NEET Chemistry: The d- and f-Block Elements questions with solutions

Q1. Identify the incorrect statement.

1. The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.
2. Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
3. The oxidation states of chromium in CrO4^2− and Cr2O7^2− are not the same.
4. Cr^2+ (d^4) is a stronger reducing agent than Fe^2+ (d^6) in water.

Answer: The oxidation states of chromium in CrO4^2− and Cr2O7^2− are not the same.

The oxidation states of chromium in CrO4^2− and Cr2O7^2− are the same (+6). Hence, statement C is incorrect.

Q2. Which one of the following ions exhibits d-d transition and paramagnetism as well?

1. CrO4²⁻
2. Cr2O7²⁻
3. MnO4²⁻
4. MnO4⁻

Answer: MnO4²⁻

MnO4²⁻ exhibits d-d transitions because Mn is in a lower oxidation state (+6), allowing partially filled d-orbitals. It is also paramagnetic due to unpaired electrons in its d-orbitals.

Q3. Magnetic moment 2.84 B.M. is given by: (At. nos, Ni = 28, Ti = 22, Cr = 24, Co = 27)

1. Ti³⁺
2. Cr²⁺
3. Co²⁺
4. Ni²⁺

Answer: Cr²⁺

The magnetic moment is calculated using the formula μ = √n(n+2), where n is the number of unpaired electrons. Cr²⁺ has an electronic configuration of [Ar] 3d⁴, with 4 unpaired electrons, giving μ = √4(4+2) = 2.84 B.M.

Q4. Magnetic moment 2.83 B.M is given by which of the following ions? (At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28)

1. Ti³⁺
2. Ni²⁺
3. Cr³⁺
4. Mn²⁺

Answer: Cr³⁺

The magnetic moment is calculated using the formula μ = √n(n+2), where n is the number of unpaired electrons. For Cr³⁺, the electronic configuration is [Ar] 3d³, with 3 unpaired electrons. Substituting n = 3 gives μ = √3(3+2) = √15 ≈ 2.83 B.M.

Q5. KMnO4 can be prepared from K2MnO4 as per the reaction: 3MnO4^2− + 2H2O → 2MnO4^− + MnO2 + 4OH^−. The reaction can go to completion by removing OH^− ions by adding:

1. KOH
2. CO2
3. SO2
4. HCl

Answer: CO2

The reaction can be driven to completion by removing OH⁻ ions. CO₂ reacts with OH⁻ to form HCO₃⁻ or CO₃²⁻, thus reducing the concentration of OH⁻ and shifting the equilibrium forward.

Q6. Which of the statements is not true?

1. On passing H2S through acidified K2Cr2O7 solution, a milky colour is observed.
2. Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis.
3. K2Cr2O7 solution in acidic medium is orange.
4. K2Cr2O7 solution becomes yellow on increasing the pH beyond 7.

Answer: Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis.

Na2Cr2O7 is not preferred over K2Cr2O7 in volumetric analysis because K2Cr2O7 is more stable and less hygroscopic, making it more suitable for precise measurements.

Q7. Copper sulphate dissolves in excess of KCN to give:

1. [Cu(CN)4]^3−
2. [Cu(CN)4]^2−
3. Cu(CN)2
4. CuCN

Answer: [Cu(CN)4]^2−

Copper sulphate reacts with excess KCN to form the complex [Cu(CN)4]^2− due to the strong complexing ability of cyanide ions with Cu(II).

Q8. German silver is an alloy of:

1. Fe, Cr, Ni
2. Cu, Zn, Ag
3. Cu, Zn, Ni
4. Cu, Sn, Al

Answer: Cu, Zn, Ni

German silver is an alloy primarily composed of copper (Cu), zinc (Zn), and nickel (Ni), and it does not contain silver despite its name.

Q9. The d-electron configurations of Cr²⁺, Mn²⁺, Fe²⁺ and Ni²⁺ are 3d⁴, 3d⁵, 3d⁶ and 3d⁸ respectively. Which one of the following aqua complexes will exhibit the minimum paramagnetic behaviour?

1. [Fe(H₂O)₆]²⁺
2. [Ni(H₂O)₆]²⁺
3. [Cr(H₂O)₆]²⁺
4. [Mn(H₂O)₆]²⁺

Answer: [Ni(H₂O)₆]²⁺

Paramagnetic behavior depends on the number of unpaired electrons. [Ni(H₂O)₆]²⁺ has a 3d⁸ configuration, resulting in 2 unpaired electrons, which is the least among the given options.

Q10. Iron carbonyl, [Fe(CO)5] is

1. Tetranuclear
2. Mononuclear
3. Dinuclear
4. Trinuclear

Answer: Mononuclear

Iron carbonyl, [Fe(CO)5], consists of a single iron atom coordinated to five carbonyl ligands, making it mononuclear.

Q11. Which of the following may be considered to be an organometallic compound?

