"Atomic Structure" — Practice

Question 1

Find the wavelength of the transition $n=4	o2$ in $	ext{He}^{+}$, and state which hydrogen transition it coincides with.

Accepted Answer

$\Delta E=13.6	imes4\left(\dfrac14-\dfrac{1}{16}ight)=54.4	imes0.1875=10.2$ eV

$\lambda=\dfrac{1240}{10.2}=121.6$ nm

This is exactly the hydrogen Lyman-alpha wavelength, because $Z^{2}\left(\dfrac{1}{n_1^{2}}-\dfrac{1}{n_2^{2}}ight)$ takes the same value for both.

Such coincidences are common between hydrogen-like species whose quantum numbers scale with $Z$.

Question 2

Hydrogen atoms are excited to $n=5$. Find the number of emission lines and how many of them fall in the Balmer series.

Accepted Answer

Total lines $=\dfrac{5	imes4}{2}=10$

Balmer lines end at $n=2$, so they come from $n=3,4,5$: **three** lines.

The remaining seven are four Lyman lines (from $n=2,3,4,5$), two Paschen lines and one Brackett line.

Adding these gives $4+3+2+1=10$, confirming the total.

Question 3

For the $5d$ orbital, find the number of radial nodes, angular nodes and total nodes, and compare with $4f$.

Accepted Answer

$5d$: $n=5$, $l=2$. Radial $=5-2-1=2$; angular $=2$; total $=4$.

$4f$: $n=4$, $l=3$. Radial $=4-3-1=0$; angular $=3$; total $=3$.

Both totals equal $n-1$, as required.

$4f$ has no radial node at all, which is characteristic of the first orbital of any new $l$ value.

Question 4

How many electrons in an atom can have $n=4$ and $m_l=-1$? How many can have $n=3$, $l=1$ and $m_s=+	frac12$?

Accepted Answer

$m_l=-1$ requires $l\ge1$, so with $n=4$ the possibilities are $l=1,2,3$: one orbital each, three orbitals, holding $6$ electrons.

$n=3$, $l=1$ is the $3p$ subshell with three orbitals. Restricting to one spin gives one electron per orbital: $3$ electrons.

The second form is the simpler one because both $l$ and the spin are pinned.

Question 5

State the nodal planes of $3d_{x^{2}-y^{2}}$ and explain why it is destabilised in an octahedral ligand field.

Accepted Answer

The function is proportional to $x^{2}-y^{2}$, which vanishes on the planes $y=x$ and $y=-x$. Those are its two angular nodes.

Its lobes therefore lie **along** the $x$ and $y$ axes.

In an octahedral field the six ligands sit on the axes, so electrons in this orbital are pushed directly against them and are repelled strongly. It rises in energy along with $d_{z^{2}}$, forming the upper pair of the octahedral splitting.

Question 6

Calculate the effective nuclear charge experienced by a $3p$ electron in chlorine ($Z=17$).

Accepted Answer

Configuration: $\left(1s^{2}ight)\left(2s^{2}2p^{6}ight)\left(3s^{2}3p^{5}ight)$

Same group, excluding the electron itself: $6$ others, contributing $0.35	imes6=2.10$

$\left(n-1ight)$ shell: $8$ electrons at $0.85$ each $=6.80$

Deeper: $2$ electrons at $1.00$ each $=2.00$

$\sigma=10.90$, so $Z_{eff}=17-10.90=6.10$

Question 7

Write the ground-state configurations of $	ext{Cr}$, $	ext{Cr}^{3+}$, $	ext{Cu}$ and $	ext{Cu}^{2+}$, and give the unpaired-electron count for each.

Accepted Answer

$	ext{Cr}$: $\left[	ext{Ar}ight]3d^{5}4s^{1}$ — **six** unpaired

$	ext{Cr}^{3+}$: $\left[	ext{Ar}ight]3d^{3}$ — three unpaired

$	ext{Cu}$: $\left[	ext{Ar}ight]3d^{10}4s^{1}$ — one unpaired

$	ext{Cu}^{2+}$: $\left[	ext{Ar}ight]3d^{9}$ — one unpaired

Chromium's six unpaired electrons make it the most paramagnetic atom in the first transition series.

