Question

Which of the following complex(s) is/are high spin?
[A] ${{\left[ Co{{F}_{6}} \right]}^{3-}}$
[B] $\left[ Co{{\left( {{H}_{2}}O \right)}_{3}}{{F}_{3}} \right]$
[C] ${{\left[ Co{{\left( CN \right)}_{6}} \right]}^{3-}}$
[D] ${{\left[ Co{{\left( {{H}_{2}}O \right)}_{3}} \right]}^{2+}}$

Answer 1

To answer this question, we have to consider the spectrochemical series. In the spectrochemical series, the ligands towards the left are weak fields and form high spin complexes. The ligands towards the right are high spin and form low spin complexes

Complete answer:
We know that ligands are the species which are attached to the central metal ion and together they form a complex. They can either form high spin or low spin complexes.
We can arrange the ligands according to their strengths in the form of a series. This series is known as the spectrochemical series-
${{I}^{-}}$ < $B{{r}^{-}}$ < $SC{{N}^{-}}$ < $C{{l}^{-}}$ < ${{F}^{-}}$ < $O{{H}^{-}}$ < ${{H}_{2}}O$ < $NC{{S}^{-}}$ < $Py$ < $N{{H}_{3}}$ < $en$ < $N{{O}_{2}}^{-}$ < $C{{N}^{-}}$ < $CO$
In this series, ligands towards the left (iodide) are weak field ligands and ligands towards right (CO) are stronger field ligands. Weak field ligands generally form high spin complexes and strong field ligands generally form low spin ligands. Ammonia is known as the ‘notorious ligand’ as it forms both high spin as well as low spin complexes.
Firstly, we have ${{\left[ Co{{F}_{6}} \right]}^{3-}}$. As we can see in the series that fluoride ion is placed towards the left in the spectrochemical series so it is a weak field ligand thus it will form high spin complexes.
Next, we have $\left[ Co{{\left( {{H}_{2}}O \right)}_{3}}{{F}_{3}} \right]$. Here also, both water and fluorine are weak field ligands thus, they will form a high spin complex.
Then we have ${{\left[ Co{{\left( CN \right)}_{6}} \right]}^{3-}}$. As we can see in the series that cyanide lies towards the extreme right it is a strong field ligand. Therefore, it will form weak field ligands.
And lastly, we have ${{\left[ Co{{\left( {{H}_{2}}O \right)}_{3}} \right]}^{2+}}$Here also we have water as ligand which is a weak field ligand therefore, it will form high spin complex.
As we can see from the above discussion that ${{\left[ Co{{F}_{6}} \right]}^{3-}},\left[ Co{{\left( {{H}_{2}}O \right)}_{3}}{{F}_{3}} \right]\text{ and }{{\left[ Co{{\left( {{H}_{2}}O \right)}_{3}} \right]}^{2+}}$form high spin complexes.

Therefore, the correct answers are options [A] ${{\left[ Co{{F}_{6}} \right]}^{3-}}$, [B] $\left[ Co{{\left( {{H}_{2}}O \right)}_{3}}{{F}_{3}} \right]\text{ }$ and [D] $\text{ }{{\left[ Co{{\left( {{H}_{2}}O \right)}_{3}} \right]}^{2+}}$

Note:
We can also answer this question on the basis of CFSE i.e. crystal field stabilisation energy. CFSE is the stability that results from placing a transition metal ion in the crystal field degenerated by a set of ligands. It is defined as the difference in energy of electronic configuration in isotropic field and the energy of electronic configuration in ligand field.
A high spin complex is formed if the pairing energy is higher than the energy of the extent of splitting i.e. more energy is required in pairing than the energy required for the electron to move to the higher energy level.