Exercises — Alkali metals (Group 1) — physical - chemical properties, anomaly of Li, diagonal Li-Mg

WHAT we do: fill orbitals in order of increasing energy until we place 19 electrons.

Order: $1s,2s,2p,3s,3p,4s$. $\text{K}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^1$ In short form, the inner $1s^{2}2s^{2}2p^{6}3s^{2}3p^6$ block is argon, so: $\text{K}:[\text{Ar}]4s^1$ The valence electron is the single $4s^1$ electron. That lonely outer electron is the reason potassium is a soft, reactive alkali metal — exactly the $n{s}^1$ pattern from the parent note.

WHY: I.E. is the energy to pull the outer electron off. Down the group the electron sits farther from the nucleus and is more shielded, so it leaves more easily → I.E. decreases down the group.

Group order top→bottom: Li > Na > K > Cs (decreasing). Increasing I.E. means we reverse it: $\boxed{\text{Cs}<\text{K}<\text{Na}<\text{Li}}$

Step 1 — balanced equation (WHY: fixes the mole ratio). $2\text{Na}+2{\text{H}}_2\text{O}\to 2\text{NaOH}+{\text{H}}_2\uparrow$ Step 2 — moles of Na. $n_{\text{Na}}=\frac{4.6}{23}=0.2\text{mol}$ Step 3 — stoichiometry. From the equation $2\text{ mol Na}\to 1\text{ mol }{\text{H}}_2$, so $n_{{\text{H}}_2}=\frac{0.2}{2}=0.1\text{mol}$ Step 4 — volume at STP. $V=0.1\times 22.4=\boxed{2.24\text{L}}$

Recall the pattern from the parent note: as the cation grows it can cradle a bigger oxygen anion.

• (a) Li → normal oxide (${\text{O}}^{2-}$): $4\text{Li}+{\text{O}}_2\to 2{\text{Li}}_2\text{O}$
• (b) Na → peroxide (${\text{O}}_2^{2-}$): $2\text{Na}+{\text{O}}_2\to {\text{Na}}_2{\text{O}}_2$
• (c) K → superoxide (${\text{O}}_2^{-}$): $\text{K}+{\text{O}}_2\to {\text{KO}}_2$

WHY: small ${\text{Li}}^{+}$ only makes a stable lattice with a small anion; large ${\text{K}}^{+}$ stabilises the big superoxide ion. This is the lattice-energy argument.

WHAT the figure shows: two bars — the mass ratio K:Na and the volume ratio K:Na. Density is mass ÷ volume, so if volume grows faster than mass, density drops.

Model each atom as a sphere of radius $r$, volume $V=\frac{4}{3}\pi r^3$, so $\rho =\frac{M}{\frac{4}{3}\pi r^3}.$ Use metallic radii $r_{\text{Na}}\approx 186\text{ pm}$, $r_{\text{K}}\approx 227\text{ pm}$ and masses $M_{\text{Na}}=23$, $M_{\text{K}}=39$.

Mass ratio: $\frac{M_{\text{K}}}{M_{\text{Na}}}=\frac{39}{23}\approx \mathrm{1.70.}$

Volume ratio: ${\left(\frac{r_{\text{K}}}{r_{\text{Na}}}\right)}^3={\left(\frac{227}{186}\right)}^3\approx (1.22{)}^3\approx \mathrm{1.82.}$

Density ratio: $\frac{{\rho }_{\text{K}}}{{\rho }_{\text{Na}}}=\frac{1.70}{1.82}\approx 0.93<1.$ Since the ratio is below 1, K is less dense than Na — the volume ballooned (extra shell, big radius jump) faster than the mass grew. That is the density anomaly, made numeric.

WHAT the figure shows: a tiny ${\text{Li}}^{+}$ hugged tightly by water dipoles vs a big ${\text{K}}^{+}$ with water held loosely far away.

Hydration enthalpy scales roughly as charge density $\propto z/r$: the smaller the ion, the closer water's negative (oxygen) end can approach the $+$ charge, and the stronger the electrostatic pull. Since $r_{{\text{Li}}^{+}}\ll r_{{\text{K}}^{+}}$, ${\text{Li}}^{+}$ releases far more energy on hydration (about $-520\text{ kJ/mol}$ vs about $-320\text{ kJ/mol}$).

Why it matters: the overall energetics of $\text{M}(s)\to {\text{M}}^{+}(aq)$ is the sum $\Delta H=\Delta H_{\text{sub}}+\text{I.E.}+\Delta H_{\text{hyd}}.$ Even though Li's I.E. is large, its enormous negative $\Delta H_{\text{hyd}}$ overcompensates, making the aqueous electron-loss most favourable. This is why Li is the strongest reducing agent in water. See hydration enthalpy and reduction potentials.

Add the three steps (WHY: enthalpy is a state function — the sum along the path equals the net change).

