3.2.9 · Chemistry › p-Block
Members: F, Cl, Br, I, At (At radioactive hai). Valence config n s 2 n p 5 — noble gas se ek electron kam, isliye ye family table mein sabse zyada non-metallic, oxidizing hai.
Ek halogen atom full shell se ek electron kam hota hai. Toh uski poori "personality" hai — "ek electron grab karo." Yeh ek akeli craving explain karti hai: high electronegativity, high electron affinity, strong oxidizing power, aur X − ions aur single covalent bonds banane ki tendency. Jaise-jaise group mein neeche jaate hain, atom bada hota hai, nucleus ki incoming electron pe pull kamzor hoti hai, toh har "grabbing" property group mein neeche jaate jaate kam hoti hai .
Property
Trend
WHY
Atomic/ionic radius
increases
naye shells add hote hain
Ionization enthalpy
decreases
electron nucleus se door
Electronegativity
decreases (F sabse zyada EN)
bonding e⁻ pe pull kamzor
Electron gain enthalpy
Cl pe sabse zyada negative, F pe nahi
F itna chhota hai ki uske e⁻–e⁻ repulsion gain ko partly cancel kar deti hai
Oxidizing power
decreases
X − mein reduce hona mushkil hota hai
Bond enthalpy X − X
Cl > Br > F > I (F anomalous hai)
chhota F atom → lone-pair repulsion F − F ko kamzor karti hai
"F ki electron gain enthalpy sabse zyada negative hai kyunki yeh sabse zyada electronegative hai."
Sahi lagta hai: F bonds mein electrons sabse zyada hard grab karta hai. Fix: Tiny F atom mein electron add karne se woh already-crowded 2 p shell mein jaata hai → strong e⁻–e⁻ repulsion. Isliye F ka Δ e g H , Cl se kam negative hota hai. Electronegativity (bonded) ≠ electron gain enthalpy (isolated atom).
"Oxidizing power" = X 2 kitna badly electrons steal karke 2 X − banna chahta hai. Isko standard reduction potential E ∘ se measure karte hain:
X 2 + 2 e − → 2 X − larger E ∘ ⇒ stronger oxidizer
Reduction ko ek Born–Haber-style cycle mein todenge (per mole of X 2 , aqueous):
2 1 X 2 atomize, 2 1 Δ d i ss H X ( g ) Δ e g H X − ( g ) Δ h y d H X − ( a q )
Release hone wali total energy E ∘ govern karti hai. F ke liye:
Δ d i ss H ( F 2 ) low hai (weak F–F bond, lone-pair repulsion) → todna aasaan.
Δ h y d H ( F − ) bahut high hai (chhota ion → strong hydration).
Yeh do terms F ki thodi si kharab electron gain enthalpy ko overpower kar dete hain, jisse F₂ sabse strong oxidizing agent ban jaata hai.
Paani mein H X ki acid strength = H − X kitni aasaani se toot ke H + deta hai . Sabse bada factor bond strength hai, electronegativity nahi. Bada X → lamba, kamzor H–X bond → ionize karna aasaan → zyada strong acid .
"HF sabse strong acid hai kyunki F sabse zyada electronegative hai."
Sahi lagta hai: zyada EN bonding pair ko H se kheench lega. Fix: F chhota hai → bahut strong, short H–F bond + solution mein strong H-bonding → HF apna proton pakde rehta hai. Isliye HF sabse weak HX acid hai (yeh ek weak acid hai; HCl, HBr, HI strong hain). Electronegativity oxoacids ke liye matter karta hai, HX ke liye nahi jahan bond enthalpy dominate karta hai.
H X ki stability (thermal) bond strength ke saath chalti hai: H F > H C l > H B r > H I (HI aasaani se decompose ho jaata hai).
Interhalogens do alag halogens ke compounds hote hain, type X X n ′ jahan n = 1 , 3 , 5 , 7 aur X bada, kam electronegative halogen (central) hota hai, X ′ chhota hota hai.
Bada central atom zyada chhote partners kyun leta hai? Geometry: bade central atom ke paas itna surface area hota hai ki wo kaafi saare chhote atoms ko apne around fit kar sake. Isliye sirf bada I/Br n = 5 , 7 tak pahunch sakta hai (e.g. I F 7 ), jabki similar-sized pairs ke liye n chhota hota hai (e.g. C l F ).
Type
Example
Hybridisation
Shape (VSEPR)
X X ′
C l F , B r F , I C l
—
linear
X X 3 ′
C l F 3 , B r F 3
s p 3 d
bent T-shape
X X 5 ′
I F 5 , B r F 5
s p 3 d 2
square pyramidal
X X 7 ′
I F 7
s p 3 d 3
pentagonal bipyramidal
Key facts (WHY/HOW):
Interhalogen bonds X − X ′ , X − X se kamzor hote hain → interhalogens, halogens se zyada reactive hote hain (F₂ ko chhodkar).
Ye covalent, diamagnetic, volatile hote hain; hydrolyse hokar halide + oxohalide dete hain.
Fluorinating agents ki tarah use hote hain (e.g. C l F 3 ).
Pseudohalogens polyatomic molecules hain jo chemically halogens (X₂) jaisi resemblance rakhte hain : e.g. cyanogen ( C N ) 2 , thiocyanogen ( S C N ) 2 , ( O C N ) 2 .
Inke anions pseudohalides hain: C N − , S C N − , O C N − , N 3 − , jo X − ki tarah behave karte hain.
