3.3.7 · Chemistry › d-Block (Transition Metals) & f-Block
Intuition Bada picture (WHY ye important hain)
Mn aur Cr dono d-block mein hain jahan metal kai electrons de sakta hai. KMnO₄ mein Mn apni sabse zyada oxidation state +7 mein hai; K₂Cr₂O₇ mein Cr +6 mein hai. Ek metal jo apne maximum tak electrons chhod chuka hai, woh electrons wapas lene ke liye bahut desperate hota hai — electrons ke liye yahi bhoook exactly hai jo ek oxidising agent hota hai. Isliye dono compounds powerful oxidisers hain, aur poora topic bas yeh bookkeeping hai ki "kitne electrons leta hai, aur kis medium mein."
Intuition WHY do-step prep hoti hai?
Natural ore mein Mn ya Cr low/intermediate oxidation state mein hota hai. +7 (Mn) ya +6 (Cr) tak pahunchne ke liye metal ko oxidise karna padta hai. Nature tumhe +7 free mein nahi dega, isliye industrial prep = (1) alkaline medium mein oxidiser ke saath fuse/roast karo taaki oxidation states badhe, phir (2) acidify / electrolyse karo taaki final salt ban sake.
Step 1 — Mn(+4) → Mn(+6) (manganate) ko alkaline fusion se oxidise karo:
2 M n O 2 + 4 K O H + O 2 fuse 2 K 2 M n O 4 + 2 H 2 O
Hawa ka O₂ (ya KNO₃) oxidiser hai. Product = green manganate M n O 4 2 − .
Step 2 — Mn(+6) → Mn(+7) (permanganate) electrolytic oxidation se (ya disproportionation/Cl₂/O₃ se):
2 M n O 4 2 − electrolytic oxidation 2 M n O 4 − + 2 e −
Product = purple permanganate M n O 4 − .
Worked example WHY electrolytic, sirf "aur zyada hawa" kyun nahi?
O₂ +6 par ruk jaata hai. Aakhri electron bahar karne ke liye (Mn⁶⁺→Mn⁷⁺) ek aur zyada strong oxidiser chahiye. Anode par electrons forcibly remove kiye jaate hain — anode sabse strongest possible oxidiser hai, isliye woh karta hai jo hawa nahi kar sakti. Yeh step kyun? Yeh cleanly green→purple convert karta hai bina koi foreign reagent daale.
Step 1 — Chromite ko alkali + air ke saath roast karo, Cr(+3) → Cr(+6) chromate:
4 F e C r 2 O 4 + 8 N a 2 C O 3 + 7 O 2 → 8 N a 2 C r O 4 + 2 F e 2 O 3 + 8 C O 2
Yellow chromate C r O 4 2 − .
Step 2 — Acidify karo (chromate → dichromate):
2 N a 2 C r O 4 + H 2 S O 4 → N a 2 C r 2 O 7 + N a 2 S O 4 + H 2 O
Step 3 — Na salt ko K salt mein convert karo (K₂Cr₂O₇ kam soluble hai, crystallise ho jaata hai):
N a 2 C r 2 O 7 + 2 K C l → K 2 C r 2 O 7 + 2 N a C l
Intuition WHY medium matter karta hai
M n O 4 − ko reduce hona hi hai, lekin kitna neeche girega yeh depend karta hai ki kitne H⁺ available hain. Acid (zyada H⁺) use 5 electrons grab karne deta hai (→Mn²⁺). Neutral/halka alkaline (kam H⁺) sirf 3 grab karta hai (→MnO₂). Strong alkali mein sirf 1 grab karta hai (→MnO₄²⁻).
Medium
Mn kya banta hai
Electrons
Colour change
Acidic
M n 2 + (+2)
5
purple → colourless
Neutral/weak alkali
M n O 2 (+4)
3
purple → brown ppt
Strong alkali
M n O 4 2 − (+6)
1
purple → green
Worked example Acidic KMnO₄ + oxalic acid (ek classic titration)
Oxidation: C 2 O 4 2 − → 2 C O 2 + 2 e − (×5)
Reduction: M n O 4 − + 8 H + + 5 e − → M n 2 + + 4 H 2 O (×2)
2 M n O 4 − + 5 C 2 O 4 2 − + 16 H + → 2 M n 2 + + 10 C O 2 + 8 H 2 O
×5 aur ×2 kyun? Electrons lost (10) = electrons gained (10) hone chahiye → 5 aur 2 ka LCM. ~60 °C par garam karo kyunki reaction shuru mein slow hoti hai (bana hua Mn²⁺ ise autocatalyse karta hai).
Worked example Acidic KMnO₄ + Fe²⁺
F e 2 + → F e 3 + + e − (×5), combine:
M n O 4 − + 5 F e 2 + + 8 H + → M n 2 + + 5 F e 3 + + 4 H 2 O
Worked example Dichromate + Fe²⁺ (iron ka volumetric estimation)
F e 2 + → F e 3 + + e − (×6):
C r 2 O 7 2 − + 6 F e 2 + + 14 H + → 2 C r 3 + + 6 F e 3 + + 7 H 2 O
×6 kyun? Ek dichromate 6 e⁻ leta hai, har Fe²⁺ 1 e⁻ deta hai → 6 Fe²⁺ chahiye.
Intuition WHY dichromate ek
secondary standard hai lekin KMnO₄ primary standard nahi hai
K₂Cr₂O₇ pure form mein mil sakta hai, non-hygroscopic hai, aur cold dilute HCl se react nahi karta → iska solution known concentration par rehta hai. KMnO₄ 100% pure milna mushkil hai aur dheere dheere paani ko oxidise karta hai, isliye use karne se pehle standardise karna padta hai.
