Carbon and its Compounds — Class 10 Science PYQs

(A) / Both A and R are true and R is the correct explanation of A.

(A) / Both A and R are true and R is the correct explanation of A. (1 Mark)

(C) Catenation (1)

(C)/4 (1 Mark)

(a)
A - $\ce{C2H5OH}$/ Ethanol ($\frac{1}{2}$)
$\ce{H \quad H}$
$\ce{| \quad |}$
$\ce{H-C-C-OH}$
$\ce{| \quad |}$
$\ce{H \quad H}$ ($\frac{1}{2}$)
B - $\ce{CH3COOH}$ /Ethanoic acid/ acetic acid ($\frac{1}{2}$)
$\ce{H \quad O}$
$\ce{| \quad ||}$
$\ce{H-C-C-OH}$
$\ce{|}$
$\ce{H}$ ($\frac{1}{2}$)
C- $\ce{CH3COOC2H5}$ /Ethyl ethanoate/ ethyl acetate/ester ($\frac{1}{2}$)
$\ce{H \quad O \quad H \quad H}$
$\ce{| \quad || \quad | \quad |}$
$\ce{H-C-C-O-C-C-H}$
$\ce{| \quad \quad \quad | \quad |}$
$\ce{H \quad \quad \quad H \quad H}$ ($\frac{1}{2}$)
Alternate answer
$\ce{C2H5OH ->[Acidified K2Cr2O7] CH3COOH}$
Ethanol(A) Ethanoic acid (B) (1+1)
$\ce{A + B ->[conc H2SO4] C (CH3COOC2H5)}$
Ester
Ethyl ethanoate/ethyl acetate (1)
OR
(b)
(i) Unsaturated hydrocarbons having double or triple bonds have a
strong tendency to become saturated. (1)
OR
Unsaturated hydrocarbons are reactive and undergo addition
reactions. (1)
(ii) Long chain of hydrocarbon in detergents does not form insoluble
precipitate with Calcium or Magnesium ions of hard water and hence
they remain effective even in hard water. (1)
(iii) Carbon is tetravalent. It can neither form $\ce{C4+}$ cation nor $\ce{C4-}$ anion.
It completes its octet by sharing its valence electrons with other atoms
of carbon or other elements forming covalent bonds. (1)

(i)
$\bullet$ 1) Carbon has unique ability of catenation in which it can form bonds with other atoms of carbon. ($\frac{1}{2}$ Mark)
2) Carbon is tetravalent and small in size, due to which it can form strong and stable bonds with other atoms. ($\frac{1}{2}$ Mark)
$\bullet$ Carbon has four electrons in its outermost shell. It cannot form $\ce{C^{4+}}$ or $\ce{C^{4-}}$ ions as they are highly unstable. So, it shares its electrons with other atoms and forms covalent bonds. (1 Mark)
(ii) $\ce{C2H5CHO}$ (1 Mark)

(a) (i)
(I) They do not give rise to charged particles/ ions. (1)
(II) Soap reacts with calcium and magnesium salts present in
hard water and forms insoluble substances called Scum. (1)
(III) C-C bonds are strong and stable whereas Si-Si bonds are
relatively weak. (1)
(ii) (I) $\ce{CH3CH2OH ->[acidified K2Cr2O7+heat] CH3COOH}$ (1)
(II) $\ce{CH2=CH2 + H2 ->[Ni] CH3-CH3}$ (1)

(a) (i) • The unique ability of carbon to form bonds with other atoms of carbon giving rise to large molecules. (1 Mark)
• Because of strong and stable C-C bond. (1 Mark)
(ii) • A group of organic compounds with the same functional group and similar chemical properties / Series of compounds in which the same functional group substitutes for Hydrogen in a carbon chain / Compounds having same general formula, successive member differ by $\ce{CH2}$ or 14u. (1 Mark)
• Example: - Methane-Ethane or any other. (1 Mark)
(iii) Because soap forms insoluble precipitates (scum) by reacting with calcium and magnesium salts present in hard water. (1 Mark)

(a) (i) To form $\ce{C^{4+}}$ ion large amount of energy is required to remove 4 electrons. For forming $\ce{C^{4-}}$ anion, it would be difficult for the nucleus with six protons to hold on to ten electrons, that is, four extra electrons. Hence energetically not feasible. Hence carbon completes its octet by sharing of electrons and forms covalent bonds. (2 Marks)
(ii) (I) Sodium ethoxide and Hydrogen gas are formed / $\ce{2Na + 2CH3CH2OH -> 2CH3CH2O-Na+ + H2}$ (1 Mark)
(II) $\ce{CH3COOC2H5}$ is formed / Ester is formed / $\ce{CH3-COOH + CH3-CH2OH ->[Acid] CH3-C-O-CH2-CH3 + H2O}$ (1 Mark)
(Either statement or equation is to be marked)
(iii) Sodium or Potassium salt of long chain carboxylic acid. ($\frac{1}{2}$ Mark)
Because soap forms scum with hard water. ($\frac{1}{2}$ Mark) (5 Marks)

