Assign oxidation number to the underlined elements in each of the following species: (a) NaH2P̲O4 (b) NaHS̲O4

Exercises · 7.1

Q. Assign oxidation number to the underlined elements in each of the following species:

(a) NaH2P̲O4

(b) NaHS̲O4

(d) K2M̲n̲O4

(e) CaO̲2

(f) NaB̲H4

(g) H2S̲2O7

(h) KAl(S̲O4)2.12 H2O

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Concept understanding — Oxidation Reduction

Let’s start with something you already know from everyday life.

The intuition: what does “oxidation” really mean?

Think of a piece of iron left out in the rain. Over time, it turns into reddish-brown rust. Or think of a slice of apple turning brown when you leave it on the table. Or a fire burning wood to ash. In all these cases, something is combining with oxygen — that’s the original meaning of “oxidation.” The iron combines with oxygen from the air to form iron oxide (rust). The apple’s chemicals react with oxygen in the air. The wood burns because carbon in the wood combines with oxygen.

So the first, simplest idea: oxidation = adding oxygen. And the reverse — taking oxygen away — was called reduction. For example, if you heat iron oxide with carbon, the carbon steals the oxygen away, leaving pure iron. That’s reduction: removing oxygen.

But chemists soon realised this was too narrow. Many reactions that look like oxidation-reduction don’t involve oxygen at all. For instance, when sodium metal reacts with chlorine gas to make table salt, no oxygen is involved — yet the sodium clearly “rusts” in a sense, and the chlorine “steals” something from it.

So the definition had to be broadened.

The precise modern definition: electron transfer

Here’s the clean, exam-ready statement:

Oxidation is the loss of electrons by a substance.

Reduction is the gain of electrons by a substance.

They always happen together — you cannot have one without the other. That’s why we call them redox reactions (short for reduction-oxidation).

Let’s see this with the sodium-chlorine example:

Sodium atom ( $\text{Na}$ ) loses one electron to become $\text{Na}^+$ . That’s oxidation.
Chlorine atom ( $\text{Cl}$ ) gains that electron to become $\text{Cl}^-$ . That’s reduction.

You can write the two halves separately:

\text{Na} \rightarrow \text{Na}^+ + e^- \quad (\text{oxidation})

\text{Cl} + e^- \rightarrow \text{Cl}^- \quad (\text{reduction})

Add them together:

\text{Na} + \text{Cl} \rightarrow \text{Na}^+ + \text{Cl}^-

That’s table salt.

Tip

A handy mnemonic: OIL RIG — Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

How to spot a redox reaction without seeing electrons

You can’t watch electrons move directly. So chemists use oxidation numbers (also called oxidation states) — a bookkeeping system that tracks electrons.

Rules (simplified for first-time learners):

An atom in its elemental form has oxidation number 0.
A monatomic ion has oxidation number equal to its charge (e.g., $\text{Na}^+$ is +1, $\text{Cl}^-$ is -1).
Oxygen is usually -2 (except in peroxides).
Hydrogen is usually +1 (except in metal hydrides).
The sum of oxidation numbers in a neutral compound is 0; in a polyatomic ion, it equals the ion’s charge.

Then:

Oxidation = increase in oxidation number.
Reduction = decrease in oxidation number.

Example: Rusting of iron.

4\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3

Fe starts at 0 (elemental). In $\text{Fe}_2\text{O}_3$ , each Fe is +3. So Fe’s oxidation number goes up from 0 to +3 → oxidation.
O starts at 0 (in $\text{O}_2$ ). In $\text{Fe}_2\text{O}_3$ , each O is -2. So O’s oxidation number goes down from 0 to -2 → reduction.

Watch out

A common mistake: thinking that “reduction” means something becomes smaller or less. It doesn’t — it’s about gaining electrons (or losing oxygen, in the old sense). The name comes from metallurgy: when you “reduce” iron ore to iron, you’re taking away oxygen, so the mass reduces.

One more way to think about it

In any redox reaction, one substance donates electrons (the reducing agent) and another accepts them (the oxidising agent).

The reducing agent gets oxidised (it loses electrons).
The oxidising agent gets reduced (it gains electrons).

This sounds backwards at first, but it’s consistent: the agent does something to the other substance. The reducing agent reduces the other substance (by giving it electrons), so the reducing agent itself gets oxidised.

Important

Reducing agent → gets oxidised (loses electrons).

Oxidising agent → gets reduced (gains electrons).

Summary for your notes

Concept	Old definition (oxygen)	Modern definition (electrons)
Oxidation	Gain of oxygen	Loss of electrons
Reduction	Loss of oxygen	Gain of electrons
Redox reaction	Both happen together	Both happen together

Final takeaway: Whenever you see a reaction where elements change their oxidation numbers, you’re looking at a redox reaction. The substance whose oxidation number increases is oxidised (loses electrons). The one whose oxidation number decreases is reduced (gains electrons). And they always come as a pair.

A quick search for "Oxidation Reduction class 11 chemistry" or "NCERT chemistry syllabus oxidation reduction" will confirm what's true here: this concept is a standard, curriculum-aligned part of Class 11 Chemistry. Given how often it's tested in JEE Main, NEET and state CET Chemistry papers, it's worth revisiting this explanation until the reasoning feels automatic, not just the final formula.

Oxidation numbers are assigned using a fixed set of rules (known oxidation states for common ions, neutral molecule sum = 0, polyatomic ion sum = charge). The answers are: (a) P = +5,

(b) S = +6,

(d) Mn = +6, (e) O = –1, (f) B = +3, (g) S = +6, (h) S = +6.

