Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
What is the energy (in J atom$^{-1}$) required for the following process?
$Li^{2+}(g) \to Li^{3+}(g) + e^-$
(Take the ionization energy for the H atom in the ground state as $2.18 \times 10^{-18}$ J atom$^{-1}$)
(1) $8.72 \times 10^{-18}$
(2) $1.962 \times 10^{-18}$
(3) $1.962 \times 10^{-17}$
(4) $6.54 \times 10^{-17}$
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Answer: Option (3)
Key idea: Ionization energy of hydrogen-like ions (one-electron species)
Identify that $Li^{2+}$ is a hydrogen-like ion (one electron) with atomic number $Z = 3$
Use the formula for ionization energy of hydrogen-like ions: $E = Z^2 \times E_H$
The surface of sodium metal is irradiated with radiation of wavelength $x$ nm. The kinetic energy of ejected electrons is $2.8 \times 10^{-20}$ J. The work function of sodium is $2.3$ eV. The value of $x$ is $\_\_\_\_\_\_\_\_\_\_ \times 10^2$ nm. (Nearest integer)
(Given : $h = 6.6 \times 10^{-34}$ J s ; $1$ eV $= 1.6 \times 10^{-19}$ J; $c = 3.0 \times 10^8$ m s$^{-1}$)
Which of the following is correct set of 4 quantum numbers of $19^{th}$ electron in Chromium (Atomic number = 24) in accordance with Aufbau principle?
(1) $n = 3, l = 2, m = +2, s = +\frac{1}{2}$
(2) $n = 3, l = 2, m = -2, s = +\frac{1}{2}$
(3) $n = 4, l = 1, m = 0, s = +\frac{1}{2}$
(4) $n = 4, l = 0, m = 0, s = +\frac{1}{2}$
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Answer: Option (4)
Key idea: Electronic Configuration of Chromium (Exception to Aufbau Principle) and Quantum Numbers
Write the electronic configuration of Chromium (Z=24). Note: Cr is an exception with configuration $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$ (not $4s^2 3d^4$)
Count electrons according to Aufbau filling order: $1s^2$ (1-2), $2s^2$ (3-4), $2p^6$ (5-10), $3s^2$ (11-12), $3p^6$ (13-18)
The 19th electron enters the $4s$ orbital (before $3d$ fills according to Aufbau principle)
For the 19th electron in $4s^1$: $n=4$ (4th shell), $l=0$ (s-orbital), $m=0$ (only value for $l=0$)
Spin quantum number $s = +\frac{1}{2}$ (first electron in orbital takes positive spin by convention)
The species having identical radii according to the Bohr's theory are:
A. H (first orbit)
B. He$^+$ (first orbit)
C. He$^+$ (Second orbit)
D. Li$^{2+}$ (first orbit)
E. Be$^{3+}$ (Second orbit)
Choose the correct answer from the options given below:
(1) A and C Only
(2) A and E Only
(3) B and E Only
(4) C and D Only
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Answer: Option (2)
Key idea: Bohr's atomic radius formula for hydrogen-like species
Use Bohr's formula for radius: $r_n = \frac{n^2}{Z} \times a_0$ where $n$ is orbit number, $Z$ is atomic number, and $a_0$ is Bohr radius
Calculate for each species: A. H: $r = \frac{1^2}{1} \times a_0 = a_0$
B. He$^+$: $r = \frac{1^2}{2} \times a_0 = 0.5a_0$; C. He$^+$: $r = \frac{2^2}{2} \times a_0 = 2a_0$
D. Li$^{2+}$: $r = \frac{1^2}{3} \times a_0 = 0.33a_0$; E. Be$^{3+}$: $r = \frac{2^2}{4} \times a_0 = a_0$
Compare all radii: Only species A and E have radius = $a_0$
Arrange the following atomic orbitals of multi electron atoms in order of increasing energy.
