Problem 10 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The number of possible optical isomers for the complexes \(\mathrm{MA}_{2} \mathrm{~B}_{2}\) with \(s p^{3}\) and \(d s p^{2}\) hybridized metal atom, respectively, is: Note: A and \(\mathrm{B}\) are unidentate neutral and unidentate monoanionic ligands, respectively. (a) 0 and 2 (b) 2 and 2 (c) 0 and 0 (d) 0 and 1

Step-by-Step Solution

Answer

(c) 0 and 0

Step 1: Understanding the Complex

We need to find the possible optical isomers for \(\mathrm{MA}_{2}\mathrm{~B}_{2}\) complexes with two different hybridizations: \(sp^3\) and \(dsp^2\). Given \(A\) and \(B\) as unidentate ligands, the spatial arrangement of these ligands around the metal atom determines their optical activity.

Step 2: Analyzing \(sp^3\) Hybridization

For an \(sp^3\) hybridized complex \(\mathrm{MA}_2\mathrm{B}_2\), the geometry is tetrahedral. A tetrahedral complex can have optical isomers if it is asymmetrical. However, in this case, since two identical \(A\) and two identical \(B\) ligands are present, they cause symmetry in the structure. Hence, no optical isomers are possible for \(sp^3\) hybridization.

Step 3: Analyzing \(dsp^2\) Hybridization

For an \(dsp^2\) hybridized complex, the geometry is square planar. Optical isomerism is typically not observed in square planar complexes because they tend to be symmetrical. Therefore, the \(\mathrm{MA}_{2}\mathrm{B}_{2}\) complex with \(dsp^2\) hybridization does not exhibit optical isomerism as well.

Step 4: Conclusion

Based on the above analysis, neither \(sp^3\) nor \(dsp^2\) hybridized \(\mathrm{MA}_{2}\mathrm{B}_{2}\) complexes exhibit optical isomerism. Thus, the number of possible optical isomers is 0 for both hybridizations.

Key Concepts

Hybridization

Hybridization is a critical concept in chemistry that explains how atomic orbitals combine to form new hybrid orbitals. This process affects the geometry and properties of molecules. It allows atoms to form more bonds than usual. For the complex \((\mathrm{MA}_2\mathrm{B}_2)\), we explore two types of hybridization: \(sp^3\) and \(dsp^2\).In \(sp^3\) hybridization, one s and three p orbitals mix to create four equivalent hybrid orbitals. This type of hybridization is associated with a tetrahedral geometry. For the complex discussed, where \((A\) and \(B\) are different ligands, \(sp^3\) hybridization leads to a symmetrical environment because the \(A\) ligands are paired, and so are the \(B\) ligands.Furthermore, \(dsp^2\) hybridization involves one d, one s, and two p orbitals. This forms four hybrid orbitals that align in a square planar arrangement. These orbitals again lead to a symmetrical configuration, especially in \(\mathrm{MA}_2\mathrm{B}_2\) type complexes, diminishing any chance of optical isomerism due to symmetry.Understanding hybridization provides a clear view of the molecular geometry and potential for isomerism in metal complexes.

Tetrahedral Geometry

Tetrahedral geometry is one of the simplest and most important geometries in chemical coordination. When a central atom undergoes \(sp^3\) hybridization, it forms a shape where the atoms or groups bonded to it are placed at the corners of a tetrahedron.This geometry is characterized by:

• Four bonds of equal length.
• Bond angles of approximately 109.5 degrees.

For the \(\mathrm{MA}_2\mathrm{B}_2\) complex, the ligands in a tetrahedral arrangement can differ, but in this specific case, two pairs of identical ligands (While tetrahedral molecules can exhibit optical isomerism if there is no center of symmetry, the presence of two identical pairs (Thus, when considering optical isomerism in the \(sp^3\) hybridized \(\mathrm{MA}_2\mathrm{B}_2\) complex, the symmetrical arrangement due to paired ligands \(A\) and \(B\) prevents any optical activity.

Square Planar Geometry

Square planar geometry typically arises in \(dsp^2\) hybridized complexes, which commonly occurs with transition metals. Here, the ligands arrange themselves in the same plane, forming a square around the central metal atom.This geometry is noted for:

• All ligands lying in the same plane.
• 90-degree bond angles between adjacent ligands.

In the \(\mathrm{MA}_2\mathrm{B}_2\) complex with \(dsp^2\) hybridization, the square planar form is symmetrical. Each pair of identical ligands (Such symmetry in a square planar arrangement usually eliminates the possibility of optical isomerism. Because optical isomers are non-superimposable mirror images, the symmetrical nature of square planar geometry does not allow for such isomers.Therefore, understanding the square planar structure helps in recognizing why certain complexes do not exhibit optical isomerism, furthering comprehension of molecular symmetry and its impact on isomerism.