CBSE Class 12 Chemistry The d and f Block Elements MCQs Set 11

Question 1. The ions of metals of Group 12 [Zn, Cd and Hg] have completely filled \( d \)-orbitals and so they
(a) behave like semiconductors.
(b) have very high melting points.
(c) do not behave like transition metals.
(d) behave like superconductors.
Answer: (c) do not behave like transition metals.
In simple words: Since these metals have all their d-orbitals completely full, they are not counted as true transition elements.

Exam Tip: Remember that transition metals are defined by having incomplete d-orbitals in their ground state or common oxidation states.

Question 2. Metallic radii of some transition elements are given below. Which of these elements will have highest density?

(a) Fe
(b) Ni
(c) Co
(d) Cu
Answer: (d) Cu
When moving from left to right across a period, the metallic radius falls while the atomic mass rises. This reduction in metallic radius along with an increase in atomic weight leads to an elevated density in metals. Thus, among the four options, Cu is located on the far right of these d-block elements and possesses the greatest density (\( 8.9\text{ g/cm}^3 \)).
In simple words: As you move right in the periodic table, atoms get heavier and slightly smaller, packing more tightly. This makes copper the densest of these four metals.

Exam Tip: Density depends on both atomic mass and atomic volume; since copper has the highest mass and a small radius among the four, its density is the highest.

Question 3. Which of the following statement is not correct?
(a) Copper liberates hydrogen from acids.
(b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine.
(c) \( \text{Mn}^{3+} \) and \( \text{Co}^{3+} \) are oxidising agents in aqueous solution.
(d) \( \text{Ti}^{2+} \) and \( \text{Cr}^{2+} \) are reducing agents in aqueous solution.
Answer: (a) Copper liberates hydrogen from acids.
Copper is positioned below hydrogen within the electrochemical series and consequently does not release \( \text{H}_2 \) gas from acidic solutions. Thus, option (a) is incorrect. The other three choices (b, c, and d) represent true statements.
In simple words: Copper is not reactive enough to push hydrogen out of acids because it sits below hydrogen in the reactivity series.

Exam Tip: Remember that metals with positive standard reduction potentials (like copper, \( E^\circ = +0.34\text{ V} \)) cannot reduce \( \text{H}^+ \) ions to hydrogen gas.

Question 4. Which of the following transition metals does not show variable oxidation state?
(a) Ti
(b) Cr
(c) Cu
(d) Sc
Answer: (d) Sc
Scandium fails to exhibit multiple oxidation states because it only shows a stable \( +3 \) oxidation state.
In simple words: Scandium has only one stable oxidation state, which is +3, because it easily loses all three of its outer electrons at once.

Exam Tip: Scandium has the electronic configuration \( [\text{Ar}]3d^1 4s^2 \); losing all three valence electrons gives a stable noble gas configuration, leaving no other stable states.

Question 5. Which of the following characteristics of transition metals is associated with their catalytic activity?
(a) Paramagnetic nature
(b) Colour of hydrated ions
(c) High enthalpy of atomisation
(d) Variable oxidation states
Answer: (d) Variable oxidation states
The catalytic behaviour of transition metals is attributed to their ability to show multiple oxidation states.
In simple words: Transition metals make great catalysts because they can easily change their charge to help chemical reactions happen faster.

Exam Tip: Catalysts function by providing an alternative reaction pathway with lower activation energy, which is facilitated by the transition metal's variable oxidation states.

Question 6. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of \( \text{Cr}^{3+} \) ion (Atomic no: Cr = 24) is...
(a) 2.87 B.M.
(b) 3.87 B.M.
(c) 3.47 B.M.
(d) 3.57 B.M.
Answer: (b) 3.87 B.M.
The configuration of chromium is \( \text{Cr} = [\text{Ar}]3d^5 4s^1 \). For the trivalent ion, it becomes \( \text{Cr}^{3+} = [\text{Ar}]3d^3 4s^0 \), giving \( n = 3 \) unpaired electrons. Thus, the spin-only magnetic moment is calculated as: \[ \mu_{\text{spin only}} = \sqrt{n(n+2)}\text{ B.M.} = \sqrt{3(3+2)}\text{ B.M.} = \sqrt{15}\text{ B.M.} \approx 3.87\text{ B.M.} \]
In simple words: Since Chromium-3+ has 3 unpaired electrons, we use the magnetic moment formula to find that its strength is 3.87 B.M.

Exam Tip: A quick shortcut: the spin-only magnetic moment value is always 'n point something' where n is the number of unpaired electrons (e.g., for \( n=3 \), the value is \( 3.87\text{ B.M.} \)).

Question 7. Which of the following characteristics make transition elements good catalysts?
P. Their tendency to form reaction intermediates with the reactants thereby reducing the activation energy.
Q. Their ability to have multiple oxidation states.
R. Their ability to form complex compounds.
(a) Only P
(b) Only Q
(c) Only Q and R
(d) P, Q and R
Answer: (d) P, Q and R
Each of the listed statements is correct.
In simple words: Transition metals are excellent catalysts because they can form temporary compounds, change their oxidation states, and bind easily with other molecules.

