Every Book Back multiple-choice question from Transition and Inner Transition Elements (12th Standard Chemistry, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
Sc( Z=21) is a transition element but Zinc (z=30) is not because
- A. both \(Sc^{3+}\) and \(Zn^{2+}\) ions are colourless and form white compounds.
- B. in case of Sc, 3d orbitals are partially filled but in Zn these are completely filledCorrect
- C. last electron as assumed to be added to 4s level in case of zinc
- D. both Sc and Zn do not exhibit variable oxidation states
Explanation. Transition elements must have an incompletely filled d subshell in their ground state or stable ionic states. Scandium has a partially filled 3d orbital, whereas zinc has a completely filled 3d orbital in both its metallic and ionic forms.
Q2
Which of the following d block element has half filled penultimate d sub shell as well as half filled valence sub shell?
- A. CrCorrect
- B. Pd
- C. Pt
- D. none of these
Explanation. Chromium exhibits an exceptional electronic configuration of \([Ar] 3d^5 4s^1\). In this state, both the penultimate 3d subshell and the valence 4s subshell are half-filled, which provides the atom with extra stability through exchange energy.
Q3
Among the transition metals of 3d series, the one that has highest negative \(M/M^{2+}\) standard electrode potential is
- A. TiCorrect
- B. Cu
- C. Mn
- D. Zn
Explanation. Titanium is identified as having the most significant negative standard oxidation potential in the 3d transition series. This indicates a high thermodynamic tendency for the metal to oxidize and form divalent cations in solution.
Q4
Which one of the following ions has the same number of unpaired electrons as present in \(V^{3+}\)?
- A. \(Ti^{3+}\)
- B. \(Fe^{3+}\)
- C. \(Ni^{2+}\)Correct
- D. \(Cr^{3+}\)
Explanation. Vanadium(III) has two unpaired electrons in its \(3d^2\) configuration. Nickel(II) also has two unpaired electrons in its \(3d^8\) configuration when placed in an octahedral field, matching the number of unpaired electrons.
Q5
The magnetic moment of \(Mn^{2+}\) ion is
- A. 5.92BMCorrect
- B. 2.80BM
- C. 8.95BM
- D. 3.90BM
Explanation. Manganese(II) possesses five unpaired electrons in its 3d subshell. Calculating the spin-only magnetic moment using the formula \(\mu = \sqrt{n(n+2)}\), where \(n\) is 5, results in approximately 5.92 Bohr Magnetons.
Q6
The catalytic behaviour of transition metals and their compounds is ascribed mainly due to
- A. their magnetic behaviour
- B. their unfilled d orbitals
- C. their ability to adopt variable oxidation statesCorrect
- D. their chemical reactivity
Explanation. The primary reason transition metals act as effective catalysts is their ability to exhibit multiple oxidation states. This allows them to form various intermediate complexes with reactants, effectively lowering the activation energy of chemical reactions.
Q7
Match items in column - I with the items of column – II and assign the correct code.
**Column-I**
A. Cyanide process
B. Froth floatation process
C. Electrolytic reduction
D. Zone refining
**Column-II**
(i) Ultrapure Ge
(ii) Dressing of ZnS
(iii) Extraction of Al
(iv) Extraction of Au
(v) Purification of Ni
- A. A-(i), B-(ii), C-(iii), D-(iv)
- B. A-(iii), B-(iv), C-(v), D-(i)
- C. A-(iv), B-(ii), C-(iii), D-(i)Correct
- D. A-(ii), B-(iii), C-(i), D-(v)
Explanation. The cyanide process is used for extracting gold, froth flotation for concentrating sulfide ores like ZnS, electrolytic reduction for reactive metals like aluminum, and zone refining is specialized for producing ultrapure semiconductors like germanium.
Q8
Wolframite ore is separated from tinstone by the process of
- A. Smelting
- B. Calcination
- C. Roasting
- D. Electromagnetic separationCorrect
Explanation. Since wolframite has magnetic properties while tinstone does not, they are separated using electromagnetic separation. The ore is passed over a magnetic roller, which attracts the magnetic wolframite while the non-magnetic tinstone falls further away.
Q9
Which one of the following is not feasible
- A. \(Zn(s) + Cu^{2+}(aq) \rightarrow Cu(s) + Zn^{2+}(aq)\)
- B. \(Cu(s) + Zn^{2+}(aq) \rightarrow Zn(s) + Cu^{2+}(aq)\)Correct
- C. \(Cu(s) + 2Ag^{+}(aq) \rightarrow 2Ag(s) + Cu^{2+}(aq)\)
- D. \(Fe(s) + Cu^{2+}(aq) \rightarrow Cu(s) + Fe^{2+}(aq)\)
Explanation. Reaction feasibility is determined by standard reduction potentials. Zinc is more reactive than copper and has a lower reduction potential, meaning copper cannot displace zinc ions from their salt solution.
