Hydrides — ionic, covalent, interstitial: Visual walkthrough

This is the visual companion to the parent topic. We build every idea from zero. If a symbol appears, it was defined the line before.

Step 1 — Meet the two players and the one question

WHAT. We have two neutral atoms: a metal atom $M$ (we will use sodium, $Na$ ) and a hydrogen atom $H$ . "Neutral" means the number of positive protons equals the number of negative electrons, so the atom has zero net charge.

WHY. Before any bonding, we must know what each atom wants. Hydrogen has one electron in its lowest shell but room for two — the same full shell as helium. So hydrogen is tempted to grab one more electron and become $H^{-}$ (the hydride ion, a hydrogen atom carrying one extra electron, net charge $- 1$ ). Sodium has one loosely-held outer electron it is tempted to lose, becoming $Na^{+}$ .

PICTURE. In the figure, the loose sodium electron (orange dot) sits far from its nucleus — easy to pull away. Hydrogen (teal) has an empty "slot" in its shell.

The one question of this whole page ::: does the electron transfer completely (ionic), or get shared (covalent)?

Step 2 — The rule that decides: who is greedier?

WHAT. Compare the two electronegativities. Define the difference $Δ χ = χ (H) - χ (M) .$ Here $Δ χ$ is the gap in greediness; $χ (H) = 2.1$ is hydrogen's pull; $χ (M)$ is the metal's pull.

WHY. A big gap means one atom overwhelmingly wins the electron — that leans toward complete transfer (ionic); a small gap means sharing (covalent).

PICTURE. The figure is a number line of $χ$ . Sodium sits far left, hydrogen middle-right; the long orange arrow between them ( $Δ χ = 1.2$ ) is wide. Beryllium ( $χ = 1.5$ ) sits close to hydrogen — a short arrow ( $Δ χ = 0.6$ ).

$Δ χ (Na) = 2.1 - 0.9 = 1.2 Δ χ (Be) = 2.1 - 1.5 = 0.6$

Here each term: $2.1$ is hydrogen's pull (same in both), $0.9$ and $1.5$ are the metals' pulls.

Step 3 — The energy tug-of-war (the Born–Haber idea)

WHAT. We break the reaction $Na(s) + \frac{1}{2} H_{2} (g) \to NaH(s)$ into small imaginary steps. Some steps cost energy (positive numbers, $+$ ) — we must pump energy in. One step pays back energy (negative, $-$ ) — energy is released. We add them all up.

WHY. Energy can't be created or destroyed, so the total change $Δ H_{f}$ (the "formation enthalpy" — net energy released or absorbed making one mole of the compound) is the same no matter what path we take. Splitting into small steps lets us find each cost from tables and sum them. A negative total means the compound forms happily (releases energy).

Before the table, one symbol to name up front: let $D$ stand for the bond dissociation energy of $H_{2}$ — the energy needed to snap one whole $H_{2}$ molecule into two separate H atoms. Since we only need one H atom per formula unit of NaH, we use half of it, written $\frac{1}{2} D$ .

PICTURE. A staircase: each step is a labelled block, upward blocks = cost (orange), the one big downward block = payback (teal). Whether the finish line is below the start decides everything.

The four costs and one giant payback, in kJ/mol:

symbol	meaning	value
$Δ H_{sub}$	rip Na out of its solid into gas	$+ 108$
$I E_{1}$	tear the first (outer) electron off Na → $Na^{+}$	$+ 496$
$\frac{1}{2} D$	snap half an $H_{2}$ into one H atom ( $D$ = H–H bond dissociation energy)	$+ 218$
$E A$	H swallows the electron → $H^{-}$ (releases)	$- 73$
$U$	ions snap into the crystal (lattice energy)	$- 808$

Step 4 — Cost side: freeing the electron

WHAT. Follow the first three uphill blocks. $Δ H_{sub} = + 108$ lifts a sodium atom out of the metal. $I E_{1} = + 496$ (the first ionization energy, energy to remove the first electron) strips that atom to $Na^{+}$ . $\frac{1}{2} D = + 218$ (half of $D$ , the H–H bond dissociation energy defined above) makes one lone H atom.

WHY. For $Na^{+}$ and $H^{-}$ to exist as separate ions, we must pay to free the electron from sodium and to free a single H atom. These are unavoidable up-front costs.

PICTURE. The figure zooms into the three orange upward arrows stacked, running total climbing to $+ 108 + 496 + 218 = + 822$ kJ/mol — high above the start line.

$Δ H_{sub} + 108 + I E_{1} + 496 + \frac{1}{2} D + 218 = + 822 kJ/mol (so far a big debt)$

At this point the reaction looks like a loser: we are $+ 822$ in the red.

Step 5 — Payback: H grabs the electron, then the crystal snaps shut

WHAT. Two downhill blocks. $E A = - 73$ ( $E A$ = electron affinity of H — energy released when H accepts an electron to become $H^{-}$ ). Then the giant one: $U = - 808$ ( $U$ = lattice energy, energy released when free ions $Na^{+}$ and $H^{-}$ slam together into a solid crystal).

WHY. The tiny $H^{-}$ (radius $\approx 146$ pm) and $Na^{+}$ pack tightly in an NaCl-type lattice; opposite charges close together release enormous energy — that is what $- 808$ represents. This single payback is what makes ionic hydrides possible.

PICTURE. The two teal downward arrows, the second one huge, dropping the running total below the start line.

$debt so far + 822 E A - 73 U - 808 = - 59 kJ/mol$

$Δ H_{f} (NaH) = + 108 + 496 + 218 - 73 - 808 = - 59 kJ/mol$

Each symbol: sum of the three costs ( $+ 822$ ), minus H's payback ( $- 73$ ), minus the crystal's giant payback ( $- 808$ ). Negative total ⇒ NaH forms and releases energy. The tug-of-war is won by the lattice.