Common mistake Common traps ko steel-man karo
Trap 1: "KMnO₄ hamesha Mn²⁺ deta hai." Kyun sahi lagta hai: acidic titrations pehle padhate hain. Fix: sirf acid mein. Neutral → MnO₂ (3e⁻), strong base → green MnO₄²⁻ (1e⁻).
Trap 2: "Chromate→dichromate ek redox/oxidation hai." Kyun sahi lagta hai: colour dramatically badalta hai, redox jaisa. Fix: Cr dono sides par +6 hai — yeh acid–base condensation hai, redox nahi.
Trap 3: KMnO₄ titrations mein HCl use karna. Kyun sahi lagta hai: HCl ek common acid hai. Fix: KMnO₄, Cl⁻ ko Cl₂ mein oxidise karta hai, oxidiser consume hota hai → galat result. Dilute H₂SO₄ use karo.
Trap 4: Electrons galat balance karna — bhool jaana ki electrons lost = electrons gained . Fix: hamesha LCM lo (jaise 5 & 2 → 10).
Recall Feynman: ek 12-saal ke bacche ko samjhao
Do laalchi bacche imagine karo. Purple baccha (KMnO₄) itna laalchi hai ki woh 5 toffiyaan (electrons) grab kar sakta hai agar room mein bahut saari toffiyaan hain (acid), lekin sirf 3 ya 1 agar room thoda khali ho. Orange baccha (K₂Cr₂O₇) hamesha 6 toffiyaan grab karta hai lekin sirf acid room mein khelta hai. Jab woh toffiyaan grab karte hain toh colour badal jaata hai: purple→clear, orange→green. Unhe banane ke liye pehle hum unhe "charge up" karte hain — dull ore ko washing-soda aur hawa ke saath pakate hain (aur purple wale ke liye thodi electricity bhi) jab tak metal super-laalchi na ho jaye.
Mnemonic Numbers yaad rakho
"7→2 give Five (acid), 7→4 give Three (neutral), 7→6 give One (base)" Mn ke liye.
"Dichromate Drinks 6, Demands acid" — 6 electrons, sirf acidic.
Colours: P ermanganate P urple, M anganate M ossy-green; C hromate C anary-yellow, D ichromate D eep-orange.
KMnO₄ — Mn ki oxidation state +7
K₂Cr₂O₇ — Cr ki oxidation state +6
KMnO₄ prep ke liye manganese ka ore Pyrolusite, MnO₂
K₂Cr₂O₇ prep ke liye chromium ka ore Chromite, FeCr₂O₄
KMnO₄ prep ka Step 1 (equation) 2 M n O 2 + 4 K O H + O 2 → 2 K 2 M n O 4 + 2 H 2 O (green manganate)
KMnO₄ prep ka Step 2 MnO₄²⁻ → MnO₄⁻ ka electrolytic oxidation (green→purple)
Manganate vs permanganate ka colour Manganate green (+6), permanganate purple (+7)
Acidic KMnO₄ half-reaction M n O 4 − + 8 H + + 5 e − → M n 2 + + 4 H 2 O (5 e⁻)
Neutral KMnO₄ product & electrons MnO₂ (Mn +4), 3 electrons, brown ppt
Strong-alkali KMnO₄ product & electrons MnO₄²⁻ (green), 1 electron
Dichromate half-reaction C r 2 O 7 2 − + 14 H + + 6 e − → 2 C r 3 + + 7 H 2 O (6 e⁻)
Chromate ⇌ dichromate equilibrium 2 C r O 4 2 − + 2 H + ⇌ C r 2 O 7 2 − + H 2 O (acid→orange, base→yellow)
Kya chromate→dichromate ek redox hai? Nahi — Cr +6 rehta hai, sirf condensation/acid-base hai
KMnO₄ ke saath dilute H₂SO₄ kyun, HCl kyun nahi? HCl ka Cl⁻ Cl₂ mein oxidise ho jaata hai, KMnO₄ consume ho jaata hai
KMnO₄ + oxalic acid balanced eqn 2 M n O 4 − + 5 C 2 O 4 2 − + 16 H + → 2 M n 2 + + 10 C O 2 + 8 H 2 O
K₂Cr₂O₇ + Fe²⁺ balanced eqn C r 2 O 7 2 − + 6 F e 2 + + 14 H + → 2 C r 3 + + 6 F e 3 + + 7 H 2 O
K₂Cr₂O₇ primary/secondary standard kyun hai Pure, stable, non-hygroscopic; KMnO₄ nahi hota
Oxalic acid–KMnO₄ titration garam kyun karte hain Shuru mein slow hoti hai; Mn²⁺ ise autocatalyse karta hai
Standard Electrode Potential & E° values — explain karta hai kyun E°(MnO₄⁻)=+1.51 > E°(Cr₂O₇²⁻)=+1.33
Oxidation States of Transition Metals — kyun +7 (Mn) aur +6 (Cr) accessible hain
Balancing Redox Reactions (ion-electron method)
Volumetric Analysis / Titrations — permanganometry & dichrometry
Le Chatelier's Principle — chromate⇌dichromate shift
Colour & d-d Transitions — kyun MnO₄⁻ itna intensely purple hai (charge transfer)
roast with alkali plus air
Na to K exchange with KCl
High oxidation state Mn +7, Cr +6
Green manganate MnO4 2-, Mn +6
Purple permanganate MnO4-, Mn +7
Yellow chromate CrO4 2-, Cr +6
Orange dichromate Cr2O7 2-