A) \[Cu{{F}_{2}}\]
B) \[CuI\]
C) \[NaCI\]
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 \[\text{NaCl}\]and \[\text{MgC}{{\text{l}}_{\text{2}}}\]are colourless compounds. (ii) \[Cu{{F}_{2}}\] oxidation state of Cu in \[Cu\,{{F}_{2}}\]is \[+\,2\] \[C{{u}^{2+}},=1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{o}},3{{d}^{9}}\] \[\therefore \]\[Cu{{F}_{2}}\] in which Cu has one unpaired electron is coloured. (iii) \[CuI\] Oxidation state of Cu in\[CuI=+1\] \[C{{u}^{+}}-1{{s}^{2}},2{{s}^{2}},2{{p}^{6}},3{{s}^{2}},3{{p}^{6}},4{{s}^{o}},3{{d}^{10}}\] It has no unpaired electron. So \[CuI\]is colourless. \[\therefore \] Only \[Cu{{F}_{2}}\]is coloured among given choices.You need to login to perform this action.
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