Question
Answers

Which of the following is a high spin complex?
A. $[Cr{{(gly)}_{3}}]$
B. ${{[CoB{{r}_{2}}C{{l}_{2}}{{(SCN)}_{2}}]}^{3-}}$
C. ${{[Fe{{(CN)}_{6}}]}^{3-}}$
D. $Na[PtBrCl{{(N{{O}_{2}})}_{2}}{{(N{{H}_{3}})}_{2}}]$

Answer Verified Verified
Hint: Think about when a complex is identified as a high spin complex and what it signifies about the electrons in that complex. Consider what role the ligands will play and how they may cause a complex to be high spin or low spin complex.

Complete step by step answer:
A high spin complex is a complex that has a significant number of unpaired electrons present in its d-orbital. If the energy that is required for the pairing of two electrons is greater than the energy that is required to place the electron in an orbital that is one energy level higher, then the system would prefer to go with the option that costs less energy. So, the electrons will fill all the orbitals first before pairing up the electrons. This will result in a large number of unpaired electrons and will cause the complex to have a high spin.
The field of the ligand will decide whether the energy required for placing the electron in an orbital is more or the energy required for the pairing is more. If the ligand is a strong field ligand, then the energy required for pairing will be less than the energy required to add the electron in a new orbital. Thus, less unpaired electrons will be left and the complex is more likely to become a low-spin complex.
A weak field ligand on the other hand will indicate that the energy required for the pairing is more than the addition into a new orbital. So, complexes with weak field ligands will have electrons placed in multiple orbits that are not placed together. These complexes will tend to have a high spin.
So, to solve the question, we have to figure out which complexes have the most weak field ligands. This will give us the complex with the highest spin. We can refer to the spectrochemical series to determine whether a ligand is a strong field ligand or a weak field ligand. The spectrochemical series from strongest to weakest is as follows:
\[CO>C{{N}^{-}}>N{{O}_{2}}^{-}>N{{H}_{3}}>Pyridine>{{H}_{2}}O>O{{H}^{-}}>{{F}^{-}}>N{{O}_{3}}^{-}>C{{l}^{-}}>B{{r}^{-}}>{{I}^{-}}\]
Usually, the divide between strong field and weak field ligands is considered to be at the point between $N{{H}_{3}}$ and pyridine. We see that the ligands $SC{{N}^{-}}$ and $gly$ are not mentioned in the series. But, we know that $SC{{N}^{-}}$ is a strong field ligand and $gly$ is a weak field one. So according to this, let us figure out which complex has the highest spin:
- Option A only has weak field ligands, so it is a high spin complex.
- Option B has 4 weak field ligands and 2 strong field ligands.
- Option C has 6 strong field ligands
- Option D has 2 weak field ligands and 4 strong field ligands
From all these observations we can state that the complex mentioned in option A i.e. $[Cr{{(gly)}_{3}}]$ has all weak field ligands and hence is the complex with the highest spin.

Note: Although weak field ligands are also present in 2 options given along with strong ligands the effect of only strong field ligands is seen since they show the dominant behaviour.