D & F-Block Elements & Their Important Compounds flash cards

1. 1.What are d-block elements?

   Elements in which the last electron enters the $(n-1)d$ subshell. They occupy groups $3$ to $12$ and lie between the $s$-block and $p$-block. Comprise four series: $3d$ (Sc–Zn), $4d$ (Y–Cd), $5d$ (La, Hf–Hg) and $6d$.

   Hint: Last electron in penultimate d-orbital.

2. 2.Define a transition element (IUPAC).

   An element whose atom has a partially filled $d$ subshell, or which can give rise to cations with an incomplete $d$ subshell. Formally $(n-1)d^{1-9}$ character in element or ions.

   Hint: Partially filled d in atom or ion.

3. 3.Why are $\text{Zn}$, $\text{Cd}$, $\text{Hg}$ not regarded as typical transition metals?

   Their configuration is $(n-1)d^{10}ns^2$; the $d$ subshell is completely filled in both the atom and their common $+2$ ions ($d^{10}$). They lack a partially filled $d$ subshell, so they don't show typical transition properties.

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

4. 4.Give the general electronic configuration of d-block elements.

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

   Hint: Penultimate d + outer s.

5. 5.Why do $\text{Cr}$ and $\text{Cu}$ have anomalous configurations?

   $\text{Cr}=[\text{Ar}]3d^5 4s^1$ and $\text{Cu}=[\text{Ar}]3d^{10}4s^1$. Exactly half-filled ($d^5$) and completely filled ($d^{10}$) subshells give extra stability (symmetrical distribution + maximum exchange energy).

   Hint: Half-filled/fully-filled stability.

6. 6.Write the electronic configuration of $\text{Fe}$, $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$.

   $\text{Fe}=[\text{Ar}]3d^6 4s^2$; $\text{Fe}^{2+}=[\text{Ar}]3d^6$; $\text{Fe}^{3+}=[\text{Ar}]3d^5$. $ns$ electrons are removed before $(n-1)d$ on ionisation.

   Hint: Remove 4s first; $\text{Fe}^{3+}$ is $d^5$.

7. 7.On ionisation of a transition metal, which electrons are lost first and why?

   The $ns$ electrons are lost before $(n-1)d$ electrons. Although $4s$ fills before $3d$, once filled the $3d$ orbitals become lower in energy, so $4s$ electrons are removed first.

   Hint: $ns$ before $(n-1)d$.

8. 8.Why do transition metals show variable oxidation states?

   Because the energies of $(n-1)d$ and $ns$ orbitals are very close, so a variable number of electrons (from both s and d) can participate in bonding, giving a range of oxidation states differing by unity.

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

9. 9.Which element of the $3d$ series shows the maximum number of oxidation states, and what are they?

   $\text{Mn}$ shows the widest range, from $+2$ to $+7$ ($+2,+3,+4,+5,+6,+7$), because it has $3d^5 4s^2$ (5 unpaired d + 2 s electrons all available).

   Hint: $\text{Mn}$: +2 to +7.

10. 10.Across the $3d$ series, how do oxidation states vary from Sc to Mn and then to Zn?

    The maximum (highest) oxidation state increases from $\text{Sc}(+3)$ up to $\text{Mn}(+7)$, then decreases as $d$-electrons pair up and become less available ($\text{Fe}$ max $+6$, ... $\text{Zn}$ only $+2$).

    Hint: Peaks at Mn.

11. 11.Why is the $+2$ oxidation state increasingly stable towards the end of the $3d$ series?

    Moving right, the increasing nuclear charge holds the $3d$ electrons more tightly; after $\text{Mn}$ removal of more than two electrons becomes difficult, so $M^{2+}$ (loss of two $4s$ electrons) is favoured (e.g. Fe, Co, Ni, Cu, Zn).

    Hint: Higher Z, d electrons held tightly.

12. 12.Why does the highest oxidation state of transition metals usually occur in oxides and fluorides?

    $\text{O}$ and $\text{F}$ are small, highly electronegative elements that can stabilise high oxidation states via strong bonding and, for oxygen, multiple ($M=O$) bonding. E.g. $\text{Mn}$ is $+7$ in $\text{Mn}_2\text{O}_7$ and $\text{MnO}_4^-$.

    Hint: Small, electronegative O and F.

13. 13.Why do the metals of the second and third transition series show higher oxidation states more readily than the first?

    Larger $4d$/$5d$ orbitals are more diffuse and their electrons are more easily involved in bonding; the energy gap between ns and (n-1)d is smaller. Hence Mo, W, Ru, Os etc. readily reach high states (e.g. $\text{OsO}_4$, $+8$).

    Hint: Bigger, diffuse 4d/5d orbitals.

14. 14.Why are most transition metals and their ions coloured?

