Chapter 8: Transition & Inner Transition Elements
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. What are transition elements (d-block elements)?
Transition elements are defined as the elements which have incompletely filled d-orbitals either in their ground state or in any of their common oxidation states. They occupy the middle portion of the periodic table, strictly spanning from Group 3 to Group 11. (Group 12 elements like Zn, Cd, Hg are strictly d-block but often excluded from being typical transition elements because they have full d-orbitals).
2. Write the general valence electronic configuration of d-block elements.
The general valence electronic configuration of d-block elements is:
$$(n-1)d^{1-10} \ ns^{1-2}$$
Where 'n' is the outermost principal quantum number, $(n-1)$ is the penultimate shell, and the d-subshell is progressively filled.
3. Why are Zinc, Cadmium, and Mercury not considered true transition elements?
Zinc ($3d^{10} 4s^2$), Cadmium ($4d^{10} 5s^2$), and Mercury ($5d^{10} 6s^2$) have completely filled d-orbitals in their ground state as well as in their common oxidation states (e.g., $Zn^{2+}$ is $3d^{10}$). Because they do not have incompletely filled d-orbitals, they do not exhibit typical transition metal properties (like showing variable oxidation states, forming colored ions, or having paramagnetic behavior). Hence, they are not considered true transition elements.
4. Write the observed electronic configurations of Chromium (Cr, Z=24) and Copper (Cu, Z=29). Why do they show anomalous configurations?
- Chromium (Z=24): Expected: $[Ar] 3d^4 4s^2$. Observed: $[Ar] 3d^5 4s^1$.
- Copper (Z=29): Expected: $[Ar] 3d^9 4s^2$. Observed: $[Ar] 3d^{10} 4s^1$.
Reason: Half-filled ($d^5$) and completely filled ($d^{10}$) d-subshells offer extra stability due to symmetrical distribution of electrons and high exchange energy. Therefore, one $4s$ electron shifts to the $3d$ subshell to attain this extra stability.
5. Discuss the trend of atomic radii in a transition series (e.g., 3d series).
In a transition series (left to right):
- Initially (Sc to Cr): Atomic radius decreases. This is because the effective nuclear charge increases while the shielding effect of the newly added $(n-1)d$ electrons is poor.
- Middle (Mn to Ni): Atomic radius remains almost constant. The increased nuclear charge is counterbalanced by the increased shielding/screening effect of the additional d-electrons.
- End (Cu to Zn): Atomic radius slightly increases. Electron-electron repulsion in the nearly filled or completely filled d-orbitals overcomes the effective nuclear charge, causing a slight expansion.
6. Why do transition metals have high melting and boiling points?
Transition metals have high melting and boiling points because they exhibit very strong metallic bonding. This strong bonding is due to the involvement of a large number of delocalized electrons from both the outermost $ns$ and the inner $(n-1)d$ orbitals. Greater the number of unpaired d-electrons, stronger the metallic bonding, and higher the melting point.
7. Explain why the ionization enthalpies of transition elements increase gradually along a series.
As we move left to right in a transition series, the nuclear charge increases. At the same time, electrons are added to the inner $(n-1)d$ subshell. These inner d-electrons offer a poor shielding effect. Therefore, the effective nuclear charge pulling on the outermost $ns$ electrons gradually increases, making it slightly more difficult to remove them. Hence, ionization enthalpy increases gradually.
8. Transition metals show variable oxidation states. Why?
Transition metals show variable oxidation states because the energy difference between the outermost $ns$ orbital and the inner $(n-1)d$ orbitals is very small. As a result, not only the $ns$ electrons but also the $(n-1)d$ electrons can participate in bond formation. Depending on the nature of the combining element, a varying number of d-electrons can be lost or shared.
9. Which transition metal of the 3d series exhibits the largest number of oxidation states and why?
Manganese (Mn) exhibits the largest number of oxidation states (from +2 to +7) in the 3d series.
Reason: Manganese ($[Ar] 3d^5 4s^2$) has the maximum number of unpaired electrons (five in the 3d subshell and two in the 4s subshell) available for bond formation.
