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: What lowers the extinction coefficient.
A transition that is both spin forbidden and Laporte forbidden is very weak.

Step 2: Examine the complexes.
The tetrahedral cobalt and nickel complexes have allowed, intense bands. The octahedral chromium complex is only Laporte forbidden.

Step 3: Pick the weakest.
[Mn(H$_2$O)$_6$]$^{2+}$ is high spin d$^5$, so every d to d band is spin and Laporte forbidden, giving the lowest extinction coefficient.

Step 4: Answer.
\[ \boxed{[\text{Mn(H}_2\text{O)}_6]\text{SO}_4} \]