A) \[{{H}_{2}}S\]and\[S{{O}_{2}}\]
B) \[N{{H}_{3}}\]and\[N{{O}_{2}}\]
C) \[N{{a}_{2}}S\]and\[N{{a}_{2}}O\]
D) \[{{N}_{2}}O\]and\[NO\]
Correct Answer: D
Solution :
Key Idea: According to law of multiple proportions when two elements combine to form more than one compound, the fixed amount of one element combining with other element have simple whole number ratio. [a] \[{{H}_{2}}S\] and \[S{{O}_{2}}\] are not example of law of multiple proportion because elements are different. [b] \[N{{H}_{3}}\] and \[N{{O}_{2}}\] are not examples of law of multiple proportion because elements are different. [c] \[N{{a}_{2}}S\] and \[N{{a}_{2}}O\] are not example of law of multiple proportion because elements are different. [d] \[{{N}_{2}}O\] and \[NO\] here \[16\] parts of oxygen react with \[14\] and \[28\] parts of nitrogen. \[14:28\] is \[1:2\] which is simple whole number ratio. \[\therefore \]It follows law of multiple proportion.
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