Question
Amongst $$TiF_6^{2 - },CoF_6^{3 - },C{u_2}C{l_2}$$ and $$NiCl_4^{2 - },$$ which are the colourless species? ( atomic number of $$Ti = 22,Co = 27,Ni = 28,Cu = 29,$$ )
A.
$$CoF_6^{3 - }\,{\text{and}}\,NiCl_4^{\,2 - }$$
B.
$$TiF_6^{2 - }\,{\text{and}}\,C{u_2}C{l_2}$$
C.
$$C{u_2}C{l_2}\,{\text{and}}\,NiCl_4^{\,2 - }$$
D.
$$TiF_6^{2 - }\,{\text{and}}\,CoF_6^{3 - }$$
Answer :
$$TiF_6^{2 - }\,{\text{and}}\,C{u_2}C{l_2}$$
Solution :
In $$TiF_6^{2 - }$$ titanium is in + 4 oxidation state. In $$C{u_2}C{l_2},$$ the copper is in + 1 oxidation state. Thus in both cases, transition from one $$d$$ - orbital to other is not possible.
$$\eqalign{
& Ti - 3{d^2}4{s^2} \to T{i^{4 + }} - 3{d^0}4{s^0} \cr
& Cu - 3{d^{10}}4{s^1} \to C{u^{1 + }} - 3{d^{10}}4{s^0} \cr} $$