A) \[{{N}_{3}}H\]
B) \[{{H}_{2}}O\]
C) \[N{{O}_{2}}\]
D) \[C{{O}_{2}}\]
Correct Answer: D
Solution :
Total no. of valence electrons = 5 x 2 + 6 = 16 Similar in \[C{{O}_{2}}=4+6\times 2=16\] Structure of \[{{N}_{2}}O\] \[N\equiv N:\to O\] \[C{{O}_{2}}\] \[O=C=O\]
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