10. Scandium forms only $+3$ oxidation state while Zinc forms only $+2$. Explain.
- Scandium ($Sc$, $[Ar] 3d^1 4s^2$): By losing three electrons (two from 4s and one from 3d), Sc acquires a highly stable noble gas configuration ($[Ar]$). Hence, it exclusively shows a +3 state.
- Zinc ($Zn$, $[Ar] 3d^{10} 4s^2$): By losing the two 4s electrons, Zn forms the $Zn^{2+}$ ion which has a highly stable, completely filled $3d^{10}$ configuration. It requires too much energy to remove a third electron from the stable d-subshell. Hence, it only shows a +2 state.
11. Why do transition metals in their higher oxidation states usually form compounds with Oxygen and Fluorine?
Oxygen and Fluorine are highly electronegative and have small atomic sizes. They can easily force transition metals to share or lose a large number of d-electrons, thereby bringing out their highest oxidation states. Additionally, Oxygen can form multiple bonds ($p\pi-d\pi$), which stabilizes high oxidation states (e.g., $Mn^{+7}$ in $MnO_4^-$).
12. What is paramagnetism? Why do most transition metal ions exhibit paramagnetism?
Paramagnetism: It is the property of substances being weakly attracted by a magnetic field.
Reason: Most transition metal ions possess one or more unpaired electrons in their $(n-1)d$ orbitals. Each unpaired electron acts as a tiny micro-magnet due to its spin and orbital motion, giving the ion a net magnetic moment and causing paramagnetism.
13. Write the "spin-only" formula to calculate the magnetic moment. What is its unit?
The "spin-only" magnetic moment ($\mu$) is calculated using the formula:
$$ \mu = \sqrt{n(n+2)} \text{ B.M.} $$
Where $n$ = number of unpaired electrons.
The unit is Bohr Magneton (B.M.).
14. Calculate the spin-only magnetic moment of $Fe^{2+}$ ion. (Atomic no. of Fe = 26).
Electronic configuration of Fe ($Z=26$) = $[Ar] 3d^6 4s^2$.
Electronic configuration of $Fe^{2+}$ = $[Ar] 3d^6 4s^0$.
In $3d^6$, there are 4 unpaired electrons and 1 paired set ($n = 4$).
Using the formula: $\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24}$.
$\mu \approx 4.90 \text{ B.M.}$
15. Why are $Sc^{3+}$, $Ti^{4+}$, and $Zn^{2+}$ ions diamagnetic and colorless?
- $Sc^{3+}$ ($3d^0$) and $Ti^{4+}$ ($3d^0$) have completely empty d-orbitals.
- $Zn^{2+}$ ($3d^{10}$) has completely filled d-orbitals.
Since they have zero unpaired electrons, they are diamagnetic. Because d-d transitions cannot occur without partially filled d-orbitals, they do not absorb visible light and are therefore colorless.
16. Why do transition metal ions form colored compounds?
The formation of color is primarily due to d-d transitions.
In a free transition metal ion, the five d-orbitals are degenerate (have equal energy). When ligands approach the central metal ion to form a complex, the d-orbitals split into two sets of different energies (e.g., $t_{2g}$ and $e_g$).
When white light falls on the complex, an unpaired electron absorbs a specific wavelength (energy) from the visible spectrum and gets excited from the lower energy d-orbital to the higher energy d-orbital. The transmitted or reflected light, which lacks the absorbed wavelength, appears as the complementary color.
17. Transition metals and their compounds are known to be good catalysts. Give reasons.
They act as excellent catalysts because:
- Variable Oxidation States: They can easily change their oxidation states and form unstable intermediate compounds with reactants, providing a lower activation energy pathway.
- Availability of vacant d-orbitals: They provide a large surface area with free valencies where reactant molecules can get adsorbed. This adsorption weakens the bonds in reactant molecules, lowering the activation energy.
18. Mention any two industrial processes that use transition metal catalysts.
- Haber's Process: Finely divided Iron ($Fe$) is used as a catalyst in the manufacture of Ammonia ($NH_3$).
