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

This page is the problem gym for the parent topic. Before we compute anything, we map out every kind of question the topic can ask. Then each example nails one cell of that map.

The scenario matrix

Every cell below is a class of question this chapter throws. Each worked example is tagged with the cell it lands in.

Cell	Class of question	What makes it tricky	Example
A	Stoichiometry — metal + water → H₂	mole ratio 2:1 (metal:H₂)	Ex 1
B	Stoichiometry — limiting reagent	two reactants, one runs out	Ex 2
C	Oxygen products (oxide / peroxide / superoxide)	which oxide? balance the odd formula	Ex 3
D	Trend ranking (radius, ionization enthalpy, density)	includes the K density anomaly	Ex 4
E	Thermodynamic cycle — why Li is strongest reducer	ionization enthalpy vs hydration enthalpy tug-of-war	Ex 5
F	Flame colour ↔ wavelength ↔ energy	tiny numbers, unit care	Ex 6
G	Diagonal Li–Mg anomaly (real-world/exam twist)	Li behaves like Mg, not Na	Ex 7
H	Degenerate / zero / limiting case	"what if the metal is Fr?" or 0 g	Ex 8

Prerequisite links you may want open: 2.5.04-Lattice-energy-Born-Haber-cycle, 6.1.05-Hydration-enthalpy, 5.3.02-Standard-reduction-potentials, 3.106-Periodic-trends-in-s-block.

Cell A — Metal + water stoichiometry

Worked example Ex 1 — Potassium in water

Statement: $3.9 g$ of potassium ( $K$ , molar mass $39 g/mol$ ) is dropped in excess water. Find the volume of $H_{2}$ produced at STP.

Forecast: Guess first — will it be more or less than the $1.12 L$ that $0.1 mol$ Na gave? (Same moles of metal, same 2:1 ratio... so guess the same volume.)

Step 1 — Balanced equation. $2 K + 2 H_{2} O \to 2 K O H + H_{2} ↑$ Why this step? The whole answer rides on the mole ratio, and only a balanced equation gives it: 2 metal atoms make 1 $H_{2}$ molecule.

Step 2 — Moles of K. $n_{K} = \frac{3.9}{39} = 0.10 mol$ Why? Mass alone is meaningless to the equation; the equation counts particles (moles).

Step 3 — Mole ratio to $H_{2}$ . $n_{H_{2}} = \frac{1}{2} n_{K} = \frac{0.10}{2} = 0.05 mol$ Why? The coefficients $2 : 1$ literally mean "for every 2 K, one $H_{2}$ ".

Step 4 — Volume at STP. $V = n \times 22.4 = 0.05 \times 22.4 = 1.12 L$ Why? At STP ( $273 K$ , $1 atm$ ) one mole of any gas occupies $22.4 L$ — that's the bridge from moles to litres.

Verify: Units: $mol \times \frac{L}{mol} = L$ ✓. And our forecast held — $0.1 mol$ of any Group-1 metal gives the same $1.12 L$ , because the ratio, not the identity, sets it.

Cell B — Limiting reagent

Worked example Ex 2 — Only a little water

Statement: $4.6 g$ of Na ( $23 g/mol$ ) is added to $1.8 g$ of water ( $18 g/mol$ ). Which runs out, and how much $H_{2}$ forms at STP?

Forecast: With so little water (only $0.1 mol$ ) versus $0.2 mol$ Na, guess: water is limiting.

Step 1 — Equation & moles. $2 N a + 2 H_{2} O \to 2 N a O H + H_{2}$ $n_{N a} = \frac{4.6}{23} = 0.20, n_{H_{2} O} = \frac{1.8}{18} = 0.10$ Why? Both reactants can now be spoken in the equation's language (moles).

Step 2 — Find the limiter. The equation needs Na and water in a $1 : 1$ ratio. We have $0.20$ Na but only $0.10$ water. Water is exhausted first ⇒ water is limiting, Na is in excess. Why? The reaction stops the instant any one reactant hits zero.