Exam Tip: Both physical properties (like surface area and complex formation) and chemical properties (like variable valency) contribute to transition metals acting as efficient catalysts.

Question 8. Two important compounds of transition element chromium are \( \text{K}_2\text{Cr}_2\text{O}_7 \) and \( \text{K}_2\text{CrO}_4 \). Compound \( \text{K}_2\text{Cr}_2\text{O}_7 \) is orange in colour and \( \text{K}_2\text{CrO}_4 \) is yellow in colour. The colour observed is because chromium ion in their compounds
(a) contain completely filled \( d \)-orbitals.
(b) contain empty \( d \)-orbitals.
(c) undergo \( d \)-\( d \) transition of electrons.
(d) undergo charge transfer between oxide ion and itself.
Answer: (d) undergo charge transfer between oxide ion and itself.
The brilliant orange shade of dichromate is caused by the transfer of charge (electrons) from the oxide ligands to the vacant \( d \)-orbitals of the central \( \text{Cr}^{6+} \) ion, which is referred to as ligand-to-metal charge transfer (LMCT).
In simple words: The bright color doesn't come from regular electron jumps inside the metal; instead, electrons temporarily jump from oxygen to chromium, absorbing light.

Exam Tip: In species like \( \text{CrO}_4^{2-} \) and \( \text{Cr}_2\text{O}_7^{2-} \), chromium has a \( d^0 \) configuration, so \( d \)-\( d \) transitions are impossible; the color is solely due to charge transfer.

Question 9. Which of the following statement is not correct?
(a) \( \text{La(OH)}_3 \) is less basic than \( \text{Lu(OH)}_3 \).
(b) La is actually an element of transition series rather than lanthanoids.
(c) Atomic radius of Zr and Hf is same.
(d) In lanthanoid series, the ionic radius of \( \text{Lu}^{3+} \) is smallest.
Answer: (a) \( \text{La(OH)}_3 \) is less basic than \( \text{Lu(OH)}_3 \).
Only option (a) is false. The accurate version of this statement is that \( \text{La(OH)}_3 \) is more basic than \( \text{Lu(OH)}_3 \).
In simple words: Lanthanum hydroxide is actually more basic than lutetium hydroxide because lanthanum has a larger size, making it lose its OH group more easily.

Exam Tip: As ionic size decreases from \( \text{La}^{3+} \) to \( \text{Lu}^{3+} \) (lanthanoid contraction), the covalent character of the \( \text{M}-\text{OH} \) bond increases, which decreases basicity.

Question 10. The silver UK coins are an alloy of
(a) Cu and Zn
(b) Cu and Fe
(c) Cu and Ni
(d) Cu and Al
Answer: (c) Cu and Ni
The silver-colored UK coins consist of a mixture of copper (\( \text{Cu} \)) and nickel (\( \text{Ni} \)).
In simple words: Even though they look silver, these British coins are actually made from cupronickel, which is a mix of copper and nickel.

Exam Tip: Cupronickel is highly resistant to corrosion and is widely used for minting silver-colored coins.

Question 11. From the elements of \( 3d \)-series given below, which element shows the maximum number of oxidation states?
(a) Scandium
(b) Manganese
(c) Chromium
(d) Titanium
Answer: (b) Manganese
Manganese (\( \text{Mn} \)) demonstrates the highest count of oxidation states, displaying levels from \( +2, +3, +4, +5, +6 \) up to \( +7 \).
In simple words: Manganese has the most unpaired electrons in its outer shells, letting it lose or share anywhere from 2 to 7 electrons.

Exam Tip: Manganese has the electronic configuration \( 3d^5 4s^2 \). Using all seven valence electrons allows it to show the maximum oxidation state of \( +7 \) in the \( 3d \)-series.

Question 12. Transition metals are known to make interstitial compounds. Formation of interstitial compounds makes the transition metal
(a) more hard
(b) more soft
(c) more ductile
(d) more metallic
Answer: (a) more hard
The creation of interstitial compounds causes the transition metal to become much harder.
In simple words: When tiny non-metal atoms fit into the gaps inside a metal's structure, they lock the metal atoms in place so they can't slide around, making it harder.

Exam Tip: Interstitial compounds retain metallic conductivity and high chemical stability, but show increased hardness and melting point compared to the host metal.

Question 13. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment?
(a) \( 3d^7 \)
(b) \( 3d^5 \)
(c) \( 3d^8 \)
(d) \( 3d^2 \)
Answer: (b) \( 3d^5 \)
An increased quantity of unpaired electrons leads to a greater magnetic moment. Because the \( 3d^5 \) system has \( 5 \) unpaired electrons, it displays the greatest magnetic moment: \[ \mu = \sqrt{5(5+2)}\text{ B.M.} = \sqrt{35}\text{ B.M.} \approx 5.95\text{ B.M.} \ ]
In simple words: The configuration with 5 single electrons has the highest number of unpaired spins, giving it the strongest magnetic pull.

Exam Tip: Always remember Hund's rule: fill orbitals singly first. A \( 3d^5 \) configuration has 5 unpaired electrons, whereas \( 3d^7 \) has only 3, and \( 3d^8 \) has 2.

