Question 1

At what wavenumber does the first line of the hydrogen Balmer series occur in the atomic spectrum?

Accepted Answer

5R / 36 cm⁻¹. Wavenumber = R(1/2^2 - 1/3^2) = R(5/36) = 5R/36. Stored 9R/400 is wrong.

Question 2

In the hydrogen emission spectrum, a spectral series has its limit at 12186.3 cm⁻¹. This series is identified as

Accepted Answer

Paschen series. 12186.3 cm^-1 = R/9 (R=109677), so n=3 -> Paschen series. Stored Balmer (R/4 = 27419) is wrong.

Question 3

Two rapidly moving particles X and Y have de Broglie wavelengths of 1 nm and 4 nm, respectively. If the mass of X is nine times the mass of Y, what is the ratio of their kinetic energies, X: Y?

Accepted Answer

16: 9. KE ratio EX/EY = (mY lambdaY^2)/(mX lambdaX^2) = (1 x 16)/(9 x 1) = 16:9. Stored 9:1 is wrong.

Question 4

Given that Planck’s constant is 6.63 × 10⁻³⁴ J s and the speed of light is 3.0 × 10⁸ m s⁻¹, which of the following is nearest to the wavelength, in nanometres, of light having frequency 8 × 10¹⁵ s⁻¹?

Accepted Answer

4 × 10¹. The wavelength of light can be calculated using the formula λ = (c)/(f), where c is the speed of light and f is the frequency. Plugging in the values, we find that the wavelength corresponding to a frequency of 8 × 10¹⁵ s⁻¹ is approximately 3.75 × 10⁻⁷ m, which converts to 375 nm, closest to 4 × 10¹ nm.

Question 5

A lithium ion and a proton are each accelerated through the same potential difference. If their de Broglie wavelengths are λ₁ and λ₂, respectively, and m_(Li) = 9mₚ, what is the ratio λ₁: λ₂?

Accepted Answer

1: 3√(3). The de Broglie wavelength is inversely proportional to the momentum of a particle, which depends on its mass and velocity. Since the lithium ion has a mass 9 times that of the proton and both are accelerated through the same potential difference, the proton will have a higher velocity and thus a shorter wavelength, leading to the ratio of their wavelengths being 1: 3√3.

Question 6

The ionisation energy of He⁺ is 19.6 × 10⁻¹⁸ J atom⁻¹. What is the energy of the first stationary state (n = 1) of Li²⁺?

Accepted Answer

4.41 × 10⁻¹⁷ J atom⁻¹. E(Li2+, n=1) = E(He+, n=1) x (3/2)^2 = 19.6e-18 x 2.25 = 4.41e-17 J/atom. The stored value 4.41e-16 is ten times too large.

Question 7

What is the kinetic energy of an electron in the second Bohr orbit of a hydrogen atom, taking a₀ as the Bohr radius?

Accepted Answer

h² / 32π²mₑa₀². The kinetic energy of an electron in the Bohr model is derived from its orbital radius and the quantization of angular momentum. In the second orbit, the radius is twice that of the first, leading to the specific factor of 1/32 in the kinetic energy formula.

Question 8

The energy of an electron is expressed as E = -2.178 × 10⁻¹⁸ (Z² / n²). For a hydrogen atom, the wavelength of radiation needed to promote an electron from n = 1 to n = 2 is (take h = 6.62 × 10⁻³⁴ Js and c = 3.0 × 10⁸ m/s):

Accepted Answer

1.214 × 10⁻⁷ m. DeltaE(1->2) = 2.178e-18 x 0.75 = 1.6335e-18 J; lambda = (6.62e-34 x 3e8)/1.6335e-18 = 1.214e-7 m. The stored 2.816e-7 m is wrong.

Question 9

Suppose a nitrogen atom were imagined to have the configuration 1s⁷. This would place all electrons in the 1s orbital and, because they would be nearer the nucleus, would seem to have lower energy than the normal ground-state arrangement 1s² 2s² 2p³. However, such a configuration does not

Accepted Answer

Pauli exclusion principle. The Pauli exclusion principle states that no two electrons can occupy the same quantum state within a quantum system simultaneously. In the hypothetical configuration 1s⁷, all seven electrons would be in the same 1s orbital, violating this principle.

Question 10

For a hydrogen atom, the energy of the electron in the ground state is 13.6 eV. What is its energy in the second excited state?

