NCERT Solutions Class 10 Science Chapter 4 Carbon and its Compounds

Dot-structure routine for the board. Whenever you draw an electron dot structure, use the same method: count outer-shell electrons, decide how many each atom must share to reach eight, then place shared pairs until every atom is happy.

Carbon brings 4 outer electrons and oxygen brings 6 each. Carbon is short by 4, so it must share 4 electrons. Each oxygen is short by 2, so each must share 2. The only way these numbers fit is carbon making two double bonds, one to each oxygen, so the molecule comes out linear as O=C=O.

Confirm by counting electrons around each atom after bonding. Carbon shows 8 shared electrons (two double bonds). Each oxygen shows 4 shared electrons in its double bond plus 4 of its own as two lone pairs, again making 8. Students lose marks when they forget the lone pairs on oxygen, so always add those two dot-pairs on each oxygen.

Answer: Carbon dioxide is O=C=O: two carbon-oxygen double bonds, each oxygen with two lone pairs, every atom completing its octet.

Why a ring and not a chain. The shape of an element's molecule is decided by how many bonds each atom must form. Sulphur needs two bonds, so each atom acts like a link in a chain with two ends. Close the two loose ends together and you get a ring, which is exactly the crown-shaped S₈ ring.

Look at the electron bookkeeping for one atom. Sulphur starts with six outer electrons. It uses one electron in each of its two bonds, so two of its electrons become bonding electrons. That leaves four electrons as two lone pairs. Add the four shared electrons it now feels to the four lone-pair electrons and you reach eight, a complete octet.

Compare elements to see the pattern. Oxygen, just above sulphur, also needs two bonds but prefers a double bond in O₂ rather than long rings. Carbon needs four bonds, which is why it builds the endless chains this whole chapter is about. The same octet rule explains very different shapes.

Answer: S₈ is a ring of eight sulphur atoms, each joined by two single bonds and holding two lone pairs, satisfying every octet.

A systematic way to find isomers. Drawing isomers by guesswork leads to repeats or misses. A safer routine is to work down from the longest carbon chain to shorter ones, attaching the leftover carbons as branches.

For pentane you have five carbons. First put all five in a straight line, which is n-pentane. Next make the main chain four carbons long and use the fifth carbon as a methyl branch. The branch can only sit on the second carbon, giving iso-pentane. Finally make the main chain three carbons long and hang both extra carbons on the central carbon, giving neo-pentane.

A point students forget is that the same structure drawn flipped or rotated is not a new isomer. n-Pentane written left to right is the same as written right to left. So even though there seem to be many drawings, only three are genuinely different arrangements of atoms. Three is the correct and complete answer for pentane.

Answer: Three structural isomers of pentane: n-pentane, iso-pentane (2-methylbutane) and neo-pentane (2,2-dimethylpropane).

Two ideas that multiply together. The key to this answer is seeing how the two properties combine, not just listing them. Catenation decides how big and varied the carbon skeleton can be, while tetravalency decides how richly that skeleton can be decorated with other atoms. Multiply the two and you get millions of carbon compounds.

Catenation is special to carbon because its carbon-carbon bond is strong and stable. The carbon atom is small, so the shared electrons sit close to both nuclei and are held firmly. This lets carbon make chains of two, ten, or even thousands of atoms, and also rings, without the structure falling apart.

Tetravalency means every carbon offers four hands for bonding. On a long chain, the inner carbons use two hands to continue the chain and have two more free for hydrogen or other atoms. This is why the same carbon skeleton can carry many different functional groups, giving alcohols, acids, aldehydes and so on.

Answer: Catenation (self-linking into chains and rings) and tetravalency (four covalent bonds) together produce the huge number of carbon compounds.

Reading the name to build the molecule. The name cyclopentane carries all the information you need. "Pent" fixes five carbons, "ane" tells you the ring is saturated, and "cyclo" tells you those five carbons form a closed loop.

