Worked examples — Position of hydrogen in the periodic table (anomalous)

This page is the workout gym for Position of hydrogen in the periodic table (anomalous). The parent note told you why hydrogen is a chameleon. Here we hit every kind of question an exam or a curious brain can throw at that idea — one worked example per scenario — so you never meet a case you have not already practiced.

Before symbols fly, three tiny reminders in plain words:

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The scenario matrix

Every question about hydrogen's placement is really one of these cells. We will kill them all.

#	Case class	The twist it tests	Example that hits it
A	Group 1 claim (lose 1 e⁻)	H⁺ vs Na⁺ size/behaviour	Ex 1
B	Group 17 claim (gain 1 e⁻)	EA & stability of H⁻	Ex 2
C	Group 14 claim (share e⁻)	EN closeness, single-bond limit	Ex 3
D	Sign of ΔEN (bond character)	0 / small / large — all three	Ex 4
E	Degenerate input (bare proton)	zero electrons — what breaks	Ex 5
F	Limiting / trend behaviour	why H sits above Li in IE trend	Ex 6
G	Real-world word problem	hydride reacting in the field	Ex 7
H	Exam-style twist (which group?)	forced single-answer with a trap	Ex 8

Read the "Forecast" line and guess before scrolling. Guessing wrong is how the idea sticks.

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[!example] Example 1 — Cell A: the Group 1 claim

Statement. Hydrogen loses one electron to give H⁺, exactly like $\text{Na}\to {\text{Na}}^{+}$. Show quantitatively why H⁺ still does not make hydrogen a true alkali metal.

Forecast: How many times smaller is H⁺ than Na⁺ — 10×, 1000×, or 100000×?

1. Write what each ion has left. ${\text{H}}^{+}:1s^0(\text{zero electrons}){\text{Na}}^{+}:1s^{2}2s^{2}2p^6(10\text{electrons})$ Why this step? The whole difference lives in "what remains after losing one electron", so we list it first.

2. Compare the two radii. A bare proton has radius $\approx 10^{-15}\text{m}$; ${\text{Na}}^{+}$ keeps a full electron cloud, radius $\approx 10^{-10}\text{m}$. $\frac{r_{{\text{Na}}^{+}}}{r_{{\text{H}}^{+}}}=\frac{10^{-10}}{10^{-15}}=10^5$ Why this step? A ratio turns "very different" into a number an examiner can grade.

3. Interpret the number physically. $10^5=100000$. A charge that concentrated cannot float free in water; it clamps onto a lone pair of ${\text{H}}_2\text{O}$: ${\text{H}}^{+}+{\text{H}}_2\text{O}\to {\text{H}}_3{\text{O}}^{+}$ Why this step? This is the behaviour that alkali cations never show — ${\text{Na}}^{+}$ does exist as a hydrated free ion.

Verify: $10^{-10}/10^{-15}=10^5=100000$. Units cancel (metre/metre), so the ratio is pure number ✓. Conclusion: same "+1 charge", wildly different physics. Group 1 membership fails.

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[!example] Example 2 — Cell B: the Group 17 claim

Statement. Halogens grab an electron eagerly ($\text{F}+e^{-}\to {\text{F}}^{-}$). Hydrogen also forms ${\text{H}}^{-}$ in $\text{NaH}$. Use electron affinity to show H is a reluctant halogen.

Forecast: By roughly what factor is fluorine's electron-grabbing energy larger than hydrogen's?

1. Write both electron affinities. $\text{EA}(\text{H})=-73\text{kJ/mol},\text{EA}(\text{F})=-328\text{kJ/mol}$ Why this step? EA is the measure of "does this atom want an extra electron?". More negative = more wanted.

2. Take the ratio of magnitudes. $\frac{|-328|}{|-73|}=\frac{328}{73}\approx 4.49$ Why this step? A ratio near $4.5$ says fluorine releases ~$4.5\times$ more energy grabbing an electron — H⁻ is far less stable.

3. Confirm with a water test. Because ${\text{H}}^{-}$ is loosely held, it is attacked by water; ${\text{Cl}}^{-}$ is not: $\text{NaH}+{\text{H}}_2\text{O}\to \text{NaOH}+{\text{H}}_2\uparrow \text{vs}\text{NaCl}+{\text{H}}_2\text{O}\to {\text{Na}}^{+}+{\text{Cl}}^{-}$ Why this step? A real reaction proves the number, not just quotes it. See Hydrides Classification for why only very electropositive metals make ionic hydrides.

Verify: $328/73=4.493\dots \approx 4.5$ ✓. H "acts halogen" only with extreme metals — not a true Group 17 member.

