NCERT Solutions Class 10 Science Chapter 4 — Carbon and its Compounds

NCERT Class 10 Science — Carbon and its Compounds, Chapter 4 Exercises. All 15 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

The exercise falls into three parts. Questions 1, 4 and 5 are about the covalent bond and how to draw it. Questions 2, 6, 12, 13 and 14 are about carbon chains — saturated and unsaturated, functional groups and homologous series. Questions 7 to 11 and 15 are about the two compounds the chapter singles out, ethanol and ethanoic acid, and about soap. Four ideas carry all of it:

• Carbon is tetravalent and catenates — it forms four bonds and joins to itself in chains and rings, which is why its compounds are so numerous.
• A covalent bond is a shared pair of electrons, formed so that both atoms complete their outermost shells. Covalent compounds have low melting points and do not conduct electricity, because they contain no ions.
• A homologous series shares a functional group, and successive members differ by $-\text{CH}_2-$.
• A soap molecule has two ends — a water-hating hydrocarbon tail and a water-loving ionic head — and every question about cleaning, scum and micelles follows from that.

Key insight. Count electrons before drawing anything. Carbon brings four valence electrons and needs four more; hydrogen brings one and needs one; oxygen brings six and needs two. Once those numbers are written down, the electron dot structure of Question 5 draws itself, and so does the answer to Question 1 — ethane has seven bonds because six hydrogens need six bonds and the two carbons need one more between them.

Question 1

Ethane, with the molecular formula C₂H₆ has

Solution. Count the bonds rather than guessing from the formula.

Each of the six hydrogen atoms forms exactly one bond — hydrogen needs one electron and can make no more. That is 6 C–H bonds.

Each carbon must form four bonds in all. One carbon uses three of them on hydrogens, leaving one; the other carbon is in the same position. Those two spare bonds join the carbons to each other, giving 1 C–C bond.

$$6 + 1 = 7 \text{ covalent bonds}$$

Question 2

Butanone is a four-carbon compound with the functional group

Solution. The name carries the answer in its ending. In the naming rules of this chapter, a functional group is shown by a suffix: -ol for an alcohol, -al for an aldehyde, -oic acid for a carboxylic acid and -one for a ketone.

“Butanone” ends in -one, so the functional group is a ketone, $>\text{C}=\text{O}$ with a carbon on each side. “Butan-” tells us the chain has four carbons, which is consistent — a ketone needs at least three carbons, since the carbonyl group must sit in the middle of the chain.

Question 3

While cooking, if the bottom of the vessel is getting blackened on the outside, it means that

Solution. The black deposit is soot, which is unburnt carbon. Soot appears only when a carbon fuel does not get enough oxygen: complete combustion gives carbon dioxide and water and a clean blue flame, while incomplete combustion gives a sooty yellow flame and leaves carbon behind on whatever the flame touches.

$$\text{CH}_4 + 2\text{O}_2 \longrightarrow \text{CO}_2 + 2\text{H}_2\text{O} \quad \text{(complete, clean blue flame)}$$

The remedy is to open the air holes of the burner so that more oxygen reaches the flame. Option (d) says the opposite of what is happening, and (a) and (c) are not what the blackening shows.

Question 4

Explain the nature of the covalent bond using the bond formation in CH₃Cl.

Solution. Start from what each atom has and needs.

Carbon has four electrons in its outermost shell and needs four more for an octet. It cannot lose four or gain four — either would take far too much energy — so it shares.

Each hydrogen has one electron and needs one more to fill its shell. Chlorine has seven and needs one more for its octet.

In chloromethane, carbon shares one electron with each of the three hydrogen atoms and one with the chlorine atom. Each shared pair belongs to both atoms at once, so:

• carbon counts all four shared pairs — eight electrons, a complete octet;
• each hydrogen counts its shared pair — two electrons, a full shell;
• chlorine counts its seven own electrons plus the shared pair — a complete octet.

Every atom is satisfied, and four single covalent bonds hold the molecule together: three C–H and one C–Cl.

That is the nature of the covalent bond: a bond formed by the sharing of a pair of electrons between two atoms, both of which count the shared pair. Because no ions are formed, covalent compounds have low melting and boiling points — the forces between separate molecules are weak — and they are poor conductors of electricity, since there are no charged particles free to move.

Question 5

Draw the electron dot structures for (a) ethanoic acid. (b) H₂S. (c) propanone. (d) F₂.

Solution. In each case, count the valence electrons first — carbon 4, hydrogen 1, oxygen 6, sulphur 6, fluorine 7 — then place shared pairs until every atom has a full shell, and show the rest as lone pairs.

(a) Ethanoic acid, $\text{CH}_3\text{COOH}$. The methyl carbon shares a pair with each of three hydrogens and one with the second carbon. That second carbon shares two pairs with one oxygen — a double bond — and one pair with the other oxygen, which in turn shares a pair with a hydrogen. Each oxygen keeps two lone pairs.

(b) Hydrogen sulphide, $\text{H}_2\text{S}$. Sulphur has six valence electrons and needs two. It shares one pair with each hydrogen, and the remaining four electrons stay as two lone pairs.

