CBSE Class 7 Science Advanced Chemistry Atomic Structure and Transformation of Matter MCQs Set 03

Answer: (b) P

Explanation:
1. Count the electrons: 2 in the first shell, 8 in the second and 5 in the third, a total of 15.
2. An atom with 15 electrons has atomic number 15, which is phosphorus (P).
3. Its configuration 2, 8, 5 matches the figure.

Teacher's Note:
1. Count electrons shell by shell and add them up.
2. Compare with S (16), Si (14) and Cl (17), which have one or two electrons different.

Answer: (d) (iii) and (iv)

Explanation:
1. In CD one atom of C combines with one atom of D, so C gives as many electrons as D needs.
2. If C has 1 valence electron it gives 1 and D with 7 valence electrons takes 1, which is (iii).
3. If C has 2 valence electrons it gives 2 and D with 6 valence electrons takes 2, which is (iv).
4. Both atoms having 1 or both 2 valence electrons would not make an ionic pair, so (i) and (ii) are wrong. The answer is (d).

Teacher's Note:
1. Total electrons given must equal total electrons taken for a 1 : 1 compound.
2. Valence electrons of the two atoms add up to 8 here.

Answer: (d) \( {}_{x}\mathrm{B}^{x+y} \)

Explanation:
1. The subscript shows the atomic number, which is the number of protons: x.
2. The superscript shows the mass number, which is protons + neutrons = x + y.
3. So the atom is written as option (d).

Teacher's Note:
1. Atomic number = number of protons; mass number = protons + neutrons.
2. Here the book writes the atomic number as a subscript and the mass number as a superscript.

Answer: (b) Mercurous

Explanation:
1. Ferrous has valency 2, stannous has 2 and cupric has 2.
2. Mercurous has valency 1.
3. The odd one is mercurous.

Teacher's Note:
1. The 'ous' form usually has the lower valency and 'ic' the higher, but check each metal.
2. Learn mercurous (1) and mercuric (2) as a pair.

Answer: (c) Calcium hydroxide

Explanation:
1. Nitrogen pentoxide \(\ce{N2O5}\) has 2 + 5 = 7 atoms.
2. Sulphuric acid \(\ce{H2SO4}\) has 2 + 1 + 4 = 7 atoms.
3. Calcium hydroxide \(\ce{Ca(OH)2}\) has 1 + 2 + 2 = 5 atoms.
4. Ferric hydroxide \(\ce{Fe(OH)3}\) has 1 + 3 + 3 = 7 atoms.
5. The odd one is calcium hydroxide.

Teacher's Note:
1. Write the formula first, then multiply the subscript with the number inside brackets.
2. Count each element, not each symbol.

Answer: (d) F–1

Explanation:
1. He has 2 electrons (a duplet), not an octet.
2. Na has configuration 2, 8, 1, so its valence shell has 1 electron.
3. K+2 is not a stable ion; K²⁺ would have 2, 8, 7, so it has no octet.
4. F⁻ has 2, 8, so its valence shell has 8 electrons: (d).

Teacher's Note:
1. An octet means 8 electrons in the outermost shell.
2. Fluorine gains 1 electron to complete its octet.

Answer: (c) Formation of green patches

Explanation:
1. Alloying, sonorousness and conduction of electricity are physical properties or processes; no new substance forms.
2. The green patches on copper are formed by a chemical reaction with air and moisture, giving basic copper carbonate.
3. So the odd one is (c).

Teacher's Note:
1. Ask whether a new substance forms.
2. Green patches on copper are a corrosion product.

Answer: (d) 4c = 3d

Explanation:
1. Balance carbon: 3a = c. Balance hydrogen: 8a = 2d, so d = 4a.
2. Take a = 1: c = 3 and d = 4.
3. Then 4c = 12 and 3d = 12, so 4c = 3d.
4. Check: oxygen needs 3 × 2 + 4 = 10 atoms, so b = 5.

