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Question: Amongst‌ ‌‌\(TiC{l_6}^{2‌ ‌-‌ ‌},Co{F_6}^{3‌ ‌-‌ ‌},C{u_2}C{l_2}\&‌ ‌NiC{l_4}^{2‌ ‌-‌ ‌}\)‌‌ ‌the‌ ‌...

Amongst‌ ‌‌TiCl62‌‌‌‌,CoF63‌‌‌‌,Cu2Cl2&‌‌NiCl42‌‌‌‌TiC{l_6}^{2‌ ‌-‌ ‌},Co{F_6}^{3‌ ‌-‌ ‌},C{u_2}C{l_2}\&‌ ‌NiC{l_4}^{2‌ ‌-‌ ‌}‌‌ ‌the‌ ‌colorless‌ ‌
species‌ ‌are:‌ ‌ ‌
A.CoF63‌‌‌‌&‌‌NiCl42‌‌‌‌Co{F_6}^{3‌ ‌-‌ ‌}\&‌ ‌NiC{l_4}^{2‌ ‌-‌ ‌}‌ ‌
B.TiCl62‌‌‌‌&‌‌CoF63‌‌‌‌TiC{l_6}^{2‌ ‌-‌ ‌}\&‌ ‌Co{F_6}^{3‌ ‌-‌ ‌}‌ ‌
C.Cu2Cl2&‌‌NiCl42‌‌‌‌C{u_2}C{l_2}\&‌ ‌NiC{l_4}^{2‌ ‌-‌ ‌}‌ ‌
D.TiCl62‌‌‌‌&‌‌Cu2Cl2TiC{l_6}^{2‌ ‌-‌ ‌}\&‌ ‌C{u_2}C{l_2}

Explanation

Solution

We know that colored species are those species which show ddd - d transitions and have unpaired electrons. So, the colorless species will be those which do not have these conditions present in the compound. So, we will determine this by writing the electronic configuration of the central atom of each compound.

Complete step-by-step answer:
(A)TiCl62TiC{l_6}^{2 - }
Here we see that TiTi is present in +4 + 4 oxidation state
So, its electronic configuration will be:
Electronic configuration of Ti+4T{i^{ + 4}} =1s22s22p63s23p63d04s0 = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^0}4{s^0} or we can simply write it as [Ar]3d04s0[Ar]3{d^0}4{s^0}
Here zero electrons are present in the dd - orbital so there will be no ddd - d transition taking place hence this compound will thus remain colorless.

(B)CoF63Co{F_6}^{3 - }
Here Cobalt is present in +3 + 3 state so its electronic configuration will be:
Electronic configuration of Co+3=[Ar]3d64s0C{o^{ + 3}} = [Ar]3{d^6}4{s^0}
Here we see that four unpaired electrons are present in the dd - subshell so these electrons will be available for transition by absorption of light in the visible region. So, this will be a colored compound.

(C)Cu2Cl2C{u_2}C{l_2}
Here CuCu is present in +1 + 1 oxidation state so its electronic configuration will be:
Electronic configuration of Cu+=[Ar]3d104s0C{u^ + } = [Ar]3{d^{10}}4{s^0}
We know that dd - orbital is completely filled with electrons and therefore no unpaired electrons will be present for the absorption of light and therefore will make this compound colorless.

(D)NiCl42NiC{l_4}^{2 - }
Here NiNi is present in +2 + 2 oxidation state so its electronic configuration will be:
Electronic configuration of Ni2+=[Ar]3d84s0N{i^{2 + }} = [Ar]3{d^8}4{s^0}
Here as two unpaired electrons are present in the dd - orbital, so the unpaired electrons will be present for the absorption of light which will make the compound colored.

Hence the correct answer is Option (D) i.e., TiC{l_6}^{2 - }\ & C{u_2}C{l_2}.

Note: The principle which is used in writing the electronic configuration is the Aufbau’s principle. In this the electrons are filled into the atomic orbitals in their increasing order of the orbital energy which means that electrons with the lowest energy levels will be filled before the higher energy levels in the available atomic orbitals.