Question 8

Explain, using exchange energy, why chromium adopts $3d^{5}4s^{1}$ rather than $3d^{4}4s^{2}$.

Accepted Answer

In $3d^{4}$ there are four parallel electrons, giving $\dfrac{4	imes3}{2}=6$ exchange pairs.

In $3d^{5}$ there are five, giving $\dfrac{5	imes4}{2}=10$ pairs.

The gain of four exchange pairs exceeds the cost of promoting one electron from $4s$ to $3d$, which are in any case very close in energy.

The half-filled set is also spherically symmetric, lowering electron-electron repulsion further.

Question 9

An octahedral complex of a first-row transition metal has a spin-only magnetic moment of $4.90$ BM. Find the number of unpaired electrons and suggest two possible ions.

Accepted Answer

$n\left(n+2ight)=4.90^{2}=24\qquad\Rightarrow\qquad n=4$

Four unpaired electrons in a $3d$ set arise from $d^{4}$ or $d^{6}$ in the high-spin case.

Candidates: $	ext{Cr}^{2+}$ ($d^{4}$) or $	ext{Fe}^{2+}$ ($d^{6}$).

The moment alone cannot distinguish them; the colour or the oxidation-state chemistry must settle it.

Question 10

Arrange $	ext{Mn}^{2+}$, $	ext{Fe}^{3+}$, $	ext{Ni}^{2+}$ and $	ext{Ti}^{4+}$ in order of increasing spin-only magnetic moment.

Accepted Answer

$	ext{Ti}^{4+}$: $3d^{0}$, $n=0$, $\mu=0$

$	ext{Ni}^{2+}$: $3d^{8}$, $n=2$, $\mu=2.83$ BM

$	ext{Mn}^{2+}$: $3d^{5}$, $n=5$, $\mu=5.92$ BM

$	ext{Fe}^{3+}$: $3d^{5}$, $n=5$, $\mu=5.92$ BM

Order: $	ext{Ti}^{4+}<	ext{Ni}^{2+}<	ext{Mn}^{2+}=	ext{Fe}^{3+}$, with the last two identical because both are $d^{5}$.

Question 11

An electron and a proton are each confined to a region of $0.1$ nm. Compare their minimum velocity uncertainties.

Accepted Answer

$\Delta v\ge\dfrac{h}{4\pi m\Delta x}$

Electron: $\dfrac{6.63	imes10^{-34}}{4\pi\left(9.11	imes10^{-31}ight)\left(10^{-10}ight)}=5.8	imes10^{5}$ m s$^{-1}$

Proton: smaller by the mass ratio $1836$, giving $3.2	imes10^{2}$ m s$^{-1}$

The electron's uncertainty is comparable with its own orbital speed, while the proton's is negligible — which is why nuclei can be treated as fixed points when describing atomic structure.

Question 12

Find the de Broglie wavelength of an electron in the second Bohr orbit of hydrogen, and verify that a whole number of wavelengths fits the orbit.

Accepted Answer

$v_2=\dfrac{2.18	imes10^{6}}{2}=1.09	imes10^{6}$ m s$^{-1}$

$\lambda=\dfrac{6.63	imes10^{-34}}{\left(9.11	imes10^{-31}ight)\left(1.09	imes10^{6}ight)}=6.68	imes10^{-10}$ m

$r_2=0.529	imes4=2.12$ Å, so the circumference is $2\pi\left(2.12	imes10^{-10}ight)=1.33	imes10^{-9}$ m

$\dfrac{1.33	imes10^{-9}}{6.68	imes10^{-10}}=2$ — exactly two wavelengths, as the quantisation condition requires.

Practice — "Atomic Structure"

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