Question 14. Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds?
(a) They have high melting points in comparison to pure metals.
(b) They are very hard.
(c) They retain metallic conductivity.
(d) They are chemically very reactive.
Answer: (d) They are chemically very reactive.
These compounds are created when tiny atoms get locked within the metallic crystal structure. Their primary traits include:
(i) They are extremely hard and sturdy.
(ii) They possess elevated melting temperatures compared to pure metallic elements.
(iii) They maintain electrical conductivity similar to the pure metal.
(iv) They gain chemical stability, meaning they are mostly unreactive, rather than highly reactive.
In simple words: Instead of being very reactive, interstitial compounds are actually very chemically stable and don't react easily with other substances.

Exam Tip: Know the key properties of interstitial compounds: hard, high melting point, conductive, and chemically inert.

Question 15. Which of the following reactions are disproportionation reactions?
(i) \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu} \)
(ii) \( 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O} \)
(iii) \( 2\text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2 \)
(iv) \( 2\text{MnO}_4^- + 3\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow 5\text{MnO}_2 + 4\text{H}^+ \)
(a) (i)
(b) (i), (ii) and (iii)
(c) (ii), (iii) and (iv)
(d) (i) and (ii)
Answer: (d) (i) and (ii)
A chemical process where both oxidation and reduction happen to the same atom at once is termed a disproportionation reaction. Both reactions (i) and (ii) display this behavior.
In (i), \( \text{Cu}^+ \) (+1) is oxidized to \( \text{Cu}^{2+} \) (+2) and reduced to \( \text{Cu} \) (0).
In (ii), \( \text{MnO}_4^{2-} \) (+6) is oxidized to \( \text{MnO}_4^- \) (+7) and reduced to \( \text{MnO}_2 \) (+4).
In simple words: Disproportionation is when one kind of ion splits into two, with some gaining charge and others losing charge. This happens to both copper and manganate ions here.

Exam Tip: To identify a disproportionation reaction, find the element that simultaneously goes to a higher and a lower oxidation state in the products.

Question 16. In which of the following oxo metal anions does the metal not exhibit an oxidation state equal to its group number?
(a) \( \text{CrO}_4^{2-} \)
(b) \( \text{MnO}_4^- \)
(c) \( \text{Cr}_2\text{O}_7^{2-} \)
(d) \( \text{MnO}_4^{2-} \)
Answer: (d) \( \text{MnO}_4^{2-} \)
In both \( \text{Cr}_2\text{O}_7^{2-} \) and \( \text{CrO}_4^{2-} \), the oxidation state of chromium is \( +6 \). For chromium (Group 6), the group number matches the total count of valence electrons in the \( (n-1)d \) and \( ns \) subshells (\( 5 + 1 = 6 \)). In \( \text{MnO}_4^- \), the oxidation state of manganese is \( +7 \), which matches its group number (Group 7, with valence electrons \( 5 + 2 = 7 \)). However, in \( \text{MnO}_4^{2-} \), the oxidation state of manganese is \( +6 \), which does not match its group number of 7.
In simple words: Manganese is in Group 7, but in the manganate ion, its charge is only +6. Since 6 does not equal 7, this is our answer.

Exam Tip: Group number for a d-block element equals the sum of its outer s and d electrons. Match this with the calculated oxidation state of the metal to quickly answer.

Question 17. Out of the following transition elements, the maximum number of oxidation states is shown by
(a) Sc
(b) Cr
(c) Mn
(d) Fe
Answer: (c) Mn
Manganese, situated in the center of the transition series, displays the greatest variety of oxidation states, and also reaches the highest oxidation level in the entire period due to having five unpaired electrons.
In simple words: Manganese can show oxidation states all the way from +2 to +7 because it has the maximum number of single electrons available for bonding.

Exam Tip: Elements in the middle of a transition series exhibit the maximum number of oxidation states because they have the most d and s electrons to share.

Question 18. Acidified \( \text{KMnO}_4 \) oxidises sulphite to
(a) \( \text{S}_2\text{O}_3^{2-} \)
(b) \( \text{S}_2\text{O}_8^{2-} \)
(c) \( \text{SO}_2(\text{g}) \)
(d) \( \text{SO}_4^{2-} \)
Answer: (d) \( \text{SO}_4^{2-} \)
Under acidic conditions, acidified \( \text{KMnO}_4 \) converts sulphite into sulphate: \[ 5\text{SO}_3^{2-} + 2\text{MnO}_4^- + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{SO}_4^{2-} + 3\text{H}_2\text{O} \ ]
In simple words: Potassium permanganate is a very strong oxidizer. It adds oxygen to sulphite ions to turn them into sulphate ions.

Exam Tip: Write down the balanced ionic equation to see that the oxidation state of sulfur increases from +4 (in sulphite) to +6 (in sulphate).

