P-Block Elements (b & C Family) flash cards

1. 1.Which groups constitute the Boron and Carbon families, and what are their general valence-shell electronic configurations?

   Group 13 (Boron family): $ns^2 np^1$. Group 14 (Carbon family): $ns^2 np^2$. Group 13 members: B, Al, Ga, In, Tl; Group 14 members: C, Si, Ge, Sn, Pb.

   Hint: Count outermost electrons: 3 vs 4.

2. 2.What are the maximum group oxidation states of Group 13 and Group 14 elements?

   Group 13 shows a maximum of $+3$ (all valence electrons used). Group 14 shows a maximum of $+4$. Lower states $+1$ (Gp 13) and $+2$ (Gp 14) also appear due to the inert pair effect.

   Hint: Equal to number of valence electrons.

3. 3.State the inert pair effect and its cause.

   The reluctance of the two $ns^2$ electrons to participate in bonding, making lower oxidation states ($+1$, $+2$) more stable down a group. Cause: poor shielding by intervening $d$ and $f$ electrons so the $ns^2$ pair is held tightly and is hard to unpair/remove.

   Hint: Why is $\text{Tl}^{+}$ more stable than $\text{Tl}^{3+}$?

4. 4.How does the stability of the $+1$ state vary down Group 13?

   Stability of $+1$ increases down the group: $\text{B, Al}$ show mainly $+3$; $\text{Ga}$ and $\text{In}$ show both; $\text{Tl}$ is most stable as $\text{Tl}^{+}$. Thus $\text{Tl}^{3+}$ is a strong oxidant.

   Hint: Inert pair effect strongest at the bottom.

5. 5.How does the stability of the $+2$ state vary down Group 14, and which ion is a strong oxidising agent?

   $+2$ stability increases down the group; $+4$ decreases. $\text{Pb}^{2+}$ is very stable while $\text{Pb}^{4+}$ (e.g. $\text{PbO}_2$) is a strong oxidising agent. Conversely $\text{Sn}^{2+}$ is a reducing agent (readily goes to $+4$).

   Hint: Compare $\text{PbO}_2$ vs $\text{SnCl}_2$.

6. 6.Why is $\text{PbCl}_4$ unstable and covalent while $\text{PbCl}_2$ is stable and ionic?

   Due to the inert pair effect, $\text{Pb}^{4+}$ is a strong oxidiser; $\text{Cl}^{-}$ reduces it, so $\text{PbCl}_4$ decomposes to $\text{PbCl}_2 + \text{Cl}_2$. $\text{PbCl}_2$ ($+2$) is the stable, more ionic form.

   Hint: $+4$ oxidises the chloride.

7. 7.How do atomic radii vary down Group 13 — is the trend regular?

   Radii generally increase B < Al, but the increase is irregular: $\text{Ga}$ is nearly equal to (even slightly smaller than) $\text{Al}$ because the poorly shielding $3d^{10}$ electrons in Ga cause extra effective nuclear charge (d-block contraction).

   Hint: The Ga anomaly.

8. 8.Explain the trend in first ionisation enthalpy down Group 13.

   $\text{IE}_1$ decreases B → Al then shows irregularities: it slightly increases at Ga and Tl due to poor shielding by $d$ and $f$ electrons (d- and lanthanoid contraction) which increases $Z_{eff}$. Order roughly $\text{Al}<\text{Ga}<\text{In}<\text{Tl}$ is not simple.

   Hint: d and f contraction raise $Z_{eff}$.

9. 9.Why is the first ionisation enthalpy of Group 13 lower than Group 2 but the trends still noteworthy?

   Group 13 has an $np^1$ electron that is easier to remove than the paired $ns^2$ of Group 2, so $\text{IE}_1$ (13) < $\text{IE}_1$ (2). Within Group 14, $\text{IE}$ values are higher than Group 13 as expected.

   Hint: Removing $p$ vs stable $s^2$.

10. 10.How does metallic/non-metallic character change down Groups 13 and 14?

    Metallic character increases down both groups. Group 13: B is a metalloid, Al–Tl are metals. Group 14: C non-metal, Si & Ge metalloids, Sn & Pb metals.

    Hint: Top = non-metal, bottom = metal.

11. 11.List three ways boron differs anomalously from the rest of Group 13.

    (1) B is a non-metal/metalloid; others are metals. (2) B is hard, high melting, forms only covalent compounds; max covalency 4 (no d-orbitals). (3) Boron hydrides & halides are electron-deficient Lewis acids; boron never forms simple $\text{B}^{3+}$ cation.

    Hint: Small size, high IE, no d-orbitals.

12. 12.Why is the maximum covalency of boron 4 whereas Al can reach 6?

