Question

The complex having the lowest molar extinction coefficient is

• A. Na$_2$[NiBr$_4$]
• B. [Cr(NH$_3$)$_6$]SO$_4$
• C. Na$_2$[CoCl$_4$]
• D. [Mn(H$_2$O)$_6$]SO$_4$

Hint:

Spin-forbidden + Laporte-forbidden transitions give extremely low intensity, hence lowest molar extinction coefficient

Answer 1

The Correct Option is D

Step 1: Understand molar extinction coefficient.
Molar extinction coefficient ($\epsilon$) depends on the intensity of electronic transitionsForbidden transitions have very low $\epsilon$ values

Step 2: Identify type of complexes.
Octahedral complexes usually show Laporte-forbidden d–d transitions, resulting in low intensityTetrahedral complexes do not have a center of symmetry, so their transitions are more allowed and intense

Step 3: Analyze Na$_2$[NiBr$_4$] and Na$_2$[CoCl$_4$].
Both are tetrahedral complexesThus, they show relatively high molar extinction coefficients

Step 4: Analyze [Cr(NH$_3$)$_6$]SO$_4$.
This is an octahedral complex with center of symmetryHence, d–d transitions are Laporte forbidden and weak, giving low $\epsilon$

Step 5: Analyze [Mn(H$_2$O)$_6$]SO$_4$.
Mn$^{2+}$ is $d^5$ (high spin)All transitions are both spin-forbidden and Laporte-forbiddenThus, intensity is extremely low

Step 6: Compare all options.
Among all, [Mn(H$_2$O)$_6$]SO$_4$ has the most forbidden transitions, hence lowest $\epsilon$

Step 7: Conclusion.
\[ \boxed{\text{[Mn(H}_2\text{O)}_6]\text{SO}_4} \]