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

You have read the parent topic and met the trends. This page does something different: it hunts down every kind of question Group 2 can hand you, sorts them into boxes, and then works one full example for each box. By the end there should be no scenario left that surprises you.

Before we touch a single example, a promise: every symbol used below is spelled out in words the moment it appears. If you have never seen $E^{\circ}$ or "lattice energy" before, you will still be able to follow.

The scenario matrix

Group 2 questions all live inside a small grid. The two axes are which property you are reasoning about and what kind of input the question hands you (a normal case, a degenerate/edge case, a limiting case, a word problem, or an exam trap).

Definition The words we will use

Radius $r$ — how far the outermost electron sits from the nucleus, measured in picometres (pm), where $1 pm = 1 0^{- 12}$ metres. Picture a fuzzy ball; $r$ is its size.
Ionization energy $I E$ — the energy in kJ/mol (kilojoules per mole) you must pay to rip one electron off a gaseous atom. Bigger $I E$ = electron held tighter. When we write $I E_{1}$ , the subscript $1$ means the first ionization energy — the cost to remove the first (easiest) electron; $I E_{2}$ would be the cost to remove the second.
Standard reduction potential $E^{\circ}$ — a voltage (in volts, V) that measures how badly an ion wants its electrons back. A more negative $E^{\circ}$ means the metal would rather stay a metal-losing-electrons — i.e. it is a stronger reducing agent (better at giving electrons away).
Lattice energy $L E$ — energy released when gaseous ions snap together into a solid crystal. Larger magnitude = more stable solid.
Charge density — charge divided by size. A tiny ion with charge $+ 2$ has a huge charge density; it pulls hard on nearby electron clouds.

Here is the full grid. Each cell names the example that fills it.

Property → / Case ↓	Radius & IE	Reactivity ( $E^{\circ}$ , water)	Compound formation (LE, peroxide)	Anomaly / diagonal
Normal in-group compare	Ex 1 (Be vs Mg IE)	Ex 3 (Ca vs Ba water)	Ex 5 (MgO vs BaO₂)	—
Degenerate / near-tie	Ex 2 (Sr vs Ba radius near-tie)	(see note ★)	—	—
Limiting / extreme end	—	Ex 4 (Be: reacts or not?)	—	—
Word problem (real world)	—	—	Ex 6 (whitewash / Ca(OH)₂)	—
Exam-style twist	—	—	Ex 7 (peroxide sizes trap)	Ex 8 (Be–Al diagonal)

Eight examples, one per filled cell. Let's clear the board.

Ex 1 — Normal compare: Be vs Mg ionization energy

Worked example Cell: Radius & IE / Normal compare

Statement: $I E_{1} (Be) = 899 kJ/mol$ but $I E_{1} (Mg) = 738 kJ/mol$ (recall $I E_{1}$ = first ionization energy, the cost to strip the first electron). Explain why Mg's is lower, and estimate the fractional drop.

Forecast: Guess first — is the drop closer to $5%$ , $18%$ , or $40%$ ? Jot it down.

Step 1. Write the electron addresses. Be $= 1 s^{2} 2 s^{2}$ ; Mg $= 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$ . Here the leading number in each term — the 1, 2, 3 — is the shell number $n$ (also called the principal quantum number): it labels which "energy layer" an electron lives in, counting outward from the nucleus. $n = 1$ is the innermost, tightest layer; each higher $n$ is a bigger, looser layer farther out. So Be's top electron sits in shell $n = 2$ , Mg's in shell $n = 3$ . Why this step? The shell number $n$ is the single biggest lever on how tightly an electron is held — the higher the layer, the farther and looser the electron — so we must read it off before anything else.

Step 2. Compare distance. Bigger $n$ (higher layer) → outer electron farther from the nucleus. Radii: Be $112$ pm, Mg $160$ pm. Now recall the physics of electrical attraction, called Coulomb's law: the pulling force between two charges weakens as they move apart, specifically $F \propto \frac{1}{r ^{2}}$ — double the distance $r$ and the force drops to a quarter. It applies here because the nucleus (positive) and the valence electron (negative) are just two charges attracting each other. Why this step? Coulomb's law is why distance matters: Mg's larger $r$ means a much weaker grip on its outer electron, so it is easier to remove.

