Members: F, Cl, Br, I, At (At is radioactive). Valence config n s 2 n p 5 ns^2np^5 n s 2 n p 5 non-metallic, oxidizing family in the table.
A halogen atom is one electron short of a full shell . So its whole personality is "grab one electron." That single craving explains: high electronegativity, high electron affinity, strong oxidizing power, and the tendency to form X − X^- X − fades down the group .
Property
Trend
WHY
Atomic/ionic radius
increases new shells added
Ionization enthalpy
decreases
electron farther from nucleus
Electronegativity
decreases (F most EN)
weaker pull on bonding e⁻
Electron gain enthalpy
most negative at Cl , not F
F is so small its e⁻–e⁻ repulsion partly cancels the gain
Oxidizing power
decreases
harder to be reduced to X − X^- X −
Bond enthalpy X − X X\!-\!X X − X
Cl > Br > F > I (F is anomalous)
small F atom → lone-pair repulsion weakens F − F F\!-\!F F − F
"F has the most negative electron gain enthalpy because it's most electronegative."
Feels right: F grabs electrons hardest in bonds. Fix: Adding an electron to the tiny F atom forces it into an already-crowded 2 p 2p 2 p Δ e g H \Delta_{eg}H Δ e g H less negative for F than Cl. Electronegativity (bonded) ≠ electron gain enthalpy (isolated atom).
"Oxidizing power" = how badly X 2 X_2 X 2 steal electrons and become 2 X − 2X^- 2 X − E ∘ E^\circ E ∘ X 2 + 2 e − → 2 X − larger E ∘ ⇒ stronger oxidizer X_2 + 2e^- \rightarrow 2X^- \qquad \text{larger } E^\circ \Rightarrow \text{stronger oxidizer} X 2 + 2 e − → 2 X − larger E ∘ ⇒ stronger oxidizer
Break the reduction into a Born–Haber-style cycle (per mole of X 2 X_2 X 2
1 2 X 2 → atomize, 1 2 Δ d i s s H X ( g ) → Δ e g H X − ( g ) → Δ h y d H X − ( a q ) \tfrac12 X_2 \xrightarrow{\text{atomize, } \frac12\Delta_{diss}H} X(g) \xrightarrow{\Delta_{eg}H} X^-(g) \xrightarrow{\Delta_{hyd}H} X^-(aq) 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 )
The total energy released governs E ∘ E^\circ E ∘ F :
Δ d i s s H ( F 2 ) \Delta_{diss}H(F_2) Δ d i ss H ( F 2 ) low (weak F–F bond, lone-pair repulsion) → easy to break.Δ h y d H ( F − ) \Delta_{hyd}H(F^-) Δ h y d H ( F − ) very high (small ion → strong hydration).
These two terms overpower F's slightly poorer electron gain enthalpy, making F₂ the strongest oxidizing agent of all halogens.
In water, acid strength of H X HX H X H − X H\!-\!X H − X breaks to give H + H^+ H + . The single biggest factor is bond strength , not electronegativity. Bigger X → longer, weaker H–X bond → easier to ionize → stronger acid .
"HF is the strongest acid because F is most electronegative."
Feels right: more EN should pull the bonding pair off H. Fix: F is small → very strong, short H–F bond + strong H-bonding in solution → HF holds onto its proton. So HF is the weakest HX acid (it's a weak acid; HCl, HBr, HI are strong). Electronegativity matters for oxoacids , not for HX where bond enthalpy dominates .