Accepted Answer

1.51 eV. Ground n=1, first excited n=2, second excited n=3. E3 = 13.6/9 = 1.51 eV. The stored 3.4 eV is the n=2 (first excited) value.

Question 11

For a hydrogen atom, the ionization enthalpy is 1.312 × 10⁶ J mol⁻¹. What amount of energy is needed to raise the electron from the ground state (n = 1) to the first excited state (n = 2)?

Accepted Answer

9.84 × 10⁵ J mol⁻¹. Energy to go n=1 to n=2 = 1.312e6 x (1 - 1/4) = 1.312e6 x 0.75 = 9.84e5 J/mol. The stored 6.56e5 is wrong.

Question 12

What is the frequency of the limiting line in the Balmer series? (Take the Rydberg constant, R = 3.29 × 10¹⁵ cycles/s)

Accepted Answer

8.22 × 10¹⁴ s⁻¹. The series limit of the Balmer series is the n=inf -> 2 transition: nu = R(1/2^2) = 3.29e15/4 = 8.22e14 s^-1. The stored 3.29e15 is the full R, not R/4.

Question 13

What is the de Broglie wavelength associated with a car of mass 1000 kg moving at 36 km/h?

Accepted Answer

6.626 × 10⁻³⁸ m. The de Broglie wavelength is calculated using the formula λ = (h)/(mv), where h is Planck's constant, m is the mass, and v is the velocity. For a car with a mass of 1000 kg moving at 36 km/h, the resulting wavelength is extremely small, specifically 6.626 × 10⁻³⁸ m, which reflects the negligible wave-like properties of macroscopic objects.

Question 14

If the radius of the first Bohr orbit of a hydrogen atom is a₀, what is the de Broglie wavelength of the electron in the third orbit?

Accepted Answer

6πa₀. The de Broglie wavelength of an electron in a Bohr orbit is determined by the orbit's radius and the quantization condition, which states that the circumference of the orbit must be an integer multiple of the wavelength. For the third orbit, the radius is 3 times the first orbit's radius (3a₀), leading to a wavelength of 6πa₀ when applying the de Broglie relation.

Question 15

If the kinetic energy of an electron is made four times larger, how will the wavelength of its associated de Broglie wave change?

Accepted Answer

become half. The de Broglie wavelength is inversely proportional to the square root of the kinetic energy. Therefore, if the kinetic energy is increased four times, the wavelength will decrease by a factor of the square root of four, which is two, resulting in the wavelength becoming half.

Question 16

Which one of the following sets of four quantum numbers represents the valence electron of a rubidium atom (Z = 37)?

Accepted Answer

5, 0, +1/2. The correct option, 5, 0, +1/2, represents the valence electron of rubidium, which is in the fifth energy level (n=5), has an angular momentum quantum number (l) of 0 corresponding to an s orbital, and a spin quantum number of +1/2, indicating the electron's spin direction.

Question 17

For a hydrogen atom, consider an electron moving in a circular orbit of radius r, with mass m and charge e. What is the total energy of this orbiting electron?

Accepted Answer

-e²/2r. The total energy of an electron in a circular orbit around a proton is the sum of its kinetic and potential energy. The kinetic energy is given by (1/2)mv², and the potential energy due to electrostatic attraction is -e²/r. When calculated for a stable orbit, the total energy simplifies to -e²/2r, which accounts for the balance between kinetic and potential energy.

Question 18

The bond dissociation energy of H is 430.53 kJ molb9. When hydrogen is split by exposure to radiation of wavelength 253.7 nm, what percentage of the incident radiant energy is transformed into kinetic energy?

Accepted Answer

8.76%. The percentage of incident radiant energy transformed into kinetic energy is calculated by comparing the energy of the absorbed photons to the bond dissociation energy. The energy of the photons at the given wavelength is lower than the bond dissociation energy, resulting in a percentage of about 8.76% being converted into kinetic energy.

Question 19

The gas-phase conversion of an oxygen atom into oxide ion, O²⁻(g), occurs in two steps: first an exothermic electron gain and then an endothermic one, as shown below:
O(g) + e⁻ → O⁻(g); ΔH° = -141 kJ mol⁻¹
O⁻(g) + e⁻ → O²⁻(g); ΔH° = +780 kJ mol⁻¹
Even though O²⁻ is isoelectronic with neon,

Accepted Answer

the repulsion between added electrons is greater than the gain in stability from attaining a noble gas arrangement. Adding a second electron to the already-negative O- ion is opposed by electron-electron repulsion that outweighs the stability gained from the noble-gas configuration (+780 kJ/mol step). That repulsion, not electronegativity, is why gaseous O2- formation is unfavourable.

