Atomic Structure — Practice Questions with Answers

Rutherford's model proposed that an atom consists of a tiny, dense, positively charged nucleus at the centre, containing almost all the atom's mass, with electrons revolving around it in the surrounding mostly-empty space, similar to planets orbiting the Sun.

Limitation: According to classical electromagnetic theory, an accelerating (orbiting) charged particle like an electron should continuously radiate energy, causing it to spiral inward and collapse into the nucleus almost instantly — this model could not explain why atoms are actually stable.

Principal quantum number ($n$): Determines the shell and overall size/energy of the orbital.

Azimuthal quantum number ($l$): Determines the subshell and shape of the orbital ($l=0,1,2,3$ for s,p,d,f).

Magnetic quantum number ($m_l$): Determines the specific orientation of the orbital in space.

Spin quantum number ($m_s$): Describes the intrinsic spin of the electron, taking value $+1/2$ or $-1/2$.

The Pauli exclusion principle states that no two electrons in the same atom can have the same set of all four quantum numbers.

Significance: This principle directly limits the maximum number of electrons that can occupy any given orbital to exactly two (with opposite spins), since $n$, $l$, and $m_l$ define a specific orbital, leaving only $m_s$ (two possible values) to distinguish two electrons within that same orbital — this is the fundamental reason each orbital holds a maximum of two electrons.

Electronic configuration of chlorine (Z=17): $1s^2\,2s^2\,2p^6\,3s^2\,3p^5$

The Aufbau principle states electrons fill available orbitals in order of increasing energy — following this order (1s before 2s, 2s before 2p, and so on, according to the $(n+l)$ rule where applicable), 17 electrons are progressively added to the lowest-energy available orbitals, resulting in this specific configuration.

• Electrons revolve around the nucleus only in specific, allowed circular orbits without radiating energy.
• Only orbits with quantised angular momentum, $L=n\dfrac{h}{2\pi}$, are permitted.
• Energy is absorbed or emitted (as a photon) only when an electron transitions between these allowed orbits, with photon energy exactly equal to the energy difference between orbits.

De Broglie proposed that particles of matter, including electrons, exhibit wave-like properties in addition to their particle-like properties — every moving particle has an associated wavelength given by $\lambda = h/mv$.

This dual nature (both particle and wave characteristics) was later experimentally confirmed for electrons through the Davisson-Germer experiment, which demonstrated electron diffraction — a phenomenon only explainable if electrons behave as waves under appropriate conditions, providing crucial support for quantum mechanical treatment of atomic structure.

An orbital is a three-dimensional region of space around the nucleus where the probability of finding an electron is significantly high, described mathematically by a wave function derived from quantum mechanics.

This differs fundamentally from Bohr's "orbit" concept, which described a fixed, well-defined circular path with a precisely known radius and electron position at any instant — an orbital, by contrast, gives only a probability distribution (electron cloud), reflecting the inherent quantum mechanical uncertainty in simultaneously knowing an electron's exact position and momentum.

A half-filled ($d^5$) or fully-filled ($d^{10}$) subshell configuration provides extra stability, due to a combination of symmetric electron distribution and favourable exchange energy between electrons with parallel spins.

For chromium, shifting one electron from the 4s orbital into the 3d subshell (achieving the extra-stable half-filled $3d^5$ configuration, at the cost of a less-stable, singly-occupied $4s^1$) results in an overall lower-energy, more stable configuration than the "expected" $3d^44s^2$ arrangement — this exception to the simple Aufbau prediction is precisely due to this additional stability associated with half-filled subshells.

Planck's quantum theory proposed that energy is not emitted or absorbed continuously, but rather in discrete packets called "quanta." The energy of each quantum is directly proportional to the frequency of the radiation: $E=h\nu$, where $h$ is Planck's constant. Energy can only be emitted or absorbed as whole-number multiples of this fundamental quantum ($E=nh\nu$), not as any arbitrary continuous value.

According to Bohr's model, electrons in a hydrogen atom can only occupy specific, discrete (quantised) energy levels — not a continuous range of energies. When an excited electron transitions from a higher energy level to a lower one, it emits a photon whose energy exactly matches this specific, discrete energy difference.

Since only certain specific energy differences (and hence specific photon frequencies/wavelengths) are possible between the atom's quantised energy levels, only a discrete, limited set of specific wavelengths are emitted — producing distinct, separated spectral lines rather than a continuous spread of all possible wavelengths.

