Atomic Structure and Periodicity -- Practice Problems

Atomic Structure and Periodicity — Practice Problems

10 MCQ questions covering quantum numbers, effective nuclear charge, periodic trends, electron configurations, and atomic properties. Select an option to check your answer.

Quantum Numbers and Electron Configuration

Q1. How many orbitals are there in the $n = 3$ shell, and what is the maximum number of electrons it can hold?

A. 9 orbitals ( $1 \\\\times s + 3 \\\\times p + 5 \\\\times d$ ); maximum 18 electrons B. 3 orbitals; maximum 6 electrons C. 5 orbitals; maximum 10 electrons D. 7 orbitals; maximum 14 electrons’,

Answer: A

Answer: A — The $n=3$ shell has $s$ ( $\\\\ell=0$ , 1 orbital), $p$ ( $\\\\ell=1$ , 3 orbitals), and $d$ ( $\\\\ell=2$ , 5 orbitals), totaling 9 orbitals. Each orbital holds 2 electrons (Pauli principle), so maximum capacity = 18 electrons. For $n$ : total orbitals = $n^2 = 9$ ; max electrons = $2n^2 = 18$ .

Q2. What is the electron configuration of $\\\\text\{Cr\}$ (atomic number 24) and why does it deviate from the expected $[\\\\text\{Ar\}]\\\\,3d^4\\\\,4s^2$ ?

A. $[\\\\text\{Ar\}]\\\\,3d^5\\\\,4s^1$ : half-filled $d$ subshell provides extra exchange energy stabilization B. $[\\\\text\{Ar\}]\\\\,3d^4\\\\,4s^2$ : no deviation occurs C. $[\\\\text\{Ar\}]\\\\,3d^6$ : the $4s$ orbital is completely empty D. $[\\\\text\{Ar\}]\\\\,4s^2\\\\,3d^4$ : no deviation; the order $4s$ before $3d$ is always followed’,

Answer: A

Answer: A — Cr has $[\\\\text\{Ar\}]\\\\,3d^5\\\\,4s^1$ instead of $[\\\\text\{Ar\}]\\\\,3d^4\\\\,4s^2$ . The half-filled $d^5$ configuration ( $d$ orbitals each with one electron, maximizing exchange energy) provides extra stability. Similarly, Cu ( $Z=29$ ) is $[\\\\text\{Ar\}]\\\\,3d^\{10\}\\\\,4s^1$ (fully filled $d$ ). These are the two main exceptions to the Aufbau order in the first transition series.

Q3. The effective nuclear charge $Z_\{\\\\text\{eff\}\}$ experienced by a valence electron in sodium ( $\\\\text\{Na\}$ ) is approximately:

A. $Z_\{\\\\text\{eff\}\} \\\\approx 2.5$ (the 3s electron is partially shielded by 10 inner electrons) B. $Z_\{\\\\text\{eff\}\} = 11$ (no shielding) C. $Z_\{\\\\text\{eff\}\} \\\\approx 10$ (almost fully shielded) D. $Z_\{\\\\text\{eff\}\} = 1$ (completely shielded)’,

Answer: A

Answer: A — $Z_\{\\\\text\{eff\}\} = Z - S$ where $S$ is the shielding constant. For Na 3s: $Z = 11$ , $S \\\\approx 8.5$ (Slater’s rules: the 10 core electrons (1s $^2$ 2s $^2$ 2p $^6$ ) contribute fully, same-shell electrons partially). So $Z_\{\\\\text\{eff\}\} \\\\approx 2.5$ . This is why valence electrons are loosely held and Na is highly reactive (low ionization energy).

Q4. Which set of four quantum numbers $(n, \\\\ell, m_\\\\ell, m_s)$ is valid for an electron in a $4d$ orbital?

A. $(4, 3, 0, -1/2)

Answer: D

Answer: D — For a $4d$ orbital: $n=4$ , $\\\\ell=2$ (d orbital), $m_\\\\ell$ ranges from $-\\\\ell$ to $+\\\\ell$ = $\\\{-2,-1,0,+1,+2\\\}$ , $m_s = \\\\pm 1/2$ . $(4,2,-1,+1/2)$ is valid. $(4,3,0,-1/2)$ has $\\\\ell=3$ (f orbital, not d). $(4,2,3,+1/2)$ has $m_\\\\ell=3$ > $\\\\ell=2$ (invalid). $(4,2,0,+1)$ has $m_s=+1$ (invalid; must be $\\\\pm 1/2$ ).