Question 20

Which one of the following species shows diamagnetic behavior?

Accepted Answer

H2. H2 is a diatomic molecule with paired electrons, resulting in no net magnetic moment, which is characteristic of diamagnetic substances. In contrast, the other options either have unpaired electrons or are ions that do not exhibit this behavior.

Question 21

In a hydrogen atom, which statement about the relative energies of these orbitals is true?

Accepted Answer

The 3s, 3p, and 3d orbitals all possess equal energy.. For the hydrogen atom the 3s, 3p and 3d orbitals are degenerate (energy depends only on n), so option 0 is correct. The stored statement that 3s is lower than 3p holds only for multi-electron atoms.

Question 22

According to Bohr's theory, the angular momentum of an electron in 5th orbit is

Accepted Answer

2.5 h/π. Bohr's theory states that the angular momentum of an electron in a given orbit is quantized and can be calculated using the formula L = nħ, where n is the principal quantum number and ħ is the reduced Planck's constant. For the 5th orbit (n=5), the angular momentum is 5ħ, which simplifies to 2.5 h/π when considering the appropriate conversion.

Question 23

Uncertainty in the position of an electron (mass = 9.1 × 10^−31 kg) moving with a velocity 300 m s^−1, accurate upto 0.001% will be

Accepted Answer

1.92 × 10^−2 m. The uncertainty in position can be calculated using the formula for uncertainty in velocity and the Heisenberg uncertainty principle. Given the velocity and its accuracy, the resulting uncertainty in position is derived to be 1.92 × 10^−2 m, making it the correct option.

Question 24

Which one of the following sets of ions represents a collection of isoelectronic species?

Accepted Answer

K⁺, Cl^−, Ca²+, Sc³+. The ions K⁺, Cl^−, Ca²+, and Sc³+ all have the same number of electrons, specifically 18, making them isoelectronic. This means they share the same electron configuration, which is a key characteristic of isoelectronic species.

Question 25

Which of the following sets of quantum numbers represents the highest energy of an atom?

Accepted Answer

n = 3, l = 2, m = 1, s = +1/2. By the (n+l) rule: 3s (n+l=3), 3p (4), 3d (5), 4s (4). The 3d set has the highest n+l = 5, so n=3, l=2 represents the highest-energy orbital.

Question 26

The ionization enthalpy of hydrogen atom is 1.312 × 10⁶ J mol^−1. The energy required to excite the electron in the atom from n = 1 to n = 2 is

Accepted Answer

9.84 × 10⁵ J mol^−1. Ionization energy corresponds to n=1 to infinity. E(1->2) = IE*(1 - 1/2^2) = 1.312e6*0.75 = 9.84e5 J/mol.

Question 27

Calculate the wavelength (in nanometre) associated with a proton moving at 1.0 × 10³ m s^−1. Mass of proton = 1.67 × 10^−27 kg and h = 6.63 × 10^−34 Js

Accepted Answer

0.40 nm. lambda = h/(m*v) = 6.63e-34 / (1.67e-27 * 1.0e3) = 3.97e-10 m = 0.40 nm.

Question 28

In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainly with which the position of the electron can be located is (h = 6.6 × 10^−34 kg m² s^−1, mass of electron, m = 9.1 × 10^−31 kg)

Accepted Answer

1.92 × 10^−3 m. Delta_v = 600 x 0.005/100 = 0.03 m/s, so Delta_p = 9.1e-31 x 0.03 = 2.73e-32. Delta_x >= h/(4*pi*Delta_p) = 6.6e-34/(4*pi*2.73e-32) = 1.92e-3 m.

Question 29

The energy required to break one mole of Cl−Cl bonds in Cl2 is 242 kJ mol^−1. The longest wavelength of light capable of breaking a single Cl−Cl bond is (c = 3 × 10⁸ ms^−1 and N_A = 6.02 × 10²³ mol^−1).

Accepted Answer

494 nm. Energy per bond = 242000/6.02e23 = 4.02e-19 J. lambda = hc/E = (6.626e-34*3e8)/4.02e-19 = 4.95e-7 m ~ 494 nm.

Question 30

Ionisation energy of He⁺ is 19.6 × 10^−18 J atom^−1. The energy of the first stationary state (n = 1) of Li²+ is

Accepted Answer

−4.41 × 10^−17 J atom^−1. For one-electron ions E(n=1) is proportional to Z^2 and is negative. E(Li2+) = -(19.6e-18)*(3^2/2^2) = -19.6e-18 * 9/4 = -4.41e-17 J/atom.

