Exercises — Alkaline earth metals (Group 2) — physical - chemical properties, anomaly of Be, diagonal Be-Al

"Atomic radius" = the size of the atom, measured tip-to-tip in picometres (pm; $1\text{pm}=10^{-12}\text{m}$). Going down Group 2 you add a whole new electron shell each step, so the atom grows.

Increasing order (smallest → largest): $\text{Be }(112)<\text{Mg }(160)<\text{Ca }(197)<\text{Sr }(215)<\text{Ba }(217)$ This is exactly top-to-bottom in the group. Look at the rising red staircase in the figure above — that IS this ordering. Answer: Be, Mg, Ca, Sr, Ba.

Beryllium (Be). Reason: Be is tiny with very high charge density (charge packed into small volume), so its surface grows a tough, sticky oxide skin ($\text{BeO}$) that seals the metal from water. This is the same "protective layer" idea that makes aluminium unreactive in air — a first hint of the diagonal Be–Al relationship.

Take the ratio so the constant of proportionality and $Z_{\text{eff}}^2$ cancel: $\frac{I{E}_{n=3}}{I{E}_{n=2}}=\frac{Z_{\text{eff}}^2/3^2}{Z_{\text{eff}}^2/2^2}=\frac{1/9}{1/4}=\frac{4}{9}\approx 0.444$ So $IE$ falls to about $44.4\%$ — roughly less than half. The drop is driven entirely by $n$ here (we held $Z_{\text{eff}}$ fixed), which is exactly why the falling-$IE$ curve in the figure above tracks rising shell number, not a falling $Z_{\text{eff}}$.

Both are $M^{2+}{O}^{2-}$, so $q_{+}{q}_{-}=(2)(2)=4$ is identical — only the size term differs. Lattice energy = the energy released when gaseous ions snap together into a crystal; smaller ions sit closer, pull harder, release more.

$L{E}_{\text{MgO}}\propto \frac{4}{72+140}=\frac{4}{212},L{E}_{\text{BaO}}\propto \frac{4}{136+140}=\frac{4}{276}$ $\frac{L{E}_{\text{MgO}}}{L{E}_{\text{BaO}}}=\frac{276}{212}\approx 1.30$ $\text{MgO}$ has the larger lattice energy (about $1.30\times$). Its small ions pack tighter — this is why $\text{MgO}$ is famously stable and high-melting.

The peroxide ion $O_2^{2-}$ is large and floppy (two oxygen atoms bonded, spreading the $-2$ charge out). To survive in a crystal it needs a partner that won't rip it apart.

• Small cation (Be²⁺, Mg²⁺): enormous charge density → it polarizes $O_2^{2-}$, tugging electron density so hard the weak O–O bond snaps, collapsing the peroxide into the compact, high-lattice-energy $O^{2-}$. So only $\text{MgO}$ forms.
• Large cation (Ba²⁺): gentle, diffuse charge → doesn't polarize $O_2^{2-}$; and a big cation size-matches the big anion, giving a decent stable lattice. So ${\text{BaO}}_2$ survives.

Rule of thumb: large cations stabilise large anions (size compatibility).

Define the key quantity first. Density $\rho =\frac{\text{mass of atoms}}{\text{volume they occupy}}$. A clean way to estimate the volume side is the molar volume $V_m=\frac{M}{\rho }({\text{cm}}^3/\text{mol}),$ where $M$ is the molar mass (g/mol). $V_m$ literally tells you "how many cm³ one mole of the metal spreads over." Plugging the data: $V_m(\text{Be})=\frac{9.0}{1.85}\approx 4.9,V_m(\text{Mg})=\frac{24.3}{1.74}\approx 14.0,V_m(\text{Ca})=\frac{40.1}{1.54}\approx \mathrm{26.0.}$ So calcium's atoms occupy far more volume per mole (26.0 vs 4.9 for Be) — its atoms are genuinely large and spread out. Its molar mass has grown too, but not fast enough to keep pace with that ballooning volume, so $\rho$ dips.

A correction on packing: both Ca and Be sit in efficient close-packed lattices (Ca is fcc, $\approx 74\%$ packing fraction; Be is hcp, also $\approx 74\%$). Packing fraction is not the reason — it is essentially the same. The real driver is the raw atomic size (huge molar volume for Ca) outrunning its modest mass gain. ${\rho }_{\text{Ca}}(1.54)<{\rho }_{\text{Mg}}(1.74)<{\rho }_{\text{Be}}(1.85)$ Answer: Calcium is the least dense. The pattern only becomes "heavier = denser" from Sr onward, once atomic mass climbs fast enough to dominate the volume growth. See the density picture below — Ca is the visible dip.

