Hydrides — ionic, covalent, interstitial — Hinglish notes

3.1.4 · Chemistry › Hydrogen and s-Block

Hydrides Kya Hain?

Key question: Kya hydrogen H⁺, H⁻, ya in dono ke beech kuch banta hai?

Jawab partner ki electronegativity pe depend karta hai:

Partner H se kam electronegative ho (metals) → H banta hai H⁻ (ionic)
Partner H se zyada electronegative ho (non-metals) → shared electrons (covalent)
Partner ek transition metal ho → H interstitial sites occupy karta hai (interstitial)

Figure — Hydrides — ionic, covalent, interstitial

1. Ionic (Saline) Hydrides

Formation — First Principles Se Derivation

Step 1: Electronegativity comparison

H: 2.1 (Pauling scale)
Na: 0.9, Ca: 1.0, Li

Step 2: Electron transfer energetics

Sodium hydride ke liye: $Na(s) + \frac{1}{2} t e x t H_{2} (g) \to NaH(s)$

Energy balance:

Na ka Sublimation: $Na(s) \to Na(g)$ ke liye +108 kJ/mol chahiye
Na ka Ionization: $Na(g) \to Na^{+} (g) + e^{-}$ ke liye +496 kJ/mol chahiye
H₂ ka Dissociation: $\frac{1}{2} H_{2} (g) \to H (g)$ ke liye +218 kJ/mol chahiye
H ki Electron affinity: $H(g) + e^{-} \to H^{-} (g)$ se −73 kJ/mol release hota hai
Lattice energy: $Na^{+} (g) + H^{-} (g) \to NaH (s)$ se −808 kJ/mol release hota hai

Ye step kyun? Badi negative lattice energy (−808) saare positive energy terms ko overcome kar leti hai. Ye tabhi possible hai jab:

H⁻ ka radius chhota ho (~146 pm in a 6-coordinate lattice)
Na⁺ aur H⁻ NaCl-type crystal mein efficiently pack ho jayein

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

Properties — Ye Aisa Behave Kyun Karte Hain

Property	Value	Kyun?
State	Crystalline solids	Ionic bonding → high lattice energy
Melting point	High (LiH: 689°C)	M⁺ aur H⁻ ke beech strong electrostatic forces
Density	Parent metal se zyada	H⁻ (~146 pm) space efficiently fill karta hai
Electrical conductivity	Molten state mein conduct karta hai	Melt hone pe mobile ions
Chemical nature	Strong reducing agents	H⁻ easily electrons donate karta hai: $H^{-} \to H + e^{-}$

Worked example Worked Example 2: Water ke saath Reaction

Question: Ionic hydrides water ke saath violently kyun react karte hain?

Step 1: Reaction likho $NaH(s) + H_{2} O(l) \to NaOH(aq) + H_{2} (g)$

Step 2: Mechanism — ye kyun hota hai

H⁻ ek strong Brønsted base hai: $p K_{a} (H_{2}) \approx 35$
H₂O ek weak acid hai: $p K_{a} = 15.7$
Acid-base: $H^{-} + H_{2} O \to H_{2} + O H^{-}$ Step 3: Energetics
H−H bond energy jo ban raha hai: 436 kJ/mol
O−H bond break hona: 463 kJ/mol
Net: bond-wise thoda endothermic, lekin entropy increase (gas produce hoti hai) aur Na⁺/OH⁻ ka solvation reaction drive karte hain

Ye step kyun? H⁻ He ke saath isoelectronic hai, jo ise ek extremely weak Brønsted acid (H₂) → strong base banata hai.

Result: Exothermic, flammable H₂ gas produce hoti hai → violent reaction.

2. Covalent (Molecular) Hydrides

Covalent kyun? Non-metals ki electronegativity (2.5-4.0) H (2.1) se comparable ya zyada hoti hai.

Group ke hisaab se examples:

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: Water bent kyun hai?

Step 1: Oxygen mein 6 valence electrons hain Step 2: H ke saath 2 bonds banata hai → 2 bonding pairs Step 3: Baaki 4 electrons → 2 lone pairs Step 4: VSEPR theory → 4 electron pairs tetrahedrally arrange hote hain Step 5: Lone pairs bonding pairs se zyada repel karte hain → bond angle 109.5° se compress hokar 104.5° ho jaata hai

Ye step kyun? Lone pairs zyada angular space occupy karte hain (nucleus ke closer) → bonding pairs ko paas kar dete hain.