Question 31

For the He⁺ ion, the light emitted in the transition from n=4 to n=2 has the same frequency as which transition in the hydrogen atom?

Accepted Answer

n=2 to n=1. Frequency is proportional to Z^2(1/n_f^2 - 1/n_i^2). For He+ (Z=2), 4->2 gives 4*(1/4 - 1/16) = 3/4. For hydrogen, 2->1 gives (1 - 1/4) = 3/4, which is the same. So it matches the n=2 to n=1 transition.

Question 32

Which of the following is the energy of a possible excited state of hydrogen?

Accepted Answer

−3.4 eV. The energy of an excited state of hydrogen corresponds to a less negative value than the ground state energy of -13.6 eV, and -3.4 eV represents the energy level of the first excited state, indicating that the electron is further from the nucleus.

Question 33

The radius of the second Bohr orbit for hydrogen atom is: (Planck's const. h = 6.6262 × 10⁻³⁴ Js; mass of electron = 9.1091 × 10⁻³¹ kg; charge of electron = 1.60210 × 10⁻¹⁹ C; permittivity of vacuum ε₀ = 8.854185 × 10⁻¹² kg⁻¹ m⁻³ A²)

Accepted Answer

2.12 Å. The radius of the second Bohr orbit for a hydrogen atom is determined by the formula rₙ = n² * (ε₀ * h²) / (π * m * e²), where n is the principal quantum number. For n=2, this calculation results in a radius of approximately 2.12 Å, making option D the correct choice.

Question 34

For emission line of atomic hydrogen from n_i = 8 to n_f = n, the plot of wave number (ν̄) against (1/n²) will be (The Rydberg constant, R_H is in wave number unit)

Accepted Answer

Linear with slope R_H. For n_i = 8 -> n_f = n, nu_bar = R_H(1/n^2 - 1/8^2). Plotting nu_bar against (1/n^2) gives a straight line of slope +R_H and intercept -R_H/64.

Question 35

For any given series of spectral lines of atomic hydrogen, let Δν̄ = ν̄_max − ν̄_min be the difference in maximum and minimum frequencies in cm⁻¹. The ratio Δν̄_Lyman / Δν̄_Balmer is:

Accepted Answer

9: 4. The ratio Δν̄_Lyman / Δν̄_Balmer is derived from the differences in energy levels involved in the transitions for each series. The Lyman series transitions occur from higher energy levels to n=1, while the Balmer series transitions go to n=2, resulting in a greater frequency range for the Lyman series, specifically leading to a ratio of 9:4 when calculated.

Question 36

The number of orbitals associated with quantum numbers n = 5, mₛ = +1/2 is:

Accepted Answer

25. The number of orbitals for a given principal quantum number n is calculated using the formula n². For n = 5, this results in 5² = 25 orbitals.

Question 37

Which of the following is the energy of a possible excited state of hydrogen ?

Accepted Answer

-3.4 eV. The energy levels of hydrogen are quantized, and the excited states have negative energy values that are less negative than the ground state. In this case, -3.4 eV corresponds to the first excited state, indicating that the electron has absorbed energy and moved to a higher energy level.

Question 38

The radius of the second Bohr orbit for hydrogen atom is - (Planck's Const. h = 6.6262 × 10⁻³⁴ Js; mass of electron = 9.1091 × 10⁻³¹ kg; charge of electron e = 1.6021 × 10⁻¹⁹ C; permittivity of vacuum ε0 = 8.854185 × 10⁻¹² kg⁻¹ m⁻³ A²)

Accepted Answer

2.12 Å. The radius of the second Bohr orbit for a hydrogen atom is determined by the formula rₙ = n² * (h² / (4 * π² * m * e²)), where n is the principal quantum number. For n=2, this calculation yields approximately 2.12 Å, making it the correct answer.

Question 39

If the series limit frequency of the Lyman series is vL, then the series limit frequency of the Pfund series is:

Accepted Answer

vL/25. The series limit frequency of the Pfund series is derived from the energy levels of the hydrogen atom, specifically the transitions to the n=5 level. Since the Lyman series corresponds to transitions to n=1 and has a series limit frequency of vL, the Pfund series, which transitions to n=5, results in a frequency that is a fraction of vL, specifically vL/25.

