Question

Which of these complexes has the lowest-energy \[d - d\] transition band in a UV-Vis spectrum?
A. \[{\left[ {CoBr{{\left( {N{H_3}} \right)}_5}} \right]^{3 + }}\]
B. \[{\left[ {Co{H_2}O{{\left( {N{H_3}} \right)}_5}} \right]^{3 + }}\]
C. \[{\left[ {Co\left( {CN - \kappa C} \right){{\left( {N{H_3}} \right)}_5}} \right]^{2 + }}\]
D. \[{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}\]

Answer 1

Hint: The splitting energy of the orbital depends on the type of ligand approaching the central metal atom. Here we will analyse each compound respectively and find which ligand is strong field and which is field ligand. Compounds having higher number of high field ligands which have high splitting energy.

Complete Answer:
When the ligand approaches a central metal atom, then depending on the nature of the ligand the splitting energy can be estimated. When a strong field ligand approaches the central metal atom then the \[d - d\] transition of electrons involves high energy. While when a weak field ligand approaches central metal atom transition energy is lesser than that of strong field ligand. We can arrange the ligands on the basic of its strength in electromagnetic spectrum as:
\[B{r^ - }{\text{ }} \ll {\text{ }}{H_2}O{\text{ }} \prec {\text{ }}N{H_3}{\text{ }} \ll {\text{ }}C{N^ - }\] , increasing order of strength of ligands.
On the basis of above order of strength of ligand we will analyse each compound as:
A. \[{\left[ {CoBr{{\left( {N{H_3}} \right)}_5}} \right]^{3 + }}\]
Here we have one \[B{r^ - }\] and five\[N{H_3}\] ligands. From the above order we can see \[B{r^ - }\] is the weakest ligand among all. Thus its energy is lowest among all.
B. \[{\left[ {Co{H_2}O{{\left( {N{H_3}} \right)}_5}} \right]^{3 + }}\]
Here we have one \[{H_2}O\] and five \[N{H_3}\] ligands. From the above order we can see \[{H_2}O\] is a stronger ligand than \[B{r^ - }\]. Thus its energy is greater than \[{\left[ {CoBr{{\left( {N{H_3}} \right)}_5}} \right]^{3 + }}\].
C. \[{\left[ {Co\left( {CN - \kappa C} \right){{\left( {N{H_3}} \right)}_5}} \right]^{2 + }}\]
Here we have one \[C{N^ - }\] and five \[N{H_3}\] ligands. From the above order we can see \[C{N^ - }\] is the strongest ligand among all. Thus its energy is greatest among all.
D. \[{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}\]
Here we have six \[N{H_3}\] ligands. From the above order we can see \[N{H_3}\] is a stronger ligand than \[B{r^ - }\]. Thus its energy is greater than \[{\left[ {CoBr{{\left( {N{H_3}} \right)}_5}} \right]^{3 + }}\].
Therefore we can say that the complex which has the lowest-energy \[d - d\] transition band in a UV-Vis spectrum is \[{\left[ {CoBr{{\left( {N{H_3}} \right)}_5}} \right]^{3 + }}\].

Hence the correct option is A.

Note:
It must be noted that whenever a strong or high field ligand approaches central metal then its splitting energy is large and pairing of electrons takes place. The splitting energy of the orbital is denoted by symbol \[{\Delta _ \circ }\]. \[C{N^ - }\] is the strongest ligand and \[B{r^ - }\] is the weakest ligand among the above given ligands.