Step 3 — $H_{2}$ from the limiter. $n_{H_{2}} = \frac{1}{2} n_{H_{2} O} = \frac{0.10}{2} = 0.05 mol$ Why? Never use the excess reagent for the yield — the limiter caps everything.

Step 4 — Volume at STP. $V = 0.05 \times 22.4 = 1.12 L$ Why? Same molar-volume bridge as Ex 1 — moles of gas times $22.4 L/mol$ gives litres at STP.

Verify: Leftover Na $= 0.20 - 0.10 = 0.10 mol = 2.3 g$ sits unreacted. Sanity: had we (wrongly) used all $0.20 mol$ Na, we'd predict $0.10 mol H_{2} = 2.24 L$ — double the truth. So choosing the limiter mattered. ✓

Cell C — Oxygen products

Worked example Ex 3 — Which oxide, and balance it

Statement: $7.8 g$ of $K$ burns in excess oxygen. Name the product and find its mass. ( $K = 39$ , $O = 16$ .)

Forecast: K is a large cation → it stabilises the large superoxide ion $O_{2}^{-}$ . Guess product = $K O_{2}$ .

Step 1 — Pick the product. Potassium forms the superoxide $K O_{2}$ . Why this step? From the lattice-energy argument (see 2.5.04-Lattice-energy-Born-Haber-cycle): a big $K^{+}$ pairs best with a big anion. Small Li gives $L i_{2} O$ , Na gives $N a_{2} O_{2}$ , but K/Rb/Cs give $M O_{2}$ .

Step 2 — Balanced equation. $K + O_{2} \to K O_{2}$ Why? Already balanced (1 K, 2 O each side) — the superoxide formula does the balancing for us.

Step 3 — Moles of K. $n_{K} = \frac{7.8}{39} = 0.20 mol$ Why? The equation speaks in moles, so we must convert the given $7.8 g$ into moles before applying the $1 : 1$ ratio.

Step 4 — Mass of $K O_{2}$ . Ratio $K : K O_{2} = 1 : 1$ , and $M (K O_{2}) = 39 + 32 = 71 g/mol$ . $m = 0.20 \times 71 = 14.2 g$ Why? Convert product moles back to grams to answer "mass".

Verify: Oxygen absorbed $= 14.2 - 7.8 = 6.4 g$ . Check independently: $K O_{2}$ has 2 O per K, so $0.20 mol K$ needs $0.20 mol O_{2}$ = $0.20 \times 32 = 6.4 g$ ✓.

Cell D — Trend ranking (with the density anomaly)

Figure s01 (below): a warm-paper line plot of density (y-axis, g/cm³) against the five Group-1 metals Li→Cs (x-axis, left to right = down the group). The teal line climbs Li→Na, then a burnt-orange highlighted point at K sits lower than Na (the "red dip", arrowed "K dips below Na — anomaly"), then rises again through Rb→Cs. A dashed plum line shows the naive "should-keep-rising" expectation for contrast.

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

Worked example Ex 4 — Order these, and spot the odd one

Statement: Arrange Li, Na, K, Rb, Cs by (a) atomic radius, (b) first ionization enthalpy, (c) density. Flag the anomaly.

Forecast: Radius and ionization enthalpy move oppositely down a group. Density "should" rise — but one metal breaks rank. Which? (In the figure, look for the burnt-orange point that dips below its left neighbour.)

Step 1 — Radius. Down the group a new shell is added each step, so $r : Li < Na < K < Rb < Cs .$ Why? Distance to the outer electron grows faster than the effective nuclear charge — see 3.106-Periodic-trends-in-s-block.

Step 2 — Ionization enthalpy. Farther electron = weaker pull = $ionization enthalpy : Li > Na > K > Rb > Cs .$ Why? Exactly the mirror of radius — the same distance effect, opposite sign.