Put the five carbons at the corners of a pentagon and connect neighbours with single bonds. Each carbon has now used two of its four bonds inside the ring. The two bonds left over on every carbon take one hydrogen each, so each ring carbon becomes a CH₂ unit. Five carbons each carrying two hydrogens give ten hydrogens, so the formula is C₅H₁₀.

Compare with open-chain pentane, C₅H₁₂. When you bend the chain round and join the two ends, those two end carbons now bond to a carbon instead of a hydrogen, so the molecule loses exactly two hydrogens. That explains the general ring formula C_nH_2n.

Answer: Cyclopentane is C₅H₁₀, a saturated five-membered carbon ring of CH₂ groups joined by single bonds.

Translate the name, then fill in hydrogens. Read the root for the carbon count and the affix for the functional group, draw the carbon skeleton with that group in place, then add hydrogens until every carbon has its four bonds.

Take butanone as a worked example. "But" fixes four carbons and "-one" fixes a ketone, a C=O carbon sitting inside the chain. Place the four carbons in a row, put the double-bonded oxygen on the second carbon, then top up hydrogens to give CH₃-CO-CH₂-CH₃.

The bromopentane part tests isomerism. A halogen is a substituent, so it can ride on different carbons of the same chain. Putting bromine on carbon 1, 2 or 3 gives different molecules with the same formula C₅H₁₁Br, so the answer to the starred question is a clear yes.

Answer: Structures as drawn; bromopentane has structural isomers because the -Br can attach at different positions.

A three-step naming routine. Always do the same three things: count the carbons in the longest chain, identify the functional group, then attach it as the correct prefix or suffix while fixing the spelling.

Step one is the carbon count. Memorise the first six: meth (1), eth (2), prop (3), but (4), pent (5), hex (6). For CH₃-CH₂-Br the chain is two carbons (eth), for HCHO just one (meth), and the last compound is six carbons (hex).

Step two is the group: bromine is the prefix "bromo", an aldehyde uses "-al", a triple bond uses "-yne". Step three is to combine and tidy the spelling, dropping the final "e" before a vowel suffix. That gives bromoethane, methanal and hexyne.

Answer: (i) Bromoethane, (ii) Methanal, (iii) Hexyne.

Track the oxygen and hydrogen. The cleanest way to prove a reaction is oxidation is to compare the starting and ending molecules atom by atom. Adding oxygen, or removing hydrogen, is oxidation.

In ethanol the key carbon carries two hydrogens and one -OH group (-CH₂OH). In ethanoic acid that same carbon carries a doubly bonded oxygen as well as an -OH (-COOH). The molecule has picked up an extra oxygen and shed hydrogen, so by the oxygen-gain rule this is oxidation.

Ethanol will not turn into the acid on its own; it needs an oxidising agent to hand over the oxygen. Alkaline potassium permanganate or acidified potassium dichromate do this job, and as they give up oxygen they are reduced and lose colour. The colour change in the test tube is a visible signal that oxidation is taking place.

Answer: The conversion adds oxygen to ethanol to form ethanoic acid, so it is an oxidation reaction driven by an oxidising agent.

Heat depends on completeness of burning. A fuel gives out its maximum heat only when it burns completely, and complete burning needs a generous oxygen supply. Welding needs that maximum heat, so the oxygen supply must be as rich as possible.

Air is mostly nitrogen, with only about a fifth being oxygen. Burning ethyne in air starves the flame of oxygen, so the carbon is not fully oxidised. The result is incomplete combustion: some carbon escapes as soot, the flame turns smoky and yellow, and much less heat is produced. Such a flame is not hot enough to melt metal.

Pure oxygen removes this limitation. With oxygen all around it, ethyne burns completely to carbon dioxide and water, releasing its full energy in a clean, intensely hot blue flame near three thousand degrees Celsius. That is more than enough to melt steel, and with no soot the joint stays clean.

Answer: Air supplies too little oxygen, so ethyne burns incompletely with a smoky, cool flame; pure oxygen gives complete combustion and a flame hot enough (about 3000°C) for welding.

Test the acidic property, since only one has it. The reliable way to tell two compounds apart is to pick a property that one has and the other does not. Here the carboxylic acid is acidic and the alcohol is neutral, so any acid test gives a clear yes or no answer.

