A) \[Z{{n}^{2+}}\]
B) \[C{{r}^{3+}}\]
C) \[{{V}^{2+}}\]
D) \[T{{i}^{3+}}\]
Correct Answer: A
Solution :
Key Idea The colour of the ion depends on the presence of unpaired electrons. In the absence of upaired electrons ion becomes colourless. \[Z{{n}^{2+}}\] Zn(30) : \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{2}}\] \[Z{{n}^{2+}}\] : \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}\] There is no unpaired electron in \[Z{{n}^{2+}}\]. So, it is colourless. \[C{{r}^{3+}}\] Cr(24) : \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{5}},4{{s}^{1}}\] \[C{{r}^{3+}}\] : \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}\] There are three unpaired electron in \[C{{r}^{3+}}\]hence, it is deep green in colour. \[{{V}^{2+}}\] V(23) : \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{3}},4{{s}^{2}}\] \[{{V}^{2+}}\] : \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}\] Coloured due to presence of three unpaired electrons. \[T{{i}^{3+}}\] Ti(22) : \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{2}},4{{s}^{2}}\] \[T{{i}^{3+}}\] : \[1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}\] Coloured due to presence of one unpaired electron.
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