Step 3. Compare shielding. Mg has $10$ inner electrons hiding its $3 s$ electron from the nucleus; Be has only $2$ . More shielding → the outer electron feels a weaker effective pull. Why this step? Shielding is the second lever; together with distance it explains the whole drop.

Step 4. Compute the fractional drop: $\frac{899 - 738}{899} = \frac{161}{899} \approx 0.179 = 17.9%.$ Why this step? Turning "lower" into a number lets us check our forecast and sanity-test the trend size.

Verify: $738 = 899 \times (1 - 0.179) \approx 899 \times 0.821 \approx 738$ . ✓ Answer $\approx 18%$ — the middle forecast. A drop of one shell giving ~one-fifth off $I E$ is physically sensible.

Ex 2 — Degenerate case: Sr vs Ba, a near-tie in radius

Worked example Cell: Radius & IE / Degenerate (near-tie)

Statement: Atomic radii: Sr $= 215$ pm, Ba $= 217$ pm. The gap is tiny even though Ba sits a whole period below. What does this near-tie mean, and why doesn't Ba jump far ahead?

Forecast: Will the Sr→Ba radius jump be bigger or smaller than the Ca→Sr jump ( $197 \to 215$ )?

Step 1. Measure both jumps. Ca→Sr $= 215 - 197 = 18$ pm. Sr→Ba $= 217 - 215 = 2$ pm. Why this step? "Near-tie" is a claim about a difference; we quantify it before explaining it.

Step 2. Explain the shrinking jump. Going from Sr's period to Ba's period, the row includes the ten transition ( $d$ -block) elements. Those added $d$ -electrons shield the nucleus poorly, so as the nucleus keeps gaining protons the outer electrons feel an extra pull that drags the whole atom back in. This poor $d$ -shielding cancels most of the size increase you'd otherwise expect from adding a new shell. Why this step? The degenerate near-tie is not an accident — it is the fingerprint of poor $d$ -shielding (a genuine effect for Ba, which sits before the lanthanide $4 f$ fill, so no $f$ -shielding is involved here), exactly the mechanism you must name in an exam.

Step 3. State the takeaway: radius trend still increases down the group, but the increments shrink, and Sr–Ba is the flattest step. Why this step? Edge cases test whether you know the trend can nearly flatten without ever reversing.

Verify: $18 pm > 2 pm$ , so the Sr→Ba jump is much smaller ✓. Radius still monotonically non-decreasing: $197 < 215 < 217$ ✓.

Ex 3 — Normal compare: Ca vs Ba with water

Worked example Cell: Reactivity / Normal compare

Statement: Both Ca and Ba react with cold water. Using $E^{\circ}$ values ( $E_{Ca}^{\circ} = - 2.87$ V, $E_{Ba}^{\circ} = - 2.91$ V), decide which reacts more vigorously and by how much (in volts).

Forecast: Which is the stronger reducing agent — the more negative or the less negative $E^{\circ}$ ?

Step 1. Recall the rule: a more negative $E^{\circ}$ means the metal gives up electrons more eagerly → stronger reducing agent → more vigorous with water. Why this step? $E^{\circ}$ is our numeric proxy for "how hard it reacts"; we must fix its sign-meaning first.

Step 2. Compare: $- 2.91 < - 2.87$ , so Ba is more negative → Ba reacts more vigorously. Difference $= ∣ - 2.91 - (- 2.87) ∣ = 0.04$ V. Why this step? Reading the sign correctly is the whole question; the gap tells us it is a modest difference.

Step 3. Sanity-link to trend: down the group $I E$ falls and $E^{\circ}$ gets more negative, so reactivity rises Ca < Sr < Ba. Ba being top matches. Why this step? Cross-checking against the master trend guards against a sign slip.