Properties — Structure Behavior Dictate Karta Hai

Property	Trend	Kyun?
State	Gases ya volatile liquids	Weak van der Waals forces (except H-bonding cases)
Boiling point	H₂O > HF > NH₃ >> CH₄	H₂O, HF, NH₃ mein Hydrogen bonding
Solubility	Polar hydrides water mein soluble	"Like dissolves like" + H-bonding
Acidity	HF > H₂O > NH₃ > CH₄	Partner ki electronegativity H−X bond ko weaken karti hai
Reducing power	Group mein neeche jaane pe ghatta hai	H−X bond strength ghatti hai

Worked example Worked Example 3: Boiling Point Anomaly

Question: H₂O (100°C) H₂S (−60°C) se bahut zyada kyun boil karta hai?

Step 1: Molecular weight consideration

H₂O: 18 g/mol
H₂S: 34 g/mol
Expected: H₂S zyada boil karna chahiye (zyada electrons → stronger dispersion)

Step 2: Electronegativity check

O: 3.5, S: 2.5
H₂O highly polar hai (δ⁺H−O^(δ⁻)) Step 3: H₂O mein Hydrogen bonding
O mein high electronegativity + small size hai
H doosre O ke paas closely approach kar sakta hai
Hydrogen bonds form hote hain: $H_{2} O \dots H-O-H$
Bond strength: ~20 kJ/mol (dispersion ke 2 kJ/mol se zyada)

Ye step kyun? H-bonds todhne mein significant energy lagti hai → elevated boiling point.

Step 4: H₂S mein kyun nahi?

Sulfur larger hai → lower electronegativity → weaker dipole
H₂S mein H-bonding negligible hai Conclusion: H₂O ko 40.7 kJ/mol chahiye (enthalpy of vaporization) vs H₂S ke liye 18.7 kJ/mol.

Worked example Worked Example 4: Electron-Deficient Hydrides

Question: B₂H₆ (diborane) kyun exist karta hai lekin BH₃ kyun nahi?

Step 1: Valence electrons count karo

B mein 3 valence electrons hain
BH₃ ko 3 B−H bonds ke liye 6 electrons chahiye honge → B ke paas empty p orbital hai Step 2: Stability issue
BH₃ electron-deficient hai → highly reactive
B₂H₆ mein dimerize ho jaata hai

Step 3: B₂H₆ ki structure

    H    H
     \  /
  H---B---BH
     /  \
    H    H

Chaar terminal B−H bonds (2c-2e normal bonds)
Do bridging H atoms (3-center-2-electron bonds) Step 4: 3c-2e bond derivation
Diborane mein har boron overall sp³-hybridized hai, lekin description simplify karke ye kehte hain: ek H1s orbital dono borons mein se ek ek sp³-hybrid orbital ke saath overlap karta hai (purane texts B ko sp² describe karte hain jo bridging ke liye ek unhybridized p orbital use karta hai — modern treatment bridge ke liye sp³ hybrids use karti hai)
3 molecular orbitals bante hain: bonding, non-bonding, antibonding
2 electrons bonding MO fill karte hain → B−H−B mein share hote hain Ye step kyun? 3c-2e bonds B ko bina full octet ke zyada stable electron configuration achieve karne dete hain.

Result: B₂H₆ stable hai; BH₃ sirf transient species ke roop mein exist karta hai.

3. Interstitial (Metalic) Hydrides

Formation — Ek Physical Process, Typical Bonding Nahi

Metals hydrogen kyun absorb karte hain:

Transition metals mein partially filled d-orbitals hote hain
Metal lattices mein tetrahedral aur octahedral holes hote hain
H atom chhota hota hai (covalent radius ~31 pm, metalic radius ~53 pm) → interstitial voids mein fit ho jaata hai

Process: $H_{2} (g) surface 2 H(adsorbed) diffusion H(interstitial)$

Step 1: H₂ metal surface pe dissociate hota hai (d-orbital electrons se catalyzed) Step 2: H atoms bulk mein diffuse karte hain Step 3: Interstitial voids occupy karte hain

Ye step kyun? Ionic/covalent se unlike, ye absorption hai, reaction nahi. H partial electron density retain karta hai.

Properties — Metal-Like, Molecule-Like Nahi

Property	Behavior	Kyun?
Appearance	Metalic luster	Metalic bonding retain rehti hai
Conductivity	Electricity conduct karta hai (pure metal se kam)	H atoms electrons scatter karte hain lekin metalic bonds nahi todhte
Hardness	Zyada hard aur brittle	Lattice mein H crystal distort karta hai → dislocation movement hindered
Density	Parent metal se kam	H light hai; lattice thoda expand hota hai
Magnetic	Reduced magnetism	H electrons metal d-orbitals ke saath interact karte hain
Heat of formation	Low	Weak bonding; easily reversible