Question 40

Among the following, the energy of 2s orbital is lowest in:

Accepted Answer

K. Orbital energy scales as -13.6*Z_eff^2/n^2, so for a fixed shell (2s) the energy becomes more negative as nuclear charge rises. Among H, Li, Na, K the 2s electrons feel the greatest effective nuclear charge in K (Z=19), so 2s is lowest in energy in K.

Question 41

For any given series of spectral lines of atomic hydrogen, let Δν = νmax − νmin be the difference in maximum and minimum frequencies in cm−1. The ratio ΔνLyman / ΔνBalmer is:

Accepted Answer

9: 4. For Lyman (n1=1): nu_max = R, nu_min = R(1-1/4) = 3R/4, so delta = R/4. For Balmer (n1=2): nu_max = R/4, nu_min = R(1/4-1/9) = 5R/36, so delta = R/9. Ratio = (R/4)/(R/9) = 9:4.

Question 42

The region in the electromagnetic spectrum where the Balmer series lines appear is-

Accepted Answer

Visible. The Balmer series (transitions to n=2) of hydrogen falls in the visible region. (Lyman -> UV, Paschen/Brackett -> infrared.)

Question 43

In the sixth period, the orbitals that are filled are

Accepted Answer

6s, 4f, 5d, 6p. By the Aufbau order, period 6 fills the 6s, then 4f, then 5d, and finally 6p subshells. So the orbitals filled in the sixth period are 6s, 4f, 5d, 6p.

Question 44

The number of orbitals associated with quantum numbers n = 5, mₛ = +1/2 is:

Accepted Answer

25. The number of orbitals in shell n=5 is n^2 = 25. Each orbital can hold one electron with ms = +1/2, so 25 orbitals are associated with n=5, ms=+1/2.

Question 45

For the Balmer series in the spectrum of H atom, ν̄ = R_H {(1)/(n₁²)-(1)/(n₂²) }, the correct statements among (I) to (IV) are:
(I) As wavelength decreases, the lines in the series converge
(II) The integer n1 is equal to 2
(III) The lines of longest wavelength corresponds to n2 = 3
(IV) T

Accepted Answer

(3) (I), (II), (III). As wavelength decreases the Balmer lines converge to the series limit (I true); n1 = 2 for Balmer (II true); the longest-wavelength line is the 3->2 transition, n2=3 (III true). The ionization energy of hydrogen requires n1=1, so it cannot be obtained from Balmer lines (IV false). Correct set: (I), (II), (III).

Question 46

The radius of the second Bohr orbit, in terms of the Bohr radius, a0, in Li2+ is:

Accepted Answer

4a0/3. The Bohr orbit radius is r = (n^2/Z)*a0. For Li2+ (Z=3) and n=2, r = (4/3)*a0 = 4a0/3.

Question 47

In which of the following pairs, the outer most electronic configuration will be the same?

Accepted Answer

Cr+ and Mn2+. Both Cr+ and Mn2+ have the same outermost electronic configuration of 3d5, as the removal of electrons from Cr and Mn leads to a similar distribution in their d orbitals.

Question 48

Given below are two statements:
Statement I: Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom.
Statement II: Bohr's model of hydrogen atom contradicts Heisenberg's uncertainty principle.
In the light of the above statements, choose the most appropriate an

Accepted Answer

Both statement I and statement II are true.. Rutherford's gold foil experiment primarily provided insights into the nuclear structure of the atom, but it does not address the quantized energy levels that explain the hydrogen line spectrum. Additionally, Bohr's model, while revolutionary, does not fully align with Heisenberg's uncertainty principle, which highlights the limitations of precisely knowing both position and momentum of particles.

Question 49

Which of the following sets of quantum numbers is not allowed?

Accepted Answer

n = 3, ℓ = 3, m = −3, s = − 1/2. For a given n, the azimuthal quantum number l ranges 0 to n-1. With n=3, l can be 0,1,2 only, so n=3, l=3, m=-3 is not allowed.

Question 50

The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is [Given The threshold frequency of platinum is 1.3 × 10¹⁵ s⁻¹ and h = 6.6 × 10⁻³⁴ J s.]

Accepted Answer

8.58 × 10⁻¹⁹ J. The correct option is derived from the equation E = hν, where E is the energy of the photon, h is Planck's constant, and ν is the threshold frequency. By substituting the given values, the energy calculated aligns with the minimum energy required to initiate the photoelectric effect in platinum.

JEE Main Chemistry: Structure of Atom questions with solutions

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