$E_2 = \dfrac{-13.6}{2^2} = \dfrac{-13.6}{4} = -3.4\text{ eV}$

Converting to joules: $E_2 = -3.4\times1.6\times10^{-19}$

$E_2 = \mathbf{-5.44\times10^{-19}\text{ J}}$

Following the Aufbau principle, filling orbitals in order of increasing energy for 26 electrons:

$1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^6$

Verification: $2+2+6+2+6+2+6 = \mathbf{26}$ electrons total ✓, which can also be written as $[Ar]3d^64s^2$.

Kinetic energy gained: $KE = qV = 1.6\times10^{-19}\times200 = 3.2\times10^{-17}\text{ J}$

Using $KE = p^2/2m$: $p = \sqrt{2mKE} = \sqrt{2\times9.1\times10^{-31}\times3.2\times10^{-17}}$

$p = \sqrt{5.824\times10^{-47}} \approx 7.632\times10^{-24}\text{ kg·m/s}$

$\lambda = \dfrac{h}{p} = \dfrac{6.63\times10^{-34}}{7.632\times10^{-24}} \approx \mathbf{8.69\times10^{-11}\text{ m} = 0.0869\text{ nm}}$

$E_4 = -13.6/16 = -0.85\text{ eV}$; $E_2 = -13.6/4 = -3.4\text{ eV}$

Energy released: $\Delta E = E_4-E_2 = -0.85-(-3.4) = 2.55\text{ eV}$

Converting to joules: $\Delta E = 2.55\times1.6\times10^{-19} = 4.08\times10^{-19}\text{ J}$

$\lambda = \dfrac{hc}{\Delta E} = \dfrac{6.63\times10^{-34}\times3\times10^8}{4.08\times10^{-19}} \approx \mathbf{4.875\times10^{-7}\text{ m} = 487.5\text{ nm}}$

$r_3 = 0.529\times3^2 = 0.529\times9$

$r_3 = \mathbf{4.761\text{ Å} = 4.761\times10^{-10}\text{ m}}$

Number of subshells = $n$ = 4 (i.e., 4s, 4p, 4d, 4f)

Number of orbitals = $n^2$ = $4^2$ = 16 (1 for s, 3 for p, 5 for d, 7 for f: $1+3+5+7=16$)

Maximum electrons = $2n^2$ = $2(16)$ = $\mathbf{32\text{ electrons}}$

At $n=\infty$, $E_\infty = 0$ (since $-13.6/\infty^2 \to 0$). Energy released: $\Delta E = E_\infty - E_1 = 0-(-13.6) = 13.6\text{ eV}$

Converting to joules: $\Delta E = 13.6\times1.6\times10^{-19} = 2.176\times10^{-18}\text{ J}$

$\lambda = \dfrac{hc}{\Delta E} = \dfrac{6.63\times10^{-34}\times3\times10^8}{2.176\times10^{-18}}$

$\lambda \approx \mathbf{9.14\times10^{-8}\text{ m} = 91.4\text{ nm}}$ (this is the Lyman series limit, in the ultraviolet region)

Nitrogen's electronic configuration is $1s^22s^22p^3$ — the last (differentiating) electron enters the 2p subshell.

By Hund's rule, the three 2p electrons individually occupy the three separate p-orbitals ($p_x, p_y, p_z$) with parallel spin — the third electron enters the last of these three orbitals.

Quantum numbers for this third 2p electron: $n=2$, $l=1$ (for p subshell), $m_l = +1$ (by convention, the third of the three orbitals, though $-1,0,+1$ ordering can vary by convention), $m_s = +1/2$ (unpaired, parallel spin).

$E_3 = -13.6/9 \approx -1.511\text{ eV}$; $E_1 = -13.6\text{ eV}$

Energy released: $\Delta E = E_3-E_1 = -1.511-(-13.6) = 12.089\text{ eV}$

Converting to joules: $\Delta E = 12.089\times1.6\times10^{-19} \approx 1.934\times10^{-18}\text{ J}$

Frequency: $f = \dfrac{\Delta E}{h} = \dfrac{1.934\times10^{-18}}{6.63\times10^{-34}} \approx \mathbf{2.92\times10^{15}\text{ Hz}}$

Element identification: Number of protons (atomic number, Z) = 12, which corresponds to magnesium (Mg).

Mass number: $A = \text{protons}+\text{neutrons} = 12+12 = \mathbf{24}$

Charge on ion: Since neutral magnesium has 12 electrons (equal to its 12 protons), but this ion has only 10 electrons, it has lost 2 electrons — giving a charge of $\mathbf{+2}$, i.e., $Mg^{2+}$.