Teacher's Note:
1. Put a = 1 and work out the rest.
2. Test each option with the numbers c = 3, d = 4.

Answer: (b) Charring

Explanation:
1. Dissolving, melting and crystallisation are physical changes, so sugar is still sugar.
2. Charring breaks sugar down into carbon, which is a chemical change.
3. So sugar cannot retain its chemical properties in charring.

Teacher's Note:
1. A chemical change makes a new substance.
2. Black carbon appears when sugar is overheated.

Answer: (d) N₂

Explanation:
1. Balance N: put 2 before NH₃ for one N₂. Then H gives 3 H₂O. Then O gives 3 CuO and 3 Cu.
2. Balanced: \(\ce{3CuO + 2NH3 -> 3Cu + 3H2O + N2}\)
3. The coefficients are 3, 2, 3, 3, 1, so the lowest is 1 for N₂.

Teacher's Note:
1. Start with the element that appears in the fewest compounds.
2. A coefficient of 1 is usually not written.

Answer: (c) ABDC

Explanation:
1. Chromate \(\ce{CrO4^2-}\) has 4 oxygen atoms.
2. Silicate \(\ce{SiO3^2-}\) has 3.
3. Nitrite \(\ce{NO2-}\) has 2.
4. Hydroxyl \(\ce{OH-}\) has 1.
5. Decreasing order: A, B, D, C, which is option (c).

Teacher's Note:
1. Write each radical's formula before comparing.
2. Nitrite (NO₂⁻) and nitrate (NO₃⁻) are easily confused.

Answer: (d) EDABC

Explanation:
1. Write the configurations: calcium 2, 8, 8, 2 (L shell has 8); oxygen 2, 6 (6); boron 2, 3 (3); lithium 2, 1 (1); helium 2 (0).
2. Decreasing: calcium, oxygen, boron, lithium, helium = E, D, A, B, C.
3. The answer is option (d).

Teacher's Note:
1. The L shell is the second shell.
2. Helium has no electron in the L shell.

Answer: (a) (i) → (D); (ii) → (A); (iii) → (E); (iv) → (C); (v) → (B)

Explanation:
1. Oxygen has 8 protons and 8 neutrons: (i) → (D).
2. Neon (2, 8) has 8 electrons in its valence shell: (ii) → (A).
3. Potassium (2, 8, 8, 1) has 8 electrons in its penultimate shell: (iii) → (E).
4. Hydride ion (H⁻) has 2 electrons in its valence shell: (iv) → (C).
5. Fluorine has 9 protons and 10 neutrons: (v) → (B).
6. This is option (a).

Teacher's Note:
1. Mass number minus atomic number gives neutrons: fluorine 19 − 9 = 10.
2. The penultimate shell is the one just before the outermost shell.

Answer: (d) B

Explanation:
1. Configurations: O is 2, 6; N is 2, 5; C is 2, 4; B is 2, 3.
2. Only boron has 3 electrons in its outer shell: B.

Teacher's Note:
1. Fill shells 2, 8, 8 in order.
2. Outer shell electrons for B are 3.

Answer: (d) BA₂

Explanation:
1. A is 2, 7, so it needs 1 electron and forms A⁻.
2. B is 2, 8, 2, so it loses 2 electrons and forms B²⁺.
3. One B²⁺ needs two A⁻, so the formula is BA₂.

Teacher's Note:
1. Cross the valencies: B has 2 and A has 1.
2. The metal is written first.

Answer: (b) n

Explanation:
1. In this notation the subscript m is the atomic number and the superscript n is the mass number.
2. So the mass number is n.

Teacher's Note:
1. Subscript means atomic number; superscript means mass number.
2. Mass number = protons + neutrons.

Answer: (d) Phosphate

Explanation:
1. Silicate \(\ce{SiO3^2-}\), zincate \(\ce{ZnO2^2-}\) and sulphate \(\ce{SO4^2-}\) all have valency 2.
2. Phosphate \(\ce{PO4^3-}\) has valency 3.
3. The odd one is phosphate.

