question_answer

In\[TiF_{6}^{2-},CoF_{6}^{3-},C{{u}_{2}}C{{l}_{2}}\]and\[NiCl_{4}^{2-}\]the colourless species is (Atomic number Ti = 22, Co = 27, Cu = 29, Ni = 28)

A)  \[TiF_{6}^{2-}\]and\[C{{u}_{2}}C{{l}_{2}}\]

B)  \[C{{u}_{2}}C{{l}_{2}}\]and\[NiCl_{4}^{2-}\]

C)  \[TiF_{6}^{2-}\]and\[CoF_{6}^{3-}\]

D)  \[CoF_{6}^{3-}\]and\[NiCl_{4}^{2-}\]

Correct Answer: A

Solution :

 \[TiF_{6}^{2-},CoF_{6}^{3-},C{{u}_{2}}C{{l}_{2}}\] and\[NiCl_{4}^{2-}\] give \[T{{i}^{4+}},C{{o}^{3+}},C{{u}^{+}}\]and\[N{{i}^{2+}}\]ions respectively.

\[T{{i}^{4+}}=3{{d}^{0}},{{45}^{0}}\]
\[C{{o}^{3+}}=3{{d}^{6}},{{45}^{0}}\]
\[C{{u}^{+}}=3{{d}^{10}},{{45}^{0}}\]
\[N{{i}^{2+}}=3{{d}^{8}},{{45}^{0}}\]

\[TiF_{6}^{2-}\]and\[C{{u}_{2}}C{{l}_{2}}\]species are colourless due to the absence to unparied electron in\[T{{i}^{4+}}\]and \[C{{u}^{+}}\]ions.