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Which of the following ions is expected to be colourless - $Ti^{4+}, Cu^{2+}, V^{3+}$ and $Fe^{2+}$?

$\begin{array}{1 1}Only \;Ti^{4+}\\ Cu^{2+}\; and \;Fe^{2+} \\ Only \;V^{3+} \\ Ti^{4+}\; and \;V^{3+} \end{array}$

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Answer: Only $Ti^{4+}$
The electronic configurations of $Ti^{4+}, Cu^{2+}, V^{3+}$ and $Fe^{2+}$ are $3d^0, 3d^9, 3d^1$ and $3d^5$ respectively.
There are no unpaired electrons in the case of $Ti^{4+}$. Hence it is colourless.
answered Jan 30, 2014 by sreemathi.v
edited Aug 5, 2014 by balaji.thirumalai

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