Class 11 Chemistry · Chapter 2
Short answer:
Class 11 Chemistry Chapter 2 Structure of Atom atom ke andar ke sub-atomic particles, atomic models (Thomson, Rutherford, Bohr) aur quantum mechanical model cover karta hai — quantum numbers, orbitals ki shape, aur electronic configuration tak. Ye chapter numerical-heavy hai isliye structure of atom class 11 formulas yaad hone chahiye. Neeche step-by-step NCERT solutions diye gaye hain jo class 11 chemistry chapter 2 structure of atom important questions revise karne aur apna khud ka class 11 chemistry ncert solutions pdf banane ke kaam aayenge.
Chapter 1 poore Class 10 Chemistry ki neev hai — waise hi Class 11 Chemistry ka Chapter 2, Structure of Atom, poori Physical Chemistry (aur agle saal Class 12 tak) ki neev hai. Yahin se quantum numbers, orbitals aur electronic configuration shuru hote hain, jo Chapter 3 (Classification of Elements and Periodicity in Properties) aur Chapter 4 (Chemical Bonding) me seedhe kaam aate hain.
Rationalised class 11 chemistry syllabus 2026-27 me is baar sirf 9 chapters hain — Part I me 1-4, Part II me 5-9 (poori list neeche FAQ me hai). Kuch pura units hata diye gaye hain (jaise States of Matter, Hydrogen, s-Block, p-Block, Environmental Chemistry) lekin Structure of Atom apni poori 6-section wali structure ke saath syllabus me hai — subatomic particles ki discovery se lekar quantum mechanical model tak koi topic delete nahi hua.
Photon energy, de Broglie wavelength, Bohr model ke radius/energy formulas, aur Heisenberg uncertainty principle jaise calculations har saal class 11 chemistry chapter 2 structure of atom important questions me repeat hote hain. Is spec me poore in-text aur exercise questions step-by-step solve kiye gaye hain, plus ek formula table jo revision ke liye quick reference ban sakta hai.
Chapter 2 Summary — 5 Minute Revision
2.1 Discovery of Sub-atomic Particles: Cathode ray experiments (J.J. Thomson) se electron discover hua — charge/mass ratio (e/m) = 1.758820 × 1011 C kg–1. Millikan ke oil drop experiment se electron ka charge = –1.602176 × 10–19 C mila, jisse mass ≈ 9.1094 × 10–31 kg nikla. Goldstein ke canal rays se proton discover hua, aur Chadwick ne neutron discover kiya.
2.2 Atomic Models: Thomson ka plum-pudding model aur Rutherford ka nuclear model (alpha-particle scattering experiment se) — dono ki apni limitations thi; Rutherford model electron stability aur atomic spectra explain nahi kar paya.
2.3 Developments Leading to Bohr's Model: Electromagnetic radiation ka dual (wave + particle) nature — Planck's quantum theory, photoelectric effect, aur hydrogen atomic spectrum ki study (Rydberg formula) ne Bohr model ki zameen taiyar ki.
2.4 Bohr's Model for Hydrogen Atom: Electron fixed circular orbits (stationary states) me ghoomta hai, jinki energy quantised hoti hai. Model ne hydrogen spectrum explain kiya lekin multi-electron atoms aur fine spectral lines (Zeeman/Stark effect) explain nahi kar paya.
2.5 Towards Quantum Mechanical Model: de Broglie ne dikhaya ki electron jaise particles ki bhi wave nature hoti hai (λ = h/mv). Heisenberg's uncertainty principle ne bataya ki electron ki exact position aur momentum ek saath precisely nahi maapi ja sakti — isi se orbit ka concept khatam hoke orbital (probability region) ka concept aaya.
2.6 Quantum Mechanical Model of Atom: Schrödinger equation electron ko wave function (ψ) se describe karti hai. Char quantum numbers (n, l, ml, ms) electron ki energy, shape, orientation aur spin define karte hain. Orbitals ki shapes (s, p, d), unki energies (Aufbau principle), Pauli exclusion principle, aur Hund's rule milke electronic configuration decide karte hain.
In-Text Questions — Solutions
2.1 (i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate the mass and charge of one mole of electrons.
(i) Mass of one electron = 9.10939 × 10–31 kg = 9.10939 × 10–28 g.