- Contact Process: Vanadium pentoxide ($V_2O_5$) is used as a catalyst in the oxidation of $SO_2$ to $SO_3$ for the manufacture of Sulfuric acid. (Another example: Nickel in the hydrogenation of oils).
19. What are interstitial compounds? Give examples.
Interstitial Compounds: These are compounds formed when small atoms like Hydrogen (H), Carbon (C), or Nitrogen (N) get trapped inside the interstitial spaces (voids) of the transition metal crystal lattice. They are non-stoichiometric.
Examples: Steel and cast iron (contain carbon), $TiC$, $VH_{0.56}$.
20. State three properties of interstitial compounds.
- They are very hard and rigid (some borides/carbides approach diamond in hardness).
- They have higher melting points than the pure parent metals.
- They retain metallic conductivity.
- They are chemically inert.
21. Why do transition metals form alloys easily?
Alloys are homogeneous solid solutions of two or more metals. Transition metals readily form alloys with each other because they have very similar atomic radii (within a 15% difference). Due to this similar size, atoms of one transition metal can easily substitute the atoms of another transition metal in its crystal lattice without severely distorting it.
22. What are the components of Brass and Bronze?
- Brass: An alloy of Copper (Cu) and Zinc (Zn).
- Bronze: An alloy of Copper (Cu) and Tin (Sn).
23. In what oxidation state does Manganese exist in Potassium Permanganate ($KMnO_4$)?
Let the oxidation state of Mn be '$x$'.
In $KMnO_4$: $K$ is $+1$, $O$ is $-2$.
$(+1) + x + 4(-2) = 0$
$1 + x - 8 = 0 \implies x = +7$.
Manganese exists in the +7 oxidation state.
24. Briefly outline the preparation of $KMnO_4$ from pyrolusite ore ($MnO_2$).
Step 1: Oxidation of $MnO_2$ to Manganate. Pyrolusite is fused with KOH in the presence of an oxidizing agent ($O_2$ or $KNO_3$) to give a green mass of potassium manganate ($K_2MnO_4$).
$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$
Step 2: Electrolytic oxidation of Manganate to Permanganate. The green aqueous solution of $K_2MnO_4$ undergoes electrolytic oxidation at the anode to form purple $KMnO_4$.
$MnO_4^{2-} \rightarrow MnO_4^- + e^-$ (at anode)
25. What is the equivalent weight of $KMnO_4$ in an acidic medium? (Given Molar Mass = 158).
In an acidic medium, $KMnO_4$ acts as a strong oxidizing agent and gets reduced from $Mn^{+7}$ to $Mn^{+2}$.
$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
Change in oxidation state = $7 - 2 = 5$ (which is the number of electrons gained).
$\text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Change in Oxidation State}} = \frac{158}{5} = \mathbf{31.6}$
26. What happens when $KMnO_4$ reacts with an acidic solution of Ferrous sulfate ($FeSO_4$)?
$KMnO_4$ acts as an oxidizing agent. It oxidizes ferrous ions ($Fe^{2+}$) to ferric ions ($Fe^{3+}$), while it gets reduced itself to $Mn^{2+}$, making the purple color of the solution disappear.
Ionic equation: $MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$
27. Describe the color and structure of the Permanganate ion ($MnO_4^-$).
The permanganate ion is intensely purple/deep violet in color. This color is not due to d-d transitions (as $Mn^{+7}$ is $d^0$), but due to Ligand to Metal Charge Transfer (LMCT) from Oxygen to Manganese.
Its structure is Tetrahedral, with Manganese at the center bonded to four Oxygen atoms.
28. In what oxidation state does Chromium exist in Potassium Dichromate ($K_2Cr_2O_7$)?
Let the oxidation state of Cr be '$x$'.
In $K_2Cr_2O_7$: $K$ is $+1$, $O$ is $-2$.
$2(+1) + 2x + 7(-2) = 0$
$2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$.
Chromium exists in the +6 oxidation state.
29. Explain the effect of pH on Chromate and Dichromate ions.
Chromate ($CrO_4^{2-}$, yellow) and Dichromate ($Cr_2O_7^{2-}$, orange) ions are interconvertible in an aqueous solution depending on the pH.
