Hydrides — ionic, covalent, interstitial: Exercises

Before we start, here is the one energy tool we reuse everywhere.

Formula Born–Haber sum (the master equation)

To decide if a metal hydride $M H_{n}$ forms from the elements, we add up every energy cost and every energy payback: $Δ H_{f} = sublime metal S + ionize metal \sum I E + split H_{2} n \cdot D_{\frac{1}{2} H_{2}} + make H^{-} n \cdot E A_{H} + lattice U$

$S$ = sublimation energy: turning solid metal into gas atoms (always costs energy, $+$ ).
$I E$ = ionization energy: ripping electrons off the metal atom ( $+$ ).
$D_{\frac{1}{2} H_{2}}$ = half the H–H bond energy: cost to make one free H atom from $\frac{1}{2} H_{2}$ ( $+$ ).
$E A_{H}$ = electron affinity of H: energy when $H + e^{-} \to H^{-}$ (here it is released, so $- 73$ ).
$U$ = lattice energy: the big payback when the gas ions crash together into a crystal (large $-$ ).

A negative $Δ H_{f}$ means "downhill" → the compound tends to form.

The picture below is that same sum drawn as a staircase — each positive step climbs, the big lattice step plunges below the start, and where you land is $Δ H_{f}$ . We will refer back to it in L3.

Figure — Hydrides — ionic, covalent, interstitial

L1 — Recognition

Problem 1.1 (L1)

Classify each hydride as ionic, covalent, or interstitial: $CaH_{2}$ , $CH_{4}$ , $TiH_{1.8}$ , $HF$ , $KH$ , $PdH_{0.6}$ .

Recall Solution 1.1

Rule of thumb — look at the partner of hydrogen. (A quick vocabulary note first: the periodic table splits into blocks by which orbital the outermost electrons fill. Group 1 & 2 metals are the "s-block". The middle rectangle — the "transition metals" like Ti, Pd, Fe — is the "d-block" because their d-orbitals are filling; you don't need any orbital detail here, just recognise them as the metals in the central block.)

$CaH_{2}$ → Ca is a Group 2 metal (not Be) → ionic (contains $H^{-}$ ).
$CH_{4}$ → C is a non-metal, similar electronegativity to H → covalent.
$TiH_{1.8}$ → Ti is a transition metal (from the central d-block), and the formula is non-stoichiometric (1.8, not a whole number) → interstitial.
$HF$ → F is a non-metal → covalent.
$KH$ → K is a Group 1 metal → ionic.
$PdH_{0.6}$ → Pd is a transition (d-block) metal, fractional formula → interstitial.

Problem 1.2 (L1)

State the electronic configuration of the hydride ion $H^{-}$ and name the noble gas it is isoelectronic with.

Recall Solution 1.2

Neutral H has configuration $1 s^{1}$ . Adding one electron gives $H^{-}$ with $1 s^{2}$ — the same two-electron closed shell as helium (He). "Isoelectronic" just means "same number of electrons": both have 2.

L2 — Application

Problem 2.1 (L2)

Complete and balance the reaction of calcium hydride with water, and state one everyday use that follows from it.

Recall Solution 2.1

$H^{-}$ is a strong base; it strips a proton from water to make $H_{2}$ : $CaH_{2} (s) + 2 H_{2} O (l) \to Ca(OH)_{2} (a q) + 2 H_{2} (g)$ Balance check (count each element on both sides):

Ca: left has 1 (in $CaH_{2}$ ); right has 1 (in $Ca(OH)_{2}$ ). ✓
O: left has 2 (from $2 H_{2} O$ ); right has 2 (the two O in $Ca(OH)_{2}$ ). ✓
H — left side: 2 H from $CaH_{2}$ , plus $2 \times 2 = 4$ H from $2 H_{2} O$ → total 6.
H — right side: $Ca(OH)_{2}$ has 2 H (one H in each of its two OH groups), plus $2 H_{2}$ has $2 \times 2 = 4$ H → total 6. ✓

Six H on each side, so the equation is balanced. Use: because it consumes water and releases $H_{2}$ , $CaH_{2}$ is a portable drying agent and a field source of hydrogen gas (weather balloons).

Problem 2.2 (L2)

Given the H–H bond energy is $436 kJ/mol$ , what is $D_{\frac{1}{2} H_{2}}$ , the cost of making one gaseous H atom from $\frac{1}{2} H_{2}$ ?

Recall Solution 2.2

Splitting one whole $H_{2}$ molecule into two H atoms costs $436 kJ/mol$ . We only want one atom (from half a molecule): $D_{\frac{1}{2} H_{2}} = \frac{1}{2} \times 436 = 218 kJ/mol .$ This is the $+ 218$ term you plug into the Born–Haber sum.

