Intuition The Core InsightHydrogen's dual nature (can gain or lose electrons) means it forms three fundamentally different types of compounds depending on its partner. Think of hydrogen as a chameleon: with reactive metals it becomes H⁻ (ionic), with non-metals it shares electrons (covalent), and with transition metals it slips into the metal's crystal lattice like a guest sneaking into gaps at crowded party (interstitial).
Why this matters : The same element (H₂) produces salt-like solids, molecular gases, and metalic alloys—all because of electronegativity differences and atomic size.
A binary compound of hydrogen with another element. Classification depends on bonding type and electronic nature of hydrogen .
The key question: Does hydrogen become H⁺, H⁻, or something in between?
Answer depends on the partner's electronegativity :
Partner less electronegative than H (metals) → H becomes H⁻ (ionic)
Partner more electronegative than H (non-metals) → shared electrons (covalent)
Partner is a transition metal → H occupies interstitial sites (interstitial)
Formed when hydrogen reacts with highly electroactive metals (Groups 1 & 2, except Be). Hydrogen exists as hydride ion H⁻ with electronic configuration 1s² (same as helium).
Step 1 : Electronegativity comparison
H: 2.1 (Pauling scale)
Na: 0.9, Ca: 1.0, Li
Step 2 : Electron transfer energetics
For sodium hydride:
Na(s) + 1 2 t e x t H 2 (g) → NaH(s) \text{Na(s) + } \frac{1}{2}text{ H}_2\text{(g) } \rightarrow \text{ NaH(s)} Na(s) + 2 1 t e x t H 2 (g) → NaH(s)
Energy balance:
Sublimation of Na : Na(s) → Na(g) \text{Na(s) → Na(g)} Na(s) → Na(g) Ionization of Na : Na(g) → Na + (g) + e − \text{Na(g) → Na}^+\text{(g) + e}^- Na(g) → Na + (g) + e − Dissociation of H₂ : 1 2 H 2 ( g ) → H ( g ) \frac{1}{2} \text{H}_2(g) \rightarrow \text{H}(g) 2 1 H 2 ( g ) → H ( g ) Electron affinity of H : H(g) + e − → H − (g) \text{H(g) + e}^- \rightarrow \text{H}^-\text{(g)} H(g) + e − → H − (g) Lattice energy : Na + ( g ) + H − ( g ) → NaH ( s ) \text{Na}^+(g) + \text{H}^-(g) \rightarrow \text{NaH}(s) Na + ( g ) + H − ( g ) → NaH ( s ) −808 kJ/mol
Why this step? The large negative lattice energy (−808) overcomes all positive energy terms. This is only possible because:
H⁻ has a small radius (~146 pm in a 6-coordinate lattice)
Na⁺ and H⁻ pack efficiently in NaCl-type crystal
Net reaction : Δ H f = + 108 + 496 + 218 − 73 − 808 = − 59 kJ/mol \Delta H_f = +108 + 496 + 218 - 73 - 808 = -59 \text{ kJ/mol} Δ H f = + 108 + 496 + 218 − 73 − 808 = − 59 kJ/mol
Property
Value
Why?
State Crystalline solids
Ionic bonding → high lattice energy
Melting point High (LiH: 689°C)
Strong electrostatic forces between M⁺ and H⁻
Density Higher than parent metal
H⁻ (~146 pm) fills space efficiently
Electrical conductivity Molten state conducts
Mobile ions when melted
Chemical nature Strong reducing agents
H⁻ easily donates electrons: H − → H + e − \text{H}^- \rightarrow \text{H} + e^- H − → H + e −
Worked example Worked Example 1: Why LiH but not BeH₂?Setup : Predict if beryllium forms ionic hydride.
Step 1 : Calculate Pauling electronegativity difference
Be: 1.5, H: 2.1
Δχ = 2.1 − 1.5 = 0.6 (covalent character likely)
Step 2 : Consider ionization energy
Be has small size → very high IE₁ (900 kJ/mol) and IE₂ (1757 kJ/mol)
Forming Be²⁺ costs 2657 kJ/mol total
Step 3 : Lattice energy compensation?
BeH₂ lattice energy ≈ −3000 kJ/mol
Not enough to compensate sublimation (324) + ionization (2657) + other terms
Why this step? High charge density of Be²⁺ causes covalent character (Fajans' rules: small cation polarizes H⁻)
Conclusion : BeH₂ is polymeric and covalent, not ionic.
Worked example Worked Example 2: Reaction with WaterQuestion : Why do ionic hydrides react violently with water?
