D&F Block flash cards

1. 1.Which blocks of the periodic table are called the d-block and f-block?

   d-block: elements in which the last electron enters the $(n-1)d$ subshell — groups 3 to 12 (transition elements). f-block: elements in which the last electron enters the $(n-2)f$ subshell — the lanthanoids and actinoids (inner transition elements).

   Hint: Which subshell fills last?

2. 2.Define a transition element according to NCERT.

   An element whose atom in the ground state or ion in a common oxidation state has a partially filled $d$ subshell ($d^{1}$ to $d^{9}$).

   Hint: Partially filled d in atom or common ion.

3. 3.Why are Zn, Cd and Hg (group 12) not regarded as typical transition elements?

   Their atoms and common ions ($M^{2+}$) have a fully filled $d^{10}$ configuration — no partially filled $d$ subshell. So they lack most characteristic transition-metal properties.

   Hint: $d^{10}$ in both atom and $M^{2+}$.

4. 4.Write the general outer electronic configuration of d-block elements.

   $(n-1)d^{1-10}\,ns^{1-2}$, where $n$ is the outermost principal quantum number.

   Hint: Two subshells: (n-1)d and ns.

5. 5.Name the four transition series and the d subshell each fills.

   3d series (1st): Sc–Zn. 4d series (2nd): Y–Cd. 5d series (3rd): La, Hf–Hg. 6d series (4th, incomplete): Ac, Rf onward.

   Hint: 3d, 4d, 5d, 6d.

6. 6.Why do Cr and Cu have anomalous configurations $3d^{5}4s^{1}$ and $3d^{10}4s^{1}$?

   Because exactly half-filled ($d^{5}$) and completely filled ($d^{10}$) $d$ subshells have extra stability (symmetry + exchange energy). One $4s$ electron shifts to $3d$ to attain this stability.

   Hint: Half-filled and fully-filled stability.

7. 7.Write the ground-state electronic configuration of $\text{Cr}$ (Z = 24) and $\text{Cu}$ (Z = 29).

   $\text{Cr}: [\text{Ar}]\,3d^{5}4s^{1}$. $\text{Cu}: [\text{Ar}]\,3d^{10}4s^{1}$.

   Hint: Both take only one 4s electron.

8. 8.When a transition metal forms a cation, which electrons are lost first?

   The $ns$ electrons are removed before the $(n-1)d$ electrons. So for $M^{n+}$ ions the configuration is $(n-1)d^{x}$ with an empty $ns$.

   Hint: 4s goes before 3d on ionisation.

9. 9.Write the electronic configuration of $\text{Fe}$, $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$ (Z = 26).

   $\text{Fe}: [\text{Ar}]\,3d^{6}4s^{2}$; $\text{Fe}^{2+}: [\text{Ar}]\,3d^{6}$; $\text{Fe}^{3+}: [\text{Ar}]\,3d^{5}$ (extra-stable half-filled).

   Hint: Remove 4s first, then a 3d for +3.

10. 10.Why are transition metals and their compounds generally studied together as a distinct group?

    Because they share characteristic properties: variable oxidation states, coloured ions, paramagnetism, catalytic activity, and a tendency to form complex, interstitial and alloy compounds, all arising from partially filled $d$ orbitals.

    Hint: Common set of d-orbital properties.

11. 11.Describe the general physical (metallic) properties of transition elements.

    They are hard, high-melting, high-density metals with high tensile strength, good conductors of heat and electricity, malleable and ductile, and form alloys readily.

    Hint: Strong metallic bonding.

12. 12.Why do transition metals have high melting and boiling points?

    Because of strong metallic bonding — both the $ns$ and unpaired $(n-1)d$ electrons participate in bonding, giving a large number of bonding electrons.

    Hint: Unpaired d electrons in metallic bonding.

13. 13.Which 3d element has the highest melting point in its series, and why?

    Chromium (and vanadium is also very high). Melting points rise to a maximum near the middle where the maximum number of unpaired $d$ electrons are available for metallic bonding.

    Hint: Maximum unpaired electrons ≈ middle of series.

14. 14.Why does Mn have an abnormally low melting point compared with its neighbours Cr and Fe?

    $\text{Mn}$ is $3d^{5}4s^{2}$; its stable half-filled $d^{5}$ configuration makes those d electrons less available for metallic bonding, weakening the bonding.

    Hint: $d^{5}$ stability lowers bonding.

15. 15.How does atomic radius vary across a transition series (e.g. Sc → Zn)?