Step 3 — Density with the anomaly. $ρ = \frac{M}{V} \approx \frac{M}{\frac{4}{3} π r ^{3}} .$ What is $r$ here? We model each metal atom as a sphere of radius $r$ (its metallic/atomic radius) packing into the solid, and $\frac{4}{3} π r^{3}$ is that sphere's volume. This is an approximation: real metals pack with gaps and specific crystal structures, so the true density differs by a constant packing factor — but that factor is roughly the same across Group 1, so the trend it predicts is trustworthy even though absolute numbers aren't exact. Result: $ρ : Li < K < Na < Rb < Cs .$ Why the odd order? Going Na→K, the radius jumps so much (3rd→4th shell) that the volume ( $\propto r^{3}$ ) outruns the mass gain, so $ρ_{K} < ρ_{N a}$ . That is the anomaly — the burnt-orange dip in figure s01.

Verify (numbers): Real data $ρ_{N a} \approx 0.97, ρ_{K} \approx 0.86 g/cm^{3}$ . Since $0.86 < 0.97$ , K really is below Na ✓. Every other adjacent pair increases ✓.

Cell E — Thermodynamic cycle (why Li is the strongest reducer)

Figure s02 (below): a grouped bar chart on warm paper. Four groups on the x-axis — $Δ H_{sub}$ , ionization enthalpy, $Δ H_{hyd}$ , and Net — each with a burnt-orange bar (Li) beside a teal bar (Na). The $Δ H_{hyd}$ bars point down (negative); Li's dives deepest, arrowed "Li's huge negative hydration wins". The Net group shows Li's bar shorter than Na's, arrowed "Net Li < Net Na → Li stronger reducer".

Worked example Ex 5 — The Li reducing-power paradox

Statement: Li has the highest ionization enthalpy in Group 1, yet the most negative standard reduction potential $E^{\circ} (- 3.04 V)$ — the strongest reducing agent in water. Resolve this with the cycle $Δ H = Δ H_{sub} + I.E. + Δ H_{hyd}$ . Given (kJ/mol): for Li, $Δ H_{sub} = 161$ , $I.E. = 520$ , $Δ H_{hyd} = - 520$ ; for Na, $Δ H_{sub} = 108$ , $I.E. = 496$ , $Δ H_{hyd} = - 406$ .

Forecast: Guess where Li "wins back" its high ionization enthalpy. (Hint: $L i^{+}$ is tiny, so which term is huge?)

Step 0 — Why enthalpy at all? $E^{\circ}$ is set by the free-energy change $Δ G^{\circ} = - n F E^{\circ}$ , and $Δ G = Δ H - T Δ S$ . So strictly we should use $Δ G$ . We use $Δ H$ as a proxy because, across Li vs Na, the entropy terms $Δ S$ (of sublimation, ionization, hydration) are similar in size and largely cancel when we take the difference between two metals at the same $T$ . What's left — the part that actually differs — is dominated by the enthalpy terms. So comparing $Δ H$ correctly predicts which $E^{\circ}$ is more negative, even though it is not the full story. Why this tool? Enthalpies are tabulated and easy; the entropy correction would shift both numbers by roughly the same amount and not change the ordering.

Step 1 — Sum the cycle for Li. $Δ H_{L i} = 161 + 520 + (- 520) = 161 kJ/mol .$ Why? $E^{\circ}$ tracks the overall enthalpy of turning solid metal into hydrated ion, not ionization enthalpy alone — see 5.3.02-Standard-reduction-potentials.

Step 2 — Sum the cycle for Na. $Δ H_{N a} = 108 + 496 + (- 406) = 198 kJ/mol .$ Why? We repeat the identical three-term sum for Na so the two totals are built the same way and can be compared fairly — comparing only becomes valid when both sides use the same recipe.