Step 6 — Edge case: turn the knob to beryllium and watch it fail

WHAT. Repeat the same accounting for $Be + H_{2} \to BeH_{2}$ . Two things change dramatically: beryllium must lose two electrons (charge $+ 2$ ), so we pay $I E_{1} + I E_{2}$ . Here $I E_{2}$ is the second ionization energy — the energy to tear off the second electron, after the first has already gone, i.e. removing an electron from the already-positive $Be^{+}$ ion. Numerically $I E_{1} + I E_{2} = 900 + 1757 = 2657$ kJ/mol.

WHY. $I E_{2}$ is huge because you are prying an electron away from an ion that is already positively charged and clings to its remaining electrons much harder. So $Be^{2 +}$ is brutally expensive to make. Even though its lattice energy is large ( $U \approx - 3000$ kJ/mol), the payback can't cover the sky-high ionization bill — and a small, highly-charged $Be^{2 +}$ distorts (polarizes) the soft $H^{-}$ cloud so strongly the electron gets shared back, turning the bond covalent (Fajans' idea: small strong cation + big squishy anion ⇒ covalent character).

PICTURE. Two staircases side by side. Left: sodium finishes below the start (ionic wins). Right: beryllium's ionization tower is so tall the crystal payback can't bring it below start — plus the arrow bends, showing the electron pulled back into a shared bond.

Now let us actually finish the sum instead of stopping halfway. Adding up all the purely ionic Born–Haber terms for BeH₂ (both bonds, both H atoms):

$Δ H_{f}^{(ionic model)} (BeH_{2}) = Δ H_{sub} + 324 + I E_{1} + I E_{2} 2657 + D 2 (218) - 2 E A 2 (73) + U (- 3000) = + 271 kJ/mol$

Term by term: $+ 324$ to sublime Be, $+ 2657$ to strip both electrons, $+ 436$ to make two H atoms, $- 146$ as both H atoms grab electrons, $- 3000$ from the crystal. The total is $+ 271$ kJ/mol — still positive. A positive $Δ H_{f}$ means the purely-ionic BeH₂ would cost energy to form, so it does not form as an ionic solid. Nature instead chooses the sharing route (covalent, polymeric BeH₂), which is downhill. Same recipe, different ingredient, opposite answer.

Step 7 — Degenerate check: what if the metal were as greedy as H?

WHAT. Imagine sliding $χ (M)$ up until $Δ χ = 0$ — the metal pulls exactly as hard as hydrogen (like a non-metal partner, $χ \approx 2.1$ ).

WHY. With no greediness gap, neither atom wins the electron. There is no $H^{-}$ , no $M^{+}$ , hence no lattice-energy payback — the entire $- 808$ -style rescue vanishes. The only stable outcome left is sharing: a covalent molecular hydride.

PICTURE. The number line from Step 2, animated to the limit: the orange transfer-arrow shrinks to a point, and a new "shared cloud" (plum) grows between the atoms — the smooth handover from ionic to covalent as $Δ χ \to 0$ .

This closes every case: big $Δ χ$ + affordable ionization ⇒ ionic (NaH); small $Δ χ$ or brutal ionization ⇒ covalent (BeH₂, CH₄); the boundary is exactly where the lattice payback stops covering the ionization bill.

The one-picture summary

One frame, whole story: the energy staircase for NaH (finishes below start → ionic, forms) overlaid on Be's (can't finish below → covalent), with the decision arrow keyed to $Δ χ$ and to the ionization cost.

Recall Feynman retelling — say it like a story

Two atoms walk up. Sodium is generous with its one loose electron; hydrogen has an empty seat and wants one. To actually hand the electron over we pay three tolls: pull sodium out of its metal ( $+ 108$ ), rip its electron off ( $+ 496$ ), and split an $H_{2}$ ( $+ 218$ ) — we're $+ 822$ in the hole. Then hydrogen pockets the electron and gives back a little ( $- 73$ ), and the finale: the tiny hydride ion and the sodium ion snap into a crystal and hand back a fortune ( $- 808$ ). Add it up: $- 59$ . Net energy released — so the salt forms. Now swap in beryllium. It must give up two electrons, and the second one ( $I E_{2}$ ) is savagely expensive, costing $2657$ total; summing the whole ionic ledger lands at $+ 271$ — still positive, so ionic BeH₂ never forms; it shares instead (covalent). That single difference — how expensive the electrons are to free versus how much the crystal pays back — is the entire reason NaH is a salt and BeH₂ is not. (And Mg sits in between: borderline.)

Recall Quick self-test

Why is the lattice energy the hero of NaH's formation? ::: It's the single most negative term ( $- 808$ kJ/mol); it alone outweighs the combined $+ 822$ cost of freeing the electron, giving a net $- 59$ . Does NaH obey the " $Δ χ ≳ 1.7$ = ionic" rule of thumb? ::: No — NaH's $Δ χ = 1.2$ is below $1.7$ yet it is genuinely ionic. The rule of thumb is only a rough screen; the true test is the energy balance (lattice payback vs ionization cost). Why does ionic BeH₂ fail to form? ::: The second ionization energy $I E_{2}$ makes stripping two electrons cost $2657$ kJ/mol; the full ionic Born–Haber sum lands at $+ 271$ kJ/mol (positive), so it never forms ionically — it shares (covalent) instead. Is Be the only Group-2 exception? ::: No — MgH₂ is borderline (significant covalent character); BeH₂ is covalent, CaH₂ and heavier are clearly ionic. What happens to bonding as $Δ χ \to 0$ ? ::: No electron transfer, no ions, no lattice payback — the atoms share electrons and form a covalent molecular hydride.