CLASS 11 · CHEMISTRY · CHAPTER 2

Structure of Atom

Simple worked solutions to all 67 end-of-chapter exercises on atomic structure, radiation, quantum numbers, and electron configurations.

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Chapter 2 Exercises

Textbook pages 70–74

Question 1

NCERT Exercise 2.1: Find how many electrons have a combined mass of 1 g, and find the mass and total charge of one mole of electrons.

Steps only

Divide 1 g by electron mass; for one mole multiply electron mass and charge by Avogadro’s constant.

Answer

(i) 1.10×10²⁷ electrons.
(ii) Mass = 5.49×10⁻⁴ g; charge = −9.65×10⁴ C.

Question 2

NCERT Exercise 2.2: Find: (i) electrons in one mole of methane; (ii) neutron count and total neutron mass in 7 mg of ¹⁴C; (iii) proton count and total proton mass in 34 mg NH₃. Would temperature or pressure change the particle counts?

Steps only

Count particles in one molecule/formula unit, find moles, and multiply by Avogadro’s constant.

Answer

(i) 1.204×10²⁵ electrons.
(ii) 7 mg ¹⁴C contains 3.01×10²⁰ neutrons, mass 5.04×10⁻⁷ kg.
(iii) 34 mg NH₃ contains about 2.41×10²² protons, mass 4.03×10⁻⁵ kg. Changing T or P does not change counts for a fixed mass.

Question 3

NCERT Exercise 2.3: For ¹³C, ¹⁶O, ²⁴Mg, ⁵⁶Fe and ⁸⁸Sr, state the number of protons and neutrons in each nucleus.

Steps only

Protons = atomic number; neutrons = mass number − atomic number.

Answer

¹³₆C: 6 p, 7 n; ¹⁶₈O: 8 p, 8 n; ²⁴₁₂Mg: 12 p, 12 n; ⁵⁶₂₆Fe: 26 p, 30 n; ⁸⁸₃₈Sr: 38 p, 50 n.

Question 4

NCERT Exercise 2.4: Write nuclide symbols for atoms with (i) Z=17,A=35; (ii) Z=92,A=233; (iii) Z=4,A=9.

Steps only

Use the element with atomic number Z; write A as the superscript and Z as the subscript.

Answer

(i) ³⁵₁₇Cl
(ii) ²³³₉₂U
(iii) ⁹₄Be.

Question 5

NCERT Exercise 2.5: Sodium-lamp yellow light has wavelength 580 nm. Find its frequency and wavenumber.

Steps only

Convert 580 nm to metres (and centimetres for wavenumber), then use ν=c/λ.

Answer

ν=c/λ=5.17×10¹⁴ s⁻¹. Wavenumber=1/λ=1.72×10⁴ cm⁻¹.

Question 6

NCERT Exercise 2.6: Find the energy of one photon for (i) frequency 3×10¹⁵ Hz and (ii) wavelength 0.50 Å.

Steps only

Use E=hν or E=hc/λ; 0.50 Å = 5.0×10⁻¹¹ m.

Answer

(i) E=hν=1.99×10⁻¹⁸ J.
(ii) E=hc/λ=3.97×10⁻¹⁶ J.

Question 7

NCERT Exercise 2.7: A light wave has period 2.0×10⁻¹⁰ s. Find its wavelength, frequency and wavenumber.

Steps only

Frequency is reciprocal of period; use c=νλ and take reciprocal wavelength.

Answer

ν=1/T=5.0×10⁹ Hz; λ=cT=6.0×10⁻² m; wavenumber=1/λ=16.7 m⁻¹ (1.67×10² cm⁻¹).

Question 8

NCERT Exercise 2.8: How many 4000 pm photons together provide 1 J of energy?

Steps only

Convert 4000 pm to metres, calculate one photon’s energy, then divide 1 J by it.

Answer

Energy per photon=hc/λ=4.97×10⁻¹⁷ J. Number=1/(4.97×10⁻¹⁷)=2.01×10¹⁶ photons.

