)=+1 \\\ & x-2=+1 \\\ & x=+3 \\\ \end{aligned}$
Thus, Electronic configuration- ${{t}_{2g}}^{3}{{e}_{g}}^{0}$
For (D)- Complex- $\left[ Mn{{\left( {{H}_{2}}O \right)}_{6}} \right]S{{O}_{4}}$
Coordination number – 6 i.e. octahedral geometry.
Oxidation number-
$\begin{aligned} & x+0=+2 \\\ & x=+2 \\\ \end{aligned}$
Thus, Electronic configuration- ${{t}_{2g}}^{3}{{e}_{g}}^{2}$

Therefore, as we can see the incorrectly matched pair is option (B). hence, the answer.

Note: In the tetrahedral field the term ‘g’ is not used as in an octahedral field, because tetrahedral geometry does not have a centre of symmetry as in octahedral geometry. Also, the oxalate ion is weak ligand generally but in our given case (in option (A)), the central metal is Co. In this complex, oxalate will behave as a strong ligand (special case) and hence the electrons will pair up in ${{t}_{2g}}$ orbital.