The oxidation state of nitrogen in Na-NO-3- is -3-5-4-5

Question

Toppr Admin · Accepted Answer

The overall charge is zero -because it is not an ion- Since sodium is a 1A family member- the charge is -1- The charge of oxygen is almost always -2-Hence- 1-OS-xA0-of-xA0-N-3-xD7-x2212-2-0So- the oxidation of nitrogen in that problem is -5-

The oxidation state of nitrogen in $N a {N O}_{3}$ is :
$+ 3$
$+ 5$
$- 4$
$- 5$

The overall charge is zero (because it is not an ion). Since sodium is a 1A family member, the charge is +1. The charge of oxygen is almost always -2.
Hence, $1 + O S o f N + (3 \times - 2) = 0$
So, the oxidation of nitrogen in that problem is +5.

The oxidation state of nitrogen in NaNO3 is :+3+5−4−5

The overall charge is zero (because it is not an ion). Since sodium is a 1A family member, the charge is +1. The charge of oxygen is almost always -2.Hence, 1+OS of N+(3×−2)=0So, the oxidation of nitrogen in that problem is +5.

The oxidation state of nitrogen in $N a {N O}_{3}$ is :
$+ 3$
$+ 5$
$- 4$
$- 5$

The overall charge is zero (because it is not an ion). Since sodium is a 1A family member, the charge is +1. The charge of oxygen is almost always -2.
Hence, $1 + O S o f N + (3 \times - 2) = 0$
So, the oxidation of nitrogen in that problem is +5.