Question 9

NCERT Exercise 2.9: A 4×10⁻⁷ m photon strikes a metal with 2.13 eV work function. Find photon energy, emitted-electron kinetic energy and electron speed.

Steps only

Subtract work function from photon energy; convert kinetic energy to joules before finding speed.

Answer

(i) Photon energy=hc/λ=3.10 eV.
(ii) KE=3.10−2.13=0.97 eV.
(iii) v=√(2KE/mₑ)=5.84×10⁵ m s⁻¹.

Question 10

NCERT Exercise 2.10: Light of wavelength 242 nm just ionises sodium. Find the ionisation energy in kJ mol⁻¹.

Steps only

Find energy per photon, then multiply by Avogadro’s constant.

Answer

Per atom E=hc/λ=8.22×10⁻¹⁹ J. Per mole = 8.22×10⁻¹⁹×Nₐ = 495 kJ mol⁻¹.

Question 11

NCERT Exercise 2.11: A 25 W bulb emits 0.57 μm yellow light. Find photons emitted each second.

Steps only

Convert 0.57 μm to metres. Divide bulb energy each second by energy per photon.

Answer

Energy per photon=hc/λ=3.49×10⁻¹⁹ J. Rate=25 J s⁻¹/(3.49×10⁻¹⁹ J)=7.16×10¹⁹ photons s⁻¹.

Question 12

NCERT Exercise 2.12: Electrons leave a metal with zero speed under 6800 Å light. Find the threshold frequency and work function.

Steps only

Zero emitted speed means the incident light is at threshold; its photon energy equals work function.

Answer

ν₀=c/λ=4.41×10¹⁴ Hz. W₀=hν₀=2.92×10⁻¹⁹ J (1.82 eV).

Question 13

NCERT Exercise 2.13: Find the wavelength emitted when a hydrogen electron drops from n=4 to n=2.

Steps only

Apply the Rydberg equation for transition n=4 to n=2.

Answer

Using 1/λ=R(1/2²−1/4²), λ=4.86×10⁻⁷ m = 486 nm.

Question 14

NCERT Exercise 2.14: Find the energy needed to ionise hydrogen when its electron is in n=5, and compare with ground-state ionisation energy.

Steps only

The energy of level n is −2.18×10⁻¹⁸/n² J; ionisation raises it to zero.

Answer

Energy=2.18×10⁻¹⁸/5²=8.72×10⁻²⁰ J per atom (0.0525 MJ mol⁻¹). It is 1/25 of ground-state ionisation energy.

Question 15

NCERT Exercise 2.15: Find the maximum number of emission lines possible as an excited hydrogen electron at n=6 drops to the ground state.

Steps only

Count every possible pair of levels among n=6, including transitions between excited levels.

Answer

Maximum lines = n(n−1)/2 = 6×5/2 = 15.

Question 16

NCERT Exercise 2.16: For hydrogen, E₁=−2.18×10⁻¹⁸ J atom⁻¹. Find (i) E₅ and (ii) the radius of the fifth Bohr orbit.

Steps only

Energy varies as 1/n²; Bohr radius varies as n².

Answer

(i) E₅=−2.18×10⁻¹⁸/25=−8.72×10⁻²⁰ J atom⁻¹.
(ii) r₅=0.529×25 pm=132 pm.

Question 17

NCERT Exercise 2.17: Find the wavenumber of the longest-wavelength transition in hydrogen’s Balmer series.

Steps only

Within Balmer series, choose the transition closest to n=2 to give the least energy and longest wavelength.

Answer

Longest wavelength is the smallest energy drop ending at n=2: n=3→2. Wavenumber=R(1/4−1/9)=1.52×10⁶ m⁻¹ (1.52×10⁴ cm⁻¹).

Question 18

NCERT Exercise 2.18: Find the energy needed to excite hydrogen from n=1 to n=5 and the wavelength emitted when it returns to the ground state. Given ground-state energy −2.18×10⁻¹¹ erg.

Steps only

Use the level-energy difference for excitation and the Rydberg relation for the return transition.

