Find out the oxidation numbers of the underlined elements in the following compounds:

(a) Na₂SO₄ (S)

• Sodium (Na) has an oxidation state of +1 (since it’s a Group 1 element).

• Oxygen (O) has an oxidation state of -2.

• Let the oxidation state of sulfur (S) be $x$.

• The sum of the oxidation states in a neutral compound is zero:

$$2(+1)+x+4(-2)=0$$

2+x−8=0

$$x=+6$$

• So, sulfur has an oxidation state of +6.

(b) AgNO₃ (Ag)

• Nitrogen (N) in NO₃⁻ has an oxidation state of +5 (since oxygen is -2 and there are three oxygens, contributing -6, so nitrogen must balance it by being +5).

• Oxygen (O) has an oxidation state of -2.

• The sum of the oxidation states in NO₃⁻ is -1 (the charge of the nitrate ion).

• Let the oxidation state of silver (Ag) be $x$.

• Since the compound is neutral:

$$x+(+5)+3(-2)=0$$

x+5−6=0

$$x=+1$$

So, silver has an oxidation state of +1.

(c) KMnO₄ (Mn)

• Potassium (K) has an oxidation state of +1.

• Oxygen (O) has an oxidation state of -2.

• Let the oxidation state of manganese (Mn) be $x$.

• The sum of the oxidation states in KMnO₄ is zero

$$(+1)+x+4(-2)=0$$

1+x−8=0

$$x=+7$$

So, manganese has an oxidation state of +7.

(d) K₂Cr₂O₇ (Cr)

• Potassium (K) has an oxidation state of +1.

• Oxygen (O) has an oxidation state of -2.

• Let the oxidation state of chromium (Cr) be $x$.

• The sum of the oxidation states in K₂Cr₂O₇ is zero:

$$2(+1)+2x+7(-2)=0$$

2+2x−14=0

2x−12=0

$$\Rightarrow x=+6$$

So, chromium has an oxidation state of +6.

(e) HNO₃ (N)

• Hydrogen (H) has an oxidation state of +1.

• Oxygen (O) has an oxidation state of -2.

• Let the oxidation state of nitrogen (N) be $x$.

• The sum of the oxidation states in HNO₃ is zero:

$$(+1)+x+3(-2)=0$$

1+x−6=0

$$x=+5$$

So, nitrogen has an oxidation state of +5.