Number of electrons = 1 g ÷ 9.10939 × 10–28 g = 1.098 × 1027 electrons
(ii) Mass of 1 mole electrons = 6.022 × 1023 × 9.10939 × 10–31 kg
= 5.48 × 10–7 kg = 0.548 mg
Charge of 1 mole electrons = 6.022 × 1023 × 1.6022 × 10–19 C
= 9.65 × 104 C (1 Faraday)
2.2 (i) Total electrons in one mole of CH4. (ii) 7 mg of 14C me total number aur total mass of neutrons. (iii) 34 mg NH3 (at STP) me total number aur total mass of protons.
(i) CH4 me electrons = C(6) + 4×H(1) = 10 electrons/molecule.
Total electrons = 10 × 6.022 × 1023 = 6.022 × 1024
(ii) 14C me neutrons/atom = 14 – 6 = 8.
Moles of C = 7×10–3 g ÷ 14 g/mol = 5×10–4 mol → atoms = 3.011×1020
Neutrons = 8 × 3.011×1020 = 2.409×1021; mass = 2.409×1021 × 1.675×10–27 kg = 4.036×10–3 g
(iii) NH3 (M=17) me protons/molecule = N(7) + 3×H(1) = 10.
Moles = 34×10–3÷17 = 2×10–3 mol → molecules = 1.2044×1021
Protons = 10 × 1.2044×1021 = 1.2044×1022; mass = 1.2044×1022 × 1.6726×10–27 kg ≈ 2.014×10–2 g
2.3 136C, 168O, 2412Mg, 5626Fe, 8838Sr me protons aur neutrons ki sankhya batao.
136C → protons = 6, neutrons = 13–6 = 7
168O → protons = 8, neutrons = 8
2412Mg → protons = 12, neutrons = 12
5626Fe → protons = 26, neutrons = 30
8838Sr → protons = 38, neutrons = 50
2.4 Diye gaye Z aur A ke liye atom ka poora symbol likho: (i) Z=17, A=35 (ii) Z=92, A=233 (iii) Z=4, A=9
(i) 3517Cl
(ii) 23392U
(iii) 94Be
2.5 Sodium lamp se nikalne wali yellow light ka wavelength (λ) 580 nm hai. Frequency (ν) aur wavenumber (ῡ) calculate karo.
ν = c/λ = (3×108)/(5.8×10–7) = 5.172×1014 Hz
ῡ = 1/λ = 1/(5.8×10–7) = 1.724×106 m–1
2.6 Har photon ki energy nikalo jo (i) 3×1015 Hz frequency ki light se ho, (ii) 0.50 Å wavelength ki ho.
(i) E = hν = 6.626×10–34 × 3×1015
E = 1.988×10–18 J
(ii) λ = 0.50 Å = 5×10–11 m
E = hc/λ = (6.626×10–34 × 3×108)/(5×10–11) = 3.976×10–15 J
2.7 Ek light wave ka period 2.0×10–10 s hai. Iski wavelength, frequency aur wavenumber calculate karo.
ν = 1/T = 1/(2.0×10–10) = 5×109 Hz
λ = c/ν = (3×108)/(5×109) = 0.06 m = 6 cm
ῡ = 1/λ = 16.67 m–1
2.8 4000 pm wavelength wali light ke kitne photons 1 J energy denge?
λ = 4×10–9 m.
E(photon) = hc/λ = (6.626×10–34 × 3×108)/(4×10–9) = 4.970×10–17 J
Number of photons = 1 ÷ 4.970×10–17 = 2.012×1016
2.9 4×10–7 m wavelength ka photon ek metal surface (work function = 2.13 eV) par strike karta hai. Photon ki energy (eV), photoelectron ki KE aur velocity nikalo.
E(photon) = hc/λ = (6.626×10–34 × 3×108)/(4×10–7) = 4.9695×10–19 J = 3.103 eV
KE = 3.103 – 2.13 = 0.973 eV = 1.558×10–19 J
v = √(2KE/m) = √(2×1.558×10–19/9.109×10–31) = 5.85×105 m/s
2.10 242 nm wavelength ki radiation sodium atom ko just ionise karti hai. Sodium ki ionisation energy kJ mol–1 me nikalo.
E(atom) = hc/λ = (6.626×10–34 × 3×108)/(242×10–9) = 8.214×10–19 J
E(mole) = 8.214×10–19 × 6.022×1023 = 4.947×105 J/mol ≈ 495 kJ/mol
2.11 100 watt ka bulb 400 nm wavelength ki monochromatic light emit karta hai. Bulb per second kitne photons emit karta hai?