The strongest test is the reaction with a carbonate or hydrogencarbonate. Acids release carbon dioxide, seen as brisk effervescence, which you can confirm by passing the gas through lime water, which turns milky. The alcohol, being neutral, does nothing. The test tube that fizzes contains the acid.

Keep a back-up ready. Blue litmus turns red in the carboxylic acid but stays blue in the alcohol. A third option is the ester test: warming the acid with an alcohol and a little concentrated sulphuric acid gives a sweet, fruity smell. Any one separates the two, but the carbonate fizz test is the quickest and most convincing.

Answer: The carboxylic acid fizzes with sodium hydrogencarbonate (releases CO₂) and reddens blue litmus; the alcohol does neither, so these tests distinguish them.

Define by what it does to the other substance. The simplest, exam-safe definition describes its effect on its partner: it is the substance that causes oxidation in something else by handing over oxygen or pulling away hydrogen.

A small but important point is what happens to the oxidising agent itself. When it donates oxygen, it ends up with less oxygen than before, which is reduction. So in every such reaction the oxidising agent is reduced while its partner is oxidised; the two changes always go together.

Connect the idea to real reagents. Alkaline potassium permanganate and acidified potassium dichromate turn alcohols into carboxylic acids by adding oxygen. Their colour change, permanganate fading from purple and dichromate turning from orange to green, is direct evidence that they are being reduced as they oxidise the alcohol.

Answer: An oxidising agent supplies oxygen (or removes hydrogen) to oxidise another substance and is itself reduced; alkaline KMnO₄ and acidified K₂Cr₂O₇ are common examples.

A test needs a difference; detergents give none. To test for hardness you need a substance that behaves differently in hard and soft water. Soap fits because it lathers freely in soft water but struggles in hard water, forming scum instead.

Soap is the sodium salt of a long-chain carboxylic acid. In hard water its carboxylate ions grab the calcium and magnesium ions and fall out as an insoluble scum, leaving little soap free to make lather. So soap foams poorly in hard water and well in soft water, a clear observable difference.

A detergent is built differently. Its charged end (usually a sulphonate group) does not form insoluble salts with calcium and magnesium, so it stays dissolved and keeps foaming whether the water is hard or soft. Because the result looks the same in both cases, a detergent cannot reveal hardness.

Answer: A detergent cannot test hardness because it lathers equally in hard and soft water; use soap, which forms scum and lathers poorly only in hard water.

Two jobs: loosening and lifting. Washing has a chemical part and a physical part, and this question is about the physical part. The soap wraps oily dirt in micelles, but those dirt-filled micelles are still stuck onto the fibres. Something has to knock them loose.

Oily grime fills the gaps between fibres and grips them. Soap molecules slide their water-hating tails into this oil and their water-loving heads out into the water, breaking the oil into tiny droplets held in micelles. So far, those micelles are merely formed in place; nothing has moved them out of the weave.

Agitation supplies the energy for that final step. Beating clothes on a stone, scrubbing with a brush, or spinning them in a machine repeatedly squeezes the fabric, shaking the soap-coated dirt free of the fibres. Once free, each dirt droplet is sealed inside a micelle and a rinse washes it away.

Answer: Agitation gives the mechanical force that frees the soap-trapped dirt from the cloth fibres, after which the micelles carry it into the water and it rinses away.

Build the molecule, then tally. The reliable method for any bond-count question is to draw the complete structure first and only then count, because a careful drawing makes the answer obvious.

Ethane is two carbons bonded to each other, and since carbon is tetravalent each carbon needs four bonds. One of those is the central carbon-carbon bond. That leaves three bonds free on each carbon, filled by hydrogen atoms, giving the familiar CH₃-CH₃.

Now tally the lines. There is one carbon-carbon bond, then three carbon-hydrogen bonds on each carbon, six in all. One plus six is seven, so ethane has seven covalent bonds and option (b) is correct. The same draw-then-count habit handles bond counts for any small molecule.

