Problem 94
Question
Which of the following will have three stereo-isomeric forms? (1) \(\left[\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]\) (2) \(\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\) (3) \(\mathrm{K}_{3}\left[\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2} \mathrm{Cl}_{2}\right]\) (4) \(\left[\mathrm{Co}\left(\mathrm{en}_{2}\right) \mathrm{ClBr}\right]\) (Here, en = ethylene diamine) (a) 1 and 2 (b) 1 and 3 (c) 1 and 4 (d) 3 and 4
Step-by-Step Solution
Verified Answer
The correct answer is (d) 3 and 4.
1Step 1: Understanding Stereoisomerism
Stereoisomerism occurs in compounds that have the same chemical formula but different arrangements in space. For a complex to have stereoisomers, it must contain chiral centers or planes, or possess different spatial arrangements of ligands that lead to non-superimposable mirror images.
2Step 2: Analyzing Option 1
In \([\mathrm{Cr}(\mathrm{NO}_{3})_{3}(\mathrm{NH}_{3})_{3}]\), Chromium can have a coordination number of 6 with octahedral geometry. Both \(\mathrm{NO}_3\) and \(\mathrm{NH}_3\) ligands are monodentate, providing symmetry in case they are coordinated alternately. However, the symmetry and identical nature of \(\mathrm{NO}_3\) ligands typically leads to only one form or two forms considering \(\Lambda\) and \(\Delta\) isomer forms.
3Step 3: Analyzing Option 2
In \([\mathrm{Co} (\mathrm{C}_2 \mathrm{O}_4)_3]^{3-}\), the \(\mathrm{C}_2 \mathrm{O}_4\) is a bidentate ligand and forms a very stable coordination sphere with possible octahedral geometry around Co. There can be enantiomers (\(\Lambda\) and \(\Delta\)), but no positional isomers due to the identical nature of each of the \(\mathrm{C}_2 \mathrm{O}_4\) ligands.
4Step 4: Analyzing Option 3
In \([\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_2 \mathrm{Cl}_2]^{3-}\), the \(\mathrm{C}_2 \mathrm{O}_4\) ligands and \(\mathrm{Cl}\) ligands can arrange themselves in a way to form cis/trans isomers (two isomers), plus additional enantiomers of each (\(\Lambda\) and \(\Delta\) forms), leading to three stereoisomeric forms.
5Step 5: Analyzing Option 4
In [\(\mathrm{Co} (\mathrm{en})_2 \mathrm{ClBr}\)], \(\mathrm{en}\) is a bidentate ligand creating a chiral center around Co in its possible octahedral geometry. The different positions of \(\mathrm{Cl}\) and \(\mathrm{Br}\) allow for \(cis/trans\) isomerism alongside \(\Lambda\) and \(\Delta\) enantiomers, resulting in three stereoisomeric forms.
6Step 6: Conclusion
Choose the options that could give three stereoisomeric forms. According to our analysis, Options 3 \([\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_2 \mathrm{Cl}_2]^{3-}\) and 4 [\(\mathrm{Co} (\mathrm{en})_2 \mathrm{ClBr}\)] can exist in three stereoisomeric forms: cis, trans, and their respective enantiomers.
Key Concepts
Chiral ComplexesOctahedral GeometryBidentate Ligands
Chiral Complexes
Chirality is an important concept in stereochemistry, especially in coordination complexes. Simply put, a chiral complex is one that cannot be superimposed on its mirror image. This is similar to how your left hand cannot perfectly overlay your right hand. Chirality occurs when there are specific arrangements of atoms that make the complex asymmetric. In coordination chemistry, this often involves a metal ion surrounded by ligands arranged in ways that create a "handedness."
- Chiral complexes often have what are known as enantiomers, which are non-superimposable mirror images of each other.
- These enantiomers are denoted by configurations such as \(\Lambda\) and \(\Delta\), which relate to the direction in which the ligands twist around the metal center.
- In the case of the coordination compound \([\mathrm{Co} (\mathrm{en})_2 \mathrm{ClBr}]\), the presence of the bidentate ligand \({\mathrm{en}}\) allows for chirality, as the ligands wrap around the cobalt center in a particular orientation.
Octahedral Geometry
One of the most common geometries found in coordination complexes is the octahedral arrangement. In an octahedral geometry, six ligands symmetrically surround a central metal ion, akin to the shape of two four-sided pyramids joined at their bases.
- The octahedral shape is especially relevant in complexes where the coordination number is six, such as for many cobalt or chromium complexes.
- This geometry allows for a rich array of isomerism, including both geometric isomers (like cis and trans) and optical isomers (\(\Lambda\) and \(\Delta\)).
- Consider the complex \([\mathrm{Co}(\mathrm{C}_2\mathrm{O}_4)_2 \mathrm{Cl}_2]^{3-}\), which can have an octahedral arrangement leading to multiple stereoisomeric possibilities.
Bidentate Ligands
Bidentate ligands play a crucial role in the formation and stability of coordination complexes. They are a special type of ligand that can bind to a central metal ion using two donor atoms. This gives rise to a more stable chelate effect, enhancing the overall stability of the complex.
- An example of a bidentate ligand is oxalate \((\mathrm{C}_2\mathrm{O}_4)\), which coordinates through two oxygen atoms to form rings with the metal center.
- Another common bidentate ligand is ethylene diamine \((\mathrm{en})\), which uses nitrogen atoms to bind, creating five-membered chelate rings with the metal.
- When bidentate ligands are used in complexes like \([\mathrm{Co} (\mathrm{en})_2 \mathrm{ClBr}]\), they have the ability to induce chirality, leading to the presence of stereoisomers in their octahedral geometry.
Other exercises in this chapter
Problem 92
The complex with spin-only magnetic moment of \(4.9\) B.M. is (a) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3+}\) (b) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2}
View solution Problem 93
The number of isomers possible for complex \(\mathrm{K}_{2}[\mathrm{Pd}\) \(\left.\mathrm{Cl} \mathrm{Br}(\mathrm{SCN})_{2}\right]\) is (a) 1 (b) 2 (c) 3 (d) 4
View solution Problem 95
The value of magnetic moment for a complex ion is 1.73 BM. The complex ion is (a) \([\mathrm{Mn}(\mathrm{CN})]^{4}\) (b) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3
View solution Problem 96
The number of unpaired electrons in the complex ion \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is (Atomic number of \(\mathrm{Co}=27\) ) (a) 4 (b) zero (c) 2 (d) 3
View solution