(B)/2 (1 Mark)

(B)/ alkene
1
1

(D)/ (ii) and (iii) (1 Mark)

(B)/-CHO (1)

(B) / -COOH (1 Mark)

(a) (i)
u2022 A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain./ Can be represented by a single general formula and successive members differ by $\ce{-CH2}$ unit or 14u.
u2022 $\ce{C2H4}$ and $\ce{C4H8}$ belong to the same homologous series/ are alkenes.
(ii) (I) $\ce{CH3CH2OH + CH3COOH ->[Acid] CH3COOC2H5 + H2O}$
(II) $\ce{C2H5OH + 3O2 ->[Burning] 2CO2 + 3H2O}$ (balancing may be ignored)
(iii) The ionic end of soap interacts with water while the carbon chain interacts with oil/dirt. The soap molecules thus form a structure called micelle.

(i) Carbon cannot form $\ce{C^4+}$ cation because removal of four electrons is energetically not possible. Carbon cannot form $\ce{C^4-}$ anion because nucleus with six protons cannot hold ten electrons. (2 Marks)
(ii) • A series of compounds in which the same functional group substitutes for hydrogen in carbon chain is called a homologous series. (1 Mark)
• Because of higher molecular mass of $\ce{C4H10}$ than $\ce{C3H8}$ or $\ce{C2H6}$. (1 Mark)
(iii) Ethanoic acid / acetic acid / $\ce{CH3COOH}$ is formed /
$\ce{CH3-CH2OH ->[Alkaline KMnO4 + Heat][Or acidified K2Cr2O7 + Heat] CH3COOH}$ (1 Mark)

(D)/ $\ce{C4H8}$ and $\ce{C5H12}$ (1 Mark)

(D)/ Propyne (1 Mark)

(A) Carboxylic acid (1)

(D) $\ce{C3H4}$
1

(D)/ Ketone (1 Mark)

(A)/ Carboxylic acid (1 Mark)

(B)/ Methanal (1 Mark)

(i) I. Propanal (1 Mark)
II. Propyne (1 Mark)
(ii) I. Ester group (1 Mark)
II. $\ce{CH3-COOH + CH3-CH2OH ->[Acid] CH3-C(=O)-O-CH2-CH3 + H2O}$ (1 Mark)
(iii) Soaps reacts with calcium and magnesium salts present in hard water to form scum (insoluble substance). (1 Mark)

(B)/ $\ce{C4H8}$

(C)/ $\ce{C3H8}$ (1 Mark)

(A) Ethane (1 Mark)

(D) Hydrogenation (1 Mark)

(a)
A - $\ce{C2H5OH}$/ Ethanol ($\frac{1}{2}$)
$\ce{H \quad H}$
$\ce{| \quad |}$
$\ce{H-C-C-OH}$
$\ce{| \quad |}$
$\ce{H \quad H}$ ($\frac{1}{2}$)
B - $\ce{CH3COOH}$ /Ethanoic acid/ acetic acid ($\frac{1}{2}$)
$\ce{H \quad O}$
$\ce{| \quad ||}$
$\ce{H-C-C-OH}$
$\ce{|}$
$\ce{H}$ ($\frac{1}{2}$)
C- $\ce{CH3COOC2H5}$ /Ethyl ethanoate/ ethyl acetate/ester ($\frac{1}{2}$)
$\ce{H \quad O \quad H \quad H}$
$\ce{| \quad || \quad | \quad |}$
$\ce{H-C-C-O-C-C-H}$
$\ce{| \quad \quad \quad | \quad |}$
$\ce{H \quad \quad \quad H \quad H}$ ($\frac{1}{2}$)
Alternate answer
$\ce{C2H5OH ->[Acidified K2Cr2O7] CH3COOH}$
Ethanol(A) Ethanoic acid (B) (1+1)
$\ce{A + B ->[conc H2SO4] C (CH3COOC2H5)}$
Ester
Ethyl ethanoate/ethyl acetate (1)
OR
(b)
(i) Unsaturated hydrocarbons having double or triple bonds have a
strong tendency to become saturated. (1)
OR
Unsaturated hydrocarbons are reactive and undergo addition
reactions. (1)
(ii) Long chain of hydrocarbon in detergents does not form insoluble
precipitate with Calcium or Magnesium ions of hard water and hence
they remain effective even in hard water. (1)
(iii) Carbon is tetravalent. It can neither form $\ce{C4+}$ cation nor $\ce{C4-}$ anion.
It completes its octet by sharing its valence electrons with other atoms
of carbon or other elements forming covalent bonds. (1)