Question 19. \( \text{KMnO}_4 \) acts as an oxidising agent in acidic medium. The number of moles of \( \text{KMnO}_4 \) that will be needed to react with one mole of sulphide ions in acidic solution is
(a) \( \frac{2}{5} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{4}{5} \)
(d) \( \frac{1}{5} \)
Answer: (a) \( \frac{2}{5} \)
The oxidation reaction where \( \text{KMnO}_4 \) serves as an oxidant in acidic conditions is: \[ 2\text{MnO}_4^- + 16\text{H}^+ + 5\text{S}^{2-} \rightarrow 2\text{Mn}^{2+} + 5\text{S} + 8\text{H}_2\text{O} \] From the balanced ionic equation, \( 5\text{ moles} \) of sulphide (\( \text{S}^{2-} \)) ions require \( 2\text{ moles} \) of \( \text{MnO}_4^- \). Consequently, \( 1\text{ mole} \) of sulphide ion will react with: \[ \frac{2}{5}\text{ moles of KMnO}_4 \ ]
In simple words: Using the balanced chemical equation, 5 sulphide ions need 2 permanganate ions to react completely. So, 1 sulphide ion needs 2/5 of a permanganate ion.

Exam Tip: Using the ion-electron method: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \) (5-electron change) and \( \text{S}^{2-} \rightarrow \text{S} + 2e^- \) (2-electron change). Equating the electrons gives the mole ratio directly: \( 5\text{ S}^{2-} : 2\text{ MnO}_4^- \).

Question 20. \( \text{KMnO}_4 \) acts as an oxidising agent in alkaline medium. When alkaline \( \text{KMnO}_4 \) is treated with KI, iodide ion is oxidised to
(a) \( \text{I}_2 \)
(b) \( \text{IO}^- \)
(c) \( \text{IO}_3^- \)
(d) \( \text{IO}_4^- \)
Answer: (c) \( \text{IO}_3^- \)
In basic conditions, \( \text{KMnO}_4 \) functions as an oxidizing agent. Upon treating potassium iodide (\( \text{KI} \)) with alkaline \( \text{KMnO}_4 \), the iodide ion gets oxidized to iodate (\( \text{IO}_3^- \)): \[ 2\text{MnO}_4^- + \text{H}_2\text{O} + \text{I}^- \rightarrow 2\text{MnO}_2 + 2\text{OH}^- + \text{IO}_3^- \ ]
In simple words: In an alkaline (basic) solution, permanganate is less aggressive, so it oxidizes iodide ions up to iodate (\( \text{IO}_3^- \)) instead of just free iodine.

Exam Tip: Note the medium: in acidic medium, iodide (\( \text{I}^- \)) is oxidized to iodine (\( \text{I}_2 \)), but in alkaline or neutral medium, it is oxidized to iodate (\( \text{IO}_3^- \)). This is a very common exam question!

Question 21. On addition of small amount of \( \text{KMnO}_4 \) to concentrated \( \text{H}_2\text{SO}_4 \), a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following
(a) \( \text{Mn}_2\text{O}_7 \)
(b) \( \text{MnO}_2 \)
(c) \( \text{MnSO}_4 \)
(d) \( \text{Mn}_2\text{O}_3 \)
Answer: (a) \( \text{Mn}_2\text{O}_7 \)
When adding \( \text{KMnO}_4 \) to concentrated sulfuric acid, a green, oily substance of \( \text{Mn}_2\text{O}_7 \) is produced, which is extremely explosive: \[ 2\text{KMnO}_4 + 2\text{H}_2\text{SO}_4\text{(conc.)} \rightarrow \text{Mn}_2\text{O}_7 + 2\text{KHSO}_4 + \text{H}_2\text{O} \ ]
In simple words: Mixing potassium permanganate with strong sulfuric acid creates manganese heptoxide, a very dangerous green oil that can explode easily.

Exam Tip: Manganese heptoxide (\( \text{Mn}_2\text{O}_7 \)) is a highly acidic, covalent oxide of manganese in its highest (+7) oxidation state, making it extremely unstable.

Question 22. Which of the following has abnormally low value of third ionisation enthalpy?
(a) Lanthanum
(b) Gadolinium
(c) Lutetium
(d) All of the options
Answer: (d) All of the options
Lanthanum, gadolinium, and lutetium all possess abnormally low third ionization energies.
In simple words: For these three elements, removing the third electron is surprisingly easy because it leaves behind a very stable empty, half-filled, or completely filled f-subshell.

Exam Tip: The unusually low third ionization energy of \( \text{La} \), \( \text{Gd} \), and \( \text{Lu} \) is due to the extra stability of \( f^0 \), \( f^7 \), and \( f^{14} \) configurations, respectively.

Question 23. Match the properties given in Column I with the metals in Column II.

(a) (i)-(C), (ii)-(B), (iii)-(A)
(b) (i)-(C), (ii)-(A), (iii)-(B)
(c) (i)-(A), (ii)-(B), (iii)-(C)
(d) (i)-(B), (ii)-(A), (iii)-(C)
Answer: (a) (i)-(C), (ii)-(B), (iii)-(A)
The accurate pairing for these columns is: (i)-(C), (ii)-(B), and (iii)-(A).
In simple words: Curium is the actinide with 7 electrons in its f-orbital. Lutetium-3+ has its 14-electron f-shell completely full, and Cerium is known to show a +4 state.

Exam Tip: Memorize key electronic configurations: Curium (\( Z = 96 \)) is \( [\text{Rn}]5f^7 6d^1 7s^2 \) (half-filled), and Lutetium (\( Z = 71 \)) in \( +3 \) state becomes \( [\text{Xe}]4f^{14} \) (completely filled).