Step 1 : Write the reaction
NaH(s) + H 2 O(l) → NaOH(aq) + H 2 (g) \text{NaH(s) + H}_2\text{O(l) → NaOH(aq) + H}_2\text{(g)} NaH(s) + H 2 O(l) → NaOH(aq) + H 2 (g)
Step 2 : Mechanism — why this happens
H⁻ is a strong Brønsted base : p K a ( H 2 ) ≈ 35 pK_a(\text{H}_2) \approx 35 p K a ( H 2 ) ≈ 35
H₂O is a weak acid: p K a = 15.7 pK_a =15.7 p K a = 15.7
Acid-base: H − + H 2 O → H 2 + O H − H^- + H_2O \rightarrow H_2 + OH^- H − + H 2 O → H 2 + O H − Step 3 : Energetics
Bond energy of H−H being formed: 436 kJ/mol
O−H bond broken: 463 kJ/mol
Net: slightly endothermic bond-wise, but entropy increase (gas produced) and solvation of Na⁺/OH⁻ drive the reaction
Why this step? H⁻ is isoelectronic with He, making it an extremely weak Brønsted acid (H₂) → strong base.
Result : Exothermic, produces flammable H₂ gas → violent reaction.
Definition Covalent HydrideFormed when hydrogen reacts with non-metals (Groups 13-17). Electrons are shared between H and the partner atom. Exist as discrete molecules .
Why covalent? Electronegativity of non-metals (2.5-4.0) is comparable to or greater than H (2.1).
Examples by group :
Group 13: B₂H₆ (electron-deficient, 3c-2e bonds)
Group 14: CH₄, SiH₄ (tetrahedral, sp³)
Group 15: NH₃, PH₃ (pyramidal, lone pair)
Group 16: H₂O, H₂S (bent, two lone pairs)
Group 17: HF, HCl (linear, polar)
Derivation : Why is water bent?
Step 1 : Oxygen has 6 valence electrons
Step 2 : Forms 2 bonds with H → 2 bonding pairs
Step 3 : Remaining 4 electrons → 2 lone pairs
Step 4 : VSEPR theory → 4 electron pairs arrange tetrahedrally
Step 5 : Lone pairs repel more than bonding pairs → bond angle compressed from 109.5° to 104.5°
Why this step? Lone pairs occupy more angular space (closer to nucleus) → push bonding pairs together.
Property
Trend
Why?
State Gases or volatile liquids
Weak van der Waals forces (except H-bonding cases)
Boiling point H₂O > HF > NH₃ >> CH₄
Hydrogen bonding in H₂O, HF, NH₃
Solubility Polar hydrides soluble in water
"Like dissolves like" + H-bonding
Acidity HF > H₂O > NH₃ > CH₄
Electronegativity of partner weakens H−X bond
Reducing power Decreases down group
H−X bond strength decreases
Worked example Worked Example 3: Boiling Point AnomalyQuestion : Why does H₂O (100°C) boil much higher than H₂S (−60°C)?
Step 1 : Molecular weight consideration
H₂O: 18 g/mol
H₂S: 34 g/mol
Expected : H₂S should boil higher (more electrons → stronger dispersion)
Step 2 : Electronegativity check
O: 3.5, S: 2.5
H₂O is highly polar (δ⁺H−O^(δ⁻))
Step 3 : Hydrogen bonding in H₂O
O has high electronegativity + small size
H can approach another O closely
Forms hydrogen bonds : H 2 O ⋯ H-O-H \text{H}_2\text{O}\cdots\text{H-O-H} H 2 O ⋯ H-O-H
Bond strength: ~20 kJ/mol (vs 2 kJ/mol for dispersion)
Why this step? Breaking H-bonds requires significant energy → elevated boiling point.
Step 4 : Why not H₂S?
Sulfur is larger → lower electronegativity → weaker dipole
H-bonding negligible in H₂S
Conclusion : H₂O requires 40.7 kJ/mol (enthalpy of vaporization) vs 18.7 kJ/mol for H₂S.
Worked example Worked Example 4: Electron-Deficient HydridesQuestion : Why does B₂H₆ (diborane) exist but not BH₃?
Step 1 : Count valence electrons
B has 3 valence electrons
BH₃ would need 6 electrons for 3 B−H bonds → B has empty p orbital
Step 2 : Stability issue
BH₃ is electron-deficient → highly reactive
Dimerizes to B₂H₆
Step 3 : Structure of B₂H₆
H H
\ /
H---B---BH
/ \
H H
Four terminal B−H bonds (2c-2e normal bonds)
Two bridging H atoms (3-center-2-electron bonds )
Step 4 : 3c-2e bond derivation
Each boron is sp³-hybridized overall in diborane, but the description simplifies as: one H1s orbital overlaps with one sp³-hybrid orbital from each of the two borons (older texts describe B as sp² using an unhybridized p orbital for bridging — modern treatment uses sp³ hybrids for the bridge)
Forms 3 molecular orbitals: bonding, non-bonding, antibonding
2 electrons fill bonding MO → shared across B−H−B
Why this step? 3c-2e bonds allow B to achieve more stable electron configuration without full octet.
Result : B₂H₆ is stable; BH₃ only exists as transient species.
