Question

Identify the optically active compounds from the following:

• A. $cis–\left[Co\right(en{)}_{2}C{l}_2{]}^{+}$
  
• B. $trans–\left[Co\right(en{)}_{2}C{l}_2{]}^{+}$
  
• C. $\left[Co\right(en{)}_3{]}^{3+}$
• D. $\left[Cr\right(N{H}_3{)}_{5}Cl]$

Answer 1

Answer: (A), (C)

$\left[Co\right(en{)}_3{]}^{3+}$

Optical isomers:

Optically active compounds rotate the plane polarized light, one of the isomers rotates the light in one direction and the other rotates in the opposite direction.

For a compound to be optically active it should have absence of plane of symmetry$\left(POS\right)$ and center of symmetry
$\left(COS\right)$.

Analysing the options

Option (A)

$\left[Co\right(en{)}_3{]}^{3+}$



It has no element of symmetry (POS & COS), so it is optically active.

Option (B)

$trans–\left[Co\right(en{)}_{2}C{l}_2{]}^{+}$


Since the given compound has two elements of symmetry, i.e. plane of symmetry and centre of symmetry, so it is optically inactive compound.

Option (C)

$cis–\left[Co\right(en{)}_{2}C{l}_2{]}^{+}$



It has no element of symmetry (POS & COS), so it is also optically active.

Option (D)

The compound $\left[Cr\right(N{H}_3{)}_{5}Cl{]}^{2+}$ does not exhibit optical isomerism. This is due to presence of POS (plane of symmetry), hence it is optically inactive compound .


Hence, the correct options are (A) and (C).