A) \[HgC{{l}_{2}}\]
B) \[{{B}_{2}}{{H}_{6}}\]
C) \[TiC{{l}_{4}}\]
D) \[S{{O}_{2}}\]
Correct Answer: A
Solution :
\[HgC{{l}_{2}}\] show dimerisation. It found in dimer form.
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