A) \[{{K}_{2}}Cr{{O}_{4}}\]
B) \[Cr{{O}_{2}}\]
C) \[CrC{{l}_{3}}\]
D) \[Cr{{F}_{2}}\]
Correct Answer: A
Solution :
Oxidation number of \[Cr\]in options a, b, c and d are +6, + 4, + 3, + 3 respectively. In given options a has high oxidation number therefore its radii will be low. Atomic radii decreases with increase in oxidation no.
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