Stability of H X HX H X H F > H C l > H B r > H I HF > HCl > HBr > HI H F > H C l > H B r > H I
Interhalogens are compounds of two different halogens , type X X n ′ XX'_n X X n ′ n = 1 , 3 , 5 , 7 n = 1,3,5,7 n = 1 , 3 , 5 , 7 X X X larger, less electronegative halogen (central), X ′ X' X ′
Why does a bigger central atom take more small partners? Geometry: a large central atom has the surface area to fit many small atoms around it. So only large I/Br can reach n = 5 , 7 n=5,7 n = 5 , 7 I F 7 IF_7 I F 7 n n n C l F ClF C l F
Type
Example
Hybridisation
Shape (VSEPR)
X X ′ XX' X X ′ C l F , B r F , I C l ClF, BrF, ICl C l F , B r F , I C l —
linear
X X 3 ′ XX'_3 X X 3 ′ C l F 3 , B r F 3 ClF_3, BrF_3 C l F 3 , B r F 3 s p 3 d sp^3d s p 3 d bent T-shape
X X 5 ′ XX'_5 X X 5 ′ I F 5 , B r F 5 IF_5, BrF_5 I F 5 , B r F 5 s p 3 d 2 sp^3d^2 s p 3 d 2 square pyramidal
X X 7 ′ XX'_7 X X 7 ′ I F 7 IF_7 I F 7 s p 3 d 3 sp^3d^3 s p 3 d 3 pentagonal bipyramidal
Key facts (WHY/HOW):
Interhalogen bonds X − X ′ X\!-\!X' X − X ′ weaker than X − X X\!-\!X X − X more reactive than halogens (except F₂).
They are covalent, diamagnetic, volatile ; hydrolyse to give halide + oxohalide.
Used as fluorinating agents (e.g. C l F 3 ClF_3 C l F 3
Pseudohalogens are polyatomic molecules that chemically resemble halogens (X₂): e.g. cyanogen ( C N ) 2 (CN)_2 ( C N ) 2 ( S C N ) 2 (SCN)_2 ( S C N ) 2 ( O C N ) 2 (OCN)_2 ( O C N ) 2 pseudohalides : C N − , S C N − , O C N − , N 3 − CN^-, SCN^-, OCN^-, N_3^- C N − , S C N − , O C N − , N 3 − X − X^- X −
Why call them "pseudo"? Because they mimic halogen reactions:
( C N ) 2 + 2 O H − → C N − + O C N − + H 2 O (CN)_2 + 2OH^- \rightarrow CN^- + OCN^- + H_2O ( C N ) 2 + 2 O H − → C N − + O C N − + H 2 O C l 2 + 2 O H − → C l − + O C l − + H 2 O \;Cl_2 + 2OH^- \rightarrow Cl^- + OCl^- + H_2O C l 2 + 2 O H − → C l − + O C l − + H 2 O H X HX H X
Q1. Will B r 2 Br_2 B r 2 C l − Cl^- C l − C l 2 Cl_2 C l 2
Compare E ∘ E^\circ E ∘ C l 2 / C l − = + 1.36 Cl_2/Cl^- = +1.36 C l 2 / C l − = + 1.36 B r 2 / B r − = + 1.07 Br_2/Br^- = +1.07 B r 2 / B r − = + 1.07 B r 2 + 2 C l − → 2 B r − + C l 2 Br_2 + 2Cl^- \rightarrow 2Br^- + Cl_2 B r 2 + 2 C l − → 2 B r − + C l 2 Δ E ∘ = 1.07 − 1.36 = − 0.29 V < 0 \Delta E^\circ = 1.07 - 1.36 = -0.29\,V < 0 Δ E ∘ = 1.07 − 1.36 = − 0.29 V < 0 Why this step? Negative cell potential ⇒ non-spontaneous.
Answer: No. Only a stronger oxidizer displaces a weaker; here Cl₂ is stronger, so the reverse happens.
Q2. Arrange H F , H C l , H I HF, HCl, HI H F , H C l , H I
Bond enthalpies: H − F ( 565 ) > H − C l ( 431 ) > H − I ( 297 ) H\!-\!F(565) > H\!-\!Cl(431) > H\!-\!I(297) H − F ( 565 ) > H − C l ( 431 ) > H − I ( 297 ) Why this step? Weaker bond = easier H⁺ release = stronger acid.
Answer: H I > H C l > H F HI > HCl > HF H I > H C l > H F
Q3. Predict shape of C l F 3 ClF_3 C l F 3
Central Cl: 7 valence e⁻; 3 bonds use 3 → 4 e⁻ left = 2 lone pairs. Total = 3 bp + 2 lp = 5 domains ⇒ s p 3 d sp^3d s p 3 d Why this step? Equatorial lone pairs minimize 90° repulsions.
Answer: Bent T-shape (≈ T-shaped, slightly < 90°).
