15 extra multiple-choice questions for Transition and Inner Transition Elements (12th Standard Chemistry, Samacheer Kalvi), beyond the ones printed in the textbook — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
According to the periodic classification, which groups in the modern periodic table are occupied by transition metals?
- A. Groups 1 to 2
- B. Groups 3 to 12Correct
- C. Groups 13 to 18
- D. Groups 1 to 18
Explanation. Transition elements are located in the center of the periodic table, spanning from group 3 to group 12, positioned between the s-block and p-block elements.
Q2
What is the general outer electronic configuration of d-block elements, where 'n' represents the principal quantum number of the outermost shell?
- A. \((n-1)d^{1-10} ns^{1-2}\)Correct
- B. \((n-1)d^5 ns^1\)
- C. \((n-1)d^{10} ns^2\)
- D. \(nf^{1-14} (n-1)d^{1-10} ns^2\)
Explanation. The general electronic configuration involves the filling of the penultimate d-subshell and the outermost s-subshell, specifically \((n-1)d\) and \(ns\) orbitals.
Q3
Which transition metal is recognized as having the highest electrical conductivity of all known elements at room temperature?
- A. Copper
- B. Gold
- C. SilverCorrect
- D. Aluminum
Explanation. While many transition metals are good conductors, silver exhibits the highest electrical conductivity of all known elements at room temperature.
Q4
In the 3d transition series, why does the atomic radius remain nearly constant from chromium to copper?
- A. Increased shielding by 4s electrons
- B. Decrease in nuclear charge
- C. Balancing of increased nuclear charge by 3d-4s electron repulsionCorrect
- D. Absence of d-electrons
Explanation. The increase in nuclear charge is countered by the strong repulsion between the added 3d electrons and the 4s electrons, leading to a stable atomic size.
Q5
Which element in the 3d series exhibits the highest number of different oxidation states?
- A. Scandium
- B. Chromium
- C. ManganeseCorrect
- D. Iron
Explanation. Manganese has the maximum number of unpaired electrons in its 3d subshell, allowing it to exhibit six different oxidation states ranging from +2 to +7.
Q6
Why are the standard reduction potential (\(E^0\)) values for Manganese (\(Mn^{2+}/Mn\)) and Zinc (\(Zn^{2+}/Zn\)) more negative than expected from the general trend?
- A. High ionization enthalpy
- B. Stability of half-filled (\(d^5\)) and completely filled (\(d^{10}\)) configurationsCorrect
- C. Low metallic bonding
- D. Small atomic size
Explanation. The extra stability associated with the half-filled \(d^5\) configuration in \(Mn^{2+}\) and the completely filled \(d^{10}\) configuration in \(Zn^{2+}\) results in more negative reduction potentials.
Q7
Using the spin-only formula, what is the calculated magnetic moment for a transition metal ion containing three unpaired electrons?
- A. 1.73 BM
- B. 2.83 BM
- C. 3.87 BMCorrect
- D. 4.90 BM
Explanation. The spin-only magnetic moment is calculated using \(\mu_s = \sqrt{n(n+2)}\). For \(n=3\), \(\mu_s = \sqrt{3(5)} = \sqrt{15}\), which is approximately 3.87 BM.
Q8
Which type of catalyst is specifically used in the industrial process to convert acetaldehyde into acetic acid?
- A. Iron
- B. Vanadium pentoxide
- C. Rhodium or Iridium complexCorrect
- D. Ziegler-Natta catalyst
Explanation. Rhodium or Iridium complexes are employed as efficient catalysts in the carbonylation of methanol or the oxidation of acetaldehyde to produce acetic acid.
Q9
According to the Hume-Rothery rules for forming a substitute alloy, the difference between the atomic radii of the solvent and solute must be less than what percentage?
- A. 5%
- B. 10%
- C. 15%Correct
- D. 25%
Explanation. To form a substitute alloy, the atomic sizes of the elements must be very similar, specifically within a 15% difference in their radii.
Q10
Which of the following properties is characteristic of interstitial compounds formed by transition metals?
- A. They are always stoichiometric
- B. They have lower melting points than pure metals
- C. They are chemically very reactive
- D. They are hard and show high electrical conductivityCorrect
Explanation. Interstitial compounds are typically non-stoichiometric, hard, possess high melting points, and maintain the electrical and thermal conductivity of the parent metal.
Q11
What is the oxidation state of manganese in the higher oxide \(Mn_2O_7\), which is known to be covalent and acidic?
- A. +2
- B. +4
- C. +6
- D. +7Correct
Explanation. In \(Mn_2O_7\), oxygen has an oxidation state of -2. To balance the neutral molecule, each manganese atom must have an oxidation state of +7.
Q12
What is the general electronic configuration for the lanthanoid series of inner transition elements?
- A. \([Xe] 4f^{1-14} 5d^{0-1} 6s^2\)Correct
- B. \([Rn] 5f^{1-14} 6d^{0-2} 7s^2\)
- C. \([Xe] 4f^1 5d^1 6s^2\)
- D. \([Xe] 5d^{1-10} 6s^2\)
Explanation. Lanthanoids are characterized by the preferential filling of the inner 4f subshell, with a general configuration involving the [Xe] core and \(6s^2\) valence electrons.
Q13
What is the primary cause of the gradual decrease in atomic and ionic radii observed across the lanthanoid series?
- A. Increase in number of shells
- B. Effective shielding of 4f electrons
- C. Poor shielding effect of 4f electronsCorrect
- D. Decrease in nuclear charge
Explanation. The diffused shape of 4f orbitals results in poor shielding, causing the increased nuclear charge to pull the valence shell closer, resulting in lanthanoid contraction.
Q14
Which of the following actinoid elements are capable of exhibiting the highest oxidation state of +7?
- A. Thorium and Uranium
- B. Neptunium and PlutoniumCorrect
- C. Americium and Curium
- D. Californium and Einsteinium
Explanation. While +3 is common, Neptunium (\(Np\)) and Plutonium (\(Pu\)) are notable for exhibiting oxidation states as high as +7.
Q15
Compared to lanthanoids, why do actinoids show a greater tendency to form coordination complexes?
- A. Larger ionic size
- B. Higher binding energy of 5f orbitals
- C. Lower binding energy of 5f orbitals and higher ionic chargeCorrect
- D. Lack of unpaired electrons
Explanation. The lower binding energy of 5f orbitals compared to 4f orbitals, along with high ionic charges, makes actinoids more prone to forming complexes.