Teacher's Note:
1. Learn radicals by their charge: –1, –2, –3.
2. Phosphate is the common –3 radical.

Answer: (d) Sulphur trioxide

Explanation:
1. Nitric acid \(\ce{HNO3}\) has 5 atoms.
2. Methane \(\ce{CH4}\) has 5 atoms.
3. Phosphorus trioxide \(\ce{P2O3}\) (as written in the book) has 5 atoms.
4. Sulphur trioxide \(\ce{SO3}\) has 4 atoms.
5. The odd one is sulphur trioxide.

Teacher's Note:
1. Count atoms for each formula.
2. In \(\ce{SO3}\) there is 1 S and 3 O.

Answer: (c) O–1

Explanation:
1. Ca²⁺ is 2, 8, 8, Ne is 2, 8 and N³⁻ is 2, 8, all with 8 electrons in the outer shell.
2. O⁻ is 2, 7, so it has only 7.
3. The answer is O⁻.

Teacher's Note:
1. Work out the electron count after adding or removing charge.
2. O needs 2 electrons for an octet, not 1.

Answer: (d) Rusting

Explanation:
1. Expansion, melting and magnetisation do not form a new substance, so they are physical changes.
2. Rusting forms iron oxide, a new substance, so it is a chemical change.
3. The odd one is rusting.

Teacher's Note:
1. Rusting needs oxygen and moisture.
2. Test: is a new substance formed?

Answer: (b) \(\ce{H2SO4}\)

Explanation:
1. Balance iron first: 2 Fe(OH)₃ gives Fe₂(SO₄)₃.
2. Sulphate needs 3 H₂SO₄.
3. Hydrogen: 6 + 6 = 12 on the left, so 6 H₂O.
4. Balanced: \(\ce{2Fe(OH)3 + 3H2SO4 -> Fe2(SO4)3 + 6H2O}\)
5. The coefficient 3 belongs to H₂SO₄.

Teacher's Note:
1. Balance the polyatomic ions as a group.
2. Confirm oxygen: 6 + 12 = 12 + 6.

Answer: (b) Curdling

Explanation:
1. Churning, boiling and refrigerating do not create new substances in milk.
2. Curdling changes lactose to lactic acid and clots the protein, so the composition changes.
3. The answer is curdling.

Teacher's Note:
1. Curdling is caused by bacteria and acid.
2. It cannot be reversed.

Answer: (d) 2a = 2b = c

Explanation:
1. Balanced equation: \(\ce{2Pb(NO3)2 -> 2PbO + 4NO2 + O2}\)
2. So a = 2, b = 2, c = 4, d = 1.
3. Check the options: 2a = 4, 2b = 4 and c = 4, so 2a = 2b = c.
4. The other options fail, for example a = b = 2c gives 2 = 8.

Teacher's Note:
1. Balance Pb and N first, then O.
2. Substitute the numbers a = 2, b = 2, c = 4 into each option.

Answer: (a) BCAD

Explanation:
1. Ammonium is +1, cupric is +2, aluminium is +3 and stannic is +4.
2. Increasing order: B, C, A, D, which is option (a).

Teacher's Note:
1. Cupric is Cu²⁺ and cuprous is Cu⁺.
2. Stannic is Sn⁴⁺ and stannous is Sn²⁺.

Answer: (b) BACEFD

Explanation:
1. Valence electrons: sodium 1, helium 2, silicon 4, phosphorus 5, fluorine 7, neon 8.
2. Increasing order: B, A, C, E, F, D, which is option (b).

Teacher's Note:
1. Write each configuration: Si is 2, 8, 4 and P is 2, 8, 5.
2. Helium's shell is full with 2.

