Question
Which of the following shows nitrogen in its increasing order of oxidation number?
A.
$${N_2}O < NO < N{O_2} < NO_3^ - < NH_4^ + $$
B.
$$NH_4^ + < {N_2}O < NO < N{O_2} < NO_3^ - $$
C.
$$NH_4^ + < {N_2}O < N{O_2} < NO_3^ - < NO$$
D.
$$NH_4^ + < NO < {N_2}O < N{O_2} < NO_3^ - $$
Answer :
$$NH_4^ + < {N_2}O < NO < N{O_2} < NO_3^ - $$
Solution :
$${N_2}O = + 1,NO = + 2,N{O_2} = + 4,$$ $$NO_3^ - = + 5,NH_4^ + = - 3$$
Increasing order of oxidation state will be
$$NH_4^ + < {N_2}O < NO < N{O_2} < NO_3^ - $$