E(photon) = hc/λ = (6.626×10–34 × 3×108)/(400×10–9) = 4.970×10–19 J
Photons/s = Power ÷ E(photon) = 100 ÷ 4.970×10–19 = 2.012×1020
2.12 Ek electron transition 1.3225 nm radius wali orbit se shuru hokar 211.6 pm radius wali orbit par khatam hoti hai. Wavelength nikalo, transition series aur spectrum region batao.
rn = 52.9 n² pm (Z=1). r=1322.5 pm → n²=25 → n=5. r=211.6 pm → n²=4 → n=2. Ye n=5→2 transition Balmer series (visible region) hai.
1/λ = RH(1/2² – 1/5²) = 1.097×107 × 0.21 = 2.304×106 m–1
λ = 1/2.304×106 = 4.341×10–7 m = 434.1 nm
2.13 Hydrogen atom me En = –2.18×10–18/n² J hai. n=2 orbit se electron poori tarah remove karne ke liye zaroori energy nikalo. Is transition ke liye longest wavelength (cm me) kya hogi?
E(required) = E∞ – E2 = 0 – (–2.18×10–18/4) = 5.45×10–19 J
λ = hc/E = (6.626×10–34 × 3×108)/(5.45×10–19) = 3.647×10–7 m = 3.647×10–5 cm
Poore Class 11 Chemistry ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q40)
Q1. Thomson's model of atom ki koi do limitations likhiye.
(i) Ye model Rutherford ke alpha-particle scattering experiment ke results (kuch particles ka bada angle se deflect hona) explain nahi kar saka.
(ii) Ye model atomic spectra (line spectra) ko bhi explain nahi kar paya.
Q2. Rutherford's model of atom ki koi do limitations likhiye.
(i) Classical electromagnetic theory ke hisaab se accelerating electron energy radiate karke nucleus me spiral hokar gir jaana chahiye — is model se atom ki stability explain nahi hoti.
(ii) Ye model hydrogen ke discrete line spectrum ko bhi explain nahi kar paya.
Q3. Atomic number aur mass number me difference batao.
Atomic number (Z) = neutral atom ke nucleus me protons ki sankhya (= electrons ki sankhya bhi); ye element ki identity decide karta hai.
Mass number (A) = protons + neutrons (nucleons) ka total. A = Z + neutrons ki sankhya.
Q4. Ek anion ka mass number 37 hai aur uspar –1 charge hai. Agar ismein electrons se 11.1% zyada neutrons hain, to iska symbol batao.
Electrons = e; protons = e – 1 (anion); neutrons = 1.111e
A = protons + neutrons = (e–1) + 1.111e = 2.111e – 1 = 37 → e = 18
protons = 17, neutrons = 1.111×18 ≈ 20 → Z = 17 (Cl)
Ion = 37Cl–
Q5. Ek cation ka mass number 56 hai aur uspar +3 charge hai. Agar ismein electrons se 30.4% zyada neutrons hain, to iska symbol batao.
Electrons = e; protons = e + 3 (cation); neutrons = 1.304e
A = (e+3) + 1.304e = 2.304e + 3 = 56 → e = 23
protons = 26, neutrons = 1.304×23 ≈ 30 → Z = 26 (Fe)
Ion = 56Fe3+
Q6. Ek shell ke s, p aur d subshells ko increasing energy order me arrange karo.
Multi-electron atoms me (same n ke liye): energy order ns < np < nd hota hai, kyunki penetration aur shielding effect ki wajah se s > p > d order me nucleus ke close hote hain. Hydrogen-like (single electron) species me teeno ki energy sirf n par depend karti hai, isliye ns = np = nd.
Q7. En = –2.18×10–18 (Z²/n²) J formula use karke, He+ ke n=2 orbit se electron remove karne ki energy aur us transition ki longest wavelength (cm me) nikalo.
E2(He+, Z=2) = –2.18×10–18×(4/4) = –2.18×10–18 J
E(required) = 0 – (–2.18×10–18) = 2.18×10–18 J
λ = hc/E = (6.626×10–34×3×108)/(2.18×10–18) = 9.118×10–8 m = 9.118×10–6 cm
Q8. He+ ki first orbit se associated energy aur radius calculate karo.