</details>,‘ $(4, 2, 3, +1/2)\} correctAnswer=\{0\} explanation="For a$ 4d $orbital:$ n=4 $,$ \ell=2 $(d orbital),$ m_\ell $ranges from$ -\ell $to$ +\ell $=$ {-2,-1,0,+1,+2} $,$ m_s = \pm 1/2 $.$ (4,2,-1,+1/2) $is valid.$ (4,3,0,-1/2) $has$ \ell=3 $(f orbital, not d).$ (4,2,3,+1/2) $has$ m_\ell=3 $>$ \ell=2 $(invalid).$ (4,2,0,+1) $has$ m_s=+1 $(invalid; must be$ \pm 1/2 $)."\} difficulty="easy" />,'$ (4, 2, 0, +1)} correctAnswer={0} explanation=“For a $4d$ orbital: $n=4$ , $\\ell=2$ (d orbital), $m_\\ell$ ranges from $-\\ell$ to $+\\ell$ = $\{-2,-1,0,+1,+2\}$ , $m_s = \\pm 1/2$ . $(4,2,-1,+1/2)$ is valid. $(4,3,0,-1/2)$ has $\\ell=3$ (f orbital, not d). $(4,2,3,+1/2)$ has $m_\\ell=3$ > $\\ell=2$ (invalid). $(4,2,0,+1)$ has $m_s=+1$ (invalid; must be $\\pm 1/2$ ).”} difficulty=“easy” />,‘ $(4, 2, -1, +1/2)\} correctAnswer=\{0\} explanation="For a$ 4d $orbital:$ n=4 $,$ \ell=2 $(d orbital),$ m_\ell $ranges from$ -\ell $to$ +\ell $=$ {-2,-1,0,+1,+2} $,$ m_s = \pm 1/2 $.$ (4,2,-1,+1/2) $is valid.$ (4,3,0,-1/2) $has$ \ell=3 $(f orbital, not d).$ (4,2,3,+1/2) $has$ m_\ell=3 $>$ \ell=2 $(invalid).$ (4,2,0,+1) $has$ m_s=+1 $(invalid; must be$ \pm 1/2 $)."\} difficulty="easy" />]\} correctAnswer=\{0\} explanation="For a$ 4d $orbital:$ n=4 $,$ \ell=2 $(d orbital),$ m_\ell $ranges from$ -\ell $to$ +\ell $=$ {-2,-1,0,+1,+2} $,$ m_s = \pm 1/2 $.$ (4,2,-1,+1/2) $is valid.$ (4,3,0,-1/2) $has$ \ell=3 $(f orbital, not d).$ (4,2,3,+1/2) $has$ m_\ell=3 $>$ \ell=2 $(invalid).$ (4,2,0,+1) $has$ m_s=+1 $(invalid; must be$ \pm 1/2$).”} difficulty=“easy” />

Q5. The first ionization energy of nitrogen is higher than that of oxygen. This is because:

A. “Nitrogen has a half-filled $2p^3$ subshell (exchange energy stabilization), while oxygen’s $2p^4$ has one paired electron”, ‘Nitrogen has a smaller atomic radius B. Oxygen has a noble gas electron configuration C. Nitrogen has fewer electrons’,

Answer: A

Answer: A — N: $2s^2 2p^3$ (half-filled $p$ subshell) is stabilized by exchange energy (electrons prefer parallel spins in different orbitals). O: $2s^2 2p^4$ has one paired electron in a $p$ orbital, adding electron-electron repulsion. The pairing energy penalty makes O’s $IE_1$ lower than N’s. This is an exception to the general trend of increasing IE across a period.

Periodic Trends and Atomic Properties

Q6. Moving down Group 17 (the halogens), which trend correctly describes atomic radius, ionization energy, and electronegativity?

A. Atomic radius increases; ionization energy decreases; electronegativity decreases B. Atomic radius decreases; ionization energy increases; electronegativity increases C. Atomic radius increases; ionization energy increases; electronegativity decreases D. All three properties increase’,

Answer: A

Answer: A — Down Group 17: each successive element adds a principal quantum shell, so atomic radius increases. The increased distance from the nucleus and additional shielding reduce the effective nuclear charge felt by valence electrons, so IE and electronegativity decrease. F (smallest radius, highest IE and EN) > Cl > Br > I > At. This trend is observed for all main groups.

Q7. Lanthanide contraction refers to the gradual decrease in ionic radii across the lanthanide series (La to Lu). This occurs because:

A. The 4f electrons provide poor shielding of the increasing nuclear charge, causing the outer electrons to be drawn closer B. The atomic mass decreases across the series C. The number of protons decreases D. The principal quantum number decreases’,

Answer: A

Answer: A — Across the lanthanides, $Z$ increases by 1 per element while electrons are added to the inner 4f subshell. 4f orbitals are diffuse and provide poor shielding. The increased $Z$ with poor shielding compresses all orbitals (including the outer 6s), causing ionic radii to decrease from La $^\{3+\}$ to Lu $^\{3+\}$ . This makes transition metals after the lanthanides (Hf onwards) have similar sizes to their period 5 counterparts.