Assertion - Reason

Directions - In the following questions an Assertion (A) is followed by a corresponding Reason (R). 
Use the following keys to choose the appropriate answer,
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A),
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A),
(c) (A) is true, but (R) is false,
(d) (A) is false, but (R) is true.

Question 24. Assertion (A) The electronic configuration of \( \text{Cu(II)} \) is \( 3d^9 \), whereas that of \( \text{Cu(I)} \) is \( 3d^{10} \).
Reason (R) \( \text{Cu(II)} \) has greater effective nuclear charge.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
Cu(II) is more stable than Cu(I). Although Cu(I) possesses a stable \( 3d^{10} \) configuration, Cu(II) has a \( 3d^9 \) configuration. However, Cu(II) exhibits greater stability because of a higher effective nuclear charge, which allows it to attract 17 electrons instead of the 18 held in Cu(I). Therefore, both the Assertion and Reason are correct, and the Reason provides the proper explanation.
In simple words: Copper-II is more stable because its nucleus has a stronger grip on its remaining electrons. Both statements are true and the reason explains the statement correctly.

Exam Tip: Even though Cu(I) has a fully filled d-orbital, Cu(II) is often more stable in aqueous solutions due to its high hydration enthalpy and stronger effective nuclear charge.

Question 25. Assertion (A) Zinc is not regarded as a transition element.
Reason (R) In zinc, \( 3d \) orbitals are completely filled in its ground state as well as in its oxidised state.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
Both the Assertion and Reason are true, and the Reason acts as the correct explanation for the Assertion.
In simple words: Zinc is not a transition metal because its d-orbitals are always completely full, both as a pure metal and in all of its compounds.

Exam Tip: Transition metals are defined as having partially filled d-orbitals in their ground state or in any of their common ions, which zinc lacks.

Question 26. Assertion (A) Copper is a non-transition element.
Reason (R) Copper has completely filled \( d \)-orbitals in its ground state.
Answer: (d) (A) is false, but (R) is true.
The Assertion (A) is false, but the Reason (R) is true. Copper is classified as a transition element since its \( \text{Cu}^{2+} \) ion possesses a partially filled \( d \)-subshell (\( 3d^9 \)).
In simple words: Copper is actually a transition metal because its common copper-II ion has an incomplete set of d-electrons, even though its pure metal state has a full d-orbital.

Exam Tip: Copper has a \( 3d^{10} 4s^1 \) configuration in its ground state, but in its stable \( +2 \) state, it has a \( 3d^9 \) configuration, which makes it a transition element.

Question 27. Assertion (A) Zn do not show any characteristic properties of transition elements except complex formation.
Reason (R) Zinc has completely filled \( 3d \)-subshell.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
The \( 3d \)-subshell in zinc remains fully occupied both in its elemental form and in its common ionic states. Consequently, it fails to display most defining features of transition metals, with the exception of forming complexes. Therefore, both Assertion (A) and Reason (R) are correct, and the Reason successfully explains the Assertion.
In simple words: Because zinc's outer d-shell is completely full, it does not act like other transition metals, except for its ability to bind with other molecules to make complex compounds.

Exam Tip: Because Zn has a stable \( 3d^{10} \) shell, it doesn't show transition properties like variable valency, catalytic activity, or colored compounds in its usual states.

Question 28. Assertion (A) Zr and Hf have almost identical radii.
Reason (R) Both Zr and Hf exhibit similar properties.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Both Zr and Hf share nearly identical atomic radii as a result of lanthanoid contraction.
In simple words: Zirconium and Hafnium have nearly the same size and show very similar chemistry, but their similar properties don't explain why they are the same size.

Exam Tip: Lanthanoid contraction is the main cause behind the identical atomic radii of Zr and Hf, which makes their chemical separation very difficult.

Question 29. Assertion (A) Highest oxidation state of manganese in fluoride is +4 (\( \text{MnF}_4 \)) but highest oxidation state in oxides is +7 (\( \text{Mn}_2\text{O}_7 \)).
Reason (R) Fluorine stabilises lower oxidation state.
Answer: (c) (A) is true, but (R) is false.
The maximum oxidation state manganese reaches in fluorides is \( +4 \) (\( \text{MnF}_4 \)), whereas in oxides it achieves \( +7 \) (\( \text{Mn}_2\text{O}_7 \)). This occurs because fluorine can only establish single bonds, while oxygen is capable of forming multiple double bonds with the metal. Consequently, Assertion (A) is true, but Reason (R) is incorrect.
In simple words: Oxygen is much better at bringing out high oxidation states because it can form double bonds, whereas fluorine can only make single bonds. So, the reason is false.

Exam Tip: Oxygen's capacity to form multiple bonds with transition metals allows it to stabilize higher oxidation states (like \( \text{Mn}_2\text{O}_7 \)) compared to fluorine.

