A) \[N{{O}_{2}}\]
B) \[NO\]
C) \[{{N}_{2}}{{O}_{5}}\]
D) \[{{N}_{2}}O\]
Correct Answer: A
Solution :
When lead nitrate is heated it gives lead oxide and nitrogen per oxide. \[2Pb{{(N{{O}_{3}})}_{2}}\xrightarrow{\Delta }2PbO+\underset{nitrogen\,per\,oxide}{\mathop{4N{{O}_{2}}+{{O}_{2}}}}\,\]
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