Step 3 — Compare. Lower (more favourable) $Δ H$ ⇒ ion forms more readily ⇒ stronger reducing agent. $Δ H_{L i} = 161 < Δ H_{N a} = 198,$ so Li is the stronger reducer, matching its more-negative $E^{\circ}$ . Why? Li's enormous (very negative) $Δ H_{hyd} = - 520$ — because $L i^{+}$ is tiny and water clamps onto it hard (see 6.1.05-Hydration-enthalpy) — overpays its high ionization enthalpy.

Verify: $161 + 520 - 520 = 161$ ✓; $108 + 496 - 406 = 198$ ✓; and $161 < 198$ ✓ — the tiny-ion hydration term is the hero.

Cell F — Flame colour ↔ energy

Worked example Ex 6 — From wavelength to energy gap

Statement: Sodium's flame is yellow at $λ = 589 nm$ . Find the energy gap $Δ E$ of the electronic jump. Use $h = 6.626 \times 1 0^{- 34} J\cdots$ , $c = 3.0 \times 1 0^{8} m/s$ .

Forecast: Visible light carries a few $\times 1 0^{- 19} J$ per photon. Guess the answer is near $3 \times 1 0^{- 19} J$ .

Step 1 — Formula. $Δ E = \frac{h c}{λ} .$ Why this tool and not $Δ E = h ν$ directly? We are given $λ$ , not frequency $ν$ ; substituting $ν = c / λ$ gives exactly this form — one step, no extra unknown.

Step 2 — Convert units. $λ = 589 nm = 589 \times 1 0^{- 9} m .$ Why? SI needs metres, or the joules come out wrong by powers of ten.

Step 3 — Plug in. $Δ E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{589 \times 1 0 ^{- 9}} \approx 3.37 \times 1 0^{- 19} J .$ Why? Now that every quantity is in SI ( $h$ in J·s, $c$ in m/s, $λ$ in m), the arithmetic delivers the energy directly in joules — no leftover conversion needed.

Verify: Units: $\frac{J\cdots \cdot m/s}{m} = J$ ✓. Magnitude $3.37 \times 1 0^{- 19}$ sits in the visible-photon band ✓ — matches the forecast.

Cell G — Diagonal Li–Mg twist

Worked example Ex 7 — "Li acts like Mg, not Na" (exam classic)

Statement: LiF is poorly soluble in water, unlike NaF, KF (all very soluble). Which property class explains it, and which Group-2 element does Li mimic?

Forecast: Guess the partner element from the diagonal rule (down-one, right-one from Li).

Step 1 — Name the diagonal partner. Down from Li is Na, right of Na is Mg ⇒ Li resembles Mg (both small, high charge density). $M g F_{2}$ is also sparingly soluble. Why? Similar charge/size ratio → similar bonding character, per the diagonal relationship.

Step 2 — The tug-of-war behind solubility. A salt dissolves when the energy released by water wrapping the ions (hydration enthalpy, $Δ H_{hyd}$ ) can pay back the energy needed to rip the crystal apart (lattice energy, $U$ ). So we compare two energies: $U \propto \frac{z ^{+} z ^{-}}{r ^{+} + r ^{-}}, Δ H_{hyd} \propto \frac{z ^{+}}{r ^{+}} + \frac{z ^{-}}{r ^{-}} .$ Why this formula for $U$ ? Two point charges of magnitude $z^{+}$ and $z^{-}$ (here both $1$ for LiF) separated by a centre-to-centre distance $\approx r^{+} + r^{-}$ attract by Coulomb's law, whose energy scales as $\frac{z ^{+} z ^{-}}{separation}$ ; the hidden constant bundles physical constants and the crystal geometry (Madelung factor), detail in 2.5.04-Lattice-energy-Born-Haber-cycle. Why the hydration form? Water pulls on each ion separately, and the pull on one ion scales as its charge over its radius, so the two ions contribute additively — see 6.1.05-Hydration-enthalpy. Why it matters here: both $U$ and $Δ H_{hyd}$ grow when the ions are small, so the winner of the tug-of-war decides solubility.