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[!example] Example 3 — Cell C: the Group 14 claim

Statement. By electronegativity, hydrogen (2.1) sits nearest carbon (2.5). Show the numeric closeness, then give the one structural reason it still is not carbon.

Forecast: Is H closer in EN to Na, C, or F?

1. List the EN gaps. $|2.1-0.9|=1.2(\text{to Na}),|2.1-2.5|=0.4(\text{to C}),|2.1-4.0|=1.9(\text{to F})$ Why this step? The smallest gap names the closest chemical cousin.

2. Pick the minimum. $0.4<1.2<1.9$, so hydrogen is closest to carbon — both make near-covalent bonds. Why this step? EN closeness predicts bond type, and covalent bonding is carbon's signature.

3. State the disqualifier. Carbon forms four bonds and catenates (C–C–C…). Hydrogen has one $1s$ electron and one empty slot, so: $\text{max bonds}(\text{H})=1,\text{max bonds}(\text{C})=4$ Why this step? A shared "covalent tendency" is not enough — bond count separates them permanently.

Verify: gaps $1.2,0.4,1.9$; $\min =0.4$ (carbon) ✓. H is chemically nearest carbon yet locked to one bond — not Group 14 either.

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[!example] Example 4 — Cell D: every sign of ΔEN

Statement. Bond character depends on $\Delta \text{EN}=|{\text{EN}}_1-{\text{EN}}_2|$. Classify H–H, C–H and Na–H, covering the zero, small, and large cases in one shot.

Forecast: Which of these three is ionic?

1. Zero case: H–H. $\Delta \text{EN}=|2.1-2.1|=0\Rightarrow \text{purely covalent (non-polar)}$ Why this step? Identical atoms pull equally — the textbook definition of a non-polar bond.

2. Small case: C–H. $\Delta \text{EN}=|2.5-2.1|=0.4\Rightarrow \text{covalent, barely polar}$ Why this step? Small but non-zero gap = slight dipole, still shared electrons.

3. Large case: Na–H. $\Delta \text{EN}=|2.1-0.9|=1.2\Rightarrow {\text{mostly ionic (Na}}^{+}{\text{H}}^{-})$ Why this step? A gap $\gtrsim 1.7$ is fully ionic; $1.2$ is "predominantly ionic", which is why $\text{NaH}$ contains real hydride ions.

Verify: $0,0.4,1.2$ increasing ✓; the largest gap (Na–H) is the ionic one ✓. One rule, all three signs of the input covered. More at Electronegativity and Bond Character.

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[!example] Example 5 — Cell E: the degenerate input (zero electrons)

Statement. What actually "breaks" when the atom has zero electrons? Show why the usual periodic properties (shielding, radius trends) go undefined for ${\text{H}}^{+}$.

Forecast: Does ${\text{H}}^{+}$ have an electronegativity? A shielding constant?

1. Count electrons left. ${\text{H}}^{+}=1$ proton, $0$ electrons, $0$ shells. Why this step? Almost every periodic property is about electrons; remove them and the property has no operand.

2. Test shielding. Shielding $S$ counts inner electrons screening the nucleus: $S({\text{H}}^{+})=0\text{because there are no electrons at all}$ Why this step? ${\text{Na}}^{+}$ still has $S\approx 8.5$ from its 10 electrons — a genuine ion with cloud. ${\text{H}}^{+}$ is a nucleus, a different object.

3. Test electronegativity. EN measures pull on a shared pair. A bare proton in solution never shares — it just latches onto whole lone pairs ($\to {\text{H}}_3{\text{O}}^{+}$). So the "atomic EN of ${\text{H}}^{+}$" is not defined. Why this step? Recognising a degenerate/undefined case is exactly what "cover all inputs" means — you must not quote a radius trend for something with no electron cloud.

Verify: electrons $=0$, shielding $=0$, and "free ${\text{H}}^{+}$ in water" concentration $\approx 0$ (it is always hydrated) ✓. The zero-electron case is genuinely degenerate.

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[!example] Example 6 — Cell F: limiting / trend behaviour

Statement. Down Group 1, IE falls (Li → Na → K). Yet H sits above Li with a much higher IE. Show the number and explain the limiting reason.

Forecast: Is H's ionization energy higher or lower than lithium's, and by how many times?

1. List the IE values. $\text{IE}(\text{H})=1312\text{kJ/mol},\text{IE}(\text{Li})=520\text{kJ/mol}$ Why this step? The claim "H ≠ alkali metal" must survive a number, not a vibe.