E1 = –2.18×10–18×(2²/1²) = –8.72×10–18 J
r1 = 0.529×(1²/2) = 0.2645 Å
Q9. 2.05×107 m/s velocity se move kar rahe ek electron ki de Broglie wavelength nikalo.
λ = h/mv = (6.626×10–34)/(9.109×10–31×2.05×107) = 3.548×10–11 m
Q10. Ek electron ki KE = 3.0×10–25 J hai (mass = 9.1×10–31 kg). Iski wavelength nikalo.
v = √(2KE/m) = √(2×3×10–25/9.1×10–31) = 812 m/s
λ = h/mv = (6.626×10–34)/(9.1×10–31×812) = 8.97×10–7 m
Q11. Inme se isoelectronic species pehchano: Na+, K+, Mg2+, Ca2+, S2–, Ar.
Electrons: Na+=10, Mg2+=10, K+=18, Ca2+=18, S2–=18, Ar=18.
Isoelectronic groups: {Na+, Mg2+} (10 electrons) aur {K+, Ca2+, S2–, Ar} (18 electrons).
Q12. Electronic configurations likho: (a) H– (b) Na+ (c) O2– (d) F–
H–: 1s²
Na+: 1s²2s²2p⁶
O2–: 1s²2s²2p⁶
F–: 1s²2s²2p⁶
Q13. Outermost electron configuration (a) 3s¹ (b) 2p³ (c) 3p⁵ wale elements ka atomic number batao.
(a) 3s¹ → 1s²2s²2p⁶3s¹ → Z=11 (Na)
(b) 2p³ → 1s²2s²2p³ → Z=7 (N)
(c) 3p⁵ → 1s²2s²2p⁶3s²3p⁵ → Z=17 (Cl)
Q14. (a) [He]2s¹ (b) [Ne]3s²3p³ (c) [Ar]4s²3d¹ — kaunse atoms ye configurations show karte hain?
(a) [He]2s¹ → Z=3 (Li)
(b) [Ne]3s²3p³ → Z=15 (P)
(c) [Ar]4s²3d¹ → Z=21 (Sc)
Q15. g orbital exist karne ke liye n ki lowest value kya hogi?
g orbital ke liye l = 4. Chunki l, 0 se (n–1) tak hota hai, minimum n = l+1 = 5.
Q16. Ek electron 3d orbital me hai. Iske n, l aur ml ke possible values do.
n = 3, l = 2, ml = –2, –1, 0, +1, +2 (in me se koi ek)
Q17. Ek element ke atom me 29 electrons aur 35 neutrons hain. (i) Protons ki sankhya (ii) electronic configuration nikalo.
(i) Neutral atom me protons = electrons = 29.
(ii) Electronic configuration: 1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰ (Copper, Cu) — fully-filled 3d ki extra stability ki wajah se 4s²3d⁹ nahi balki 4s¹3d¹⁰ banta hai.
Q18. H2+, H2 aur O2+ me electrons ki sankhya batao.
H2+ = 1 electron
H2 = 2 electrons
O2+ = 15 electrons (O2 ke 16 me se ek nikal gaya)
Q19. n=3 wale atomic orbital ke liye l aur ml ke possible values do.
l = 0, 1, 2
l=0 → ml=0; l=1 → ml=–1,0,+1; l=2 → ml=–2,–1,0,+1,+2
Q20. 3d orbital ke electrons ke l aur ml quantum numbers list karo.
l = 2; ml = –2, –1, 0, +1, +2 (5 orbitals)
Q21. Inme se kaunse orbitals possible hain: 1p, 2s, 2p, 3f?
1p possible nahi (n=1 ke liye l sirf 0 ho sakta hai). 3f possible nahi (n=3 ke liye max l=2, f ke liye l=3 chahiye jo n≥4 maangta hai). 2s aur 2p possible hain.
Q22. s, p, d, f notation me orbital describe karo: (a) n=1,l=0 (b) n=3,l=1 (c) n=4,l=2 (d) n=4,l=3
(a) 1s (b) 3p (c) 4d (d) 4f
Q23. Batao kaunse quantum number sets possible nahi hain aur kyun: (a) n=0,l=0,ml=0,ms=+½ (b) n=1,l=0,ml=0,ms=–½ (c) n=1,l=1,ml=0,ms=+½ (d) n=2,l=1,ml=0,ms=–½ (e) n=3,l=3,ml=–3,ms=+½ (f) n=3,l=1,ml=0,ms=+½
(a) Not possible — n kabhi 0 nahi ho sakta.
