UV-Vis spectroscopy — Beer-Lambert law, conjugation and λ_max: Question bank

4.8.2 · D5 · Chemistry › Spectroscopy & Analysis (Intro) › UV-Vis spectroscopy — Beer-Lambert law, conjugation and λ_ma

Yahan sab kuch teen ideas pe tika hai jo tumne parent note mein already build kiye hain:

Absorption ek jump hai. Ek photon tabhi "khaya" jaata hai jab uski energy $\frac{h c}{λ}$ orbital gap $Δ E$ ke barabar ho (dekho Planck relation E = hν).
Zyada conjugation = bada "box" = chhota gap = bada $λ_{ma x}$ (Particle in a box picture).
Beer-Lambert $A = ε c l$ , jahan $A = - lo g_{10} T$ .

Traps se pehle, teen quick pictures un words ko pin down karti hain jo ye page baar baar use karta hai.

Figure — UV-Vis spectroscopy — Beer-Lambert law, conjugation and λ_max

Definition "Allowed" vs "forbidden" transitions —

$ε$ asal mein kya count karta hai Har jump equally likely nahi hoti. Quantum selection rules (do orbitals ki symmetry aur spin ke basis pe) decide karti hain ki ek transition easy ("allowed") hai ya almost blocked ("forbidden"). Molar absorptivity $ε$ us likelihood ka direct measure hai:

Allowed (symmetry- aur spin-permitted, jaise zyaadatar $π \to π^{*}$ ): $ε \sim 1 0^{3}$ to $1 0^{5}$ → tall, strong peaks.
Forbidden (symmetry rule torti hai, jaise kai $n \to π^{*}$ carbonyl bands): $ε \sim 10$ to $1 0^{2}$ → short, weak peaks jo sirf isliye dikhte hain kyunki real molecules thoda hilte-dulte hain aur rule thoda relax ho jaata hai.

Toh $ε$ tumhe nahi batata ki peak kahaan hai (woh $λ_{ma x}$ ka kaam hai) — woh batata hai ki peak kitna tall hai. Figure 2 dekho: ek hi wavelength pe do peaks, tall red wala allowed, short black wala forbidden.

True or false — justify

Neeche har item ek statement hai. True/false decide karo aur reason do — sirf verdict se kuch nahi milega.

Woh molecule jo longer wavelength pe absorb karta hai uska HOMO–LUMO gap bada hota hai.

False.

λ = \frac{h c}{Δ E}

, toh longer

λ

ka matlab hai photon energy chhoti hai, yani gap

Δ E

(HOMO→LUMO) chhota hai, bada nahi.

Ek colourless organic solution mein definitely koi electronic transitions nahi hote.

False. Bas uski koi transitions visible 400–700 nm window mein nahi hain; uska

π \to π^{*}

absorption UV mein ho sakta hai (jaise 400 nm se neeche), jo humari aankhein nahi dekh sakti.

Concentration double karne se absorbance double ho jaata hai (valid range mein).

True.

A = ε c l

mein

c

ke saath linear hai, toh fixed

ε

aur

l

pe,

c

double karne se

A

double hota hai — bas itna ki tum low-concentration linear region mein raho.

Concentration double karne se transmittance double ho jaata hai.

False.

c

double karne se absorbance

A_{n e w} = 2 A

ho jaata hai; phir

T_{n e w} = 1 0^{- A_{n e w}} = 1 0^{- 2 A} = (1 0^{- A})^{2} = T^{2}

. Toh

T

squared hota hai, double nahi — bahut steeper drop (jaise

T = 0.5 \to 0.25

ε = 20, 000

wali transition "allowed" hai aur

ε = 20

wali "forbidden" hai.

True (roughly). Bada

ε

(

\sim 1 0^{3}

–

1 0^{5}

) symmetry-allowed transition ko mark karta hai (jaise

π \to π^{*}

); tiny

ε

(

\sim 10

–

1 0^{2}

) ek weak, symmetry-forbidden transition ko mark karta hai (jaise kai

n \to π^{*}

carbonyl bands) — bilkul Figure 2 wala selection-rule split.

Agar sample 100% light transmit karta hai, toh uska absorbance 100% hai.

False.

T = 1

deta hai

A = - lo g_{10} (1) = 0

. Full transmittance matlab zero absorbance — dono inverse hain, equal nahi.

Same

λ_{ma x}

wale do solutions zaroor same compound hote hain.

False.

