NCERT Solutions for Class 9th Science Chapter 9 .5 Writing Chemical Formulae — Pause and Ponder

Book page 1779 Updated on2026-09-08

Q16.

Name the following: (i) CO2 (ii) NO2 (iii) SF6 (iv) PCl3

Answer

These are all covalent compounds, so they are named with the prefix system: the first element keeps its ordinary name, the second ends in -ide, and prefixes (mono-, di-, tri-, tetra-, penta-, hexa-) give the number of atoms. Mono- is dropped for the first element.

	Formula	Atoms of the second element	Name
(i)	CO₂	2 oxygen → di- + oxide	Carbon dioxide
(ii)	NO₂	2 oxygen → di- + oxide	Nitrogen dioxide
(iii)	SF₆	6 fluorine → hexa- + fluoride	Sulfur hexafluoride
(iv)	PCl₃	3 chlorine → tri- + chloride	Phosphorus trichloride

Why it happens: two different elements can form more than one compound — CO and CO₂, NO and NO₂ and N₂O₄. The prefix is what tells them apart, so it cannot be dropped from the second element. Note also that CO is carbon monoxide, not monooxide: when a prefix ending in ‘o’ or ‘a’ meets an element starting with a vowel, the last vowel of the prefix is dropped.

Did you know? SF₆ is one of the most powerful greenhouse gases known and is used as an insulating gas in high-voltage electrical switchgear. NO₂ is the reddish-brown gas responsible for the haze over busy traffic junctions.

Q17.

Write the formula for the following: (i) Sodium hydrogencarbonate (ii) Sulfur dioxide (iii) Ferric chloride (iv) Cuprous oxide

Answer

Use the criss-cross method with the charges from Table 9.1: write cation first, put the charge numbers below the symbols, swap them as subscripts, then divide by any common factor.

	Compound	Ions / valencies	Criss-cross	Formula
(i)	Sodium hydrogencarbonate	Na⁺ (1), HCO₃⁻ (1)	1 and 1 → both dropped	NaHCO₃
(ii)	Sulfur dioxide	S (valency 4), O (valency 2)	S₂O₄ ÷ 2	SO₂
(iii)	Ferric chloride	Fe³⁺ (3), Cl⁻ (1)	Fe₁Cl₃	FeCl₃
(iv)	Cuprous oxide	Cu⁺ (1), O²⁻ (2)	Cu₂O₁	Cu₂O

Why it happens: the criss-cross is a quick way of making the total positive charge cancel the total negative charge. In FeCl₃, (+3) + 3(−1) = 0; in Cu₂O, 2(+1) + (−2) = 0. Every ionic formula must come out electrically neutral.

Tip: “-ous” always means the lower charge and “-ic” the higher one — cuprous is Cu⁺ and cupric is Cu²⁺; ferrous is Fe²⁺ and ferric is Fe³⁺. Get that wrong and the whole formula changes: cupric oxide is CuO, not Cu₂O.

Q18.

Write the formulae for the compounds formed from the following pairs of ions: (i) Fe3+ and OH‒ (ii) K+ and CO3 2–

Answer

(i) Fe³⁺ and OH⁻ → Fe(OH)₃

Charges: Fe is 3+, OH is 1−
Criss-cross: Fe₁(OH)₃ → Fe(OH)₃ — ferric hydroxide
Charge check: (+3) + 3(−1) = 0 ✓

Brackets are needed because there are three hydroxide ions. Fe(OH)₃ means one iron and three OH groups; FeOH₃ would wrongly read as one O and three H.

(ii) K⁺ and CO₃²⁻ → K₂CO₃

Charges: K is 1+, CO₃ is 2−
Criss-cross: K₂(CO₃)₁ → K₂CO₃ — potassium carbonate
Charge check: 2(+1) + (−2) = 0 ✓

No brackets here, because only one carbonate ion is present. Brackets are used only when two or more identical polyatomic ions appear.

Why it happens: a polyatomic ion such as OH⁻ or CO₃²⁻ behaves as a single unit — its own atoms are covalently bonded to each other and the whole group carries one charge. So in criss-crossing you treat it exactly as you would a single symbol, and the bracket is what keeps that unit together.

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