Answer: (b) (i) → (E); (ii) → (D); (iii) → (A); (iv) → (C); (v) → (B)

Explanation:
1. Sulphur has 16 electrons: (i) → (E).
2. Phosphorus has 15 protons: (ii) → (D).
3. Calcium (2, 8, 8, 2) has four shells: (iii) → (A).
4. Nitrate \(\ce{NO3-}\) has valency 1: (iv) → (C).
5. Nitrogen has 7 protons + 7 neutrons, mass number 14: (v) → (B).
6. This is option (b).

Teacher's Note:
1. Atomic number of S is 16 and of P is 15.
2. Nitrogen-14 is the common isotope.

Answer: (c) CEADB

Explanation:
1. Aluminium phosphate \(\ce{AlPO4}\): 6 atoms.
2. Ammonium carbonate \(\ce{(NH4)2CO3}\): 2 × 5 + 4 = 14 atoms.
3. Magnesium sulphide \(\ce{MgS}\): 2 atoms.
4. Sodium bisulphate \(\ce{NaHSO4}\): 7 atoms.
5. Silver nitrate \(\ce{AgNO3}\): 5 atoms.
6. Increasing: C (2), E (5), A (6), D (7), B (14): CEADB, option (c).

Teacher's Note:
1. Count atoms inside brackets by multiplying.
2. Ammonium carbonate has the most atoms.

Answer: (d) BDAC

Explanation:
1. A metal replacing hydrogen from an acid is displacement: B.
2. Heating potassium nitrate to give oxygen is decomposition: D.
3. Formation of a precipitate by exchange of ions is double displacement: A.
4. Nitrogen and hydrogen joining to form ammonia is combination: C.
5. The order is B, D, A, C: option (d).

Teacher's Note:
1. Name the type of each reaction before arranging.
2. A precipitate reaction is always double displacement.

Answer: (b) Melting of wax involves change in molecular composition.

Explanation:
1. Blackening of silver forms silver sulphide, which is permanent, so (a) is true.
2. Electrolysis forms hydrogen and oxygen, so (c) is true. Ripening is chemical, so (d) is true.
3. Melting of wax is a physical change, with no change in molecular composition, so (b) is false.

Teacher's Note:
1. The question asks for the false statement.
2. Burning of wax is chemical, but melting is physical.

Answer: (c) \(\ce{X2 + Y2 -> 2XY}\)

Explanation:
1. Synthesis means two elements combine to form one compound.
2. The reaction \(\ce{X2 + Y2 -> 2XY}\) does this, so the answer is (c).
3. (a) is double displacement, (b) is decomposition, and (d) is a combination of a compound with an element.

Teacher's Note:
1. Synthesis is element + element → compound.
2. Option (d) is also combination but not between two elements.

Answer: (a) 4

Explanation:
1. Total mass of protons = 7348 × mass of electron = 7348 × 9.1 × 10⁻³¹ kg.
2. Number of protons = (7348 × 9.1 × 10⁻³¹) ÷ (1.67 × 10⁻²⁷) ≈ 4.
3. The number of protons is the atomic number, so it is 4.
4. Check: a proton is about 1837 times an electron, and 7348 ÷ 1837 = 4.

Teacher's Note:
1. Divide total positive mass by the mass of one proton.
2. Atomic number equals number of protons.

Answer: (d) \( {}_{18}\mathrm{X}^{40} \)

Explanation:
1. Neutrons = mass number – atomic number.
2. (a) 16 – 8 = 8, so 0 more. (b) 39 – 20 = 19, so 1 less. (c) 40 – 19 = 21, so 2 more.
3. (d) 40 – 18 = 22, which is 4 more than the 18 protons.
4. The answer is (d).

Teacher's Note:
1. Compute neutrons for each option.
2. Subtract the subscript from the superscript.

Answer: (c) 18

Explanation:
1. The maximum electrons in a shell is 2n².
2. For the M shell n = 3, so 2 × 3² = 18.
3. The answer is 18.