λ_{ma x}

chromophore/conjugation length ko constrain karta hai lekin unique nahi hai; kai alag molecules ke paas similar-length π-system hota hai aur woh similar wavelength ke paas absorb karte hain.

Benzene ring ek chromophore hai chahe usme koi isolated C=C double bond na ho jise tum point kar sako.

True. Uska delocalised aromatic π-system

π \to π^{*}

transitions karta hai, toh woh absorb karta hai (around 255 nm) — exactly yahi ise chromophore banata hai.

Benzene mein –OH group add karne se

λ_{ma x}

unchanged rehta hai kyunki OH ka koi double bond nahi hai.

False. Oxygen ka lone pair ring mein conjugate ho jaata hai (ek auxochrome), π-system extend ho jaata hai aur

λ_{ma x}

red-shift ho jaata hai (phenol benzene se longer wavelength pe absorb karta hai).

Molar absorptivity

ε

is baat pe depend karta hai ki solution kitna concentrated banaya.

False.

ε

molecule ka ek fixed property hai (aur solvent/wavelength ka); concentration ek alag variable

c

hai. Dono ko confuse karna Beer-Lambert ka poora point defeat kar deta hai.

Spot the error

Har line mein flawed reasoning hai. Batao kya galat hai aur kyun.

"Ye solution red dikhta hai, toh zaroor red light strongly absorb kar raha hoga."

Galat: ek solution jo absorb karta hai uska complement dikhata hai (Figure 4). Red dikhna matlab green/blue-green light absorb ho rahi hai aur red pass ho raha hai — dekho Complementary colours.

A = 0.30

matlab 30% light absorb hui."

Galat:

A

ek logarithm hai, percentage nahi.

A = 0.30

deta hai

T = 1 0^{- 0.30} \approx 0.50

, toh ~50% transmitted hai aur ~50% absorbed — absorbance fraction nahi hai.

"β-carotene orange hai kyunki UV absorb karne ke baad orange light emit karta hai."

Galat mechanism: uska colour blue light ke selective absorption se aata hai, toh transmitted mix orange dikhta hai. Ye subtractive hai, emission/fluorescence nahi.

"Kyunki conjugation

Δ E

kam karta hai, infinitely long polyene infinitely long wavelengths absorb karega."

Galat: real chains eventually deviate karti hain — box bilkul perfectly free-electron nahi hai, aur

λ_{ma x}

ek limit ki taraf converge karta hai rather than bina bound ke badhta rahe.

n \to π^{*}

ka gap

π \to π^{*}

se chhota hai, toh hamesha stronger, taller peak deta hai."

Galat: chhota gap wavelength set karta hai, intensity nahi.

n \to π^{*}

usually symmetry-forbidden hota hai, toh uska

ε

low hota hai — longer wavelength ke bawajood ek weak peak (Figure 2).

"Zyada accurate measurement ke liye

A

ko 2 ya 3 ke paas rakhna chahiye — zyada absorbance, zyada signal."

Galat: high

A

pe almost koi light detector tak nahi pahunchti, toh noise dominate karta hai aur Beer-Lambert curve curve kar jaata hai. Reliable window roughly

A = 0.2

1.0

hai.

"Biphenyl ko twist karna taaki dono rings perpendicular ho jaayein

λ_{ma x}

ko red-shift karega kyunki ab ye ek bada molecule hai."

Galat: size matter nahi karta, overlap karta hai. 90° twist rings ke beech p-orbital overlap tod deta hai, conjugation khatam kar deta hai aur blue (hypsochromic) shift laata hai.

Why questions

Causal chain ke saath jawab do, sirf keyword se nahi.

UV-Vis

σ \to σ^{*}

ki jagah

π \to π^{*}

transitions pe focus kyun karta hai?

σ \to σ^{*}

gaps itne huge hote hain ki woh vacuum UV (<150 nm) mein absorb karte hain, jahan normal instruments nahi pahunch sakte;

π \to π^{*}

gaps moderate hote hain aur measurable 200–800 nm window mein aate hain.

Longer conjugated chain lower-energy photons kyun absorb karti hai?

Particle in a box picture (Figure 3) mein, longer chain matlab longer box

L

; energy levels

1/ L^{2}

scale karte hain, toh woh squeeze ho jaate hain,

Δ E

chhota ho jaata hai aur lower-energy (longer-

λ

) light ki zaroorat padti hai.

Absorbance define karne ke liye logarithm kyun lete hain instead of sirf

I_{0} / I

use karne ke?