Answer: (b) 7 covalent bonds.

Let the suffix decide. Multiple-choice questions on functional groups are usually solved by spotting the name ending. Each functional group owns a unique suffix, so once you recognise the ending you can pick the answer.

Butanone ends in "-one", which belongs to ketones, a carbon double-bonded to oxygen sitting on a carbon in the middle of the chain. Writing out the molecule confirms it: CH₃-CO-CH₂-CH₃, where the second carbon carries the C=O.

Rule the others out to be sure. A carboxylic acid ends in "-oic acid" with -COOH; an aldehyde ends in "-al" with a terminal -CHO; an alcohol ends in "-ol" with an -OH. None match butanone, so the functional group is a ketone.

Answer: (c) ketone.

Black means soot means too little oxygen. The clue is the black colour. Black soot on cookware is a sure sign of unburnt carbon, and carbon goes unburnt only when there is not enough oxygen for complete combustion.

Compare the two ways a fuel can burn. With plenty of oxygen the carbon is fully oxidised to carbon dioxide, the flame is blue, and no soot forms. With too little oxygen some carbon survives as fine black particles that settle on the nearest cool surface, the bottom of the vessel.

The other options do not explain a black deposit. Whether the food is fully cooked or the fuel is slightly wet does not by itself blacken the vessel, and complete burning would leave it clean, ruling out option (d). Only (b) links the soot to its cause.

Answer: (b) the fuel is not burning completely.

Show every shared pair. The examiner wants you to show that a covalent bond is a shared pair, and CH₃Cl is tidy for showing exactly that, because it uses all four of carbon's bonds in two ways: three to hydrogen and one to chlorine.

Start from the electron needs. Carbon is short of four electrons, each hydrogen short of one, chlorine short of one. None can comfortably lose or gain whole electrons, so they pool electrons in pairs. Carbon contributes one electron to each bond and its partner contributes the other, making four shared pairs.

After bonding, check each atom. Carbon counts eight electrons (four shared pairs), a full octet; each hydrogen counts two; chlorine counts eight using the one shared pair plus its three lone pairs. No atom has gained or lost an electron outright, so no ions appear. That stability achieved purely by sharing is the nature of the covalent bond.

Answer: CH₃Cl has four covalent bonds (three C-H, one C-Cl), each a shared electron pair; sharing completes every atom's outer shell without forming ions.

Same recipe for every molecule. All four parts use one routine: decide how many bonds each atom needs from its valency, connect the atoms with that many shared pairs, then add lone pairs until every atom reaches eight (or two for hydrogen).

For the carbon-oxygen compounds, ethanoic acid and propanone, the trick is the carbonyl. Oxygen needs two bonds, so at a chain carbon it forms a double bond (C=O); in ethanoic acid the second oxygen takes one bond to carbon and one to hydrogen (O-H). Give the end carbons their three hydrogens and the structures are complete.

The two small molecules are simpler. In H₂S, sulphur forms two single bonds to hydrogen and keeps two lone pairs, a bent H-S-H. In F₂, each fluorine needs one bond, so the atoms share one pair and each keeps three lone pairs. Showing the lone pairs is essential; leaving them out is the most common reason these answers lose marks.

Answer: Dot structures as drawn: ethanoic acid CH₃-C(=O)-O-H, H₂S bent with two lone pairs on S, propanone CH₃-C(=O)-CH₃, and F₂ with a single shared pair and three lone pairs per fluorine.

One family, one functional group. The idea is that chemistry is decided mainly by the functional group, not by chain length. So compounds sharing the same functional group form a family whose members behave alike chemically.

The defining numerical feature is the constant difference between neighbours. Move from one member to the next and you always add a single -CH₂- unit, which weighs 14 u. This is why the molecular masses go up in steps of 14 and why one general formula, like C_nH_2n+2 for alkanes, captures every member.

The alkanes methane, ethane, propane and butane each differ from the previous by -CH₂-, all fit C_nH_2n+2, and all burn and undergo substitution in similar ways. Their boiling points rise steadily as the chain lengthens. Alcohols form another homologous series in the same way.