Teacher's Note:
1. K shell holds 2, L holds 8, M holds 18 and N holds 32.
2. The outer shell of an atom never holds more than 8.

Answer: (b) \(\ce{2Pb(NO3)2 ->[\Delta] 2PbO + 2NO2 + O2}\)

Explanation:
1. (a) Fe 2 = 2, S 3 = 3, H 12 = 12, O 18 = 18, balanced.
2. (b) Left has 4 N and 12 O; right has 2 N and 2 + 4 + 2 = 8 O. Not balanced: it needs 4NO₂.
3. (c) Fe 3 = 3, O 4 = 4, H 8 = 8, balanced. (d) Balanced.
4. So the unbalanced reaction is (b).

Teacher's Note:
1. Count each atom on both sides.
2. Nitrogen gives it away: 4 on the left, 2 on the right.

Answer: (d) M(HCO₃)₂

Explanation:
1. In \(\ce{M3(PO4)2}\), 3 M ions balance 2 phosphate ions: 3x = 2 × 3, so x = 2, M²⁺.
2. Bicarbonate is \(\ce{HCO3-}\), so two bicarbonate ions are needed per M²⁺.
3. The formula is M(HCO₃)₂.

Teacher's Note:
1. Cross the valencies of metal and radical.
2. Bicarbonate is the common name for hydrogen carbonate.

Answer: (d) 5

Explanation:
1. Balanced: \(\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\)
2. Oxygen on the right: 3 × 2 + 4 = 10, so O₂ needs 5.
3. The coefficient is 5.

Teacher's Note:
1. Balance C, then H, then O last.
2. Oxygen is the last element to balance in combustion.

Answer: (c) permanganate

Explanation:
1. Bisulphite is \(\ce{HSO3-}\), with charge –1.
2. Permanganate is \(\ce{MnO4-}\), also –1.
3. Magnesium is +2, sulphate is –2 and potassium is +1.
4. The answer is permanganate.

Teacher's Note:
1. Compare both sign and size of charge.
2. Potassium has size 1 but the opposite sign.

Answer: (c) Ag < Sn < Ni < Mg < Na

Explanation:
1. Reactivity series, from high to low: Na, Mg, Ni, Sn, Ag in this set.
2. Increasing order is the reverse: Ag < Sn < Ni < Mg < Na.
3. This is option (c).

Teacher's Note:
1. Learn: K, Na, Ca, Mg, Al, Zn, Fe, Ni, Sn, Pb, H, Cu, Ag, Au.
2. Increasing means start with the least reactive.

Answer: (b) (i) → (B); (ii) → (E); (iii) → (D); (iv) → (C); (v) → (A)

Explanation:
1. Carbonate \(\ce{CO3^2-}\) is –2: (i) → (B).
2. Ammonium \(\ce{NH4+}\) is +1: (ii) → (E).
3. Zinc ion \(\ce{Zn^2+}\) is +2: (iii) → (D).
4. Phosphate \(\ce{PO4^3-}\) is –3: (iv) → (C).
5. Bisulphide \(\ce{HS-}\) is –1: (v) → (A).
6. This is option (b).

Teacher's Note:
1. Bisulphide and bisulphite differ only in oxygen; both are –1.
2. Learn the charge of each radical along with its formula.

Answer: (a) CEBAD

Explanation:
1. (A) \(\ce{MXO3}\): 5 atoms. (B) \(\ce{MXO4}\): 6 atoms.
2. (C) \(\ce{M2(XO3)3}\): 2 + 12 = 14 atoms. (D) \(\ce{M3X}\): 4 atoms.
3. (E) \(\ce{M3(XO4)2}\): 3 + 10 = 13 atoms.
4. Decreasing: C (14), E (13), B (6), A (5), D (4): CEBAD, option (a).

Teacher's Note:
1. Cross the charges to get each formula.
2. Take care to multiply the atoms inside brackets.