The core idea is simple: oxidation numbers are a bookkeeping tool. They don't represent real charges in covalent compounds, but they let us track electron movement in redox reactions. The rules are hierarchical — you apply them in order, and the most electronegative element usually gets the negative number.

Let’s go through each species one by one.

1. NaH₂PO₄ (sodium dihydrogen phosphate)

Step 1: Sodium (Na) is always +1 in its compounds. Hydrogen (H) is usually +1, except in metal hydrides. Oxygen (O) is almost always –2. The molecule is neutral, so the sum of all oxidation numbers = 0.

Step 2: Let the oxidation number of phosphorus (P) be $x$ .

We have: 1 Na (+1) + 2 H (+1 each) + 1 P ( $x$ ) + 4 O (–2 each) = 0

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

Simplify: $+1 + 2 + x - 8 = 0$ → $x - 5 = 0$ → $x = +5$

Tip

In oxyacids of phosphorus, the oxidation state of P is often +5 (as in H₃PO₄). Here, NaH₂PO₄ is just a salt of that acid.

Answer for (a): P = +5

2. NaHSO₄ (sodium bisulphate)

Step 1: Na = +1, H = +1, O = –2. Neutral molecule.

Step 2: Let S = $x$ .

$+1 + 1 + x + 4(-2) = 0$ → $2 + x - 8 = 0$ → $x = +6$

Watch out

A common mistake is to treat HSO₄⁻ as having S = +4 because of the "bisulphite" confusion. But in bisulphate (HSO₄⁻), sulphur is in its highest common oxidation state, +6.

Answer for (b): S = +6

3. H₄P₂O₇ (pyrophosphoric acid)

Step 1: H = +1, O = –2. Neutral molecule.

Step 2: Let P = $x$ . There are 2 P atoms.

$4(+1) + 2x + 7(-2) = 0$ → $4 + 2x - 14 = 0$ → $2x - 10 = 0$ → $x = +5$

Answer for (c): P = +5

4. K₂MnO₄ (potassium manganate)

Step 1: K = +1, O = –2. Neutral molecule.

Step 2: Let Mn = $x$ .

$2(+1) + x + 4(-2) = 0$ → $2 + x - 8 = 0$ → $x = +6$

Note

Don’t confuse this with KMnO₄ (permanganate), where Mn is +7. In manganate, Mn is +6 — a green ion.

Answer for (d): Mn = +6

5. CaO₂ (calcium peroxide)

Step 1: Ca is an alkaline earth metal, always +2. The molecule is neutral.

Step 2: Let O = $x$ . There are 2 O atoms.

$+2 + 2x = 0$ → $2x = -2$ → $x = -1$

Watch out

This is a peroxide! Oxygen in peroxides (like H₂O₂, Na₂O₂) has oxidation number –1, not –2. Many students default to –2 and get Ca = +4, which is impossible for calcium.

Answer for (e): O = –1

6. NaBH₄ (sodium borohydride)

Step 1: Na = +1. Here, hydrogen is bonded to boron, which is less electronegative than hydrogen. So H takes –1 (metal hydride rule). Neutral molecule.

Step 2: Let B = $x$ .

$+1 + x + 4(-1) = 0$ → $1 + x - 4 = 0$ → $x = +3$

Tip

In borohydride, boron is in +3 state, and each hydrogen is hydridic (–1). This is a common reducing agent.

Answer for (f): B = +3

7. H₂S₂O₇ (pyrosulphuric acid / disulphuric acid)

Step 1: H = +1, O = –2. Neutral molecule.

Step 2: Let S = $x$ . There are 2 S atoms.

$2(+1) + 2x + 7(-2) = 0$ → $2 + 2x - 14 = 0$ → $2x - 12 = 0$ → $x = +6$

Answer for (g): S = +6

8. KAl(SO₄)₂·12H₂O (potassium alum)

Step 1: This is a hydrated salt. The water molecules are neutral, so they contribute 0 to the overall oxidation sum. We only need to consider the anhydrous part: KAl(SO₄)₂.

Step 2: K = +1, Al = +3 (always +3 in compounds). O = –2. Let S = $x$ . There are 2 sulphate ions, each with 1 S and 4 O.

For the whole formula unit:

$+1 + 3 + 2[x + 4(-2)] = 0$ → $4 + 2(x - 8) = 0$ → $4 + 2x - 16 = 0$ → $2x - 12 = 0$ → $x = +6$

Alternatively, treat each SO₄²⁻ ion separately: charge on sulphate is –2, so $x + 4(-2) = -2$ → $x - 8 = -2$ → $x = +6$ .

Note

The 12 water molecules are just spectators — they don't affect the oxidation numbers of the atoms in the salt.

Answer for (h): S = +6

✓Final answer

The oxidation numbers are: (a) P = +5, (b) S = +6, (c) P = +5, (d) Mn = +6, (e) O = -1, (f) B = +3, (g) S = +6, (h) S = +6.

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Q.Assign oxidation number to the underlined elements in each of the following species:

1. NaH₂PO₄ (sodium dihydrogen phosphate)

2. NaHSO₄ (sodium bisulphate)

3. H₄P₂O₇ (pyrophosphoric acid)

4. K₂MnO₄ (potassium manganate)

5. CaO₂ (calcium peroxide)

6. NaBH₄ (sodium borohydride)

7. H₂S₂O₇ (pyrosulphuric acid / disulphuric acid)

8. KAl(SO₄)₂·12H₂O (potassium alum)

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