Question 30. Assertion (A) When \( \text{KMnO}_4 \) solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time.
Reason (R) When \( \text{KMnO}_4 \) solution is added to oxalic acid solution \( \text{CO}_2 \) is formed as product.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
When a \( \text{KMnO}_4 \) solution is mixed with oxalic acid, the loss of color is initially slow but turns very rapid after a short while because the produced \( \text{Mn}^{2+} \) ions act as an autocatalyst. Therefore, both Assertion (A) and Reason (R) are correct, but the Reason does not explain the Assertion.
In simple words: The reaction starts slowly but speeds up quickly because one of the products, manganese-II, acts as a catalyst to speed up the rest of the reaction.

Exam Tip: This is a classic example of autocatalysis, where the catalyst (\( \text{Mn}^{2+} \)) is generated as a product during the reaction itself.

Question 31. Assertion (A) As the atomic number of lanthanoids increases, their atomic radius decreases.
Reason (R) From Eu to Yb, as the atomic number increases atomic radius decreases.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Both statements are true, but the Reason is not the correct explanation for the Assertion.
In simple words: While both statements about sizes shrinking are true, saying that sizes shrink from europium to ytterbium doesn't explain why they shrink across the whole series.

Exam Tip: Lanthanoid contraction is caused by the poor shielding effect of 4f electrons, which fails to offset the growing nuclear charge as atomic number increases.

Question 32. Assertion (A) \( \text{Cr}^{2+} \) is reducing, while \( \text{Mn}^{3+} \) is oxidising even when both have \( d^4 \)-configuration.
Reason (R) Configuration of Cr changes from \( d^3 \) to \( d^4 \).
Answer: (c) (A) is true, but (R) is false.
The Assertion is correct, but the Reason is false. \( \text{Cr}^{2+} \) acts as a reducing agent because its configuration transforms from \( d^4 \) to \( d^3 \), where the \( d^3 \) state features a stable, half-filled \( t_{2g} \) subshell in an octahedral field. Conversely, \( \text{Mn}^{3+} \) is oxidizing because gaining an electron converts it from \( d^4 \) to \( d^5 \), which offers a stable, half-filled \( d \)-shell configuration. Since the Reason claims the configuration changes from \( d^3 \) to \( d^4 \), it is false.
In simple words: Chromium-2+ wants to lose an electron to get a very stable half-filled level, while Manganese-3+ wants to gain one for the same reason. The reason statement is incorrect.

Exam Tip: In aqueous medium, \( \text{Cr}^{2+} \) is oxidized to \( \text{Cr}^{3+} \) to achieve the stable half-filled \( t_{2g}^3 \) configuration, making it a strong reducing agent.

Question 33. Assertion (A) \( \text{KMnO}_4 \) oxidises oxalic acid to \( \text{CO}_2 \) and itself changes to \( \text{Mn}^{2+} \) ion.
Reason (R) \( \text{KMnO}_4 \) acts as an oxidising agent.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
\( \text{KMnO}_4 \) oxidizes oxalic acid into carbon dioxide while itself converting to the nearly colorless \( \text{Mn}^{2+} \) ion, owing to the fact that \( \text{KMnO}_4 \) serves as a potent oxidizing agent. Thus, both statements are correct, and the Reason provides the right explanation.
In simple words: Permanganate is an excellent oxidizing agent, which means it easily oxidizes the oxalic acid while getting reduced to colorless manganese-II ions.

Exam Tip: In acidic medium, Mn in \( \text{MnO}_4^- \) (+7) gains 5 electrons to become \( \text{Mn}^{2+} \) (+2), acting as a powerful oxidizing agent.

Question 34. Assertion (A) The highest oxidation state of osmium is +8.
Reason (R) Osmium is a \( 5d \)-block element.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Both the Assertion and Reason are correct, but the Reason does not explain the Assertion. Osmium reaches a maximum oxidation state of \( +8 \) because it can utilize all of its 8 valence electrons (\( 2 \) from the \( 6s \) orbital and \( 6 \) from the \( 5d \) orbitals) for bonding.
In simple words: Osmium is indeed a 5d metal, and its highest oxidation state is +8. However, being in the 5d block doesn't automatically mean a metal has an oxidation state of +8.

Exam Tip: Ruthenium and Osmium can display the maximum oxidation state of \( +8 \) in their tetroxides (\( \text{RuO}_4 \) and \( \text{OsO}_4 \)) by utilizing all outer s and d electrons.

Question 35. Assertion (A) \( \text{Ce}^{4+} \) ion is good analytical reagent.
Reason (R) \( \text{Ce}^{4+} \) has a stable electronic configuration.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Both the Assertion and Reason are correct, but the Reason is not the correct explanation for the Assertion. \( \text{Ce}^{4+} \) is widely used as an analytical reagent because it acts as a strong oxidant, easily reducing to the highly stable \( \text{Ce}^{3+} \) state.
In simple words: Although cerium-IV has a stable noble-gas electron setup, it still prefers the +3 state common to all lanthanides, making it a great oxidizer for chemical analysis.

Exam Tip: \( \text{Ce}^{4+} \) is a powerful oxidizing agent in aqueous solution with an \( E^\circ \) of about \( +1.74\text{ V} \), which makes it excellent for volumetric analysis (cerimetry).