(c) Propanone, $\text{CH}_3\text{COCH}_3$. The middle carbon shares a double bond with the oxygen and a single pair with each of the two methyl carbons; each methyl carbon completes itself with three hydrogens.

(d) Fluorine, $\text{F}_2$. Each fluorine has seven valence electrons and needs one. The two atoms share one pair, and each keeps three lone pairs.

Question 6

What is an homologous series? Explain with an example.

Solution. A homologous series is a series of carbon compounds in which the same functional group replaces a hydrogen in a chain, and in which each member differs from the next by a $-\text{CH}_2-$ unit.

Take the alkanes, the simplest example:

$$\text{CH}_4 \quad \text{C}_2\text{H}_6 \quad \text{C}_3\text{H}_8 \quad \text{C}_4\text{H}_{10} \quad \dots \quad \text{C}_n\text{H}_{2n+2}$$

Each one has one more carbon and two more hydrogens than the last — one $-\text{CH}_2-$ group — so the molecular mass rises by 14 u each time.

The alcohols do the same with an $-\text{OH}$ group throughout:

$$\text{CH}_3\text{OH} \quad \text{C}_2\text{H}_5\text{OH} \quad \text{C}_3\text{H}_7\text{OH} \quad \dots$$

Two things follow, and they are why the idea is useful. All members have similar chemical properties, because the functional group is what reacts. And they show a gradation in physical properties — melting and boiling points rise steadily along the series as the chain lengthens.

Question 7

How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties?

Solution. Take the two kinds of property in turn, and give a test that can actually be performed.

Physical properties.

• Smell. Ethanoic acid has a sharp, pungent smell of vinegar; ethanol has a pleasant, characteristic smell.
• Melting point. Pure ethanoic acid freezes at 290 K, which is a cold winter’s day, so it often solidifies like ice — hence the name glacial acetic acid. Ethanol, freezing at 156 K, never does.

Chemical properties, which give the more reliable tests.

• Litmus test. Ethanoic acid is an acid and turns blue litmus red. Ethanol is neutral and leaves litmus unchanged.
• Sodium hydrogencarbonate test. This is the decisive one. Add a pinch of sodium hydrogencarbonate to each. With ethanoic acid there is brisk effervescence as carbon dioxide is released; with ethanol nothing happens at all.

$$\text{CH}_3\text{COOH} + \text{NaHCO}_3 \longrightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2$$

The gas can be confirmed by passing it through lime water, which turns milky.

Question 8

Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?

Solution. A soap molecule has two very different ends: a long hydrocarbon tail, which will not mix with water, and an ionic head ($-\text{COO}^-\text{Na}^+$), which dissolves in water readily.

Dropped into water, the molecule cannot satisfy both ends at once. The ionic heads are pulled into the water while the hydrocarbon tails are pushed out of it. The molecules resolve this by clustering: they arrange themselves into a sphere with all the tails pointing inwards, away from the water, and all the heads pointing outwards, into it. That cluster is a micelle.

In ethanol, no. Micelles form only because the hydrocarbon tail is insoluble in the solvent. Ethanol is an organic solvent and the hydrocarbon chain dissolves in it perfectly well, so there is nothing for the molecules to hide from and no reason to cluster. The soap simply dissolves as separate molecules.

Question 9

Why are carbon and its compounds used as fuels for most applications?

Solution. Several properties come together, and all of them matter in practice.

They burn in air releasing a great deal of heat — carbon compounds have high calorific values, so a small mass of fuel gives a large amount of energy.

They burn cleanly when the air supply is sufficient, giving a blue flame and only carbon dioxide and water, without smoke — as Question 3 showed, the soot appears only when combustion is incomplete.

They leave little or no ash, so nothing has to be cleared away.

And they are easily available and easy to store, transport and handle — as solids, liquids or gases, whichever the application needs.

Question 10

Explain the formation of scum when hard water is treated with soap.

Solution. Hard water is hard because it contains dissolved calcium and magnesium salts.

Soap is the sodium salt of a long-chain carboxylic acid, and that sodium salt is soluble. But when soap meets calcium or magnesium ions, those ions replace the sodium — and the calcium and magnesium salts of the same acid are insoluble.

$$2\text{C}_{17}\text{H}_{35}\text{COONa} + \text{CaCl}_2 \longrightarrow (\text{C}_{17}\text{H}_{35}\text{COO})_2\text{Ca} \downarrow + 2\text{NaCl}$$

The insoluble salt comes out as a curdy white precipitate that sticks to the clothes and the basin — this is scum.

Two things follow. The soap used up in forming scum does no cleaning at all, so a great deal of soap is wasted before any lather appears; and the scum itself has to be washed off. This is why detergents, whose calcium and magnesium salts are soluble, are used with hard water instead.

Question 11

What change will you observe if you test soap with litmus paper (red and blue)?

Solution. Soap is the salt of a strong base (sodium hydroxide) and a weak acid (a long-chain carboxylic acid). A salt of that combination gives a solution that is basic, not neutral.