Q10
Electrochemical process is used to extract
- A. Iron
- B. Lead
- C. SodiumCorrect
- D. silver
Explanation. Electrochemical methods are required for extracting highly reactive metals like sodium from their molten salts. These metals have very stable compounds that common chemical reducing agents cannot effectively reduce to the pure metallic state.
Q11
How many moles of \(I_2\) are liberated when 1 mole of potassium dichromate react with potassium iodide?
- A. 1
- B. 2
- C. 3Correct
- D. 4
Explanation. The reaction between one mole of potassium dichromate and potassium iodide in an acidic environment involves a six-electron transfer process. This stoichiometric relationship results in the production of exactly three moles of iodine for every mole of dichromate consumed.
Q12
The number of moles of acidified \(KMnO_4\) required to oxidize 1 mole of ferrous oxalate (\(FeC_2O_4\)) is
- A. 5
- B. 3
- C. 0.6Correct
- D. 1.5
Explanation. In an acidic medium, the permanganate ion acts as a five-electron oxidant. Since the complete oxidation of one mole of ferrous oxalate involves the release of three electrons, the number of moles of permanganate required is three divided by five.
Q13
Which one of the following statements related to lanthanons is incorrect?
- A. Europium shows +2 oxidation state.
- B. The basicity decreases as the ionic radius decreases from Pr to Lu.
- C. All the lanthanons are much more reactive than aluminium.Correct
- D. Ce4+ solutions are widely used as oxidising agents in volumetric analysis.
Explanation. While lanthanoids are chemically reactive metals, their properties are described as being similar to aluminium rather than significantly more reactive. The other statements accurately reflect the oxidation states, basicity trends, and analytical applications of specific lanthanoid elements.
Q14
Which of the following lanthanoid ion is diamagnetic?
- A. \(Eu^{2+}\)
- B. \(Yb^{2+}\)Correct
- C. \(Ce^{2+}\)
- D. \(Sm^{2+}\)
Explanation. The ytterbium(II) ion has an electronic configuration of xenon followed by a completely filled four-f subshell containing fourteen electrons. Because there are no unpaired electrons present in this configuration, the ion exhibits diamagnetic behavior.
Q15
Which of the following oxidation state is most common among the lanthanoids?
- A. +4
- B. +2
- C. +5
- D. +3Correct
Explanation. The plus three oxidation state is the most prevalent and stable state for all elements in the lanthanoid series. While other states like plus two or plus four exist, they are less common and often revert to the stable plus three state.
Q16
Assertion: \(Ce^{4+}\) is used as an oxidizing agent in volumetric analysis.
Reason: \(Ce^{4+}\) has the tendency of attaining +3 oxidation state.
- A. Both assertion and reason are true and reason is the correct explanation of assertion.Correct
- B. Both assertion and reason are true but reason is not the correct explanation of assertion.
- C. Assertion is true but reason is false.
- D. Both assertion and reason are false.
Explanation. Cerium in its plus four state is a powerful oxidant because it easily gains an electron to reach the plus three state. This allows it to attain the most stable and common oxidation state for lanthanoid elements.
Q17
The most common oxidation state of actinoids is
- A. +2
- B. +3Correct
- C. +4
- D. +6
Explanation. Similar to the lanthanoids, the plus three oxidation state is the most frequent baseline for elements in the actinoid series. While actinoids display a wider variety of oxidation states due to closer electronic subshell energy levels, plus three remains standard.
Q18
The actinoid elements which show the highest oxidation state of +7 are
- A. Np, Pu, AmCorrect
- B. U, Fm, Th
- C. U, Th, Md
- D. Es, No, Lr
Explanation. Neptunium, plutonium, and americium are the specific members of the actinoid series capable of reaching the maximum observed oxidation state of plus seven. This ability reflects the exceptionally small energy gap between their valence electronic subshells.
Q19
Which one of the following is not correct?
- A. \(La(OH)_3\) is less basic than \(Lu(OH)_3\)Correct
- B. In lanthanoid series ionic radius of \(Ln^{3+}\) ions decreases
- C. La is actually an element of transition metal series rather than lanthanoid series
- D. Atomic radii of Zr and Hf are same because of lanthanoid contraction
Explanation. Basic strength decreases as the size of the lanthanoid ion decreases. Since the lanthanum ion is larger than the lutetium ion due to lanthanoid contraction, lanthanum hydroxide is more basic than lutetium hydroxide, making the first statement incorrect.