Definition Interstitial HydrideFormed when hydrogen is absorbed by transition metals and some lanthanides/actinides . Hydrogen atoms occupy interstitial sites (holes) in the metal lattice without disrupting the metalic structure.
Why metals absorb hydrogen :
Transition metals have partially filled d-orbitals
Metal lattices have tetrahedral and octahedral holes
H atom is small (covalent radius ~31 pm, metalic radius ~53 pm) → fits into interstitial voids
Process :
H 2 (g) → surface 2 H(adsorbed) → diffusion H(interstitial) \text{H}_2\text{(g)} \xrightarrow{\text{surface}} 2\text{H(adsorbed)} \xrightarrow{\text{diffusion}} \text{H(interstitial)} H 2 (g) surface 2 H(adsorbed) diffusion H(interstitial)
Step 1 : H₂ dissociates on metal surface (catalyzed by d-orbital electrons)
Step 2 : H atoms diffuse into the bulk
Step 3 : Occupy interstitial voids
Why this step? Unlike ionic/covalent, this is absorption not reaction. H retains partial electron density.
Property
Behavior
Why?
Appearance Metalic luster
Metalic bonding retained
Conductivity Conducts electricity (lower than pure metal)
H atoms scatter electrons but don't break metalic bonds
Hardness Harder and more britle
H in lattice distorts crystal → dislocation movement hindered
Density Lower than parent metal
H is light; lattice expands slightly
Magnetic Reduced magnetism
H electrons interact with metal d-orbitals
Heat of formation Low
Weak bonding; easily reversible
Worked example Worked Example 5: Hydrogen StorageQuestion : Why is LaNi₅H₆ used for hydrogen storage?
Step 1 : Absorption capacity
LaNi₅ can absorb up to H/M ratio = 1.2
Forms LaNi₅H₆ (6H per formula unit)
Step 2 : Thermodynamics
LaNi 5 (s) + 3 H 2 (g) ⇌ LaNi 5 H 6 (s) \text{LaNi}_5\text{(s)} + 3\text{ H}_2\text{(g)} \rightleftharpoons \text{LaNi}_5\text{H}_6\text{(s)} LaNi 5 (s) + 3 H 2 (g) ⇌ LaNi 5 H 6 (s)
Δ H ° ≈ − 30 kJ/mol H 2 \Delta H° \approx -30 \text{ kJ/mol H}_2 Δ H ° ≈ − 30 kJ/mol H 2 Equilibrium pressure at 298 K: ~2 bar
Step 3 : Why reversible?
Low enthalpy → heating to 350K releases H₂
No strong M−H bonds (unlike covalent hydrides)
Why this step? The balance of "strong enough to store H₂ at room temp" but "weak enough to release on mild heating" is perfect for fu
Partner electronegativity
Large lattice energy -808
Crystalline solids high mp
Intuition Hinglish mein samjhoDekho, hydrogen ki khaas baat yeh hai ki uski dual nature hoti hai — woh electron gain bhi kar sakta hai (H⁻ ban ke) aur lose bhi (H⁺ ban ke). Isi wajah se jab hydrogen alag-alag elements ke saath react karta hai, toh teen bilkul alag types ke compounds bante hain. Ise ek chameleon ki tarah socho: reactive metals (jaise Na, Ca) ke saath yeh H⁻ ban jaata hai aur ionic hydride banta hai; non-metals ke saath electrons share karke covalent hydride banata hai; aur transition metals ke saath crowded party mein ghuse hue guest ki tarah metal ke crystal ke gaps (interstitial sites) mein baith jaata hai. Yeh sab decide hota hai partner ki electronegativity aur atomic size se.
Ab yeh important kyun hai? Kyunki wahi ek hi element H₂ se aap salt-jaise solids (ionic), molecular gases (covalent), aur metallic alloys (interstitial) — teeno bana sakte ho, sirf electronegativity difference ki wajah se. Ionic hydrides mein, jaise NaH, energy balance dekho toh sublimation, ionization, dissociation — sab positive energy maangte hain, lekin lattice energy (−808 kJ/mol) itni bada negative term hai ki poora reaction exothermic ho jaata hai. Isliye yeh solids high melting point wale, dense, aur strong reducing agents hote hain — kyunki H⁻ apna electron easily donate kar deta hai.
Ek mazedaar cheez samajh lo — Be ionic hydride kyun nahi banata? Kyunki Be²⁺ ka size bahut chhota hota hai aur charge density zyada, jisse woh H⁻ ko polarize kar deta hai (Fajans' rules yaad rakho). Isliye BeH₂ covalent aur polymeric ban jaata hai, ionic nahi. Yeh pattern samajhna zaroori hai kyunki exam mein aksar poocha jaata hai ki "kaunsa element ionic hydride banayega aur kaunsa nahi" — aur iska logic hamesha electronegativity difference plus cation ki polarizing power par tika hota hai.