Verify: $- 2.91 - (- 2.87) = - 0.04$ , magnitude $0.04$ V ✓. Ordering $E_{Ba}^{\circ} < E_{Ca}^{\circ}$ confirms Ba more reactive ✓.

Ex 4 — Limiting case: does Be react with water at all?

Worked example Cell: Reactivity / Limiting (extreme end of the group)

Statement: Be sits at the top of the group — the extreme small-atom end. Its $E^{\circ} = - 1.85$ V, still negative, so thermodynamics says it could reduce water. Yet Be does not react with water even when hot. Resolve this apparent contradiction.

Forecast: Is the block on the reaction thermodynamic (energy says no) or kinetic (energy says yes, but something physically stops it)?

Step 1. Check thermodynamics. $E^{\circ} = - 1.85$ V is negative → the reaction $Be + 2 H_{2} O \to Be(OH)_{2} + H_{2}$ is favourable in principle. Why this step? We must rule out "energy forbids it" before invoking anything subtler.

Step 2. Identify the real block: Be's tiny, high-charge-density surface instantly forms a tight, impermeable oxide/hydroxide layer that seals the metal off from water. Why this step? This is a kinetic barrier — a physical shield — not an energy verdict; naming it correctly is the point of the limiting case.

Step 3. Contrast the other end: Ba's larger, lower-charge-density surface forms a loose, water-permeable coat → reaction proceeds. So the group's extreme small end (Be) is passivated; the extreme large end (Ba) is violent. Why this step? The limiting case is only illuminating when placed against the opposite limit.

Verify: $E_{Be}^{\circ} = - 1.85 < 0$ V (thermodynamically favourable) yet observed rate ≈ 0 → contradiction is resolved by kinetics, not thermodynamics ✓.

Ex 5 — Normal compare: MgO vs BaO₂ (which oxide, and why)

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

The figure above is split by a dashed lavender line into two halves. On the left, a small coral circle labelled Mg²⁺ sits next to a green two-lobed "dumbbell" — that dumbbell is the peroxide ion $O_{2}^{2 -}$ (two oxygen atoms sharing a $2 -$ charge). An arrow shows the tiny Mg²⁺ pulling hard on the dumbbell and tearing it apart into two separate round $O^{2 -}$ ions, so the product is simple $MgO$ . On the right, a much larger lavender circle labelled Ba²⁺ sits beside a yellow dumbbell whose central bond is drawn intact — the big, gentle Ba²⁺ leaves the peroxide whole, giving $BaO_{2}$ . Keep this picture in mind through the steps.

Worked example Cell: Compound formation / Normal compare

Statement: In excess oxygen, Mg forms the simple oxide $MgO$ (containing the $O^{2 -}$ ion) while Ba forms the peroxide $BaO_{2}$ (containing the $O_{2}^{2 -}$ ion). Explain using ion sizes and the lattice-energy rule.

Forecast: Which ion — small or large — is better at keeping the peroxide $O_{2}^{2 -}$ in one piece?

Step 1. Meet the two anions (left side of the figure): $O^{2 -}$ is a small round ion; $O_{2}^{2 -}$ (peroxide) is the larger green dumbbell of two oxygens sharing charge. Why this step? You cannot reason about which fits which lattice until you see the size mismatch drawn out.

Step 2. Apply charge density. Mg²⁺ is tiny (72 pm) → huge charge density → it polarizes and rips apart the fragile $O_{2}^{2 -}$ dumbbell into $O^{2 -}$ (the tearing arrow on the left of the figure). So Mg is stuck with the simple oxide. Why this step? High charge density is Be/Mg's signature move; it decides the outcome here.