    Boron (2nd period) has no $d$-orbitals, so it uses only $2s$ and $2p$ (max 4 bonds, e.g. $\text{BF}_4^{-}$). Aluminium has available $3d$ orbitals, allowing covalency up to 6, e.g. $[\text{AlF}_6]^{3-}$.

    Hint: Availability of d-orbitals.

13. 13.Why does carbon show anomalous behaviour compared with the rest of Group 14?

    Small size, high electronegativity, high IE, and absence of $d$-orbitals. Consequences: maximum covalency 4, strong tendency for catenation and $p\pi{-}p\pi$ multiple bonding (e.g. $\text{C=C}$, $\text{C=O}$), which heavier members can't do effectively.

    Hint: No d-orbitals; forms $p\pi$–$p\pi$ bonds.

14. 14.Why does $\text{CO}_2$ exist as a discrete gas while $\text{SiO}_2$ is a giant covalent solid?

    Carbon forms strong $p\pi{-}p\pi$ double bonds ($\text{O=C=O}$ discrete molecules). Silicon is larger and cannot form effective $p\pi{-}p\pi$ bonds, so it forms four Si–O single bonds giving a 3D network solid.

    Hint: Multiple bonding ability of C vs Si.

15. 15.Define catenation and give the catenation order in Group 14.

    Catenation is the self-linking of like atoms into chains/rings via covalent bonds. Order: $\text{C} \gg \text{Si} > \text{Ge} \approx \text{Sn} > \text{Pb}$. It follows the M–M bond strength, which falls as atomic size increases down the group.

    Hint: Depends on X–X bond enthalpy.

16. 16.Why does carbon show the greatest catenation of all elements?

    The C–C bond enthalpy ($\approx 348\ \text{kJ mol}^{-1}$) is exceptionally high due to small size and good orbital overlap, and it is comparable to/greater than C–O, so long stable C–C chains form. Si–Si is much weaker, limiting Si catenation.

    Hint: Strong, short C–C bond.

17. 17.Why are boron trihalides good Lewis acids, and give their acid strength order.

    BX$_3$ has only 6 electrons on B (electron-deficient, empty $2p$), so it accepts a lone pair. Lewis acid strength: $\text{BF}_3 < \text{BCl}_3 < \text{BBr}_3 < \text{BI}_3$ — reverse of expected because back-bonding ($p\pi{-}p\pi$) is strongest in BF$_3$, reducing its acidity most.

    Hint: Back-bonding weakest for the biggest halogen.

18. 18.Explain why $\text{BF}_3$ is a weaker Lewis acid than $\text{BCl}_3$ despite F being more electronegative.

    F is small and its $2p$ orbital overlaps well with boron's empty $2p$, giving strong $p\pi{-}p\pi$ back-donation that partially fills boron's octet. This reduces its electron deficiency more than in $\text{BCl}_3$ (poorer $2p{-}3p$ overlap), so $\text{BF}_3$ is the weaker acid.

    Hint: Effective back-bonding needs size match.

19. 19.Give the molecular formula and structure of borax.

    Borax = $\text{Na}_2\text{B}_4\text{O}_7\cdot 10\text{H}_2\text{O}$. Its correct structural formula is $\text{Na}_2[\text{B}_4\text{O}_5(\text{OH})_4]\cdot 8\text{H}_2\text{O}$; the anion contains two 3-coordinate ($sp^2$) and two 4-coordinate ($sp^3$) boron atoms.

    Hint: Tetranuclear anion, 2 triangular + 2 tetrahedral B.

20. 20.Why is an aqueous solution of borax basic (alkaline)?

    Borax is the salt of a strong base (NaOH) and a weak acid (boric acid), so it hydrolyses: $\text{B}_4\text{O}_7^{2-} + 7\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{BO}_3 + 2\text{OH}^{-}$, releasing $\text{OH}^{-}$.

    Hint: Salt of weak acid + strong base.

21. 21.What happens when borax is strongly heated (borax bead test basis)?

    It loses water, swells, then melts to a clear glass of sodium metaborate + boric anhydride: $\text{Na}_2\text{B}_4\text{O}_7 \rightarrow 2\text{NaBO}_2 + \text{B}_2\text{O}_3$. The $\text{B}_2\text{O}_3$ dissolves coloured metal oxides to give characteristic bead colours.

    Hint: Gives $\text{NaBO}_2$ and $\text{B}_2\text{O}_3$.

22. 22.Give the formula and structure of orthoboric acid.

    $\text{H}_3\text{BO}_3$ (or $\text{B(OH)}_3$). Boron is $sp^2$, trigonal planar; the solid is a layered structure of planar $\text{BO}_3$ units held together by hydrogen bonds, giving its soapy, slippery, flaky feel.

    Hint: Planar $sp^2$ B, H-bonded 2D sheets.

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