Recall Quick self-test (hide answers)
Which halogen has most negative Δ e g H \Delta_{eg}H Δ e g H Cl (not F)
Strongest oxidizing halogen? → F₂
Strongest HX acid? → HI
Why HF weak acid? → strong short H–F bond + H-bonding
Shape of I F 7 IF_7 I F 7
General formula of interhalogen? → X X n ′ XX'_n X X n ′
Recall Feynman: explain to a 12-year-oldImagine each halogen is a kid who needs exactly one more sticker to fill their book. Fluorine is the smallest, hungriest kid — he'll snatch a sticker from anyone (strongest oxidizer). When they hold hands with hydrogen (HX), the bigger kids hold on loosely , so they drop the hydrogen easily in water — that's why HI is a strong acid but tiny tight-gripping HF won't let go. Interhalogens are just two different-sized kids holding hands, and a big kid can hold several small kids at once (like 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" (larger halogen central).
Which halogen is the strongest oxidizing agent and why? F₂; low F–F bond dissociation enthalpy + very high hydration enthalpy of F⁻ outweigh its modest electron gain enthalpy.
Why is Cl, not F, having the most negative electron gain enthalpy? F is so small that adding an electron causes strong e⁻–e⁻ repulsion in the compact 2p shell, making ΔegH less negative than Cl.
Order of HX acid strength and the governing factor? HF < HCl < HBr < HI; governed by H–X bond dissociation enthalpy (weaker bond → stronger acid).
Why is HF a weak acid despite F being most electronegative? Very strong short H–F bond plus strong hydrogen bonding hold the proton, so HF ionizes poorly.
General formula and central atom rule for interhalogens? XX'ₙ (n=1,3,5,7); the larger, less electronegative halogen is central.
Shape and hybridisation of ClF₃? Bent T-shape; sp³d (3 bond pairs + 2 lone pairs equatorial).
Shape of IF₇? Pentagonal bipyramidal, sp³d³.
Why are interhalogens more reactive than halogens? X–X' bond is weaker (and more polar) than X–X, so it breaks more easily (except F₂).
Define pseudohalogens with an example. Polyatomic species resembling X₂ chemically, e.g. cyanogen (CN)₂; their anions (CN⁻, SCN⁻, OCN⁻, N₃⁻) are pseudohalides.
Will Cl₂ displace Br⁻? Justify with E°. Yes; E°(Cl₂/Cl⁻)=+1.36 > E°(Br₂/Br⁻)=+1.07, ΔE°>0 so Cl₂+2Br⁻→2Cl⁻+Br₂ is spontaneous.
Trend in X–X bond enthalpy and the anomaly? Cl > Br > F > I; F–F is anomalously low due to lone-pair repulsion in the small F atoms.
low F-F bond plus high hydration
increasing size down group
not equal to electron gain enthalpy
Intuition Hinglish mein samjhoHalogens (F, Cl, Br, I) ka pura nature ek hi baat se samajh aata hai: inko sirf ek electron chahiye apna shell full karne ke liye. Isiliye ye sabse strong oxidizing agents hote hain — dusron se electron cheen lete hain. Group me niche jaate jaate atom bada hota hai, nucleus ki pakad kamzor hoti hai, to oxidizing power ghat-ti hai: F₂ sabse strong, I₂ sabse weak. Yaad rakho — electron gain enthalpy sabse negative Cl ka hota hai, F ka nahi , kyunki F itna chhota hai ki naya electron daalne par repulsion ho jaata hai.
HX acid strength me ek twist hai jahan students fas jaate hain. Log sochte hain "F sabse electronegative, to HF sabse strong acid" — galat! Asli factor hai bond strength . Bada atom (I) ke saath H–X bond lamba aur kamzor hota hai, to H⁺ aasani se release hota hai. Isliye order hai HI > HBr > HCl > HF (sabse weak) . HF apne chhote, mazboot bond aur hydrogen bonding ki wajah se proton chhodta hi nahi.
Interhalogens matlab do alag halogens ka combination, formula X X n ′ XX'_n X X n ′ bada halogen center me baithta hai aur chhote uske around (jaise IF₇ me Iodine center). Ye halogens se zyada reactive hote hain kyunki X–X' bond weak hota hai. Pseudohalogens jaise cyanogen ( C N ) 2 (CN)_2 ( C N ) 2