Step 3. Apply size-matching / lattice energy. The rule $L E \propto \frac{q _{+} q _{-}}{r _{+} + r _{-}}$ says a big cation + big anion still gives a survivable lattice. Here $q_{+}$ and $q_{-}$ are the charges (in electron-charge units) of the cation and anion, and — the key definitions — $r_{+}$ is the cation radius (size of the positive ion) and $r_{-}$ is the anion radius (size of the negative ion), both in pm; their sum $r_{+} + r_{-}$ is roughly the distance between the two ion centres. Ba²⁺ is large (136 pm) → low charge density → it leaves $O_{2}^{2 -}$ intact (right side of the figure, bond kept), and the big-big pairing packs stably. So Ba keeps the peroxide. Why this step? This is the quantitative backbone: peroxides need large, gentle cations.

Verify: Charge densities are $charge \div radius$ . Writing $e$ for the elementary charge (the charge on one electron/proton — our unit of "one charge"), the densities in $e / pm$ are: Mg²⁺ $= 2/72 \approx 0.0278$ , Ba²⁺ $= 2/136 \approx 0.0147$ . Mg²⁺ density is $\approx 1.89 \times$ Ba²⁺'s → Mg is the polarizer, Ba the peroxide-keeper ✓.

Ex 6 — Word problem: slaked lime for a whitewash

Worked example Cell: Compound formation / Real-world word problem

Statement: A builder burns limestone ( $CaCO_{3}$ ) to make quicklime ( $CaO$ ), then adds water to make slaked lime $Ca(OH)_{2}$ for whitewashing. Starting from $100 g$ of pure $CaCO_{3}$ , how many grams of $Ca(OH)_{2}$ can be made? (Molar masses: $CaCO_{3} = 100$ , $CaO = 56$ , $Ca(OH)_{2} = 74$ g/mol.)

Forecast: Will the product mass be more or less than the $100$ g you started with?

Step 1. Write the two reactions: $CaCO_{3} heat CaO + CO_{2} ↑, CaO + H_{2} O \to Ca(OH)_{2} .$ Why this step? Mass follows the atoms; we need the balanced path from start to product.

Step 2. Moles of $CaCO_{3} = \frac{100 g}{100 g/mol} = 1.00 mol$ . Why this step? Stoichiometry runs on moles, not grams — convert first.

Step 3. The Ca atom is conserved: $1$ mol $CaCO_{3} \to 1$ mol $CaO \to 1$ mol $Ca(OH)_{2}$ . Why this step? Mole ratios along the chain are all $1 : 1$ , so tracking calcium settles the amount.

Step 4. Convert product moles back to grams: $m_{Ca(OH)_{2}} = 1.00 mol \times 74 g/mol = 74 g .$ Why this step? The builder weighs out grams, not moles, so this is the number actually usable.

Verify: Product $74$ g $< 100$ g start — matches the forecast that mass drops. Cross-check by atom bookkeeping: we vented $CO_{2}$ ( $44$ g) then added $H_{2} O$ ( $18$ g), so $100 - 44 + 18 = 74$ g ✓. Units g ✓.

Ex 7 — Exam twist: the "peroxide from Mg" trap

Worked example Cell: Compound formation / Exam-style twist

Statement: An exam claims: "Burning Mg in excess O₂ gives magnesium peroxide MgO₂, the way Ba gives BaO₂." True or false, and pinpoint the single sentence that kills the claim.

Forecast: Trap or truth? Predict before reading on.

Step 1. Recall the peroxide-stability rule from Ex 5: peroxides need a large, low-charge-density cation to leave $O_{2}^{2 -}$ intact. Why this step? The twist is testing whether you'll blindly generalize; the rule is your anchor.

Step 2. Test Mg²⁺ against the rule. Mg²⁺ is small (72 pm), high charge density → it polarizes and shreds $O_{2}^{2 -}$ → only $MgO$ (simple oxide) forms. So $MgO_{2}$ is not the stable product. Why this step? Directly applying the rule to the trapped ion exposes the false generalization.

Step 3. The killer sentence: "Small high-charge-density Mg²⁺ cannot stabilize the peroxide ion; only the large cations (Ba²⁺, and partly Sr²⁺) can." Verdict: FALSE. Why this step? Exams reward naming the exact mechanism, not just "wrong".