    It decreases at first, stays almost constant in the middle, and rises slightly at the end. Increasing nuclear charge is largely offset by the screening of added $d$ electrons.

    Hint: Small decrease, then nearly flat, slight rise at end.

16. 16.Why are the atomic radii of the 4d and 5d series nearly equal?

    Because of the lanthanoid contraction — the steady size decrease across the 4f lanthanoids that precede the 5d series cancels the expected increase, making 4d and 5d radii almost the same.

    Hint: 4f contraction offsets the extra shell.

17. 17.Why do transition metals show a large number of oxidation states?

    Because the energies of the $(n-1)d$ and $ns$ orbitals are very close, so both sets of electrons can take part in bonding, giving many oxidation states differing by 1.

    Hint: Close (n-1)d and ns energies.

18. 18.Which oxidation state is common to almost all first-row transition metals, and why?

    The +2 state, formed by loss of the two $4s$ electrons. It becomes more stable (relative to +3) on moving from left to right across the series.

    Hint: Loss of the 4s pair.

19. 19.What is the maximum oxidation state shown in the 3d series and by which element?

    +7, shown by manganese (e.g. in $\text{KMnO}_4$), corresponding to loss/involvement of all $3d^{5}4s^{2}$ electrons.

    Hint: Mn: 5 + 2 = 7.

20. 20.Which element shows the greatest number of oxidation states in the first transition series?

    Manganese, showing states from +2 up to +7.

    Hint: Middle of series, maximum d + s electrons.

21. 21.Why do the higher oxidation states of transition metals occur mainly in oxides and fluorides?

    Because oxygen and fluorine are small, highly electronegative and strong oxidising — they stabilise high oxidation states (e.g. $\text{Mn}$ is +7 in $\text{Mn}_2\text{O}_7$ and $\text{MnO}_4^-$).

    Hint: Small, very electronegative O and F.

22. 22.How does the stability of the +2 versus +3 state change across the 3d series?

    The +2 state becomes progressively more stable and +3 less stable from left to right, as increasing nuclear charge makes it harder to remove the third electron.

    Hint: Left→right: +2 favoured.

23. 23.Why are most transition metal ions coloured?

    Because of $d$–$d$ transitions: in a ligand field the $d$ orbitals split, and an electron absorbs visible light to jump from the lower to higher $d$ level; the complementary colour is seen.

    Hint: d-d electronic transition, partly filled d.

24. 24.Why are $\text{Sc}^{3+}$, $\text{Ti}^{4+}$, $\text{Zn}^{2+}$ and $\text{Cu}^{+}$ ions colourless?

    Because they have empty ($d^{0}$) or completely filled ($d^{10}$) $d$ subshells — no $d$–$d$ transition is possible, so no visible light is absorbed.

    Hint: $d^{0}$ or $d^{10}$ = no d-d transition.

25. 25.What determines the particular colour observed for a given transition-metal complex?

    The magnitude of the $d$-orbital splitting ($\Delta$), which depends on the metal ion, its oxidation state, and the ligands. This fixes the wavelength absorbed; the complementary colour is seen.

    Hint: Depends on Δ set by ligand and metal.

26. 26.Define paramagnetism and its origin in transition metals.

    Paramagnetism is attraction into a magnetic field, caused by unpaired electrons. Transition metal ions often have unpaired $d$ electrons and are therefore paramagnetic.

    Hint: Unpaired electrons → attracted to field.

27. 27.Give the spin-only magnetic moment formula and its unit.

    $\mu = \sqrt{n(n+2)}$ BM (Bohr magneton), where $n$ = number of unpaired electrons.

    Hint: Depends only on number of unpaired e⁻.

28. 28.Calculate the spin-only magnetic moment of $\text{Mn}^{2+}$ ($3d^{5}$).

    $\text{Mn}^{2+}$ has 5 unpaired electrons: $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM.

    Hint: n = 5 unpaired.

29. 29.Why do transition metals and their compounds act as good catalysts?

    Because of their variable oxidation states (they can accept/donate electrons via intermediates) and their ability to adsorb reactants on the metal surface using partially filled d orbitals.

    Hint: Variable oxidation state + surface adsorption.

30. 30.Give two important industrial catalysts that are transition metals/compounds and their processes.

    Iron in the Haber process ($\text{N}_2 + \text{H}_2 \to \text{NH}_3$); $\text{V}_2\text{O}_5$ in the Contact process (manufacture of $\text{H}_2\text{SO}_4$). Also Ni for hydrogenation.

    Hint: Haber (Fe), Contact ($V_2O_5$).

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