    In the complex/ion the $d$ orbitals split into two sets ($t_{2g}$ and $e_g$). Electrons undergo $d$–$d$ transitions absorbing visible light of energy $\Delta$; the complementary colour of the absorbed wavelength is seen.

    Hint: $d$–$d$ transition, complementary colour.

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

    They have either empty ($d^0$: $\text{Sc}^{3+}$, $\text{Ti}^{4+}$) or completely filled ($d^{10}$: $\text{Zn}^{2+}$, $\text{Cu}^{+}$) $d$ subshells, so no $d$–$d$ electronic transitions are possible.

    Hint: $d^0$ or $d^{10}$ → no d–d transition.

16. 16.What determines the magnitude of crystal-field splitting $\Delta$ and hence the colour of a complex?

    $\Delta$ depends on the nature and field strength of the ligands (spectrochemical series), the oxidation state of the metal, geometry, and the metal itself. A larger $\Delta$ shifts absorption to shorter wavelength.

    Hint: Ligand strength, oxidation state, geometry.

17. 17.What is the origin of paramagnetism in transition metal compounds?

    Presence of unpaired electrons in the $d$ orbitals. These are attracted into a magnetic field; the more unpaired electrons, the greater the paramagnetism.

    Hint: Unpaired d-electrons.

18. 18.State the spin-only magnetic moment formula and its unit.

    $\mu = \sqrt{n(n+2)}$ Bohr magnetons (BM), where $n$ = number of unpaired electrons. It neglects orbital contribution.

    Hint: $\sqrt{n(n+2)}$ BM.

19. 19.Across the $3d$ series, why do atomic radii first decrease, then stay nearly constant, then increase slightly?

    Left→middle: increasing nuclear charge dominates, radius decreases. Middle: added $d$-electrons shield the nucleus, roughly balancing the increased charge → nearly constant. End (Cu, Zn): $d^{10}$ electron–electron repulsion causes slight increase.

    Hint: Nuclear charge vs d-shielding.

20. 20.Why are the atomic radii of the second and third (4d and 5d) transition series nearly the same?

    Because of the lanthanoid contraction: the poor shielding by $4f$ electrons (filled before the 5d series) causes a size decrease that almost exactly cancels the expected increase from the extra shell. E.g. $r(\text{Zr}) \approx r(\text{Hf})$.

    Hint: Lanthanoid contraction.

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

    They possess strong metallic bonding due to a large number of unpaired $d$ (and $s$) electrons available for interatomic (covalent-type) bonding. Maximum where the number of unpaired electrons is greatest.

    Hint: Unpaired d electrons → strong metallic bonds.

22. 22.Why does $\text{Zn}$, $\text{Cd}$, $\text{Hg}$ have low melting points among the d-block?

    They have $(n-1)d^{10}ns^2$ configuration with no unpaired $d$ electrons; only the two $s$ electrons contribute to metallic bonding, so bonding is weak. $\text{Hg}$ is even liquid at room temperature.

    Hint: $d^{10}$: no unpaired electrons for bonding.

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

    They have (i) variable oxidation states allowing them to form intermediate compounds/provide alternative low-energy paths, and (ii) large surface area with the ability to adsorb reactants and form weak bonds using $d$-orbitals.

    Hint: Variable oxidation states + surface adsorption.

24. 24.Name industrial catalysts: Haber process, Contact process, and hydrogenation of oils.

    Haber (NH$_3$): finely divided iron (with Mo promoter). Contact (H$_2$SO$_4$): $\text{V}_2\text{O}_5$. Hydrogenation of vegetable oils: finely divided nickel.

    Hint: Fe, $\text{V}_2\text{O}_5$, Ni.

25. 25.Why do transition metal ions readily form complexes?

    Because they have (i) small size and high charge density (high charge/radius ratio) and (ii) vacant $d$ orbitals of suitable energy to accept lone pairs from ligands.

    Hint: Small, highly charged + vacant d-orbitals.

26. 26.What are interstitial compounds?

    Compounds formed when small atoms such as $\text{H}$, $\text{C}$, $\text{N}$, or $\text{B}$ are trapped in the interstitial (void) sites of the transition metal lattice. They are usually non-stoichiometric, e.g. $\text{TiC}$, $\text{Fe}_3\text{H}$, $\text{VH}_{0.56}$.

    Hint: Small atoms in lattice voids.

27. 27.List characteristic properties of interstitial compounds.

    They are (i) very hard, (ii) have high melting points (higher than the parent metal), (iii) retain metallic conductivity, and (iv) are chemically rather inert. Often non-stoichiometric.

    Hint: Hard, high m.p., conducting, inert.

28. 28.Why do transition metals readily form alloys?

    They have nearly identical atomic radii, so atoms of one metal can readily replace atoms of another in the crystal lattice (substitutional alloys), e.g. brass, bronze, steel.

    Hint: Similar atomic radii → substitution.

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