Lithium: $\Delta H_{\text{Li}}=161+520+(-520)=+161\text{kJ/mol}$

Sodium: $\Delta H_{\text{Na}}=108+496+(-406)=+198\text{kJ/mol}$

Comparison: $\Delta H_{\text{Li}}=+161<\Delta H_{\text{Na}}=+198$. Lithium's aqueous ionisation costs less energy — it is more favourable.

Match with $E^{\circ }$: a more favourable (less positive) $\Delta H$ for $\text{M}\to {\text{M}}^{+}(aq)$ corresponds to a more negative $E^{\circ }$. Indeed $E_{\text{Li}}^{\circ }=-3.04\text{ V}$ is more negative than $E_{\text{Na}}^{\circ }=-2.71\text{ V}$. Consistent — the hydration term, not I.E. alone, drives Li's champion status.

Chain of ideas:

1. Diagonal relationship: $\text{Li}$ and $\text{Mg}$ share similar charge density ($z/r$) because moving right increases charge and moving down increases size, and these cancel along the diagonal. See diagonal relationship.
2. A small, high-charge-density cation strongly polarises the large carbonate anion, weakening the C–O bonds (polarising power / covalent character).
3. Therefore ${\text{Li}}_2{\text{CO}}_3$, unlike the very stable ${\text{Na}}_2{\text{CO}}_3$, decomposes on heating, just like ${\text{MgCO}}_3$: ${\text{Li}}_2{\text{CO}}_3\stackrel{\Delta }{\to }{\text{Li}}_2\text{O}+{\text{CO}}_2\uparrow$ Conclusion: ${\text{Li}}_2{\text{CO}}_3$ is thermally unstable — the lone anomaly among Group 1 carbonates, and a direct fingerprint of the Li–Mg diagonal link.

(a) Formation: $\text{K}+{\text{O}}_2\to {\text{KO}}_2(1\text{ mol K}\to 1\text{ mol }{\text{KO}}_2)$

(b) Mass of ${\text{KO}}_2$. $n_{\text{K}}=\frac{1.40}{39}=0.0359\text{mol}$ One-to-one, so $n_{{\text{KO}}_2}=0.0359\text{mol}$, and $m_{{\text{KO}}_2}=0.0359\times 71=2.55\text{g}.$

(c) ${\text{O}}_2$ from the scrubber reaction. Ratio $4{\text{KO}}_2\to 3{\text{O}}_2$: $n_{{\text{O}}_2}=0.0359\times \frac{3}{4}=0.0269\text{mol}.$

Answers: $m_{{\text{KO}}_2}\approx 2.55\text{ g}$, $n_{{\text{O}}_2}\approx 0.0269\text{ mol}$. The neat trick: potassium stores oxygen as superoxide and hands it back on demand — why ${\text{KO}}_2$ is a genuine breathing-apparatus chemical.

Reactivity with water (how violent): $\text{Li}<\text{Na}<\text{K}$. Down the group I.E. falls, the electron is given up faster, more heat evolves and the reaction becomes explosive.

Reducing strength in aqueous solution (by $E^{\circ }$): $\text{Li}>\text{K}>\text{Na}$ (from $-3.04,-2.93,-2.71\text{ V}$).

Why they disagree — the key contrast:

• Reactivity/violence is a kinetic + heat-release story: how fast the electron leaves and how much energy is dumped locally. Bigger, softer, lower-I.E. metals react faster ⇒ Cs/K most violent.
• Reducing power is a thermodynamic story set by $E^{\circ }$, which includes $\Delta H_{\text{hyd}}$. Tiny ${\text{Li}}^{+}$ has by far the largest hydration enthalpy, dragging $E^{\circ }$ to the most negative value ⇒ Li strongest reducer even though it reacts least violently.

One-line takeaway: speed of reaction $\ne$ thermodynamic driving force. Li wins on thermodynamics (hydration); K wins on kinetics (low I.E., big soft atom).

Q1.1 ::: $[\text{Ar}]4s^1$, valence $=4s^1$ Q1.2 ::: Cs < K < Na < Li Q2.1 ::: $2.24\text{ L }{\text{H}}_2$ Q2.2 ::: ${\text{Li}}_2\text{O}$, ${\text{Na}}_2{\text{O}}_2$, ${\text{KO}}_2$ Q3.1 ::: ${\rho }_{\text{K}}/{\rho }_{\text{Na}}\approx 0.93$ (K less dense) Q4.1 ::: $\Delta H_{\text{Li}}=+161$, $\Delta H_{\text{Na}}=+198\text{ kJ/mol}$; matches $E^{\circ }$ Q4.2 ::: ${\text{Li}}_2{\text{CO}}_3\to {\text{Li}}_2\text{O}+{\text{CO}}_2$ (decomposes) Q5.1 ::: $2.55\text{ g }{\text{KO}}_2$; $0.0269\text{ mol }{\text{O}}_2$