Answer

Energy supplied=2.18×10⁻¹⁸(1−1/25)=2.09×10⁻¹⁸ J. Return wavelength for 5→1: λ=9.50×10⁻⁸ m (95.0 nm).

Question 19

NCERT Exercise 2.19: For hydrogen, Eₙ=−2.18×10⁻¹⁸/n² J. Find the energy to remove an electron from n=2 and the longest wavelength that can cause this transition.

Steps only

Remove the binding energy of the n=2 level; threshold wavelength has exactly that photon energy.

Answer

Energy=2.18×10⁻¹⁸/4=5.45×10⁻¹⁹ J. Longest wavelength=hc/E=3.65×10⁻⁷ m=3.65×10⁻⁵ cm.

Question 20

NCERT Exercise 2.20: Find the de Broglie wavelength of an electron moving at 2.05×10⁷ m s⁻¹.

Steps only

Substitute electron mass and speed into λ=h/mv.

Answer

λ=h/(mₑv)=3.55×10⁻¹¹ m.

Question 21

NCERT Exercise 2.21: An electron has mass 9.1×10⁻³¹ kg and kinetic energy 3.0×10⁻²⁵ J. Find its de Broglie wavelength.

Steps only

Find momentum from kinetic energy, then use de Broglie relation.

Answer

p=√(2mK)=7.39×10⁻²⁸ kg m s⁻¹. λ=h/p=8.96×10⁻⁷ m.

Question 22

NCERT Exercise 2.22: Which of Na⁺, K⁺, Mg²⁺, Ca²⁺, S²⁻ and Ar have the same number of electrons?

Steps only

Count electrons by adjusting atomic number for ionic charge.

Answer

Na⁺ and Mg²⁺ each have 10 electrons. K⁺, Ca²⁺, S²⁻ and Ar each have 18 electrons. These are the two isoelectronic groups.

Question 23

NCERT Exercise 2.23: Give configurations for H⁻, Na⁺, O²⁻ and F⁻; identify atomic numbers for outer configurations 3s¹, 2p³ and 3p⁵; name atoms with [He]2s¹, [Ne]3s²3p³ and [Ar]4s²3d¹.

Steps only

Add or remove electrons for ions. For neutral atoms, infer the element from the configuration and count electrons.

Answer

(i) H⁻: 1s²; Na⁺: 1s²2s²2p⁶; O²⁻ and F⁻: 1s²2s²2p⁶.
(ii) Z=11, 7, 17.
(iii) Li, P, Sc.

Question 24

NCERT Exercise 2.24: What is the smallest principal quantum number n that can contain a g orbital?

Steps only

For a g subshell l=4; allowed l values are 0 to n−1, so n must be at least 5.

Answer

Lowest n=5.

Question 25

NCERT Exercise 2.25: For an electron in a 3d orbital, state all possible n, l and mₗ values.

Steps only

The “3” gives n; d means l=2; mₗ ranges from −l to +l.

Answer

n=3, l=2, and mₗ=−2,−1,0,+1,+2.

Question 26

NCERT Exercise 2.26: A neutral atom has 29 electrons and 35 neutrons. Find its proton number and electronic configuration.

Steps only

A neutral atom has equal proton and electron counts; 29 electrons identify copper.

Answer

(i) 29 protons.
(ii) Configuration: 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹ ([Ar]3d¹⁰4s¹).

Question 27

NCERT Exercise 2.27: Find the number of electrons in H₂⁺, H₂ and O₂⁺.

Steps only

Add atomic numbers of the atoms, then subtract one for a positive charge.

Answer

H₂⁺: 1 electron; H₂: 2 electrons; O₂⁺: 15 electrons.

Question 28

NCERT Exercise 2.28: For n=3, list possible l and mₗ values; give l and mₗ for 3d electrons; and decide which of 1p, 2s, 2p and 3f orbitals can exist.

Steps only

Allowed l values are 0…n−1; allowed mₗ values are −l…+l.

Answer

(i) l=0,1,2. For each l, mₗ runs from −l to +l.
(ii) 3d: l=2, mₗ=−2 to +2.
(iii) Possible: 2s and 2p; impossible: 1p and 3f.