Har thin slice light ka ek fixed fraction remove karta hai, toh attenuation multiplicative (path mein exponential) hoti hai.

lo g_{10}

lena us exponential ko ek quantity mein convert karta hai jo

c

aur

l

mein linear hai — easy calibration curves (dekho Quantitative analysis & calibration curves).

ε

ki value tumhe transition ke type ke baare mein kuch kyun batati hai?

ε

measure karta hai ki transition kitni likely hai (selection rules ke under uski "allowedness"); high

ε

(

1 0^{3}

–

1 0^{5}

) strongly allowed

π \to π^{*}

flag karta hai, jabki

ε \sim 10

–

1 0^{2}

weak, symmetry-forbidden

n \to π^{*}

flag karta hai.

–NH₂ jaisa auxochrome

λ_{ma x}

shift kyun karta hai jabki uska khud ka koi C=C nahi hai?

Uska lone pair adjacent π-system mein delocalise ho jaata hai, effectively conjugation lengthen karta hai aur

Δ E

kam karta hai — ek red shift, bina koi formal double bond contribute kiye (dekho Conjugation and resonance).

Do conjugation-equivalent dyes equal concentration pe kitne deeply coloured dikhte hain mein alag kyun ho sakte hain?

Colour ki depth

ε

pe depend karti hai (

A = ε c l

ke through), sirf

λ_{ma x}

pe nahi; zyada molar absorptivity wala dye zyada strongly absorb karta hai aur same

c

pe zyada intensely coloured dikhta hai.

HOMO–LUMO language box model ke saath bilkul align kyun hoti hai?

Electrons box levels mein pairs mein bharte hain, toh sabse upar wala filled level HOMO hai aur pehla empty wala LUMO; sabse chhota jump HOMO→LUMO hai, exactly wahi gap

Δ E

jo photon match kare (dekho HOMO and LUMO orbitals).

Edge cases

Boundary aur degenerate scenarios — woh jo exams ko pasand hain.

Jab cuvette mein bilkul clear (non-absorbing) solvent bhara ho toh

A

kya hoga?

A = 0

, kyunki

I = I_{0}

deta hai

T = 1

aur

A = - lo g_{10} (1) = 0

. Isliye tum instrument ko pure solvent pe "blank" karte ho.

Jab

T \to 0

(essentially koi light pass nahi hoti) toh

A

ka kya hoga?

A = - lo g_{10} T \to + \infty

; absorbance diverge karta hai. Practice mein detector noise aur stray light law ko isse bahut pehle tod dete hain, toh aisi readings meaningless hain.

Ek isolated single double bond (jaise ethene, ek C=C) ke liye, woh kahaan absorb karta hai aur kya normal UV-Vis instrument use dekh sakta hai?

Around 170 nm — vacuum UV mein, standard instruments ke ~190 nm cutoff se neeche, toh ye usually observe nahi hota. Band ko range mein laane ke liye conjugation chahiye.

Agar path length aadha kar do lekin concentration double kar do, toh

A

ka kya hoga?

Woh same rehta hai.

A = ε c l

mein product

c \cdot l

unchanged hai (

2 c \times \frac{l}{2} = c l

), toh absorbance identical hai.

Bahut high concentration pe, kya Beer-Lambert true absorbance ko over- ya under-estimate karta hai?

Woh

A

ko over-predict karta hai (

A

c

ka plot neeche bend karta hai / line ke neeche): solute–solute interactions aur refractive-index changes actual absorption reduce karte hain, toh measured

A

linear prediction se kam padta hai.

Ek solution ka

λ_{ma x}

350 nm (pure UV) pe hai aur koi visible absorption nahi hai — uska colour kya hoga?

Colourless / clear. 400–700 nm mein kuch bhi remove nahi hota, toh saare visible wavelengths equally pass hote hain aur aankhein koi colour nahi dekhti.

Agar kisi molecule mein odd number of π-electrons hain, toh kya simple "HOMO = level

N /2

" filling rule cleanly apply hoti hai?

Nahi — odd

N

ke saath top level half-filled hota hai (ek radical), toh pairs-fill assumption toot jaati hai; clean gap formula sirf even

N

ke liye valid hai.

Recall One-line survival kit

Colour seen ::: jo absorb hua uska complement
$A$ aur $T$ ::: inverse hain, $A = - lo g_{10} T$ ke through
Longer conjugation ::: chhota $Δ E$ , bada $λ_{ma x}$ (red shift)
$ε$ ::: molecule ka fixed fingerprint, peak ki height set karta hai, position nahi
Valid $A$ range ::: roughly 0.2–1.0