- In an acidic medium (low pH), yellow chromate changes to orange dichromate.
$2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O$ - In a basic medium (high pH), orange dichromate changes to yellow chromate.
$Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O$
30. Write the reaction of acidic $K_2Cr_2O_7$ with Hydrogen sulfide ($H_2S$).
$K_2Cr_2O_7$ acts as a strong oxidizing agent. It oxidizes $H_2S$ to elemental sulfur, while it gets reduced to $Cr^{3+}$ ions (green color).
Ionic equation: $Cr_2O_7^{2-} + 3H_2S + 8H^+ \rightarrow 2Cr^{3+} + 3S_{(s)} + 7H_2O$
31. What are inner transition elements (f-block elements)?
Inner transition elements are the elements in which the last differentiating electron enters the anti-penultimate $(n-2)f$ subshell. They are placed at the bottom of the periodic table in two separate rows.
32. Write the general electronic configuration of f-block elements.
The general valence electronic configuration of f-block elements is:
$$(n-2)f^{1-14} \ (n-1)d^{0-1} \ ns^2$$
For Lanthanoids, $n=6$ (4f series). For Actinoids, $n=7$ (5f series).
33. What are Lanthanoids? Why are they called rare earth elements?
The series of 14 elements following Lanthanum ($La$, Z=57), from Cerium ($Ce$, Z=58) to Lutetium ($Lu$, Z=71), in which the 4f subshell is progressively filled, are called Lanthanoids.
Historically, they were found as oxides in rare minerals (earths), hence they were called rare earth elements. (Note: they are not actually that rare in the earth's crust).
34. What is the most common oxidation state of lanthanoids?
The principal and most common stable oxidation state of all lanthanoids is +3. This is due to the loss of two 6s electrons and one 5d (or 4f) electron. Some elements also show +2 or +4 states (like $Ce^{4+}$ or $Eu^{2+}$) to attain stable $f^0, f^7, \text{ or } f^{14}$ configurations.
35. Define Lanthanoid Contraction.
Lanthanoid Contraction: It is defined as the steady and continuous decrease in the atomic and ionic radii (of $M^{3+}$ ions) of lanthanoid elements with the increasing atomic number from Cerium to Lutetium.
36. What is the cause of Lanthanoid contraction?
As we move along the lanthanoid series, the nuclear charge increases by one unit at each step, and the corresponding new electron enters the inner 4f subshell. The shielding (or screening) effect of 4f electrons is very poor due to the highly diffused shape of f-orbitals. Therefore, the effective nuclear charge pulling the outermost electrons inwards increases steadily, causing the atomic/ionic size to shrink.
37. Mention two important consequences of Lanthanoid Contraction.
- Similarity of 2nd and 3rd transition series: Because of the contraction, the atomic radii of elements in the 3rd transition series (e.g., Hf, Ta, W) are almost identical to their corresponding elements in the 2nd transition series (e.g., Zr, Nb, Mo), making them chemically very similar and difficult to separate (chemical twins like Zr-Hf).
- Decrease in basic strength of hydroxides: As size decreases from $La^{3+}$ to $Lu^{3+}$, the covalent character of the $M-OH$ bond increases (Fajans' rule). Therefore, $La(OH)_3$ is the most basic, and $Lu(OH)_3$ is the least basic.
38. Why are Zr and Hf called chemical twins?
Zirconium (Zr, 4d series) and Hafnium (Hf, 5d series) occur in the same group. Usually, size increases down the group. However, due to the Lanthanoid Contraction that occurs between them, the atomic radius of Hf (159 pm) is almost identical to that of Zr (160 pm). Because they have the same size and the same valence electronic configuration, their chemical properties are nearly identical, making them "chemical twins" which are very hard to separate.
39. Why are most lanthanoid ions ($M^{3+}$) colored?
Most lanthanoid ions are colored in solid state and aqueous solutions because they have partially filled 4f orbitals. The color is due to the absorption of visible light, which excites an electron from a lower energy f-orbital to a higher energy f-orbital. This is called an f-f transition. (Ions with $f^0$ like $La^{3+}$ or $f^{14}$ like $Lu^{3+}$ are colorless).
