Hydrides — ionic, covalent, interstitial: Worked examples

Before anything, three plain-word reminders (each earned in the parent note, restated so line one makes sense):

Recall Three words you must already own

Electronegativity (χ) ::: a number (Pauling scale, 0.7 → 4.0) measuring how strongly an atom pulls shared electrons toward itself. Big χ = greedy for electrons. Hydride ion H⁻ ::: a hydrogen atom that has gained one extra electron, so it has 2 electrons (config $1 s^{2}$ , same as helium). It carries a full negative charge. Lattice energy ::: the energy released when gaseous positive and negative ions snap together into a solid crystal. Always a large negative (releasing) number for ionic solids.

The scenario matrix

Every hydride question falls into one of these cells. The point of this page: leave no cell untouched.

Cell	What the question throws at you	Example that covers it
A — low χ partner	Metal much less electronegative than H → predict ionic	Ex 1 (KH)
B — high χ partner	Non-metal ≥ H in χ → predict covalent	Ex 2 (HCl)
C — d-block partner	Transition metal → predict interstitial, non-stoichiometric	Ex 3 (Pd–H)
D — the borderline / degenerate case	χ difference is small, or an element the "rule" excludes → must reason, not memorise	Ex 4 (BeH₂ vs MgH₂)
E — sign of energy (is it exothermic?)	Full Born–Haber sum: does the negative lattice term win?	Ex 5 (LiH $Δ H_{f}$ )
F — limiting / trend behaviour	"Which is bigger?" along a group or period (bond angle, boiling point)	Ex 6 (bond-angle order), Ex 7 (H₂O vs H₂S)
G — real-world word problem	A gas-volume / stoichiometry story	Ex 8 (CaH₂ hydrogen generator)
H — exam twist	Looks like one family, secretly behaves like another	Ex 9 (B₂H₆ electron count)

Cell A — Low-χ partner → ionic

Worked example Ex 1 · Potassium + hydrogen

Statement: Potassium (χ = 0.8) is heated with H₂. Name the product, the bonding type, and what charge hydrogen carries.

Forecast: Guess now — does H become H⁺ or H⁻ here?

Step 1. Compare electronegativities: H = 2.1, K = 0.8, difference $Δ χ = 2.1 - 0.8 = 1.3$ . Why this step? The sign of the difference decides who keeps the shared electrons. K is far less greedy, so electrons flow toward H.

Step 2. Since electrons move onto H, hydrogen gains one electron → H⁻ (config $1 s^{2}$ ). Why this step? A partner poorer at holding electrons than H forces H into its negative costume — this is the definition of an ionic (saline) hydride.

Step 3. Group 1 stoichiometry is $2 M + H_{2} \to 2 M H$ , so: $2 K(s) + H_{2} (g) \to 2 KH(s)$ Why this step? K loses one electron (it's group 1), H⁻ takes one, so they pair 1:1.

Verify: Charge balance $= (+ 1) + (- 1) = 0$ ✓. KH is a crystalline solid that conducts when molten (mobile ions) and reacts with water to give H₂ — all ionic-hydride fingerprints. ✓

Cell B — High-χ partner → covalent

Worked example Ex 2 · Chlorine + hydrogen

Statement: H₂ reacts with Cl₂ (χ of Cl = 3.0). Classify HCl and state hydrogen's role.

Forecast: H⁺, H⁻, or shared?

Step 1. $Δ χ = 3.0 - 2.1 = 0.9$ , and crucially Cl is more electronegative than H. Why this step? Now the greedier atom is the partner. Electrons are pulled toward Cl but not fully transferred — a partial shift, not a full one.

Step 2. A partial shift = a shared (covalent) bond with a dipole: $H^{δ +} - Cl^{δ -}$ . Why this step? Full ionic transfer needs a huge χ gap and lattice-energy payoff; here neither exists, so nature settles for sharing.

Step 3. HCl exists as discrete molecules, a gas at room temperature. Why this step? Molecules attract each other only weakly (dipole + dispersion), so no giant crystal forms → low boiling point.

Verify: HCl gas dissolves in water to give an acid (H⁺ donor) — the opposite of ionic hydrides which give bases (H⁻). Different family, different behaviour. ✓

Cell C — d-block partner → interstitial

Worked example Ex 3 · Palladium soaking up hydrogen

Statement: Palladium metal absorbs H₂ to form roughly $PdH_{0.6}$ . Why is the formula not a whole-number ratio, and what type of hydride is this?

Forecast: Can a real compound honestly have "0.6" atoms?

Step 1. Pd is a transition (d-block) metal. Its atoms already sit in a close-packed crystal with tiny empty gaps (interstitial sites) between them. See the figure.