2. Ratio and gap. $\frac{1312}{520}\approx 2.52,1312-520=792\text{kJ/mol}$ Why this step? H needs ~$2.5\times$ more energy to ionize — it clings to its electron like a non-metal, not a metal.

3. Explain the limiting cause. Hydrogen has no inner shell ($S=0$), so its single electron feels the full nuclear charge. Lithium's $2s^1$ electron is screened by two $1s$ electrons, so it leaves easily. As you go down Group 1 the outer electron gets farther and better shielded → IE keeps dropping → H is the "wrong-way" endpoint. Why this step? This links the anomaly directly to Ionization Energy Trends and Electronic Configuration of Elements.

Verify: $1312/520=2.523\dots$; $1312-520=792$ ✓; and $\text{IE}(\text{H})>\text{IE}(\text{Li})$ ✓ — opposite of true alkali behaviour.

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[!example] Example 7 — Cell G: real-world word problem

Statement. A field lab stores ${\text{CaH}}_2$ ("hydrolith") as an emergency hydrogen source. A technician drops 2.0 mol of ${\text{CaH}}_2$ into excess water. How many moles of ${\text{H}}_2$ gas are produced, and which of hydrogen's "personalities" does this reaction reveal?

Forecast: Will 2 mol of ${\text{CaH}}_2$ give 2, 4, or 8 mol of ${\text{H}}_2$?

1. Write the balanced reaction. ${\text{CaH}}_2+2{\text{H}}_2\text{O}\to {\text{Ca(OH)}}_2+2{\text{H}}_2\uparrow$ Why this step? Stoichiometry needs a balanced equation before any mole arithmetic.

2. Read the mole ratio. $1{\text{CaH}}_2:2{\text{H}}_2$. So: $n({\text{H}}_2)=2.0\text{mol}\times \frac{2}{1}=4.0\text{mol}$ Why this step? The ratio is the conversion factor turning "what we added" into "what we get".

3. Identify the personality. The hydride ${\text{H}}^{-}$ (from Ca–H, the ionic/Group-17-like behaviour) meets ${\text{H}}^{+}$ from water; they combine to ${\text{H}}_2$. One $\text{H}$ was $-1$, one was $+1$ → both landed at $0$ in ${\text{H}}_2$. That is hydrogen playing both roles in a single reaction. Why this step? The word problem's payoff is seeing the "chameleon" live, not just a number.

Verify: $2.0\times 2=4.0$ mol ${\text{H}}_2$ ✓. Atom check: left has $2\text{H}$ (from ${\text{CaH}}_2$) $+4\text{H}$ (from $2{\text{H}}_2\text{O}$) $=6\text{H}$; right has $2\text{H}$ (in ${\text{Ca(OH)}}_2$) $+4\text{H}$ (in $2{\text{H}}_2$) $=6\text{H}$ ✓.

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[!example] Example 8 — Cell H: the exam-style twist

Statement. "On the basis of electronic configuration alone, hydrogen is placed in Group ___." An exam demands one answer with a one-line reason — and it is trying to make you write "17" or "14".

Forecast: Which single group, and why does configuration outrank chemistry here?

1. Re-read the constraint. The key words are "electronic configuration alone." Why this step? Exam traps hide in qualifiers. The word "alone" bans chemical (EA/EN/bonding) arguments.

2. Match the configuration. $\text{H}=1s^1=n{s}^1$ with $n=1$. The $n{s}^1$ family is Group 1. $\text{H}:1s^1⟷\text{Li}:1s^{2}2s^1,\text{Na}:[\text{Ne}]3s^1$ Why this step? Configuration matching is a pattern match, and $n{s}^1$ points uniquely at Group 1.

3. Guard against the trap. Group 17 ($n{s}^{2}n{p}^5$) and Group 14 ($n{s}^{2}n{p}^2$) are argued from chemistry (gaining/sharing), which the qualifier forbids. So the single defensible answer is Group 1 — with the footnote that this is conventional, not chemical. Why this step? Distinguishing "what the question restricts you to" from "the full truth" is the exam skill being tested. See Alkali Metals Properties for why it still misbehaves once chemistry is allowed.

Verify: $1s^1$ fits $n{s}^1$ (Group 1); it does not fit $n{s}^{2}n{p}^5$ (17) nor $n{s}^{2}n{p}^2$ (14) ✓. Configuration-only ⇒ Group 1.

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[!recall]- Rapid self-test

Which cell does each phrase belong to?

Back to the map: Position of hydrogen in the periodic table (anomalous) · related traps in Diagonal Relationship and Hydrogen Bonding.