Question
The correct order of the oxidation states of nitrogen in $$NO,{N_2}O,N{O_2}\,{\text{and}}\,{N_2}{O_3}$$ is:
A.
$$N{O_2} < NO < {N_2}{O_3} < {N_2}O$$
B.
$$N{O_2} < {N_2}{O_3} < NO < {N_2}O$$
C.
$${N_2}O < \,{N_2}{O_3} < NO < N{O_2}$$
D.
$${N_2}O < \,NO < {N_2}{O_3} < N{O_2}$$
Answer :
$${N_2}O < \,NO < {N_2}{O_3} < N{O_2}$$
Solution :
| (oxide) |
(oxidation state) |
| $${N_2}O$$ |
$$ + 1$$ |
| $$NO$$ |
$$ + 2$$ |
| $${N_2}{O_3}$$ |
$$ + 3$$ |
| $$N{O_2}$$ |
$$ + 4$$ |
So, $${N_2}O < NO < {N_2}{O_3} < N{O_2}$$