Inhe "pseudo" kyun kehte hain? Kyunki ye halogen reactions mimic karte hain:
( C N ) 2 + 2 O H − → C N − + O C N − + H 2 O
compare karo C l 2 + 2 O H − → C l − + O C l − + H 2 O (same disproportionation!).
Ye H X -jaisi acids (HCN), silver-insoluble salts (AgCN), aur interpseudohalogen species banate hain.
Q1. Kya B r 2 , C l − ko C l 2 mein oxidize karega?
E ∘ compare karo: C l 2 / C l − = + 1.36 , B r 2 / B r − = + 1.07 .
Reaction B r 2 + 2 C l − → 2 B r − + C l 2 mein Δ E ∘ = 1.07 − 1.36 = − 0.29 V < 0 .
Yeh step kyun? Negative cell potential ⇒ non-spontaneous.
Answer: Nahi. Sirf ek stronger oxidizer, weaker wale ko displace karta hai; yahan Cl₂ stronger hai, toh ulta hoga.
Q2. H F , H C l , H I ko acid strength se arrange karo aur explain karo.
Bond enthalpies: H − F ( 565 ) > H − C l ( 431 ) > H − I ( 297 ) .
Yeh step kyun? Kamzor bond = aasaan H⁺ release = stronger acid.
Answer: H I > H C l > H F .
Q3. C l F 3 ki shape predict karo.
Central Cl: 7 valence e⁻; 3 bonds 3 use karte hain → 4 e⁻ bache = 2 lone pairs. Total = 3 bp + 2 lp = 5 domains ⇒ s p 3 d , trigonal bipyramidal arrangement. Lone pairs equatorial jaate hain.
Yeh step kyun? Equatorial lone pairs 90° repulsions minimize karte hain.
Answer: Bent T-shape (≈ T-shaped, thoda < 90°).
Recall Quick self-test (answers chhupao)
Kis halogen ka Δ e g H sabse zyada negative hai? → Cl (F nahi)
Sabse strong oxidizing halogen? → F₂
Sabse strong HX acid? → HI
HF weak acid kyun hai? → strong short H–F bond + H-bonding
I F 7 ki shape? → pentagonal bipyramidal
Interhalogen ka general formula? → X X n ′ , bada X central
Recall Feynman: ek 12-saal ke bacche ko explain karo
Socho har halogen ek aisa bachcha hai jise apni book complete karne ke liye exactly ek aur sticker chahiye . Fluorine sabse chhota, sabse bhookha bachcha hai — woh kisi se bhi sticker jhapat lega (strongest oxidizer). Jab ye hydrogen ke saath haath pakdte hain (HX), toh bade bacche dheelay pakdte hain , isliye paani mein hydrogen aasaani se chhhod dete hain — isliye HI strong acid hai lekin chhota tight-gripping HF nahi chhodta. Interhalogens bas do alag sizes ke bacche hain jo haath pakde hain, aur bada bachcha kaafi saare chhote bacchon ko ek saath pakad sakta hai (jaise IF₇).
Oxidizing power: "F ast Cl owns Br ing I ce" → F > Cl > Br > I.
HX acidity (reverse): "HI HBr o, HCl ub, HF lop" → strongest HI → weakest HF.
Interhalogen central atom: "B ig B oss in the middle" (bada halogen central hota hai).
Konsa halogen sabse strong oxidizing agent hai aur kyun? F₂; low F–F bond dissociation enthalpy + F⁻ ki bahut high hydration enthalpy uski modest electron gain enthalpy ko outweigh kar deti hai.
Cl ka, F ka nahi, electron gain enthalpy sabse zyada negative kyun hai? F itna chhota hai ki electron add karne se compact 2p shell mein strong e⁻–e⁻ repulsion hoti hai, jisse ΔegH, Cl se kam negative hoti hai.
HX acid strength ka order aur governing factor? HF < HCl < HBr < HI; H–X bond dissociation enthalpy se govern hota hai (kamzor bond → stronger acid).
HF weak acid kyun hai jabki F sabse zyada electronegative hai? Bahut strong short H–F bond plus strong hydrogen bonding proton ko pakde rehte hain, isliye HF zyada ionize nahi hota.
Interhalogens ka general formula aur central atom rule? XX'ₙ (n=1,3,5,7); bada, kam electronegative halogen central hota hai.
ClF₃ ki shape aur hybridisation? Bent T-shape; sp³d (3 bond pairs + 2 lone pairs equatorial).
IF₇ ki shape? Pentagonal bipyramidal, sp³d³.
Interhalogens, halogens se zyada reactive kyun hote hain? X–X' bond, X–X se kamzor (aur zyada polar) hota hai, isliye aasaani se toot jaata hai (F₂ ko chhodkar).
Pseudohalogens ko define karo ek example ke saath. Polyatomic species jo chemically X₂ jaisi resemblance rakhte hain, e.g. cyanogen (CN)₂; inke anions (CN⁻, SCN⁻, OCN⁻, N₃⁻) pseudohalides hain.
Kya Cl₂, Br⁻ ko displace karega? E° se justify karo. Haan; E°(Cl₂/Cl⁻)=+1.36 > E°(Br₂/Br⁻)=+1.07, ΔE°>0 isliye Cl₂+2Br⁻→2Cl⁻+Br₂ spontaneous hai.
X–X bond enthalpy ka trend aur anomaly? Cl > Br > F > I; F–F anomalously low hai kyunki chhote F atoms mein lone-pair repulsion hoti hai.
low F-F bond plus high hydration
increasing size down group
not equal to electron gain enthalpy