40. What is Misch metal? State one of its uses.
Misch metal is a well-known alloy consisting mainly of Lanthanoid metals (about 95%, primarily Cerium and Lanthanum), ~5% Iron, and traces of S, C, Ca, and Al.
Use: It is highly pyrophoric (emits sparks when struck), so it is used in making bullets, shells, and flints for gas lighters.
41. What are Actinoids?
The series of 14 elements following Actinium ($Ac$, Z=89), from Thorium ($Th$, Z=90) to Lawrencium ($Lr$, Z=103), in which the 5f subshell is progressively filled, are called Actinoids. They are all radioactive elements.
42. Define Transuranic elements.
Elements with atomic numbers strictly greater than 92 (Uranium) are known as transuranic elements. They are entirely man-made (synthetic) and highly radioactive. Examples: Neptunium (Np), Plutonium (Pu).
43. What is Actinoid Contraction? Is it greater or smaller than Lanthanoid contraction?
Actinoid Contraction: It is the gradual decrease in the atomic and ionic radii of actinoid elements with increasing atomic number from Thorium to Lawrencium.
The contraction is greater from element to element in actinoids than in lanthanoids. This is because the 5f electrons provide even poorer shielding from nuclear charge than the 4f electrons do, leading to a stronger inward pull.
44. Why do Actinoids show a much larger variety of oxidation states than Lanthanoids?
Actinoids exhibit variable oxidation states (from +3 up to +7) because the energy difference between the 5f, 6d, and 7s orbitals is extremely small. Therefore, electrons from all these orbitals can easily participate in chemical bonding, unlike lanthanoids where 4f electrons are deeply buried.
45. Compare Lanthanoids and Actinoids on the basis of Radioactivity and Oxidation states.
| Property | Lanthanoids | Actinoids |
|---|---|---|
| Radioactivity | Only Promethium (Pm) is radioactive; the rest are non-radioactive. | All actinoids are radioactive elements. |
| Oxidation States | Mostly show +3. Rarely show +2 and +4. | Show +3, but frequently show higher states (+4, +5, +6, +7). |
46. Mention any two uses of Actinoids.
- Uranium ($^{235}U$): Used as a fuel in nuclear power reactors to generate electricity.
- Plutonium ($^{239}Pu$): Used as a fuel in nuclear reactors and in the manufacture of nuclear weapons.
47. Which element of the 3d series has the lowest enthalpy of atomization and why?
Zinc (Zn) has the lowest enthalpy of atomization.
Reason: Enthalpy of atomization depends on the strength of metallic bonds. Strong metallic bonds require many unpaired d-electrons. Zinc ($3d^{10} 4s^2$) has completely filled d-orbitals and zero unpaired electrons, resulting in very weak metallic bonding. Hence, very little energy is needed to atomize it.
48. The $E^\circ$ value for $Cu^{2+}/Cu$ is positive (+0.34 V). What does this imply about copper's reactivity with acids?
The positive reduction potential means Copper is a weaker reducing agent than Hydrogen (since $E^\circ$ for $H^+/H_2$ is 0.00 V). Therefore, Copper cannot displace hydrogen gas from non-oxidizing dilute acids (like dilute HCl or $H_2SO_4$). It only reacts with oxidizing acids like Nitric acid.
49. Why do Actinoids form complexes more readily than Lanthanoids?
Actinoid ions are generally smaller in size and possess higher charge densities compared to lanthanoid ions. Consequently, they attract ligands much more strongly, leading to a much greater tendency to form coordination complexes.
50. Predict whether $Ti^{3+}$ will be coloured or colorless. (Atomic number Ti = 22).
Electronic configuration of Ti ($Z=22$) = $[Ar] 3d^2 4s^2$.
Electronic configuration of $Ti^{3+}$ = $[Ar] 3d^1 4s^0$.
Since $Ti^{3+}$ has one partially filled d-orbital ($n=1$), d-d electronic transition is possible. Therefore, $Ti^{3+}$ compounds will be colored (typically violet in aqueous solution).
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