"Exothermic" = net energy released, so a negative $\Delta H$. Add the pieces (positive = costs, negative = pay-backs): $\Delta H_{\text{rxn}}=\underset{\text{ionize}}{\underbrace{+1735}}\underset{{\text{hydrate Ca}}^{2+}}{\underbrace{-1577}}\underset{\text{rest}}{\underbrace{-588}}=-430\text{kJ/mol}$ Negative → exothermic, matching the note. The single biggest pay-back is the huge hydration energy of the small, doubly-charged ${\text{Ca}}^{2+}$: water molecules crowd around it and release energy, more than covering the $1735$ kJ ionization bill. This is why Ca reacts vigorously with cold water while Be (whose tiny ion would hydrate even harder, but whose oxide seals the surface) does not.

More negative $E°$ = stronger reducer, so sort by most-negative first: $\text{Ba}(-2.91)>\text{Ca}(-2.87)>\text{Mg}(-2.37)>\text{Be}(-1.85)$ Reducing power increases down the group — consistent with falling ionization energy. The "odd one out" is Be at $-1.85$ V: it is far less negative than a smooth extrapolation of Mg→Ba would predict, because Be's small size makes its two electrons unusually hard to strip (high $I{E}_1+I{E}_2$). Yet another anomaly-of-Be fingerprint. In the figure below, Be sits well above the near-straight Mg–Ca–Ba line.

Metallic melting point is set by the strength of metallic bonding = how strongly a shared "sea" of delocalised electrons glues the positive metal ions together. Two things control that glue: (i) how close the ions sit (smaller ion → sea closer to nuclei → stronger glue) and (ii) how effectively the valence electrons actually delocalise into that sea.

• Be highest: tiny ions → the electron sea sits extremely close to the nuclei → strongest glue → melts highest. Size (i) dominates.
• Ca → Sr → Ba genuinely falls (1115 → 1050 → 1000 K): as ions swell, the sea sits farther out, glue weakens — the naive size rule works here.
• Mg is the misfit (lowest at 923 K). Concretely: magnesium's two $3s$ valence electrons delocalise less effectively into the conduction sea than the heavier metals' valence electrons do — the heavier atoms have more diffuse valence orbitals that overlap and share more freely, feeding a stronger sea despite their size. Combine that weaker delocalisation with Mg's particular hexagonal-close-packed geometry (in which the number of strong nearest-neighbour bonding contacts contributing to the sea is comparatively low), and the net glue drops below even large Ba. So two factors — poorer $3s$ delocalisation and unfavourable hcp bonding contacts — together push Mg to the group minimum.

Verdict: the naive size rule captures Ca→Ba but is overruled at Mg, where weak $3s$-electron delocalisation plus its hcp bonding geometry, not size, decide the melting point. Never quote a Group 2 melting-point order from size alone.

Fajans' idea: a cation with high charge density (large charge $÷$ small size) pulls, or polarizes, the electron cloud of its anion so strongly that the "shared" electrons stop being fully transferred — the bond acquires covalent character.

• ${\text{Be}}^{2+}$: radius only $\approx 31$ pm, charge $+2$ → extreme charge density → it distorts ${\text{Cl}}^{-}$ heavily → covalent ${\text{BeCl}}_2$.
• ${\text{Mg}}^{2+}$: radius $\approx 72$ pm → much gentler → keeps electrons on chlorine → ionic ${\text{MgCl}}_2$.

Diagonal link to Al: ${\text{Al}}^{3+}$ (one period down, one group right of Be) has a comparable charge density to ${\text{Be}}^{2+}$ — small ion, high charge. So ${\text{AlCl}}_3$ is also covalent, volatile, dimeric (${\text{Al}}_2{\text{Cl}}_6$). Be and Al match on charge density even though they aren't in the same group — the essence of the diagonal relationship: similar polarizing power → similar chemistry.

Q: Why does Be not react with water? A: its protective oxide layer ($\text{BeO}$) seals the surface. Q: Strongest reducing agent in Group 2 (most negative $E°$)? A: Ba. Q: Group 2 metal with the lowest melting point? A: Mg. Q: Why does ${\text{BaO}}_2$ form? A: large cations stabilise large ==peroxide anions ($O_2^{2-}$)==. Q: Why is ${\text{BeCl}}_2$ covalent? A: ${\text{Be}}^{2+}$ has extreme charge density. Q: The diagonal partner of Be? A: Al (aluminium).