A) \[Cu{{F}_{2}}\]
B) \[Cul\]
C) \[NaCl\]
D) \[MgC{{l}_{2}}\]
Correct Answer: A
Solution :
Key Idea: The d block elements form coloured compounds. These compounds have ions with unpaired electron in d- subshell. (i) Na and Mg belong to s-block, so NaCI and\[MgC{{l}_{2}}\] are colourless compounds. (ii) \[Cu{{F}_{2}}\] oxidation state of Cu in \[Cu\,\,{{F}_{2}}\] is + 2 \[C{{u}^{2+}},=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{0}},3{{d}^{9}}\] \[\therefore \] \[Cu{{F}_{2}}\] in which Cu has one unpaired electron is coloured. (iii) CuI Oxidation state of Cu in \[Cul=+1\] \[C{{u}^{+}}-1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{0}},3{{d}^{10}}\] It has no unpaired electron. So Cul is colourless. \[\therefore \] Only \[Cu{{F}_{2}}\] is coloured among given choices.You need to login to perform this action.
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