Question 29

NCERT Exercise 2.29: Name the subshells represented by (a) n=1,l=0 (b) n=3,l=1 (c) n=4,l=2 (d) n=4,l=3.

Steps only

Map l=0,1,2,3 to s,p,d,f, while n gives the leading number.

Answer

(a) 1s (b) 3p (c) 4d (d) 4f.

Question 30

NCERT Exercise 2.30: For each set (a)–(f), decide whether the given four quantum numbers are allowed and explain using their rules.

Steps only

Check each set against the four allowed-value rules. Set (b) is valid: it describes a 1s electron.

Answer

Not possible: (a) n=0 (b)–(e) invalid; (f) valid. Reasons: n≥1; l=0…n−1; mₗ=−l…+l; mₛ=±½.

Question 31

NCERT Exercise 2.31: Find the maximum number of electrons with (a) n=4,mₛ=−½ and (b) n=3,l=0.

Steps only

A shell has n² orbitals and can hold one electron of each spin in every orbital; an s subshell has one orbital.

Answer

(a) n=4 shell has 16 orbitals, so up to 16 electrons can have mₛ=−½. (b) n=3,l=0 is 3s and holds 2 electrons.

Question 32

NCERT Exercise 2.32: Show that the circumference of a Bohr orbit equals an integer multiple of the electron’s de Broglie wavelength.

Steps only

Substitute mv=h/λ into the quantisation rule and rearrange.

Answer

Bohr’s angular-momentum rule is mvr=nh/(2π). Since λ=h/(mv), this gives 2πr=nλ.

Question 33

NCERT Exercise 2.33: Which hydrogen transition has the same wavelength as the n=4→2 Balmer transition of He⁺?

Steps only

For hydrogen-like ions, spectral wavenumber scales with Z². Match the energy gap to H’s 2→1 transition.

Answer

For He⁺, 4→2 has wavenumber 4R(1/2²−1/4²)=3R/4. In hydrogen the matching transition is 1→2 absorption (same energy gap); as an emission it is 2→1.

Question 34

NCERT Exercise 2.34: Find the energy needed for He⁺(g)→He²⁺(g)+e⁻, given hydrogen ground-state ionisation energy 2.18×10⁻¹⁸ J atom⁻¹.

Steps only

Hydrogen-like ionisation energy scales as Z².

Answer

He⁺ is hydrogen-like with Z=2. Ionisation energy=Z²×2.18×10⁻¹⁸=8.72×10⁻¹⁸ J per ion.

Question 35

NCERT Exercise 2.35: Carbon atoms have diameter 0.15 nm. How many fit side by side across a 20 cm ruler?

Steps only

Convert both lengths to metres and divide ruler length by atom diameter.

Answer

Number=20 cm/(0.15 nm)=0.20/(1.5×10⁻¹⁰)=1.33×10⁹ atoms.

Question 36

NCERT Exercise 2.36: If 2×10⁸ carbon atoms side by side span 2.4 cm, find the radius of one carbon atom.

Steps only

Divide the row length by number of atoms for diameter, then halve it.

Answer

Diameter=2.4 cm/(2×10⁸)=1.2×10⁻¹⁰ m. Radius=6.0×10⁻¹¹ m=0.060 nm.

Question 37

NCERT Exercise 2.37: Zinc atom diameter is 2.6 Å. Find its radius in pm and the number fitting side by side across 1.6 cm.

Steps only

Radius is half the diameter. Divide total length by atomic diameter for the count.

Answer

(a) Radius=1.3 Å=130 pm. (b) Count=1.6 cm/2.6 Å≈6.15×10⁷ atoms.

Question 38

NCERT Exercise 2.38: A particle has charge magnitude 2.5×10⁻¹⁶ C. Estimate how many excess electrons it carries.

Steps only

Divide charge magnitude by the elementary charge.

Answer

Number=|q|/e=(2.5×10⁻¹⁶)/(1.602×10⁻¹⁹)=1.56×10³, about 1,560 electrons.

Question 39

NCERT Exercise 2.39: An oil drop carries −1.282×10⁻¹⁸ C. How many excess electrons does it have?