Answer: A homologous series is a family with the same general formula and functional group whose members differ by -CH₂-; the alkanes are a standard example.

Lean on the acid character. The single biggest difference is that one is acidic and the other is neutral. Every reliable test exploits this, so organise your answer around the acidity of ethanoic acid versus the neutrality of ethanol.

On the chemical side, blue litmus turns red in ethanoic acid but stays blue in ethanol, a quick first check. The decisive test is the reaction with sodium hydrogencarbonate: ethanoic acid releases carbon dioxide with brisk effervescence, while ethanol does nothing. Passing the gas through lime water, which turns milky, confirms the CO₂.

A couple of physical clues round out the answer. Ethanoic acid smells sharply of vinegar and tastes sour, whereas ethanol has a pleasant smell. Pure ethanoic acid also freezes into an ice-like solid near seventeen degrees Celsius, earning it the name glacial acetic acid, while ethanol stays liquid in ordinary cold.

Answer: Ethanoic acid is acidic (reddens litmus, fizzes CO₂ with carbonate, glacial form freezes near 17°C); ethanol is neutral (no litmus change, no fizz).

The solvent decides whether tails must hide. A micelle forms only when the soap's hydrocarbon tails are unwelcome in the solvent and have to bunch together to escape it. So the answer depends entirely on how the solvent treats those tails.

Water is the classic case. Its molecules attract the ionic heads strongly but reject the long hydrocarbon tails. Caught between a head that wants to stay in water and a tail that wants out, the molecules compromise by forming a sphere, tails tucked inside, heads exposed to the water. That sphere is the micelle.

Ethanol changes the story. Its molecule has a short hydrocarbon part of its own, so it can dissolve the hydrocarbon tails rather than rejecting them. With the tails comfortably dissolved, there is no driving force to cluster, and soap dissolves in ethanol as separate molecules. So micelle formation is a response to a hostile solvent, not an automatic property of soap.

Answer: Micelles form in water because the hydrocarbon tails avoid water by clustering inward; in ethanol, which dissolves those tails, no micelle forms.

What makes a substance a good fuel. The question is really asking what qualities a fuel needs, and then noting that carbon and its compounds have them. A useful fuel must release plenty of heat, ignite easily, burn cleanly, and be available cheaply, and carbon compounds tick every box.

The heart of it is energy. When carbon and the hydrogen in its compounds combine with oxygen, they form carbon dioxide and water and pour out a large amount of heat, because combustion is strongly exothermic. Their high calorific value means even a small quantity delivers a lot of useful energy.

The practical qualities seal the case. Carbon fuels catch fire readily, so they are easy to use, and with a good air supply they burn with a clean flame and leave little residue. They are also abundant as wood, coal, petroleum and natural gas, and convenient to store and carry.

Answer: Because their combustion is strongly exothermic with high calorific value, they ignite and burn easily and cleanly, and they are abundant and convenient, carbon and its compounds make excellent fuels.

An ion swap that ruins the soap. Scum forms because the metal ions in hard water replace the sodium in soap, and the new salts they make refuse to dissolve. Following this ion swap step by step makes the answer clear.

Soap is sodium stearate or a similar sodium salt of a long-chain carboxylic acid, and it is soluble in water, which is why ordinary soap lathers. Hard water is loaded with calcium and magnesium ions, which are more strongly attracted to the carboxylate part than sodium is, so they push the sodium out.

The catch is that the resulting calcium and magnesium salts are insoluble. Instead of dissolving and lathering, they clump into a sticky white solid, the scum, that clings to clothes and the bath. Because this consumes soap without cleaning, you must add extra soap before any lather appears.

Answer: Calcium and magnesium ions in hard water replace the sodium in soap to form insoluble calcium/magnesium salts of the carboxylic acid, which separate as the white curdy scum.

Predict the nature, then read the litmus. The fastest way to answer a litmus question is to first work out whether the substance is acidic, basic or neutral, because litmus simply reports that nature.