L3 — Analysis

Problem 3.1 (L3)

Using the master sum, compute $Δ H_{f}$ for $NaH (s)$ from these values (kJ/mol): sublimation of Na $= + 108$ ; $I E_{1} (Na) = + 496$ ; $D_{\frac{1}{2} H_{2}} = + 218$ ; $E A_{H} = - 73$ ; lattice energy $U = - 808$ . Is formation favourable?

Recall Solution 3.1

Add every term straight down the master sum (follow the staircase figure at the top of the page — each number is one step): $Δ H_{f} = 108 + 496 + 218 - 73 - 808.$ Step by step: $108 + 496 = 604$ ; $604 + 218 = 822$ ; $822 - 73 = 749$ ; $749 - 808 = - 59$ . $Δ H_{f} = - 59 kJ/mol$ Negative → favourable (exothermic). Notice the single term that rescues it: the lattice energy $- 808$ (the big downward plunge in the figure) is bigger in magnitude than the whole positive pile ( $+ 822$ ). Ionic hydrides only exist because $H^{-}$ packs efficiently and releases huge lattice energy.

Problem 3.2 (L3)

$H_{2} O$ boils at $10 0^{\circ} C$ but $H_{2} S$ (heavier, $34 g/mol$ vs $18$ ) boils at $- 6 0^{\circ} C$ . Explain the reversal, and estimate how many times more energy per mole H₂O needs to vaporize than H₂S given $Δ H_{v a p} (H_{2} O) = 40.7$ and $Δ H_{v a p} (H_{2} S) = 18.7 kJ/mol$ .

Recall Solution 3.2

Reasoning: By mass alone, heavier $H_{2} S$ should boil higher (more electrons → stronger dispersion forces). It doesn't — so a stronger force must be at work in water. Oxygen is small and very electronegative ( $3.5$ vs S at $2.5$ ), so the O–H bond is strongly polar and each molecule forms hydrogen bonds ( $\sim 20 kJ/mol$ each) to neighbours. S is larger, less electronegative → negligible H-bonding, only weak dispersion. Breaking the H-bond network in water needs far more heat → higher boiling point. Ratio: $\frac{40.7}{18.7} \approx 2.18.$ Water needs about 2.2× the vaporization energy — consistent with the extra hydrogen-bond cost.

L4 — Synthesis

Problem 4.1 (L4)

Beryllium is a Group 2 metal, yet $BeH_{2}$ is not ionic. Build the argument from energy and from Fajans' rules. Data (kJ/mol): sublimation Be $= + 324$ ; $I E_{1} + I E_{2} (Be) = 900 + 1757 = 2657$ ; lattice energy of hypothetical ionic $BeH_{2} \approx - 3000$ ; plus $2 \times 218$ (two H atoms) $- 2 \times 73$ (two electron affinities).

Recall Solution 4.1

Energy test. For an ionic $Be^{2 +} (H^{-})_{2}$ we would need: $Δ H_{f} = 324 + 2657 + 2 (218) + 2 (- 73) + (- 3000) .$ Compute: $324 + 2657 = 2981$ ; $2 (218) = 436 \Rightarrow 3417$ ; $2 (- 73) = - 146 \Rightarrow 3271$ ; $3271 - 3000 = + 271$ . $Δ H_{f} \approx + 271 kJ/mol (positive \Rightarrow unfavourable) .$ The huge double ionization ( $2657$ ) is not paid back even by a big lattice energy. Contrast with Na (Problem 3.1) where only one modest ionization ( $496$ ) had to be covered. Fajans' reasoning. $Be^{2 +}$ is tiny with charge $+ 2$ → enormous charge density → it strongly polarizes (distorts) the $H^{-}$ electron cloud, dragging electron density into the bond region. That is exactly what a covalent bond is. Result: $BeH_{2}$ is a polymeric, electron-deficient, covalent solid (bridging 3-centre-2-electron B-like bonds), not a salt.

Problem 4.2 (L4)

$BH_{3}$ does not exist as a stable molecule; instead we get diborane $B_{2} H_{6}$ . Using an electron count, show why, and state how many electrons the two bridge bonds share.

Recall Solution 4.2

Electron count for $BH_{3}$ : B contributes 3 valence electrons; three B–H bonds need $3 \times 2 = 6$ electrons but B only supplies 3 and the three H's supply 3 → exactly 6, so all three bonds form — but boron is left with an empty p orbital (only 6 electrons around B, not the octet's 8). "Electron-deficient" → very reactive → it grabs a partner. Dimerization: two $BH_{3}$ units join into $B_{2} H_{6}$ . Read the figure below carefully: the two black borons in the centre are joined only through the red bridging H atoms — there is no line drawn directly between the borons, because there is no B–B bond. Follow the black lines out to the four terminal H atoms (ordinary 2-centre-2-electron bonds), then look at how each red bridge H connects to both borons at once. That double connection is the picture of a 3-centre-2-electron bond: one electron pair smeared across three atoms.