(b) Possible.
(c) Not possible — n=1 ke liye l sirf 0 ho sakta hai, 1 nahi.
(d) Possible.
(e) Not possible — n=3 ke liye max l = 2, l=3 allowed nahi.
(f) Possible.
Q24. Kitne electrons in quantum numbers ko satisfy karenge: (a) n=4, ms=–½ (b) n=3, l=0?
(a) n=4 ke liye total electrons = 2n² = 32; inme se aadhe (ms=–½) → 16 electrons.
(b) n=3,l=0 matlab 3s subshell → max 2 electrons.
Q25. Dikhao ki hydrogen atom ke Bohr orbit ki circumference, us orbit me ghoomne wale electron ki de Broglie wavelength ka integral multiple hoti hai.
Bohr postulate: mvr = nh/2π
2πr = nh/(mv) = n × (h/mv) = nλ (kyunki de Broglie λ = h/mv)
Isliye circumference (2πr) hamesha λ ka integral (n) multiple hota hai.
Q26. Hydrogen spectrum ka kaunsa transition He+ spectrum ki n=4 se n=2 Balmer transition ke barabar wavelength ka hoga?
He+ (Z=2): 1/λ = RH×4×(1/4 – 1/16) = RH×(3/4)
Hydrogen (Z=1) me same λ ke liye: (1/n1² – 1/n2²) = 3/4 → n1=1, n2=2 satisfy karta hai (1 – 1/4 = 3/4).
Isliye hydrogen ka n=2 → n=1 (Lyman series) transition same wavelength ka hoga.
Q27. 5×1014 Hz frequency wale radiation ke ek mole photons ki energy calculate karo.
E(photon) = hν = 6.626×10–34×5×1014 = 3.313×10–19 J
E(mole) = 3.313×10–19×6.022×1023 ≈ 1.995×105 J/mol ≈ 199.5 kJ/mol
Q28. 25 watt ka bulb 0.57 μm wavelength ki yellow light emit karta hai. Quanta emission ki rate per second nikalo.
E(photon) = hc/λ = (6.626×10–34×3×108)/(0.57×10–6) = 3.487×10–19 J
Rate = 25 ÷ 3.487×10–19 = 7.17×1019 photons/s
Q29. 6800 Å wavelength ki radiation se metal surface se zero velocity ke saath electrons emit hote hain. Threshold frequency (ν0) aur work function (W0) nikalo.
ν0 = c/λ = (3×108)/(6.8×10–7) = 4.412×1014 Hz
W0 = hν0 = 6.626×10–34×4.412×1014 = 2.924×10–19 J
Q30. Hydrogen atom ki first orbit se associated energy –2.18×10–18 J hai. Fifth orbit ki energy kya hogi?
E5 = E1/n² = –2.18×10–18/25 = –8.72×10–20 J
Q31. Hydrogen atom ki Bohr fifth orbit ka radius calculate karo.
r5 = 0.529×5² = 13.225 Å
Q32. 1000 V potential difference se accelerate hui proton ki velocity 4.37×105 m/s hai. Agar 0.1 kg ki ek hockey ball isi velocity se move kare, uski de Broglie wavelength nikalo.
λ = h/mv = (6.626×10–34)/(0.1×4.37×105) = 1.516×10–38 m
Ye value itni chhoti hai ki macroscopic objects ki wave nature practically observable nahi hoti.
Q33. Agar electron ki position ±0.002 nm accuracy se maapi jaaye, momentum me uncertainty nikalo. Kya electron ka momentum precisely define karna possible hai?
Δx = 2×10–12 m
Δp ≥ h/(4πΔx) = (6.626×10–34)/(4×3.1416×2×10–12) = 2.637×10–23 kg m/s
Ye uncertainty electron ke typical momentum ke comparable/zyada hai — isliye momentum ko exactly define karna possible nahi, jo Heisenberg's uncertainty principle ko confirm karta hai aur Bohr ke fixed-orbit concept ki limitation dikhata hai.
Q34. 3p orbital me angular aur radial nodes ki total sankhya nikalo.
Angular nodes = l = 1
Radial nodes = n – l – 1 = 3 – 1 – 1 = 1
Total nodes = n – 1 = 2
Q35. Bromine atom (35 electrons) me 2p me 6, 3p me 6 aur 4p me 5 electrons hain. Inme se kaunse electrons lowest effective nuclear charge experience karenge?
