Question
Which of the following ions will exhibit colour in aqueous solutions?
A.
$$L{a^{3 + }}\left( {Z = 57} \right)$$
B.
$$T{i^{3 + }}\left( {Z = 22} \right)$$
C.
$$L{u^{3 + }}\left( {Z = 71} \right)$$
D.
$$S{c^{3 + }}\left( {Z = 21} \right)$$
Answer :
$$T{i^{3 + }}\left( {Z = 22} \right)$$
Solution :
Key Idea Colour is obtained as a consequence of $$d-d$$ ( or $$f-f$$ ) transition, and for $$d-d$$ ( or $$f-f$$ ) transition,
presence of unpaired electrons is the necessary condition.
Electronic configuration of
$$L{a^{3 + }}\left( {Z = 57} \right) = \left[ {Xe} \right]4{f^0}5{d^0}6{s^0}$$ (no unpaired electron)
$$T{i^{3 + }}\left( {Z = 22} \right) = \left[ {Ar} \right]3{d^1}4{s^0}$$ (one unpaired electron)
$$L{u^{3 + }}\left( {Z = 71} \right) = \left[ {Xe} \right]4{f^{14}}5{d^0}6{s^0}$$ (no unpaired electron)
$$S{c^{3 + }}\left( {Z = 21} \right) = \left[ {Ar} \right]3{d^0}4{s^0}$$ (no unpaired electron)
Hence, due to the presence of unpaired electron in $$T{i^{3 + }},$$ it exhibit colour in aqueous solution.