Why this step? H atoms are small enough to slip into those gaps without forming a fixed ionic or covalent bond — like guests filling empty seats.

Step 2. Because H just fills available gaps, not every gap need be filled → the ratio is variable (non-stoichiometric), here 0.6 H per Pd. Why this step? No fixed valence is being satisfied, so no fixed formula is forced.

Step 3. The metal keeps its metallic conductivity and lustre; H sits as neutral-ish atoms in the lattice. Why this step? This is exactly the "guest sneaking into gaps" picture — the interstitial family.

Verify: $0 < 0.6 < 1$ , a fraction — impossible for ionic/covalent (whole-number valence) but normal for interstitial hydrides. ✓

Cell D — Borderline / degenerate case

Worked example Ex 4 · Be vs Mg — the exception the rule warns about

Statement: The rule says "Groups 1 & 2 form ionic hydrides." Yet BeH₂ is covalent and MgH₂ is (mostly) ionic. Explain the split.

Forecast: Same group — why does one break the rule?

Step 1. Electronegativity gaps: Be = 1.5 → $Δ χ = 0.6$ ; Mg = 1.2 → $Δ χ = 0.9$ . Why this step? A small $Δ χ$ (Be) is a first warning that sharing, not transfer, may win.

Step 2. Ionization cost to reach $M^{2 +}$ : Be²⁺ needs $I E_{1} + I E_{2} = 900 + 1757 = 2657$ kJ/mol. Why this step? Be is tiny → its electrons are held very tightly → ripping off two is extremely expensive.

Step 3. Fajans' logic: tiny, highly-charged Be²⁺ severely polarises the soft H⁻ cloud, dragging electron density back into the bond → covalent character. So BeH₂ is polymeric/covalent. Why this step? A high charge density cation distorts the anion so much that the "ionic" picture collapses.

Step 4. Mg²⁺ is bigger, lower charge density → far less polarising → MgH₂ stays essentially ionic. Why this step? This is why the parent note excludes only Be, not Mg.

Verify: The trend is monotonic — going down group 2 (Be → Mg → Ca) cation size grows, polarising power falls, ionic character rises. Consistent. ✓

Cell E — Sign of the energy (Born–Haber)

Worked example Ex 5 · Is LiH really exothermic?

Statement: Using the cycle, compute $Δ H_{f}$ for $Li(s) + \frac{1}{2} H_{2} (g) \to LiH(s)$ from: sublimation $+ 161$ , ionization $+ 520$ , $\frac{1}{2}$ H–H dissociation $+ 218$ , electron affinity of H $- 73$ , lattice energy $- 906$ (all kJ/mol).

Forecast: Positive or negative overall?

Step 1. Add all the "costs" (positive terms): $161 + 520 + 218 = 899$ kJ/mol. Why this step? You must pay to vaporise Li, strip its electron, and split H₂ before any crystal forms.

Step 2. Add the "payouts" (negative terms): $- 73 + (- 906) = - 979$ kJ/mol. Why this step? Making H⁻ and slamming Li⁺ + H⁻ into a crystal releases energy.

Step 3. Sum: $Δ H_{f} = 899 - 979 = - 80$ kJ/mol. Why this step? The whole point — the giant negative lattice energy overpowers the positive costs, so formation is exothermic. See the ladder figure.

Verify: $161 + 520 + 218 - 73 - 906 = - 80$ kJ/mol, and $- 80 < 0$ → exothermic ✓. (Mirrors the parent's NaH result of $- 59$ kJ/mol — both negative, driven by lattice energy.) ✓

Cell F — Limiting / trend behaviour

Worked example Ex 6 · Order the bond angles (covalent hydrides)

Statement: Arrange $CH_{4}, NH_{3}, H_{2} O$ by H–X–H bond angle, largest first, and justify each drop.

Forecast: Which has the smallest angle?

Step 1. Count lone pairs on the central atom: C has 0, N has 1, O has 2. Why this step? Lone pairs are the invisible hands that squeeze the bond angle.

Step 2. All start from a tetrahedral $109.5°$ skeleton (4 electron pairs). Each lone pair pushes bonding pairs closer because it hugs the nucleus more tightly and takes up more angular room. Why this step? More lone pairs → more compression → smaller angle. See the figure.

Step 3. Result: $CH_{4} (109.5°) > NH_{3} (107°) > H_{2} O (104.5°)$ .

Verify: Angle drops by ~ $2.5°$ per added lone pair ( $109.5 \to 107 \to 104.5$ ) — a consistent, monotone limiting trend. ✓

Worked example Ex 7 · The boiling-point anomaly (limiting comparison)

Statement: H₂O boils at $100° C$ but H₂S at $- 60° C$ , even though H₂S is heavier (34 vs 18 g/mol). Explain.