Case Study 36

The elements which show the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example exhibits all the oxidation states from (+ 2) to (+ 7). The lesser number of oxidation states at the extreme ends is due to either too few electrons to lose or share (Sc, Ti) or too many d electrons (hence fewer orbitals available in which to share electrons with others) for higher valence (Cu, Zn). Thus early in the series scandium (II) is virtually unknown and titanium (IV) is more stable than Ti (III) or Ti (II). On the other end, the only oxidation state of zinc is + 2 (no d electrons are involved). The maximum oxidation states of reasonable stability correspond in value to the sum of the s and d electrons upto manganese (\( \text{Ti}^{\text{IV}}\text{O}_2, \text{V}^{\text{V}}\text{O}_2^+, \text{Cr}^{\text{VI}}\text{O}_4^{2-}, \text{Mn}^{\text{VII}}\text{O}_4^- \)) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are \( \text{Fe}^{\text{II, III}}, \text{Co}^{\text{II, III}}, \text{Ni}^{\text{II}}, \text{Cu}^{\text{I, II}}, \text{Zn}^{\text{II}} \).
The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., V(II), V(III), V(IV), V(V). This is in contrast with the variability of oxidation states of non-transition elements where, oxidation states normally differ by a unit of two. An interesting feature in the variability of oxidation states of the d-block elements is noticed among the groups. Although in the p-block, the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the group of d-block.
For example in group 6, Mo (VI) and W (VI) are found to be more stable than Cr (VI). Thus Cr (VI), in the form of dichromate in acidic medium is a strong oxidising agent, whereas \( \text{MoO}_3 \) and \( \text{WO}_3 \) are not. Low oxidation states are found when a complex compound has ligands capable of \( \pi \)-acceptor character in addition to the \( \sigma \)-bonding. For example, in \( \text{Ni(CO)}_4 \) and \( \text{Fe(CO)}_5 \), the oxidation state of nickel and iron is zero.

Question 36. (i) Why \( \text{Fe}^{2+} \) compounds are less stable than \( \text{Mn}^{2+} \) compounds towards +3 oxidation states?
Answer: The \( \text{Fe}^{2+} \) ion has a \( d^6 \) electronic configuration, whereas the \( \text{Mn}^{2+} \) ion has a \( d^5 \) configuration, which is half-filled and therefore highly stable. Consequently, \( \text{Mn}^{2+} \) compounds are much more stable than \( \text{Fe}^{2+} \) compounds, meaning \( \text{Fe}^{2+} \) oxidizes more easily to the \( +3 \) state than \( \text{Mn}^{2+} \) does.
In simple words: Manganese-2+ already has a perfectly half-filled outer shell, which is very stable, while Iron-2+ has one extra electron it is eager to lose to get that same stable half-filled state.

Exam Tip: A half-filled d-subshell (\( d^5 \)) has high exchange energy and symmetrical charge distribution, giving it exceptional stability.

Question 36. (ii) Calculate the magnetic moment of (a) \( \text{Ti}^{3+} \) and (b) \( \text{Ni}^{2+} \)
Answer: For (a) \( \text{Ti}^{3+} \): The configuration is \( 3d^1 \), which gives \( n = 1 \) unpaired electron. \[ \mu = \sqrt{n(n+2)}\text{ B.M.} = \sqrt{1(1+2)}\text{ B.M.} = \sqrt{3}\text{ B.M.} \approx 1.73\text{ B.M.} \] For (b) \( \text{Ni}^{2+} \): The configuration is \( 3d^8 \), which gives \( n = 2 \) unpaired electrons. \[ \mu = \sqrt{n(n+2)}\text{ B.M.} = \sqrt{2(2+2)}\text{ B.M.} = \sqrt{8}\text{ B.M.} \approx 2.83\text{ B.M.} \ ]
In simple words: Titanium-3+ has 1 unpaired electron, giving a magnetic strength of 1.73 B.M. Nickel-2+ has 2 unpaired electrons, resulting in 2.83 B.M.

Exam Tip: Ensure you show the formula \( \mu = \sqrt{n(n+2)} \) and state the correct units (B.M. or Bohr Magneton) to secure full marks.

Question 36. (iii) Why \( \text{Cu}^+ \) ion is not stable in aqueous solution?
Answer: In aqueous systems, \( \text{Cu}^+ \) ions are unstable because their hydration enthalpy is less exothermic (less negative) compared to that of the \( \text{Cu}^{2+} \) ion, which more than offsets the second ionization energy of copper.
In simple words: Water molecules bind much more tightly to copper-II ions than copper-I ions, releasing lots of energy that makes copper-II far more stable in water.

Exam Tip: In water, \( \text{Cu}^+ \) undergoes disproportionation: \( 2\text{Cu}^+(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + \text{Cu}(\text{s}) \), driven by the highly favorable hydration energy of \( \text{Cu}^{2+} \).

Question 36. (iii) Or Why transition metal exhibit highest oxidation states with oxide or fluoride only?
Answer: Alternatively, transition elements display their maximum oxidation states when bonded with oxides and fluorides due to the small atomic size and exceptionally high electronegativity of oxygen and fluorine.
In simple words: Oxygen and fluorine are tiny atoms that love electrons so much they can pull lots of electrons away from transition metals, bringing out their highest possible charges.

Exam Tip: Both small size and high electronegativity are crucial because they enable oxygen and fluorine to stabilize high oxidation states via strong covalent or ionic bonding.

