Question

Which of the following ions is expected to be coloured?

• A. ${\left[Ti{\left(H_{2}O\right)}_6\right]}^{3+}$
• B. ${\left[Sc{\left(H_{2}O\right)}_6\right]}^{3+}$
• C. ${\left[Zn{\left(H_{2}O\right)}_6\right]}^{2+}$
• D. ${\left[Ti{\left(H_{2}O\right)}_6\right]}^{4+}$

Answer 1

Answer: (A)

${\left[Ti{\left(H_{2}O\right)}_6\right]}^{3+}$

A) Ti has the electronic configuration $\left[Ar\right]3d^2$. In this complex, it is present as $T{i}^{3+}$ which has one unpaired electron in 3d orbital. It undergoes $d^{2}s{p}^3$ (octahedral) hybridization to accommodate 6 pairs of water.
So, the complex is octahedral in shape, will be coloured and paramagnetic due to the presence of one unpaired electron.
B) Sc has the electronic configuration $\left[Ar\right]3d^1$. Since, it is present in the complex as $S{c}^{3+}$, it does not have any unpaired electron. It undergoes $d^{2}s{p}^3$ hybridization to accommodate 6 pairs of electrons from water. So, it is octahedral in shape, colourless and diamagnetic due to the absence of unpaired electrons.
C) In this complex, $Zn$ is present as $Z{n}^{2+}$ and there is no unpaired electron in the complex. So, it is colourless.
D) In this complex, $Ti$ is present as $T{i}^0$ and there is no unpaired electron in the complex. So, it is colourless.