Forecast: By mass alone, which "should" boil higher?

Step 1. By molecular mass, heavier H₂S should have stronger dispersion → higher boiling point. It doesn't → something extra is happening in H₂O. Why this step? Spotting that the naive prediction fails is the whole exam trick.

Step 2. O is small and very electronegative (3.5) → strong $O^{δ -} - H^{δ +}$ dipole; H can reach a neighbouring O closely → hydrogen bonds (~20 kJ/mol each). Why this step? These extra bonds must be broken to boil the liquid → high boiling point.

Step 3. S is larger, χ only 2.5 → weak dipole → negligible H-bonding in H₂S. Why this step? No extra glue → boils low despite extra mass.

Verify: Enthalpies of vaporisation: H₂O $= 40.7$ kJ/mol vs H₂S $= 18.7$ kJ/mol — H₂O needs ~ $2.2 \times$ more energy, matching the huge boiling-point gap. ✓

Cell G — Real-world word problem

Worked example Ex 8 · A calcium-hydride hydrogen generator

Statement: A weather balloon is filled by dropping $42 g$ of $CaH_{2}$ into excess water. How many litres of H₂ gas form at STP ( $22.4$ L/mol)? ( $M_{CaH_{2}} = 42 g/mol$ .)

Forecast: Guess the order of magnitude — a few litres, or tens of litres?

Step 1. Balanced reaction (ionic hydride + water → base + H₂): $CaH_{2} (s) + 2 H_{2} O(l) \to Ca(OH)_{2} (aq) + 2 H_{2} (g)$ Why this step? Each H⁻ grabs a proton from water and leaves as H₂; there are two H⁻ per CaH₂ → 2 mol H₂ per mol CaH₂.

Step 2. Moles of CaH₂ $= 42/42 = 1$ mol. Why this step? Convert given mass to moles before using the ratio.

Step 3. Moles of H₂ $= 1 \times 2 = 2$ mol. Why this step? Apply the 1:2 stoichiometry from step 1.

Step 4. Volume $= 2 \times 22.4 = 44.8$ L. Why this step? Convert moles of gas to volume at STP.

Verify: Units: $mol \times L/mol = L$ ✓. Note each water contributes one H and each hydride the other — that's why H⁻ from the solid, not O–H from water, sets the count. Answer $= 44.8$ L. ✓

Cell H — Exam twist (looks like one thing, behaves like another)

Worked example Ex 9 · Why B₂H₆, and where do the electrons go?

Statement: Diborane B₂H₆ looks like a simple covalent hydride, yet you cannot draw it with ordinary 2-electron bonds. Count the electrons and explain the twist.

Forecast: How many valence electrons does B₂H₆ have, and how many does a "normal" structure demand?

Step 1. Valence electrons available: each B gives 3, each H gives 1: $2 (3) + 6 (1) = 12$ electrons. Why this step? This is the honest electron budget you must spend.

Step 2. A "normal" structure needs a 2-electron bond for every connection. B₂H₆ has 6 B–H links + 1 B–B link $= 7$ bonds $\Rightarrow$ would need $7 \times 2 = 14$ electrons. Why this step? $14 > 12$ — the molecule is electron-deficient; it is short by 2 electrons. That's the twist.

Step 3. Nature's fix: two of the H atoms bridge the borons using 3-centre–2-electron (3c-2e) bonds — one electron pair glues three atoms (B–H–B) at once. Four terminal B–H bonds use $8$ electrons; the two bridge bonds use $2 \times 2 = 4$ electrons: total $8 + 4 = 12$ . ✓ Why this step? Spreading two electrons over three atoms lets 12 electrons hold the whole cage together.

Verify: Electron accounting: $8 (4 terminal B-H) + 4 (2 bridge 3c-2e) = 12$ , exactly the electrons available ✓. So BH₃ (electron-deficient monomer) dimerises to the stable B₂H₆. ✓

Recall Rapid self-test

A partner with χ = 0.8 gives which hydride type? ::: Ionic — H becomes H⁻. Why can PdHₓ have x = 0.6? ::: H fills gaps (interstitial), so the ratio is non-stoichiometric. Which single group-2 element is excluded from ionic hydrides? ::: Beryllium (Be²⁺ is too polarising → covalent BeH₂). What drives LiH formation to be exothermic? ::: The large negative lattice energy outweighs all positive cost terms. 1 mol CaH₂ + water gives how many mol H₂? ::: 2 mol (two H⁻ per formula unit).