Question

the oxidation number of n in nano₃ is
a 2+
b 4+
c 6+
d 5+
e 3+
f 3-
question 4
for the chlorate ion, clo₃⁻, what are the oxidation states of the cl and o, respectively?
a +6, -2
b +5, -2

Explanation:

Step1: Determine the oxidation number of Na and O

In \(NaNO_3\), the oxidation number of \(Na\) is \(+ 1\) (since it is in group 1 of the periodic table), and the oxidation number of \(O\) is \(-2\) (a common oxidation state for oxygen in compounds).

Step2: Set up the equation for the sum of oxidation numbers

Let the oxidation number of \(N\) be \(x\). For a neutral compound \(NaNO_3\), the sum of oxidation numbers is \(0\). So, \((+1)+x + 3\times(-2)=0\).

Step3: Solve the equation for \(x\)

$$
LATEXBLOCK0
$$

For \(ClO_3^{-}\):

Step1: Determine the oxidation number of O

The oxidation number of \(O\) is \(-2\).

Step2: Set up the equation for the sum of oxidation numbers

Let the oxidation number of \(Cl\) be \(y\). For the ion \(ClO_3^{-}\), the sum of oxidation numbers is \(-1\). So, \(y+3\times(-2)=-1\).

Step3: Solve the equation for \(y\)

$$
LATEXBLOCK1
$$

Answer:

For \(NaNO_3\), D. \(5+\)
For \(ClO_3^{-}\), B. \(+5,-2\)