A. $n=3, l=2, m=+1$
B. $n=4, l=0, m=0$
C. $n=6, l=1, m=0$
D. $n=5, l=1, m=+1$
E. $n=2, l=1, m=+1$
Choose the correct answer from the options given below:
(1) $C < D < B < A < E$
(2) $B < A < E < C < D$
(3) $E < C < D < B < A$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Aufbau Principle and (n+l) rule for orbital energy ordering
Identify the correct statements from the following:
A. Heisenberg uncertainty principle is applicable to electrons.
B. The size of $2p_{xy}$ orbital is less than the size of $3p_{xy}$ orbital.
C. The energy of $2s$ orbital of H atom is equal to the energy of $2s$ orbital of Li.
D. The electronic configuration of Cr is [Ar] $3d^5$ $4s^1$
Choose the correct answer from the options given below:
(1) A, B and C Only
(2) A, B and D Only
(3) B, C and D Only
(4) A, C and D Only
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Answer: Option (2)
Key idea: Atomic Structure: Heisenberg Uncertainty Principle, Orbital Size Comparison, Energy Levels in Different Atoms, and Electronic Configuration Anomalies
**Statement A**: Heisenberg uncertainty principle ($\Delta x \cdot \Delta p \geq \hbar/2$) applies to all quantum particles, especially electrons. ✓ CORRECT
**Statement B**: Orbital size increases with principal quantum number $n$. Since $n=3 > n=2$, the $3p$ orbital is larger than $2p$ orbital. ✓ CORRECT
**Statement C**: Energy of $2s$ orbital depends on effective nuclear charge. For H: $E = -13.6/n^2 = -3.4$ eV. For Li (Z=3): $2s$ electron experiences higher $Z_{eff}$, so energy is more negative (lower) than H. ✗ INCORRECT
**Statement D**: Cr (Z=24) has anomalous configuration [Ar] $3d^5$ $4s^1$ instead of expected [Ar] $3d^4$ $4s^2$ due to extra stability of half-filled $d^5$ subshell. ✓ CORRECT
Which of the following statement(s) is/are true?
A. If two orbitals have the same value of $(n+l)$, the orbital with lower value of n will have lower energy.
B. Energies of the orbitals in the same subshell increase with increase in atomic number.
C. The size of 2p$_z$ orbital is less than the size of 3p$_z$ orbital.
D. Among 5f, 6s, 4d, 5p and 5d orbitals, none of the orbitals have 2 radial nodes.
Choose the correct answer from the options given below:
(1) A, B and C only
(2) A and C only
(3) C and D only
(4) A only
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Answer: Option (2)
Key idea: Aufbau Principle, Orbital Energy, Orbital Size, and Radial Nodes
**Statement A**: According to Aufbau principle (n+l rule), when two orbitals have same $(n+l)$ value, the one with lower $n$ has lower energy. TRUE ✓
**Statement B**: As atomic number increases, effective nuclear charge increases, making orbital energies MORE NEGATIVE (lower, not higher). FALSE ✗
**Statement C**: For same orbital type, size increases with principal quantum number $n$. Since $n=3 > n=2$, size of $3p_z > 2p_z$. TRUE ✓
**Statement D**: Calculate radial nodes using formula: $n - l - 1$. For $5d$: $5 - 2 - 1 = 2$ radial nodes. So the statement 'none have 2 radial nodes' is FALSE ✗
Match the LIST-I with LIST-II
| List-I<br>Orbital | List-II<br>Radial nodes and nodal plane |
|---|---|
| A. 2s | I. 1 Radial node + two nodal planes |
| B. 3s | II. 1 Radial node + one nodal plane |
| C. 3p | III. 2 Radial nodes + No nodal plane |
| D. 4d | IV. 1 Radial node + No nodal plane |
Choose the *correct* answer from the options given below:
