Question

Which of the following is a colorless complex ion?

• A. $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$
• B. $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$
• C. $[Fe{F}_6{]}^{3-}$
• D. $\left[Mn\right(H_{2}O{)}_6{]}^{3+}$

Answer 1

The correct option is C $[Fe{F}_6{]}^{3-}$

Generally the symmetrical complexes having $d^{10}$ elctronic configuration or high spin $d^5$ complexes are colourless.
(a) In $\left[Cu\right(N{H}_3{)}_4{]}^{2+}$, the electronic configuration of : $C{u}^{2+}=t_{2g}^6{e}_g^3$. So it will show colour.

(b) In $\left[Cr\right(H_{2}O{)}_6{]}^{3+}$, the electronic configuration of
$C{r}^{3+}=t_{2g}^3{e}_g^0$. So it will also show colour.

(c) In $[Fe{F}_6{]}^{3-}$ electronic configuration of $F{e}^{3+}=t_{2g}^3{e}_g^2$
Hence it will form colourless complex.
(d) In $\left[Mn\right(H_{2}O{)}_6{]}^{3+}$ the electronic configuration of $M{n}^{3+}=t_{2g}^3{e}_g^1$. So this complex is expected to be coloured.