Verify: Same charge-density numbers as Ex 5 (with $e$ = one electron-charge unit): Mg²⁺ $\approx 0.0278 e / pm$ (high) vs Ba²⁺ $\approx 0.0147 e / pm$ (low). Mg²⁺ high → shreds peroxide → claim false ✓.

Ex 8 — Exam twist: the Be–Al diagonal relationship

Worked example Cell: Anomaly / diagonal — exam twist

Statement: State two concrete ways Be resembles Al (its diagonal neighbour) more than it resembles Mg (its own group), and explain the single physical quantity that drives the resemblance.

Forecast: Guess the one driving quantity — is it atomic mass, charge density, or electronegativity?

Step 1. Compute the shared quantity: charge density = charge ÷ ionic-radius (writing $e$ for one electron-charge unit). Be²⁺: $2/31 \approx 0.0645 e / pm$ . Al³⁺: $3/54 \approx 0.0556 e / pm$ . These are close; Mg²⁺'s $0.0278 e / pm$ is roughly half. Why this step? The diagonal likeness isn't magic — Be and Al happen to have similar charge densities, and that one number drives everything that follows. It settles the Forecast: the answer is charge density, not mass or electronegativity.

Step 2. List the first resemblance — covalent chlorides. Both $BeCl_{2}$ and $AlCl_{3}$ are covalent and dissolve in organic solvents, whereas $MgCl_{2}$ is ionic. High charge density lets Be²⁺/Al³⁺ pull on the chloride electron cloud so strongly that the bond turns covalent. Why this step? A concrete, testable similarity beats a vague "they act alike" — and it flows straight from the charge-density number.

Step 3. List the second resemblance — amphoteric oxides/hydroxides. $Be(OH)_{2}$ and $Al(OH)_{3}$ both dissolve in both acid and alkali (amphoteric = reacts as both acid and base), whereas $Mg(OH)_{2}$ is only basic. Again the high charge density is what lets these hydroxides behave acid-like toward strong alkali. Why this step? Two independent similarities (bonding and acid–base behaviour) trace to the same cause, which is exactly the point examiners want made.

Step 4. Name the mechanism explicitly: high charge density strongly polarizes neighbouring anions (Fajans' idea), pushing bonds toward covalent and oxides toward amphoteric. Because Be²⁺ and Al³⁺ share this trait but Mg²⁺ does not, Be lines up with its diagonal neighbour rather than its group neighbour. Why this step? The examiner wants the cause, not just the list — this sentence is the mark-scoring core.

Verify: Charge densities $e / pm$ : Be²⁺ $\approx 0.0645$ , Al³⁺ $\approx 0.0556$ , Mg²⁺ $\approx 0.0278$ . Distance Be→Al $= ∣0.0645 - 0.0556∣ = 0.0089$ is far smaller than Be→Mg $= ∣0.0645 - 0.0278∣ = 0.0367$ , so Be²⁺ genuinely sits nearer Al³⁺ ✓. Two named similarities delivered (covalent chloride, amphoteric hydroxide) ✓.

Recall Quick self-test (reveal after answering)

Which reacts more vigorously with water, Ca or Ba? ::: Ba — its $E^{\circ}$ ( $- 2.91$ V) is more negative than Ca's ( $- 2.87$ V). Why does Ba form a peroxide but Mg does not? ::: Ba²⁺ is large with low charge density and leaves $O_{2}^{2 -}$ intact; small high-charge-density Mg²⁺ shreds it into $O^{2 -}$ . From 100 g of pure CaCO₃, how much Ca(OH)₂ forms? ::: 74 g (1 mol Ca conserved × 74 g/mol). What single quantity makes Be resemble Al, and name two similarities? ::: Charge density — Be²⁺ (≈0.065 e/pm) ≈ Al³⁺ (≈0.056 e/pm), far from Mg²⁺ (≈0.028 e/pm). Similarities: covalent chlorides (BeCl₂, AlCl₃) and amphoteric hydroxides (Be(OH)₂, Al(OH)₃).