(1) A-IV, B-I, C-III, D-II
(2) A-IV, B-II, C-III, D-I
(3) A-III, B-I, C-IV, D-II
(4) A-IV, B-III, C-II, D-I
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Answer: Option (4)
Key idea: Radial and Angular Nodes in Atomic Orbitals
Use formulas: Radial nodes = $n - l - 1$ and Angular nodes (nodal planes) = $l$
For 2s: $n=2, l=0$ → Radial nodes = $2-0-1=1$, Nodal planes = $0$ → Matches IV
For 3s: $n=3, l=0$ → Radial nodes = $3-0-1=2$, Nodal planes = $0$ → Matches III
For 3p: $n=3, l=1$ → Radial nodes = $3-1-1=1$, Nodal planes = $1$ → Matches II
For 4d: $n=4, l=2$ → Radial nodes = $4-2-1=1$, Nodal planes = $2$ → Matches I
First and second ionization enthalpies of lithium are 520 kJ mol$^{-1}$ and 7297 kJ mol$^{-1}$ respectively. Energy required to convert 3.5 mg lithium (g) into Li$^{2+}$(g) [Li(g) $\to$ Li$^{2+}$(g)] is _______ kJ mol$^{-1}$. (nearest integer) [Molar mass of Li = 7 g mol$^{-1}$]
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Answer: 4
Key idea: Ionization Enthalpy and Energy Calculations
Process: $\text{Li}(g) \to \text{Li}^{2+}(g) + 2e^-$ requires both first and second ionization enthalpies
Total ionization energy per mole = $IE_1 + IE_2 = 520 + 7297 = 7817$ kJ/mol
Convert mass to moles: $n = \frac{3.5 \times 10^{-3}\text{ g}}{7\text{ g/mol}} = 5 \times 10^{-4}$ mol
Calculate total energy: $E = n \times 7817 = 5 \times 10^{-4} \times 7817 = 3.9085$ kJ
The Bohr radius of a hydrogen like species is 70.53 pm. The species and the stationary state (n) are respectively
(Given : Hydrogen atom Bohr radius is 52.9 pm)
(1) Li$^{2+}$, 3
(2) He$^+$, 3
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Bohr's Model - Radius of hydrogen-like species
Match List - I with List - II.
List - I
Electronic configuration of neutral atom (where n=2)
A. $ns^2$
B. $ns^2np^1$
C. $ns^2np^3$
D. $ns^2np^6$
List - II
1st Ionization Energy (kJ mol$^{-1}$)
I. 2080
II. 899
III. 800
IV. 1402
Choose the correct answer from the options given below:
(1) A-II, B-III, C-IV, D-I
(2) A-IV, B-III, C-II, D-I
(3) A-III, B-II, C-IV, D-I
(4) A-III, B-II, C-I, D-IV
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Answer: Option (1)
Key idea: Periodic trends in ionization energy and electronic configuration stability
Identify elements for n=2: A: $2s^2$ (Be), B: $2s^2 2p^1$ (B), C: $2s^2 2p^3$ (N), D: $2s^2 2p^6$ (Ne)
Recall ionization energy trend: Generally increases across period, but with exceptions for filled/half-filled subshells
Neon (D) has highest IE = 2080 kJ/mol due to complete noble gas configuration
Nitrogen (C) has IE = 1402 kJ/mol due to stable half-filled $p^3$ configuration
Beryllium (A) has IE = 899 kJ/mol with stable filled $s^2$ configuration
Boron (B) has lowest IE = 800 kJ/mol as $p^1$ electron is easier to remove
Consider two radiations of wavelengths
1. $\lambda_1 = 2000$ Å
2. $\lambda_2 = 6000$ Å
The ratio of the energies of these two radiations $\left(\frac{E_1}{E_2}\right)$ is __________ (Nearest integer).
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Answer: 3
Key idea: Energy-wavelength relationship for electromagnetic radiation (E = hc/λ)
Energy of a photon is given by $E = h\nu = \frac{hc}{\lambda}$
For the two radiations: $E_1 = \frac{hc}{\lambda_1}$ and $E_2 = \frac{hc}{\lambda_2}$
Taking the ratio: $\frac{E_1}{E_2} = \frac{hc/\lambda_1}{hc/\lambda_2} = \frac{\lambda_2}{\lambda_1}$