(b) (i) (I) Propanone/acetone
(½)
(II) Propyne
(½)
(ii) Hydrogenation/addition reaction.
(1)
(iii) $\ce{C4H10}$ will have higher melting point, as it has a higher molecular mass than $\ce{C3H8}$ and Melting point increases with the increase in molecular mass.
(1)
(iv) Sodium ethoxide ($\ce{C2H5ONa}$) is formed and hydrogen gas is evolved / $\ce{2Na + 2CH3CH2OH -> 2CH3CH2O-Na+ + H2}$ (balancing may be ignored)
(1)
(v) $\ce{CH3COON C2H5 + NaOH -> CH3COONa + C2H5OH}$
(½+½)

(b) (i) (I) Hydrocarbons having double bond ( $\ce{C=C}$) or triple bond ( $\ce{C \equiv C}$) are called unsaturated hydrocarbons ($\frac{1}{2}$ Mark)
Example alkene / alkyne. ($\frac{1}{2}$ Mark)
(any other example)
(II) A reaction in which one type of atom or a group of atoms take the place of another. ($\frac{1}{2}$ Mark)
Example $\ce{CH4 + Cl2 ->[sunlight] CH3Cl + HCl}$ ($\frac{1}{2}$ Mark)
(any other example)
(III) The reaction of an ester with an alkali to give alcohol and sodium salt of carboxylic acid is called saponification. ($\frac{1}{2}$ Mark)
$\ce{CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH}$ ($\frac{1}{2}$ Mark)
(ii) (5 Marks)

(i) I. Propanal (1 Mark)
II. Propyne (1 Mark)
(ii) I. Ester group (1 Mark)
II. $\ce{CH3-COOH + CH3-CH2OH ->[Acid] CH3-C(=O)-O-CH2-CH3 + H2O}$ (1 Mark)
(iii) Soaps reacts with calcium and magnesium salts present in hard water to form scum (insoluble substance). (1 Mark)

(a) • $\ce{CH3-CH2-CH2-CH2OH}$ /
H H H H
H-C-C-C-C-O-H
H H H H
• Butanol (1/2)
(b) • Ethanol gets oxidised to form ethanoic acid. (1/2)
(c) • alkaline $\ce{KMnO4}$+heat
$\ce{CH3-CH2-OH -> CH3COOH}$ (1/2)
• $\ce{CH3-COOH + CH3-CH2OH <=>[Acid] CH3-C-O-CH2-CH3 + H2O}$ (1)
• Esterification reaction. (1)
OR
(c) • Ethanol will get dehydrated to form ethene. (1/2)
• alkaline $\ce{KMnO4}$+heat
$\ce{CH3-CH2-OH -> CH3COOH}$ (1/2)
(Award full marks if equation is written with condition)
• Dehydrating agent/ Catalyst (1)

(i) X - $\ce{CH3COOH}$/ethanoic acid /acetic acid
$\frac{1}{2}$
Y - $\ce{CH3COOC2H5}$/$\ce{CH3COOCH2CH3}$/ ester/ ethyl ethanoate
$\frac{1}{2}$
Z - $\ce{CH3COONa}$/sodium ethanoate/sodium acetate
$\frac{1}{2}$
(ii) Catalyst / dehydrating agent
$\frac{1}{2}$
(iii)
$\ce{CH3-COOH + CH3-CH2OH <=>[Acid] CH3-C-O-CH2-CH3 + H2O}$
• Esterification reaction
$\frac{1}{2}$
• $\ce{CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH}$
1
• Saponification reaction/ De-esterification reaction
$\frac{1}{2}$
5

(A) / tomato, curd, ant-sting (1 Mark)

(A)/ Both A and R is true and R is correct explanation of A.

(i)
$\bullet$ X - Ethanol/ $\ce{C2H5OH}$ ($\frac{1}{2}$ Mark)
$\bullet$ Y - Ethanoic acid/ acetic acid/ $\ce{CH3COOH}$ ($\frac{1}{2}$ Mark)
(ii) Carbon dioxide/ $\ce{CO2}$ (1 Mark)
(iii)
$\bullet$ $\ce{C2H5OH ->[Acidified K2Cr2O7] CH3COOH}$ ($\frac{1}{2}$ Mark)
$\bullet$ $\ce{2CH3COOH + Na2CO3 -> 2CH3COONa + H2O + CO2}$ ($\frac{1}{2}$ Mark) (Don't deduct marks if balancing is not done)

(A) / The hydrophobic tail of the soap molecule is in the interior of the cluster, whereas hydrophilic end is on the surface of the cluster. (1 Mark)

(A) Hydrophilic head and a hydrophobic tail. (1)