Steps only

Divide charge magnitude by one electron’s charge.

Answer

n=1.282×10⁻¹⁸/1.602×10⁻¹⁹=8 electrons. The negative sign shows excess electrons.

Question 40

NCERT Exercise 2.40: In Rutherford scattering, what would change if a light-element foil such as aluminium were used instead of gold or platinum?

Steps only

The nuclear charge is smaller in light elements, so the repulsive force on alpha particles is weaker.

Answer

Light nuclei repel alpha particles less strongly than heavy nuclei, so deflections would generally be smaller and fewer particles would bounce back.

Question 41

NCERT Exercise 2.41: Why are ⁷⁹₃₅Br and ⁷⁹Br acceptable symbols, while ³⁵₇₉Br and ³⁵Br are not acceptable for bromine-79?

Steps only

For bromine-79, 79 is the mass number and 35 is bromine’s atomic number.

Answer

⁷⁹₃₅Br and ⁷⁹Br both identify bromine-79. ³⁵₇₉Br gives bromine the impossible atomic number 79, while ³⁵Br names bromine-35, not bromine-79.

Question 42

NCERT Exercise 2.42: An element has mass number 81 and 31.7% more neutrons than protons. Identify its nuclide.

Steps only

Solve N=1.317Z together with N+Z=81, then identify element by Z.

Answer

N=1.317Z and N+Z=81, so Z≈35 and N=46. Symbol: ⁸¹₃₅Br.

Question 43

NCERT Exercise 2.43: A mass-37 ion has charge −1 and 11.1% more neutrons than electrons. Identify the ion.

Steps only

Use the mass-number sum and the charge relation to solve for Z.

Answer

Charge −1 means electrons=Z+1. Neutrons are 1.111 times electron count. Solving N+Z=37 gives Z≈17, N=20, e=18. Symbol: ³⁷₁₇Cl⁻.

Question 44

NCERT Exercise 2.44: A mass-56 ion has charge +3 and 30.4% more neutrons than electrons. Identify it.

Steps only

Write electron count from the +3 charge, then solve the neutron percentage and mass-number equations.

Answer

Electrons=Z−3; neutrons=1.304(Z−3); and N+Z=56. This gives Z≈26, N=30, e=23. Symbol: ⁵⁶₂₆Fe³⁺.

Question 45

NCERT Exercise 2.45: Arrange microwave, amber light, FM radio, cosmic rays and X-rays in increasing frequency.

Steps only

Radio waves have the lowest frequency; cosmic rays have the highest energy and frequency.

Answer

FM radio < microwave < amber light < X-rays < cosmic rays.

Question 46

NCERT Exercise 2.46: A nitrogen laser emits 337.1 nm photons; 5.6×10²⁴ photons are emitted. Find total energy and power if this count is emitted each second.

Steps only

Multiply photon energy by photon count. Power is energy per time; the question does not specify duration, so only energy (or a per-second assumption) can be stated.

Answer

Energy per photon=hc/λ=5.90×10⁻¹⁹ J. Total energy=3.30×10⁶ J. A time interval is needed for power; if emission is over one second, power=3.30 MW.

Question 47

NCERT Exercise 2.47: Neon emits at 616 nm. Find (a) frequency (b) distance travelled in 30 s (c) one-photon energy (d) photon count in 2 J.

Steps only

Use c=λν and E=hν; divide total energy by energy per photon for the count.

Answer

(a) ν=4.87×10¹⁴ Hz. (b) Distance in 30 s=9.0×10⁹ m. (c) E=3.23×10⁻¹⁹ J. (d) 2/E=6.19×10¹⁸ quanta.

Question 48

NCERT Exercise 2.48: A detector receives 3.15×10⁻¹⁸ J of 600 nm light. Estimate the number of photons received.

Steps only

Calculate energy of one 600 nm photon, then divide detected energy by it.

Answer

E per photon=hc/λ=3.31×10⁻¹⁹ J. Count=3.15×10⁻¹⁸/3.31×10⁻¹⁹≈9.51 photons, about 10 photons.