Q8. Which property of an atom cannot be predicted from the hydrogen wavefunctions alone (requires consideration of electron-electron interactions)?

A. The exact energy levels of multi-electron atoms (the hydrogen solution only gives approximate values for $Z_\{\\\\text\{eff\}\}$ ) B. The number of orbitals in each shell C. The allowed quantum numbers D. The number of electrons in each subshell’,

Answer: A

Answer: A — The hydrogen atom is exactly solvable because it has only one electron. For multi-electron atoms, the Schrodinger equation cannot be solved analytically due to electron-electron repulsion. Approximations (Hartree-Fock, density functional theory) are used. Quantum numbers and orbital counts follow from the hydrogen solution, but energy levels depend on $Z_\{\\\\text\{eff\}\}$ .

Q9. The electron affinity of fluorine is less negative than that of chlorine. This anomaly is attributed to:

A. The small size of F causing significant electron-electron repulsion in the compact $2p$ subshell B. Fluorine having a higher ionization energy C. Chlorine being a liquid at room temperature D. Fluorine having fewer valence electrons’,

Answer: A

Answer: A — Electron affinity is the energy released when an atom gains an electron. F has a very compact $2p$ subshell (no $d$ orbitals); adding an electron causes significant inter-electron repulsion. Cl has a larger $3p$ subshell that better accommodates the extra electron with less repulsion. Thus $\\\\text\{EA\}(\\\\text\{Cl\}) < \\\\text\{EA\}(\\\\text\{F\})$ (more negative = more energy released).

Q10. Penetration and shielding effects explain why the $4s$ orbital is filled before the $3d$ orbital despite having a higher principal quantum number. Which orbital penetrates closer to the nucleus?

A. $4s$ (lower angular momentum $\\\\ell=0$ , no radial nodes, higher probability near nucleus) B. $3d$ (higher angular momentum, deeper penetration) C. Both penetrate equally D. Neither penetrates the core; they are both outer orbitals’,

Answer: A

Answer: A — Lower $\\\\ell$ orbitals (s < p < d < f) have less angular momentum and penetrate closer to the nucleus (spend more time near the nucleus). The $4s$ orbital has no angular nodes ( $\\\\ell=0$ ) and penetrates through inner shells better than $3d$ ( $\\\\ell=2$ , 2 angular nodes). This gives $4s$ lower energy than $3d$ for $Z \\\\approx 20$ , but for higher $Z$ , $3d$ drops below $4s$ .

Intuition

Atomic structure is the foundation of all chemistry: The arrangement of electrons in atoms determines chemical bonding, reactivity, and the properties of every element. Periodic trends (electronegativity, ionisation energy, atomic radius) emerge directly from electron configuration.

Why it matters: Understanding periodic trends lets you predict how elements will behave without memorising each one individually. It explains why halogens are reactive, why noble gases are inert, and why transition metals form coloured compounds.

The key insight: The periodic table is not just a chart — it is a map of electron configurations. Each period adds a new electron shell, and each group has the same valence electron count.

Common Mistakes

Assuming orbital filling order is always 1s, 2s, 2p, 3s, 3p, 4s, 3d: The Madelung rule (n + ℓ filling) works for most elements but fails for Cr ([Ar] 3d⁵ 4s¹, not 3d⁴ 4s²) and Cu ([Ar] 3d¹⁰ 4s¹, not 3d⁹ 4s²). These exceptions arise because half-filled and fully-filled d subshells have extra stability.

Confusing penetration with shielding: Penetration is how close an orbital gets to the nucleus (s > p > d > f). Shielding is how effectively inner electrons reduce the effective nuclear charge felt by outer electrons. Both affect orbital energy, but they are different phenomena. Lower ℓ orbitals penetrate better AND are less shielded.

Using $n$ as the sole determinant of orbital energy: For multi-electron atoms, orbital energy depends on both $n$ and ℓ due to shielding and penetration effects. The 4s orbital is lower in energy than 3d for Z ≈ 20 because 4s penetrates closer to the nucleus. For higher Z, 3d drops below 4s. Do not assume energy increases strictly with $n$ .

Cross-References

Atomic Structure and Periodicity: Detailed notes on electron configuration and periodic trends.
Coordination Chemistry: Covers transition metal complexes that build on atomic structure concepts.
Physical Chemistry: Quantum Chemistry: Quantum mechanical treatment of atomic and molecular orbitals.

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