Soap is made by reacting fats or oils (a weak, long-chain carboxylic acid) with sodium hydroxide, a strong base, in a process called saponification. A salt of a weak acid and a strong base is mildly basic, because the strong base dominates. So a soap solution is alkaline.

Knowing it is basic, the litmus results follow at once. Red litmus turns blue in the presence of a base, so red litmus dipped in soap solution turns blue. Blue litmus is already blue and a base cannot change it further, so it stays blue. Reporting both observations gives a complete answer.

Answer: Soap, being basic, turns red litmus blue and keeps blue litmus blue.

Adding hydrogen across double bonds. Hydrogenation is a specific kind of addition reaction: the guest being added is hydrogen, and the host that accepts it is an unsaturated carbon-carbon bond. Wherever there is a double or triple bond, hydrogen can slot in and saturate it.

The reaction needs a catalyst, usually finely divided nickel or palladium, to bring the hydrogen and the unsaturated molecule together on its surface. With the catalyst present, hydrogen atoms add across each C=C bond, converting it to a single bond. Ethene becoming ethane is the textbook example.

The industrial payoff is in food processing. Vegetable oils are liquids because their long chains are highly unsaturated. Hydrogenating them removes the double bonds, the chains straighten and pack together, and the liquid oil sets into a solid fat. This is how vanaspati ghee is made from cheap oils.

Answer: Hydrogenation adds hydrogen across the double or triple bonds of an unsaturated hydrocarbon using a Ni/Pd catalyst, making it saturated; industrially it turns liquid vegetable oils into solid vanaspati ghee.

Hydrogen count reveals unsaturation. Sorting a list of hydrocarbons into those that add and those that do not is really a test of saturation, and saturation can be read straight off the hydrogen count using C_nH_2n+2.

Ethane C₂H₆, propane C₃H₈ and methane CH₄ all satisfy C_nH_2n+2 exactly, so they are saturated alkanes with only single bonds. With no multiple bond to open up, they cannot add; they react by substitution.

The two that fall short are the reactive ones. Propene C₃H₆ has two hydrogens fewer than propane, the sign of one C=C bond, and ethyne C₂H₂ is short by four, indicating a C≡C bond. These readily add hydrogen and halogens.

Answer: C₃H₆ and C₂H₂ undergo addition; the rest are saturated and do not.

A colour change that reports unsaturation. The cleverness of this test is that it turns an invisible structural feature, a double or triple bond, into a colour change you can see. Bromine water is coloured, and it loses that colour only when consumed by an addition reaction.

When you shake an unsaturated hydrocarbon with bromine water, bromine adds across the C=C or C≡C bond to form a colourless product. As the orange-brown bromine is used up, the solution goes from coloured to clear. That decolourisation is your positive signal.

A saturated hydrocarbon behaves differently. With only single bonds it has nowhere for bromine to add, so the orange-brown colour persists. Colour gone means unsaturated, colour stays means saturated. An alternative is dilute alkaline potassium permanganate (Baeyer's reagent), decolourised only by unsaturated compounds.

Answer: Shake the hydrocarbon with bromine water: decolourisation means unsaturated; no colour change means saturated.

Bridging oil and water. The cleaning action of soap solves a basic problem: dirt is mostly oily, and oil will not mix with water. Soap acts as a bridge by having one end that likes oil and another that likes water.

When soap dissolves in the wash water, its hydrocarbon tails seek out the oily grime and dig into it, while its ionic heads remain dissolved in the surrounding water. As many soap molecules gather round a single oil droplet, they wrap it completely, tails inward in the oil and heads outward in the water, building the spherical micelle.

This micelle carries the dirt away. Its outer heads are all negatively charged, so neighbouring micelles repel each other and stay separated rather than clumping back onto the cloth, giving a stable emulsion. A bit of agitation shakes the dirt-filled micelles loose, and a rinse floats them off.

Answer: Soap's hydrophobic tails embed in the oily dirt and its hydrophilic heads face the water, forming charged micelles that emulsify and suspend the dirt; agitation and rinsing then carry the dirt away, cleaning the cloth.