Question 49

NCERT Exercise 2.49: A 2 ns pulse contains 2.5×10¹⁵ photons. Find its energy; state what extra information is needed if the photon wavelength is unspecified.

Steps only

Energy per photon depends on wavelength; multiply by photon count once wavelength is known.

Answer

The question does not give the wavelength, so the energy is not uniquely determined. For wavelength λ, E(total)=2.5×10¹⁵hc/λ. A numerical answer needs λ.

Question 50

NCERT Exercise 2.50: Sodium’s two absorption lines are 589 nm and 589.6 nm. Find each frequency and the energy gap between the excited states.

Steps only

Use ν=c/λ for each line; the energy spacing is h times their frequency difference.

Answer

ν₁=c/(589 nm)=5.09×10¹⁴ Hz; ν₂=c/(589.6 nm)=5.09×10¹⁴ Hz. ΔE=hc(1/589 nm−1/589.6 nm)=3.45×10⁻²² J.

Question 51

NCERT Exercise 2.51: Caesium work function is 1.9 eV. Find threshold wavelength and frequency; for 500 nm light find emitted-electron kinetic energy and speed.

Steps only

Use hc≈1240 eV·nm; subtract work function from photon energy, then use KE=½mv².

Answer

Threshold λ=hc/W=653 nm; ν₀=W/h=4.59×10¹⁴ Hz. At 500 nm, KE=2.48−1.9=0.58 eV=9.29×10⁻²⁰ J; v=4.52×10⁵ m s⁻¹.

Question 52

NCERT Exercise 2.52: Sodium is irradiated with 500, 450 and 400 nm light. The table lists photoelectron speeds as v×10⁻⁵ (cm s⁻¹) = 2.55, 4.35 and 5.35. Estimate the threshold wavelength and Planck’s constant.

Steps only

The printed speed unit is inconsistent. With the likely intended scale, v=(2.55, 4.35, 5.35)×10⁵ m/s, the 500 nm and 400 nm readings give λ₀≈540 nm and h≈6.7×10⁻³⁴ J s.

Answer

Assuming the intended speed scale is ×10⁵ m/s: λ₀≈540 nm and h≈6.7×10⁻³⁴ J s.

Question 53

NCERT Exercise 2.53: Silver photoelectrons from 256.7 nm light are stopped by 0.35 V. Find silver’s work function.

Steps only

Subtract maximum electron kinetic energy eVₛ from incident photon energy.

Answer

Photon energy=hc/λ=4.83 eV. Stopping energy=0.35 eV. Work function=4.83−0.35=4.48 eV.

Question 54

NCERT Exercise 2.54: A 150 pm photon ejects an inner electron at 1.5×10⁷ m s⁻¹. Find the electron’s binding energy.

Steps only

The photon energy supplies both binding energy and kinetic energy: Eγ=Eb+KE.

Answer

Photon energy=hc/λ=8.27×10⁻¹⁶ J. Electron KE=½mₑv²=1.02×10⁻¹⁶ J. Binding energy≈7.25×10⁻¹⁶ J.

Question 55

NCERT Exercise 2.55: A Paschen transition at 1285 nm follows ν=3.29×10¹⁵[1/3²−1/n²] Hz. Find n and the spectral region.

Steps only

Substitute observed frequency in the given equation and solve for n. Paschen lines end at n=3.

Answer

ν=c/λ=2.33×10¹⁴ Hz. Solving ν=3.29×10¹⁵(1/9−1/n²) gives n≈5. The transition is 5→3, in the infrared.

Question 56

NCERT Exercise 2.56: An emission begins at orbit radius 1.3225 nm and ends at 211.6 pm. Find its wavelength, series and spectral region.

Steps only

Convert radii to Bohr orbit numbers using rₙ=a₀n²; use the Rydberg equation for 5→2.

Answer

rₙ=0.0529n² nm. The radii give nᵢ=5 and n_f=2; transition 5→2 is Balmer. λ≈434 nm, visible light.

Question 57

NCERT Exercise 2.57: An electron microscope electron travels at 1.6×10⁶ m s⁻¹. Find its de Broglie wavelength.