Step 3 — Do the comparison for LiF. With the small $F^{-}$ and the small $L i^{+}$ , the denominator $r^{+} + r^{-}$ in $U$ is tiny, so lattice energy $U$ is very large. Because $U$ depends on the sum $r^{+} + r^{-}$ but hydration depends on each radius separately, shrinking both ions pumps $U$ up faster than it pumps hydration up. So here $U$ wins the tug-of-war: the crystal holds together and LiF barely dissolves — exactly like $M g F_{2}$ . Why? When lattice energy out-muscles hydration, the salt stays solid.

Step 4 — Contrast with big anions. For big anions ( $C l^{-}$ , $I^{-}$ ) the denominator $r^{+} + r^{-}$ grows, $U$ drops sharply, hydration now wins, and even LiCl, LiI dissolve well (LiI is very soluble). So only the small-anion salt (LiF) is the odd one out. Why? This size-sensitivity is the charge-density fingerprint of Li — the same fingerprint that makes it behave like Mg. Conclusion: LiF and $M g F_{2}$ are both sparingly soluble, both because a tiny cation + tiny fluoride gives a lattice energy too strong for water to overcome.

Verify: Prediction check — of {LiF, LiCl, LiI}, only LiF should be poorly soluble; experimentally LiF $\approx 0.13 g /100 mL$ while LiCl $\approx 83 g /100 mL$ . Ratio $≫ 100 \times$ ✓, consistent with the lattice-energy story.

Cell H — Degenerate / zero / limiting case

Worked example Ex 8 — Zero grams and the francium edge

Statement: (a) $0 g$ of Cs is dropped in water — volume of $H_{2}$ ? (b) Extrapolate: does francium (Fr) have higher or lower ionization enthalpy than Cs, and is it a stronger reducer in the gas phase?

Forecast: (a) is a trick (guess $0$ ). (b) push the trend one more shell.

Step 1 — Zero-mass limit. $n_{C s} = \frac{0}{133} = 0 \Rightarrow n_{H_{2}} = 0 \Rightarrow V = 0 L .$ Why? Stoichiometry is linear in moles; feed it zero and it must return zero — a sanity anchor for every formula above.

Step 2 — Trend to Fr (ionization enthalpy). Fr sits below Cs → one more shell → outer electron even farther → ionization enthalpy(Fr) < ionization enthalpy(Cs) (gas-phase trend continues). Why? The distance effect from Ex 4 doesn't stop at Cs — adding a shell keeps pushing the outer electron away and weakening the pull.

Step 3 — Gas-phase reducing power caveat. In the gas phase (no hydration), lowest ionization enthalpy = easiest to ionise, so Fr would look strongest. But in water, Li's giant hydration energy (Ex 5) makes Li the champion — the ranking depends on the phase. Why? This reminds us that "strongest reducer" is meaningless without stating the medium — the hydration term we met in Ex 5 can flip the order.

Verify: (a) $0 \times 22.4 = 0$ ✓. (b) Consistent with the monotone ionization-enthalpy decrease Li→Fr and the aqueous reversal already proven numerically in Ex 5 ( $161 < 198$ ) ✓.

Recall Quick self-test —

cover the answer after the ::: and say it aloud, then reveal These are active-recall prompts: read the question, answer from memory, then check the text after the triple colon. Ratio of Na to H₂ in the water reaction ::: $2 : 1$ Which alkali metal is less dense than the one above it ::: K (breaks the density trend) Oxygen product of K ::: superoxide $K O_{2}$ Why is Li the strongest reducer in water despite highest ionization enthalpy ::: its huge (very negative) hydration enthalpy overcompensates Li's diagonal partner ::: Mg $Δ E$ formula from wavelength ::: $Δ E = h c / λ$