1. Nickel tetracarbonyl
2. Chlorophyll
3. K3[Fe(C2O4)3]
4. [Co(en)3]Cl3

Answer: Nickel tetracarbonyl

Nickel tetracarbonyl (Ni(CO)4) is an organometallic compound because it contains a metal (nickel) directly bonded to carbon atoms of the CO ligands. The other options do not meet this criterion.

Q12. Among the following, which is not the π-bonded organometallic compound?

1. (CH3)4Sn
2. K[PtCl3(η2–C2H4)]
3. Fe(η5–C5H5)2
4. Cr(η6–C6H6)2

Answer: (CH3)4Sn

The compound (CH3)4Sn does not involve π-bonding as it contains only σ-bonds between tin and methyl groups, unlike the other options which involve π-bonded ligands like ethylene, cyclopentadienyl, or benzene.

Q13. Lesser is the number of unpaired electrons, smaller will be the paramagnetic behavior. As Cr2+, Mn2+, Fe2+ & Ni2+ contains:

1. Cr2+ (3d4) = 4 unpaired electrons
2. Mn2+ (3d5) = 5 unpaired electrons
3. Fe2+ (3d6) = 4 unpaired electrons
4. Ni2+ (3d8) = 2 unpaired electrons

Answer: Ni2+ (3d8) = 2 unpaired electrons

Paramagnetic behavior depends on the number of unpaired electrons. Ni2+ (3d8) has the least number of unpaired electrons (2), making it the least paramagnetic among the given ions.

Q14. The magnetic moment μ = √(n(n+2)) is given. On solving for n = 3, Cr3+ will have configuration as 1s2 2s2 2p6 3s2 3p6 3d3. What is the number of unpaired electrons?

1. 3 unpaired electrons
2. 4 unpaired electrons
3. 5 unpaired electrons
4. 6 unpaired electrons

Answer: 3 unpaired electrons

Cr3+ has a configuration of 3d3, meaning there are 3 electrons in the d-orbital. These electrons occupy separate orbitals due to Hund's rule, resulting in 3 unpaired electrons.

Q15. India has the world’s largest deposits of thorium in the form of

1. rutile
2. magnesite
3. lignite
4. monazite

Answer: monazite

Thorium is primarily found in monazite sand, which is abundant in India, especially along its coastal regions. Monazite is a phosphate mineral containing thorium and rare earth elements.

Q16. Among K, Ca, Fe and Zn, the element which can form more than one binary compound with chlorine is:

1. Fe
2. Zn
3. K
4. Ca

Answer: Fe

Iron (Fe) can exhibit multiple oxidation states, such as +2 and +3, allowing it to form more than one binary compound with chlorine, like FeCl2 and FeCl3. The other elements do not exhibit such variability in oxidation states.

Q17. Which of the following does not represent the correct order of the properties indicated?

1. Ni2+ > Cr2+ > Fe2+ > Mn2+ (size)
2. Sc > Ti > Cr > Mn (size)
3. Mn2+ > Ni2+ > Co2+ > Fe2+ (unpaired electron)
4. Fe2+ > Co2+ > Ni2+ > Cu2+ (unpaired electron)

Answer: Fe2+ > Co2+ > Ni2+ > Cu2+ (unpaired electron)

Option D is incorrect because the number of unpaired electrons decreases as we move from Fe2+ to Cu2+ in the d-block due to increasing electron pairing in the d-orbitals.

Q18. Among the given options, only Fe shows variable oxidation states so it can form two chlorides, viz. FeCl₂ and FeCl₃.

1. Among the given options, only Fe shows variable oxidation states so it can form two chlorides, viz. FeCl₂ and FeCl₃.
2. This is because of inter-electronic repulsions between lone pairs.
3. None of the above.
4. None of the above.

Answer: Among the given options, only Fe shows variable oxidation states so it can form two chlorides, viz. FeCl₂ and FeCl₃.

Iron (Fe) exhibits variable oxidation states (+2 and +3), allowing it to form FeCl₂ and FeCl₃. Other elements in the options do not show such variability in oxidation states.

Q19. The manganate and permanganate ions are tetrahedral, due to:

1. The π-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese
2. There is no π-bonding
3. The π-bonding involves overlap of p-orbital of oxygen with p-orbitals of manganese
4. The π-bonding involves overlap of d-orbital of oxygen with d-orbitals of manganese

Answer: The π-bonding involves overlap of p-orbitals of oxygen with d-orbitals of manganese

The tetrahedral shape of manganate and permanganate ions arises because π-bonding occurs via the overlap of p-orbitals of oxygen with d-orbitals of manganese, stabilizing the structure.

Q20. Which of the following statements about the interstitial compounds is incorrect?

1. They are chemically reactive.
2. They are much harder then the pure metal.
3. They have higher melting points than the pure metal.
4. They retain metallic conductivity.

Answer: They are chemically reactive.

Interstitial compounds are generally chemically inert due to the strong metallic bonding and the small non-metal atoms fitting into the metal lattice without disrupting it. Thus, option A is incorrect.