JEE Chemistry The D and F Block Elements MCQs Set 17

Answer: (c) \( \ce{Zn} \)

Explanation:
1. \( \ce{Zn + 2NaOH -> Na2ZnO2 + H2} \).
2. Copper, silver and nickel do not react with alkali.

Teacher's Note:
1. Link each compound to its colour, oxidation state and reaction medium; these facts are asked repeatedly.
2. For redox questions, count the electrons exchanged per formula unit before choosing an option.

Answer: (c) Wrought iron less than 0.15% carbon; steel 0.15 to 2.0%; pig iron over 2%

Explanation:
1. Wrought iron is nearly carbon free, steel has an intermediate carbon content and pig iron has the most (over 2 percent).

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (c) 1. Fe(III) gives a brown colour with potassium ferricyanide; 2. Fe(II) gives a blue precipitate with potassium ferricyanide; 3. Fe(III) gives a red colour with potassium thiocyanate. Statements 1, 2 and 3

Explanation:
1. Fe(III) with ferricyanide gives a brown solution, Fe(II) with ferricyanide gives Turnbull's blue, and Fe(III) with thiocyanate gives blood-red \( \ce{[Fe(SCN)]^{2+}} \).
2. Fe(II) does not give a brown colour with thiocyanate (statement 4 is false), so statements 1, 2 and 3 are correct.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (b) Both A and R are true but R is not the correct explanation of A

Explanation:
1. Both statements are true.
2. The yellow colour of chromate arises from ligand-to-metal charge transfer (the metal is d0, so there is no d-d transition); the oxidation state alone does not explain the colour.

Teacher's Note:
1. Link each compound to its colour, oxidation state and reaction medium; these facts are asked repeatedly.
2. For redox questions, count the electrons exchanged per formula unit before choosing an option.

Answer: (a) Ni

Explanation:
1. German silver is a copper-zinc-nickel alloy.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (b) 50 mL of 0.1 M \( \ce{FeSO4} \)

Explanation:
1. Moles of permanganate = 10 x 0.1 = 1 mmol.
2. It oxidises 5 mmol of \( \ce{Fe^{2+}} \).
3. 5 mmol of 0.1 M solution = 50 mL.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (a) steel is heated in an atmosphere of ammonia

Explanation:
1. Nitriding hardens the steel surface by forming iron nitrides when it is heated in ammonia.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (c) L.D. process

Explanation:
1. The Linz-Donawitz (basic oxygen) process, which blows pure oxygen onto molten pig iron, is the modern method.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (c) reduction of copper ions

Explanation:
1. \( \ce{Zn + Cu^{2+} -> Zn^{2+} + Cu} \): copper ions gain electrons and are reduced.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (c) \( \ce{Fe2O3 + SO2 + SO3} \)

Explanation:
1. \( \ce{2FeSO4 -> Fe2O3 + SO2 + SO3} \).

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (d) hard and brittle

Explanation:
1. Quenching traps carbon in a strained lattice, making the steel hard and brittle.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (c) 14%

Explanation:
1. Stainless steel contains 12 to 20 percent chromium; among the options, 14 percent is the only one in this range.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (d) \( \ce{Zn(NO3)2 + NH4NO3} \)

Explanation:
1. \( \ce{4Zn + 10HNO3 -> 4Zn(NO3)2 + NH4NO3 + 3H2O} \): nitrate is reduced all the way to ammonium.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (d) pig iron

Explanation:
1. The crude iron tapped from the blast furnace is pig iron (about 4 percent carbon).

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (d) laterite

Explanation:
1. Laterite is a soil rich in hydrated oxides of aluminium and iron.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (a) \( \ce{Zn} \)

Explanation:
1. Zinc displaces silver from the cyanide complex, and the process is the cyanide process.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (a) iron

Explanation:
1. Iron displaces copper from solution: \( \ce{Fe + Cu^{2+} -> Fe^{2+} + Cu} \).
2. Sodium reacts with water instead, and Ag and Hg are less reactive than Cu.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (d) \( \ce{Cr2O3} \)

Explanation:
1. Balance chromium: 8 on the left, 4 in chromate, so 4 Cr go into 2X: X = \( \ce{Cr2O3} \).
2. Check oxygen: 28 = 16 + 6 + 6.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (d) a thin oxide layer coating the metal

Explanation:
1. Conc. nitric acid forms a thin, unreactive oxide film on iron that stops further reaction.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (d) \( \ce{AgNO3} \)

Explanation:
1. Silver nitrate decomposes to molten silver metal, a globule; lead and copper nitrates leave oxides and sodium nitrate leaves nitrite.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (b) zinc

Explanation:
1. Galvanisation is the coating of iron with zinc; zinc also protects iron sacrificially.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (a) \( \ce{[Cu(NH3)4]SO4} \)

Explanation:
1. Ammonia (not ammonium ion) is the ligand: \( \ce{Cu^{2+} + 4NH3 -> [Cu(NH3)4]^{2+}} \).

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (b) \( \ce{K2CrO4} \)

Explanation:
1. Silver chromate \( \ce{Ag2CrO4} \) is brick red; AgI is yellow and AgBr pale yellow.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (d) colourless and diamagnetic

Explanation:
1. Sc(III) is 3d0: no d electrons, so no d-d transition (colourless) and no unpaired electrons (diamagnetic).

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (b) passive

Explanation:
1. Iron becomes passive in conc. nitric acid because of a protective oxide film.

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.

Answer: (b) 3

Explanation:
1. \( \ce{Cr2O7^{2-} + 14H^+ + 6e^- -> 2Cr^{3+} + 7H2O} \).
2. Six electrons per two Cr atoms, so 3 per Cr atom (+6 to +3).

Teacher's Note:
1. Write the electronic configuration of the species first; most d-block questions follow from it.
2. Compare options using trends (oxidation states, size, magnetism) rather than memorising isolated facts.