4p electrons lowest effective nuclear charge experience karenge, kyunki ye nucleus se sabse door hain aur inner electrons (1s se lekar 3p tak) unhe zyada shield karte hain.
Q36. In pairs me se kaunsa orbital zyada effective nuclear charge experience karega: (i) 2s aur 3s (ii) 4d aur 4f (iii) 3d aur 3p?
(i) 2s (nucleus ke zyada close, kam shielding)
(ii) 4d (lower l → zyada penetration → higher Zeff than 4f)
(iii) 3p (lower l → zyada penetration → higher Zeff than 3d)
Q37. Al aur Si me unpaired electrons kramshah 1 aur 2 hote hain. Electronic configuration ke aadhar par justify karo.
Al (Z=13): 1s²2s²2p⁶3s²3p¹ → 3p me sirf 1 electron → 1 unpaired electron.
Si (Z=14): 1s²2s²2p⁶3s²3p² → Hund's rule ke according 3p ke 2 electrons alag-alag orbitals me parallel spin ke saath rehte hain → 2 unpaired electrons.
Q38. Sodium lamp se emit hui light ka wavelength 590 nm hai. Frequency (ν) aur wavenumber (ῡ) nikalo.
ν = c/λ = (3×108)/(5.9×10–7) = 5.085×1014 Hz
ῡ = 1/λ = 1.695×106 m–1
Q39. H atom ke excited electron n=6 se ground state me aane par maximum kitni emission lines ban sakti hain?
Number of lines = n(n–1)/2 = 6×5/2 = 15
Q40. (a) s, p aur d orbitals ki shapes describe karo. (b) Aufbau principle aur Hund's rule ko ek-ek example ke saath samjhaao.
(a) s-orbital spherical (non-directional) hota hai; p-orbital dumbbell-shaped hota hai (px, py, pz — teeno axes ke along, beech me ek nodal plane); d-orbitals me char (dxy, dyz, dzx, dx²–y²) double-dumbbell (cloverleaf) shape ke hote hain aur dz² ek dumbbell + donut-shaped ring ka combination hota hai.
(b) Aufbau principle: electrons increasing energy order me orbitals fill karte hain — carbon (Z=6) ka configuration 1s²2s²2p² isi order me banta hai.
(c) Hund's rule: degenerate orbitals (jaise 2p) me electrons pehle akele-akele parallel spin ke saath fill hote hain, phir pairing hoti hai — nitrogen (Z=7) me 2p³ configuration 2px¹2py¹2pz¹ hoti hai, kisi bhi orbital me pairing nahi.
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Energy of photon | E = hν = hc/λ |
| Wavenumber | ῡ = 1/λ = ν/c |
| de Broglie wavelength | λ = h/mv = h/p |
| Heisenberg uncertainty principle | Δx · Δp ≥ h/4π |
| Bohr radius (nth orbit) | rn = 0.529 × (n²/Z) Å |
| Energy of electron in nth orbit | En = –2.18 × 10–18 × (Z²/n²) J = –13.6 (Z²/n²) eV |
| Velocity of electron in nth orbit | vn = 2.188 × 106 × (Z/n) m/s |
| Rydberg formula (H-spectrum) | 1/λ = RH (1/n1² – 1/n2²), RH = 1.097 × 107 m–1 |
| Number of subshells in a shell | = n |
| Number of orbitals in a subshell | = 2l + 1 |
| Max. electrons in a subshell | = 2(2l + 1) |
| Max. electrons in a shell | = 2n² |
| Angular / radial nodes | Angular nodes = l ; Radial nodes = n – l – 1 ; Total nodes = n – 1 |
| Orbital angular momentum | = [l(l+1)]1/2 · h/2π |
| Spin quantum number | ms = +½ or –½ |
| Maximum emission lines from nth excited state to ground state | = n(n–1)/2 |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Orbit vs orbital ko same maan lena: Bohr ka "orbit" ek fixed circular path hai, jabki quantum mechanical "orbital" electron milne ki probability region hai — exams me ye galti bahut common hai.
- Bohr energy formula me negative sign bhool jaana: En hamesha negative hoti hai (bound electron); ionisation energy nikalte waqt E∞ – En = 0 – (negative) karna hota hai, seedha En ka magnitude nahi.