Bridge electron count: each bridge is a 3-centre-2-electron (3c-2e) bond — one pair of electrons (

2

electrons) is smeared over three atoms (B, H, B). So the two bridges share

2 + 2 = 4

electrons total among the bridging region. (Full molecule: 12 valence electrons = 4 terminal bonds + 2 bridge bonds, with none left over for a B–B line.)

L5 — Mastery

Problem 5.1 (L5)

A titanium hydride sample has formula $TiH_{1.73}$ . (a) Explain why the subscript is not a whole number. (b) If you have $1.00 mol$ of Ti fully loaded to $TiH_{1.73}$ , how many moles of $H_{2}$ gas were absorbed? (c) Why does this make interstitial hydrides useful for hydrogen storage?

Recall Solution 5.1

(a) Interstitial hydrides are non-stoichiometric: H atoms slot into the gaps (interstitial sites) between the metal atoms in the crystal lattice — see the interstitial-site figure below, where the red H atom sits in a hole between four black metal atoms. Not every gap need be filled, and the filling depends on temperature and pressure — so the H:Ti ratio is a continuous variable, not a fixed integer. This is impossible for ionic/covalent hydrides where whole-number valence rules apply.

(b)

1.73

mol of H atoms per mol Ti. Since

H_{2}

contains 2 atoms:

n (H_{2}) = \frac{1.73}{2} = 0.865 mol H_{2} .

(c) The metal soaks up hydrogen at moderate pressure and releases it on gentle heating — reversibly and at high volumetric density (H atoms packed into lattice gaps beat compressed gas). So interstitial hydrides act like a hydrogen sponge, ideal for safe storage/transport in fuel-cell technology.

Problem 5.2 (L5)

Rank $NH_{3}$ , $PH_{3}$ , and $CH_{4}$ by H–X–H bond angle, largest first, and justify each value ( $109. 5^{\circ}$ , $10 7^{\circ}$ , $93. 5^{\circ}$ available) using lone pairs and central-atom size.

Recall Solution 5.2

Ranking: $CH_{4} (109. 5^{\circ}) > NH_{3} (10 7^{\circ}) > PH_{3} (93. 5^{\circ})$ . How to read the figure below: three molecules side by side, each drawn as a central atom with two H bonds so you can see the angle between them. The middle one, $NH_{3}$ , is highlighted in red because it is the pivotal case — one lone pair pushing the angle down from the perfect tetrahedral value. Trace how the "V" of the two bonds gets narrower as you move left→right: wide for $CH_{4}$ , a little tighter for $NH_{3}$ , sharply pinched for $PH_{3}$ .

$CH_{4}$ : carbon has 4 bonding pairs, 0 lone pairs → perfect tetrahedron → $109. 5^{\circ}$ .
$NH_{3}$ : nitrogen has 3 bonds + 1 lone pair. A lone pair takes more angular room and squeezes the bonds together, dropping the angle to $10 7^{\circ}$ .
$PH_{3}$ : phosphorus is in the same group as N but much larger. The big P atom holds the H's far apart, orbital overlap is poor, and the bonds sit almost at $9 0^{\circ}$ (pure p-orbitals) → $93. 5^{\circ}$ . Rule captured: more lone pairs and larger central atoms both shrink the bond angle.

Self-test recap — how to use it

Recall Self-test prompts (reveal after answering aloud)

An ionic hydride example, and what ion it contains (Problem 1.1) ::: NaH / KH / CaH₂ (Group 1 & 2 except Be); contains the hydride ion H⁻ The single energy term that makes ionic-hydride formation exothermic (Problem 3.1) ::: the large negative lattice energy — it outweighs the whole positive pile of sublimation + ionization + bond-splitting Why $TiH_{1.73}$ has a fractional formula (Problem 5.1) ::: interstitial hydrides are non-stoichiometric — H atoms fill lattice gaps, and how many gaps fill depends on temperature/pressure Name and electron count of the bridge bonds in $B_{2} H_{6}$ (Problem 4.2) ::: 3-centre-2-electron (3c-2e) bonds — one electron pair shared across B–H–B; two bridges = 4 bridging electrons, no B–B bond Why $H_{2} O$ boils far above $H_{2} S$ despite being lighter (Problem 3.2) ::: hydrogen bonding in water (O is small and very electronegative), which is absent in the larger, less electronegative H₂S Why $BeH_{2}$ is covalent, not ionic (Problem 4.1) ::: forming Be²⁺ costs too much (double ionization 2657 kJ/mol) and tiny high-charge Be²⁺ polarizes H⁻ (Fajans' rules) → covalent character