Steps only

Use de Broglie relation with electron mass and the given speed.

Answer

λ=h/(mₑv)=4.55×10⁻¹⁰ m=455 pm.

Question 58

NCERT Exercise 2.58: A neutron has wavelength 800 pm. Find its speed.

Steps only

Rearrange λ=h/(mₙv) and substitute λ=800 pm.

Answer

v=h/(mₙλ)=4.95×10² m s⁻¹.

Question 59

NCERT Exercise 2.59: An electron in hydrogen’s first Bohr orbit travels at 2.19×10⁶ m s⁻¹. Find its de Broglie wavelength.

Steps only

Substitute the electron’s orbital speed into λ=h/mv.

Answer

λ=h/(mₑv)=3.32×10⁻¹⁰ m=332 pm.

Question 60

NCERT Exercise 2.60: A hockey ball of mass 0.1 kg moves at 4.37×10⁵ m s⁻¹. Find its de Broglie wavelength.

Steps only

Use the hockey ball’s mass and speed in de Broglie’s equation.

Answer

λ=h/(mv)=6.63×10⁻³⁴/(0.1×4.37×10⁵)=1.52×10⁻³⁸ m.

Question 61

NCERT Exercise 2.61: If electron position uncertainty is 0.002 nm, find minimum momentum uncertainty. Can its momentum also be specified as h/(4π×0.05 nm)?

Steps only

Use ΔxΔp≥h/(4π), then compare the minimum momentum uncertainty with the proposed momentum.

Answer

Δp≥2.64×10⁻²³ kg m s⁻¹. The proposed momentum is about 1.05×10⁻²⁴ kg m s⁻¹, smaller than this uncertainty, so it cannot be specified that precisely.

Question 62

NCERT Exercise 2.62: Arrange the six listed electron states by increasing orbital energy using the n+l rule; identify equal-energy states.

Steps only

For multi-electron atoms, lower n+l means lower energy; for ties, lower n has lower energy. Spin and magnetic quantum numbers do not change orbital energy here.

Answer

Increasing energy: 5 (3p) < 2 and 4 (3d, equal energy) < 3 and 6 (4p, identical quantum sets) < 1 (4d). For equal n+l, the smaller n has lower energy.

Question 63

NCERT Exercise 2.63: For bromine, which electrons in 2p, 3p and 4p experience the lowest effective nuclear charge?

Steps only

Outer 4p electrons are farther from the nucleus and more shielded than 2p and 3p electrons.

Answer

The 2p electron experiences the greatest effective nuclear charge; 4p experiences the least.

Question 64

NCERT Exercise 2.64: In each pair, identify the orbital with greater effective nuclear charge: (i) 2s/3s (ii) 4d/4f (iii) 3d/3p.

Steps only

Within a subshell, lower n is less shielded; within one shell, penetration generally follows s>p>d>f.

Answer

(i) 2s > 3s
(ii) 4d > 4f
(iii) 3p > 3d.

Question 65

NCERT Exercise 2.65: Which has greater effective nuclear charge: the unpaired 3p electron in Al or the unpaired 3p electron in Si?

Steps only

Si has one more proton than Al, while shielding changes little across the same subshell.

Answer

The 3p electron in Si experiences greater effective nuclear charge.

Question 66

NCERT Exercise 2.66: Find unpaired-electron counts in P, Si, Cr, Fe and Kr.

Steps only

Write ground-state configurations and apply Hund’s rule; Cr is [Ar]3d⁵4s¹.

Answer

(a) P: 3 (b) Si: 2 (c) Cr: 6 (d) Fe: 4 (e) Kr: 0.

Question 67

NCERT Exercise 2.67: For n=4, find (a) number of subshells and (b) maximum number of electrons with mₛ=−½.

Steps only

The shell has l=0,1,2,3 and 1+3+5+7=16 orbitals; one spin-down electron can occupy each.

Answer

(a) Four subshells: 4s, 4p, 4d, 4f. (b) There are 16 orbitals, so 16 electrons can have mₛ=−½.