- Azimuthal quantum number (l) ki range galat likhna: l ki value 0 se (n–1) tak hoti hai, "1 se n" nahi — isse orbital ka naam bhi galat nikal jaata hai.
- Orbital naam aur l value mix karna: s orbital ke liye l = 0 hota hai, na ki l = 1; ye confusion p aur d orbitals tak bhi chal jaata hai.
- Numericals me units convert na karna: nm, Å, pm, cm jaise alag-alag units ko metre me convert kiye bina seedha E = hc/λ me daal dena — answer 10 ke power se hi galat ho jaata hai.
- Aufbau principle ko exceptions ke bina apply karna: Cr (4s¹3d⁵) aur Cu (4s¹3d¹⁰) jaise elements me half-filled/fully-filled d-subshell ki extra stability ki wajah se "expected" configuration (4s²3d⁴ / 4s²3d⁹) galat ho jaati hai.
Board-Style Important Questions
- Very Short Answer: Aufbau principle ka statement likhiye aur ek example se samjhaiye.
- Short Answer: Bohr's model of hydrogen atom ki koi do limitations likhiye.
- Short Answer: Char quantum numbers ke naam likhiye aur har ek ka significance ek-ek line me samjhaiye.
- Short Answer: de Broglie relation likhiye aur ye batao ki macroscopic objects me wave nature observable kyun nahi hoti.
- Short Answer: Aufbau principle, Pauli exclusion principle aur Hund's rule of maximum multiplicity — teeno ko ek-ek example ke saath samjhaiye.
- Long Answer: Quantum mechanical model of atom ke main postulates likhiye aur s, p, d orbitals ki shapes describe kijiye.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Chemistry 2026-27 rationalised syllabus me total kitne chapters hain? (class 11 chemistry all chapters list ncert)
Rationalised session me total 9 chapters hain, 2 parts me: Part I — 1. Some Basic Concepts of Chemistry, 2. Structure of Atom, 3. Classification of Elements and Periodicity in Properties, 4. Chemical Bonding and Molecular Structure. Part II — 5. Thermodynamics, 6. Equilibrium, 7. Redox Reactions, 8. Organic Chemistry – Some Basic Principles and Techniques, 9. Hydrocarbons.
Structure of Atom (Chapter 2) me koi topic delete hua hai? (class 11 chemistry deleted syllabus topics)
Nahi — class 11 chemistry deleted syllabus topics me sirf 5 poore units/chapters hate gaye hain: States of Matter (Gases and Liquids), Hydrogen, s-Block Elements, Some p-Block Elements, aur Environmental Chemistry. Structure of Atom apne sabhi 6 sections (2.1 se 2.6) ke saath poora syllabus me hai.
Structure of Atom padhne se pehle Chapter 1 revise karna zaroori hai kya?
Haan — some basic concepts of chemistry class 11 notes (mole concept, molar mass, Avogadro number) is chapter ke electrons/protons/neutrons counting wale numericals me seedhe use hote hain, isliye Chapter 1 clear hona zaroori hai.
Structure of Atom ke baad agla important chapter kaunsa hai?
Chapter 3 — Classification of Elements and Periodicity in Properties. Yahan periodic trends explain karne ke liye electronic configuration aur quantum numbers ka wahi gyaan use hota hai jo Structure of Atom me seekha, isliye classification of elements and periodicity in properties important questions attempt karne se pehle ye chapter solid honi chahiye.
Structure of Atom ke concepts Part II ke chapters me bhi kaam aate hain?
Haan, indirectly — energy calculations ka base class 11 chemistry chapter 5 thermodynamics numericals me, electron-transfer ka concept redox reactions class 11 important questions me, orbital overlap chemical bonding and molecular structure class 11 notes pdf me, aur electron-pushing mechanisms class 11 chemistry organic chemistry basic principles and techniques notes tatha hydrocarbons class 11 ncert solutions me use hote hain.
Structure of Atom ka NCERT solutions PDF kahan milega?
Neeche diye gaye poore in-text aur exercise solutions ko print/save karke apna khud ka class 11 chemistry ncert solutions pdf revision ke liye bana sakte ho — saath me formula table aur mistakes list zaroor dekh lena.
Class 11 Chemistry — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Chemistry ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
NCERT Kaksha ek swatantra shaikshik platform hai aur NCERT ya CBSE se aadhikarik roop se sambaddh (officially affiliated) nahi hai. Kisi galti ki jaankari dene ke liye contact kijiye.