Question

The complex that can show optical activity

• A. cis– ${\left[{\mathrm{CrCl}}_2{\left(\mathrm{ox}\right)}_2\right]}^{3-}$(ox – oxalate)
• B. trans –${\left[\mathrm{Fe}{\left({\mathrm{NH}}_3\right)}_2{\left(\mathrm{CN}\right)}_4\right]}^{-}$
• C. trans –${\left[\mathrm{Cr}\left({\mathrm{Cl}}_2\right){\left(\mathrm{ox}\right)}_2\right]}^{-3}$
• D. cis– ${\left[\mathrm{Fe}{\left({\mathrm{NH}}_3\right)}_2{\left(\mathrm{CN}\right)}_4\right]}^{-}$

Answer 1

The correct option is A

cis– ${\left[{\mathrm{CrCl}}_2{\left(\mathrm{ox}\right)}_2\right]}^{3-}$(ox – oxalate)

The correct option is option A

The explanation for the correct option:

1. The ability of a compound to rotate the plane of polarised light is known as optical activity, also referred to as optical rotation. Compounds with this property are referred to as optically active materials.
2. The three conditions for a coordination complex to be optically active are It should have two ambident ligands, only the cis-isomer exhibits optical activity if it exhibits geometric isomerism, and the mirror image of the chemical must not be super impossible. i.e., Chiral should be used.
3. Also, oof plane of symmetry exists in a compound i.e. an imaginary plane that bisects a molecule into two halves such that they are mirror images of each other then a compound is optically inactive.
4. In cis– ${\left[{\mathrm{CrCl}}_2{\left(\mathrm{ox}\right)}_2\right]}^{3-}$(ox – oxalate) there is no plane of symmetry and also the mirror images are not super impossible.
5. Also, there is no plane of symmetry.
6. Hence, it is optically active.

The explanation for the incorrect option:

Option (B): trans –${\left[\mathrm{Fe}{\left({\mathrm{NH}}_3\right)}_2{\left(\mathrm{CN}\right)}_4\right]}^{-}$

1. In trans –${\left[\mathrm{Fe}{\left({\mathrm{NH}}_3\right)}_2{\left(\mathrm{CN}\right)}_4\right]}^{-}$ the plane of symmetry exists thus it is optically inactive.
2. Hence, it is an incorrect option.

Option (C): trans –${\left[\mathrm{Cr}\left({\mathrm{Cl}}_2\right){\left(\mathrm{ox}\right)}_2\right]}^{-3}$

1. In trans –${\left[\mathrm{Cr}\left({\mathrm{Cl}}_2\right){\left(\mathrm{ox}\right)}_2\right]}^{-3}$ the plane of symmetry exists thus it is optically inactive.
2. Hence, it is an incorrect option.

Option (D): cis– ${\left[\mathrm{Fe}{\left({\mathrm{NH}}_3\right)}_2{\left(\mathrm{CN}\right)}_4\right]}^{-}$

1. In cis– ${\left[\mathrm{Fe}{\left({\mathrm{NH}}_3\right)}_2{\left(\mathrm{CN}\right)}_4\right]}^{-}$the plane of symmetry exists thus it is optically inactive.
2. Hence, it is an incorrect option.

Final answer: The complex that can show optical activity is cis– ${\left[{\mathrm{CrCl}}_2{\left(\mathrm{ox}\right)}_2\right]}^{3-}$(ox – oxalate).