Case Study 37

The first ionisation enthalpies of the lanthanoids are around 600 kJ mol^-1, the second about 1200 kJ mol^-1 comparable with those of calcium. A detailed discussion of the variation of the third ionisation enthalpies indicates that the exchange enthalpy considerations appear to impart a certain degree of stability of empty, half-filled and completely filled orbitals f level. This is indicated from the abnormally low value of the third ionisation enthalpy of lanthanum, gadolinium and lutetium. In their chemical behaviour, in general, the earlier members of the series are quite reactive similar to calcium but, with increasing atomic number, they behave more like aluminium. Values for \( \text{E}^\circ \) for the half-reaction: \( \text{Ln}^{3+}(\text{aq}) + 3e^- \rightarrow \text{Ln}(\text{s}) \) are in the range of -2.2 to -2.4 V except for Eu for which the value is -2.0 V. This is, of course, a small variation. The metals combine with hydrogen when gently heated in the gas. The carbides, \( \text{Ln}_3\text{C} \), \( \text{Ln}_2\text{C}_3 \), and \( \text{LnC}_2 \) are formed when the metals are heated with carbon. They liberate hydrogen from dilute acids and burn in halogens to form halides. They form oxides \( \text{M}_2\text{O}_3 \) and hydroxides \( \text{M(OH)}_3 \). The hydroxides are definite compounds, not just hydrated oxides. They are basic like alkaline earth metal oxides and hydroxides. The best single use of the lanthanoids is for the production of alloy steels for plates and pipes. A well known alloy is mischmetal which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al. A good deal of mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.

Question 37. (i) (a) What is the nature of bonding in \( \text{La}_2\text{O}_3 \) and \( \text{Lu}_2\text{O}_3 \)?
Answer: With a reduction in ionic size, the covalent character of bonds rises. Consequently, \( \text{La}_2\text{O}_3 \) exhibits an ionic nature, whereas \( \text{Lu}_2\text{O}_3 \) is covalent in nature.
In simple words: As lanthanide atoms get smaller from lanthanum to lutetium, they share electrons more instead of completely giving them up. This makes lanthanum oxide ionic and lutetium oxide covalent.

Exam Tip: According to Fajan's rules, smaller cations polarize anions more effectively, leading to increased covalent character in their compounds.

Question 37. (i) (b) What is the trend in the stability of oxo salts of lanthanoid from La to Lu?
Answer: As the atomic radius shrinks because of lanthanoid contraction, the stability of the corresponding oxo salts also declines from La to Lu.
In simple words: The stability of oxo salts decreases as you go from lanthanum to lutetium because the metal ions get smaller and smaller.

Exam Tip: Lanthanoid contraction increases the charge-to-size ratio, which increases polarising power and decreases the thermal stability of their oxo salts.

Question 37. (ii) Among lanthanoids, Ln (III) compounds predominate over Ln (II) and Ln (IV).
Answer: Virtually all lanthanoids display a highly stable \( +3 \) oxidation state as their dominant state. However, certain elements like Ce(IV), Nd(II), Nd(IV), Sm(II), Eu(II), Tm(II), Yb(II), Tb(IV), and Dy(IV) also display \( +2 \) or \( +4 \) states in order to reach stable \( f^0 \), \( f^7 \), or \( f^{14} \) configurations.
In simple words: The +3 oxidation state is the standard, most stable state for all lanthanides, though a few can temporarily use +2 or +4 to get a stable, half-filled, or completely empty f-orbital.

Exam Tip: The +3 state is dominant because the combined sum of the first three ionization enthalpies is favorable for all lanthanoids under normal conditions.

Question 37. (iii) Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U. Why?
Answer: This is because as the \( 5f \) subshell starts getting filled, its orbitals penetrate less into the inner electron core. Consequently, \( 5f \) electrons are much more efficiently shielded from the nucleus compared to the \( 4f \) electrons of their lanthanoid counterparts. Thus, the outermost valence electrons feel a weaker nuclear pull and are more easily removed or shared for bonding in actinides.
In simple words: The outer f-orbitals in actinides are shielded much better from the nucleus than those in lanthanides, so their outer electrons are held more loosely and have lower ionization energies.

Exam Tip: The larger size and poorer penetration of \( 5f \) orbitals relative to \( 4f \) orbitals result in more effective shielding, lowering the ionization energies of early actinides.

Question 37. (iii) Or Why irregularities in electronic configuration of actinoids are more than lanthanoids?
Answer: Alternatively, actinides display greater irregularities in their electron configurations because of the comparable energies of the \( 5f \), \( 6d \), and \( 7s \) subshells, combined with the stabilization gained from empty (\( f^0 \)), half-filled (\( f^7 \)), and fully filled (\( f^{14} \)) occupancies of the \( 5f \)-orbital.
In simple words: Because the energies of 5f, 6d, and 7s orbitals are extremely close in actinides, electrons easily hop between them to find the most stable configurations like half-filled or empty f-shells.

Exam Tip: The energy difference between \( 5f \) and \( 6d \) subshells in actinides is much smaller than that between \( 4f \) and \( 5d \) in lanthanides, leading to greater configurational variability.