So a soap solution behaves as a mild alkali:

• Red litmus turns blue.
• Blue litmus stays blue — no change.

This mild alkalinity is also why soap feels slippery on the skin and why strongly alkaline soaps can irritate it.

Question 12

What is hydrogenation? What is its industrial application?

Solution. Hydrogenation is the addition of hydrogen to an unsaturated hydrocarbon in the presence of a catalyst — nickel or palladium — to give a saturated one. The double or triple bond opens and a hydrogen atom attaches to each of the carbons that shared it:

$$\text{C}_2\text{H}_4 + \text{H}_2 \xrightarrow{\ \text{Ni}\ } \text{C}_2\text{H}_6$$

The industrial application is the hardening of vegetable oils. Vegetable oils contain long unsaturated carbon chains and are liquid at room temperature; animal fats have saturated chains and are solid. Hydrogenating a vegetable oil in the presence of nickel saturates those chains and turns the oil into a solid fat — vanaspati ghee.

It is worth noting what the chapter says about this: vegetable oils are generally the healthier choice, precisely because they are unsaturated.

Question 13

Which of the following hydrocarbons undergo addition reactions: C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄?

Solution. Only unsaturated hydrocarbons undergo addition reactions, because addition needs a double or triple bond that can open up. Saturated hydrocarbons have only single bonds and undergo substitution instead.

So the question is which of these five are unsaturated, and the general formulae settle it. A saturated hydrocarbon fits $\text{C}_n\text{H}_{2n+2}$.

Compound	$2n+2$	Saturated?
$\text{C}_2\text{H}_6$	6	Yes (ethane)
$\text{C}_3\text{H}_8$	8	Yes (propane)
$\text{C}_3\text{H}_6$	8, but has 6	No — propene
$\text{C}_2\text{H}_2$	6, but has 2	No — ethyne
$\text{CH}_4$	4	Yes (methane)

Propene has a double bond and ethyne a triple bond, so those two undergo addition.

Question 14

Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.

Solution. The standard test is bromine water, and it works because only unsaturated compounds undergo addition.

Add a few drops of bromine water — which is reddish-brown — to each hydrocarbon and shake.

• An unsaturated hydrocarbon adds bromine across its double or triple bond, and the colour disappears: the bromine water is decolourised.

$$\text{C}_2\text{H}_4 + \text{Br}_2 \longrightarrow \text{C}_2\text{H}_4\text{Br}_2 \quad \text{(colourless)}$$

• A saturated hydrocarbon cannot add bromine, so the reddish-brown colour remains.

A second, rougher test is the flame. Saturated hydrocarbons burn with a clean blue flame; unsaturated ones, having a higher proportion of carbon, burn with a yellow sooty flame.

Question 15

Explain the mechanism of the cleaning action of soaps.

Solution. Most dirt clings to clothes in an oily film, and oil will not dissolve in water — which is why water alone does not clean. Soap works by getting between the two.

Step 1 — the tails go into the oil. A soap molecule has a hydrocarbon tail that cannot mix with water but mixes readily with oil. Its ionic head is the opposite. When soap is added, the tails bury themselves in the oily droplet while the heads stay out in the water.

Step 2 — a micelle forms around the dirt. Many molecules do this at once, so the oil droplet ends up surrounded by soap molecules, tails inwards and heads outwards — a micelle, with the dirt trapped at its centre.

Step 3 — the droplet is lifted away. Agitating the clothes helps the micelles, with their dirt inside, to lift off the fibres and disperse into the water. Because the outside of every micelle is ionic, the droplets stay suspended in the water as an emulsion and, being all of the same charge, repel one another rather than joining back together.

Step 4 — rinsing. The suspended micelles are simply carried away with the rinse water, and the dirt goes with them.

Common mistakes

• Counting 6 bonds in ethane because the formula shows six hydrogens. The C–C bond has to be counted as well: $6 + 1 = 7$.
• Forgetting the lone pairs in Question 5. An electron dot structure must show every valence electron — both oxygens in ethanoic acid keep two lone pairs, sulphur keeps two, and each fluorine keeps three.
• Saying a homologous series is “compounds with the same general formula”. The two defining features are the same functional group and a difference of one $-\text{CH}_2-$ between successive members.
• Answering Question 8 with “yes, micelles form in ethanol too”. They do not — the hydrocarbon tail dissolves in ethanol, so there is nothing for the molecules to escape from and no reason to cluster.
• Explaining scum as “the soap does not dissolve in hard water”. The soap dissolves perfectly well; it is the calcium and magnesium salts formed from it that are insoluble.
• Choosing $\text{C}_2\text{H}_6$ in Question 13 because it has two carbons like ethyne. Compare each formula with $\text{C}_n\text{H}_{2n+2}$: ethane fits it and is saturated, ethyne falls four hydrogens short and is not.

Practise next

• The in-text questions of this chapter, thirteen across five boxes, which test the same bonding, the same functional groups and the same soap chemistry in shorter form.
• Chapter 3, Metals and Non-metals, whose ionic bond is the exact contrast to the covalent bond of Questions 4 and 5 — worth reading side by side.