DSE Chemistry Diagnostic: Atomic Structure

Unit Test 1: Transition Metal Electron Configuration

Question

The element chromium (atomic number 24) has the electron configuration $[Ar]\,3d^{5}4s^{1}$ rather than $[Ar]\,3d^{4}4s^{2}$.

(a) Explain why chromium adopts this configuration. [3 marks]

(b) Write the electron configuration of $Cr^{3+}$. [1 mark]

(c) A student claims that $Cr^{3+}$ has the same electron configuration as the scandium atom ($Sc$, $Z = 21$). Determine whether this claim is correct, and explain your reasoning. [2 marks]

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Worked Solution

(a) The half-filled $3d$ subshell ($3d^{5}$) and the singly-occupied $4s$ orbital provide extra stability due to symmetrical electron distribution and exchange energy (the energy lowering from parallel spins in degenerate orbitals). The energy gained from having five unpaired electrons with parallel spins in the $3d$ subshell outweighs the small energy difference between $3d$ and $4s$ orbitals.

(b) When forming $Cr^{3+}$, electrons are removed from the $4s$ orbital first (since $4s$ is at a higher energy than $3d$ once the atom is ionised), then from $3d$:

$Cr^{3+}: [Ar]\,3d^{3}$

(c) Scandium atom: $Sc = [Ar]\,3d^{1}4s^{2}$ (atomic number 21).

$Cr^{3+} = [Ar]\,3d^{3}$.

These are not the same. The student's claim is incorrect. While both are $3d$ transition metal species, $Cr^{3+}$ has three $3d$ electrons whereas $Sc$ has one $3d$ electron and two $4s$ electrons.

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Unit Test 2: Successive Ionisation Energies

Question

The table below shows the first six successive ionisation energies of an element $X$.

Ionisation Energy	Value (kJ/mol)
1st	578
2nd	1817
3rd	2745
4th	11577
5th	14842
6th	18379

(a) Identify element $X$ and explain your reasoning. [3 marks]

(b) Explain why there is a large jump between the 3rd and 4th ionisation energies. [2 marks]

(c) The 4th ionisation energy of $X$ (11577 kJ/mol) is lower than the 4th ionisation energy of silicon (16091 kJ/mol). Explain this observation in terms of nuclear charge and electron shielding. [3 marks]

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Worked Solution

(a) Element $X$ is aluminium ($Al$, $Z = 13$).

Reasoning: There are three relatively low ionisation energies (578, 1817, 2745) followed by a large jump to 11577. This indicates that $X$ has three valence electrons in the outermost shell. The electron configuration is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}$, which corresponds to aluminium.

(b) The first three electrons are removed from the third shell ($3s$ and $3p$). The 4th electron must be removed from the second shell ($2p$), which is much closer to the nucleus and experiences a much greater effective nuclear charge (significantly less shielding from inner electrons). Hence the large jump.

(c) Aluminium's 4th IE removes an electron from $2p$ ($1s^{2}2s^{2}2p^{5}$ after removal). Silicon's 4th IE also removes from $2p$ ($1s^{2}2s^{2}2p^{3}$ after removal).

Silicon ($Z = 14$) has a greater nuclear charge than aluminium ($Z = 13$). Both have the same number of inner shell electrons ($1s^{2}2s^{2}$), so shielding is essentially identical. Since silicon's $2p$ electrons experience a greater effective nuclear charge, they are held more tightly, requiring more energy to remove.

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Unit Test 3: Periodic Trend Explanation

Question

(a) Explain why the first ionisation energy of phosphorus ($P$) is higher than that of sulphur ($S$), even though sulphur has a greater nuclear charge. [3 marks]

(b) Explain why the first ionisation energy generally increases across Period 3 (Na to Ar). [3 marks]

(c) The atomic radius of aluminium is 143 pm and that of silicon is 117 pm. Explain the large difference in atomic radius between these two elements despite their similar nuclear charges ($Z = 13$ and $Z = 14$). [2 marks]

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Worked Solution

(a) Phosphorus has electron configuration $[Ne]\,3s^{2}3p^{3}$. The three $3p$ electrons occupy separate orbitals (each $3p$ orbital has one electron), minimising electron-electron repulsion.

Sulphur has configuration $[Ne]\,3s^{2}3p^{4}$. The fourth $3p$ electron must pair up in an already-occupied orbital, introducing extra inter-electronic repulsion. This repulsion offsets the effect of sulphur's greater nuclear charge, making the 1st IE of $P$ higher than $S$.

(b) Across Period 3, the nuclear charge increases by one proton per element. The additional electrons enter the same principal energy level ($n = 3$), so they provide poor shielding of each other from the nucleus. The effective nuclear charge therefore increases steadily, pulling electrons closer and making them harder to remove.

(c) The difference arises because:

• Both have the same number of electron shells.
• Silicon has one more proton in the nucleus.
• The shielding is almost identical (same inner electrons).
• The significantly greater effective nuclear charge in Si pulls all electrons inward more strongly, resulting in a notably smaller atomic radius.

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Integration Test 1: Ionisation Energy + Periodic Position

Question

Element $Y$ is in Period 4. The ratio of its 3rd ionisation energy to its 2nd ionisation energy is approximately 8.5, and its 4th ionisation energy is only about 1.4 times its 3rd.

(a) Identify the group of element $Y$. [2 marks]

(b) The first ionisation energy of $Y$ is 419 kJ/mol and that of the element immediately to its right in the periodic table is 403 kJ/mol. Explain why the ionisation energy decreases across this particular boundary. [3 marks]

(c) $Y$ forms a compound $YCl$ that is ionic. Explain why $YCl$ has a higher melting point than $SCl_{2}$ (where $S$ is sulphur), even though both contain chlorine. [3 marks]

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Worked Solution

(a) The large jump from 2nd to 3rd IE (ratio 8.5) indicates that the 3rd electron is removed from a new inner shell. The 4th IE is only 1.4 times the 3rd, meaning the 4th electron is still in the same shell as the 3rd. Therefore $Y$ has two valence electrons and is in Group 2.

(b) $Y$ is in Group 2, Period 4: $Y = Ca$ ($Z = 20$). The element to its right is $Sc$ ($Z = 21$, Group 3, Period 4).

The decrease from Ca ($419$) to Sc ($403$) occurs because:

• Ca has the electron configuration $[Ar]\,4s^{2}$: both $4s$ electrons are paired in the same orbital.
• Sc has the configuration $[Ar]\,3d^{1}4s^{2}$: the $3d$ electron provides poor shielding of the outer $4s$ electrons. However, the key factor is that the 1st IE of Sc removes a $4s$ electron from an atom where the $3d$ subshell is being filled. The $4s$ electron in Sc is effectively at a similar or slightly lower effective nuclear charge than in Ca because the $3d$ electron shields poorly, and the atomic radius of Sc is slightly larger than Ca. The result is a small decrease in 1st IE.

(c) $CaCl_{2}$ (ionic) has strong electrostatic forces between $Ca^{2+}$ and $Cl^{-}$ ions throughout the lattice, requiring a large amount of energy to overcome. $SCl_{2}$ is a simple covalent molecule with only weak London dispersion forces (van der Waals forces) between molecules. Ionic lattice energy far exceeds intermolecular forces, so $CaCl_{2}$ has a much higher melting point.

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Integration Test 2: Electron Configuration + Properties

Question

Consider the following species: $K^{+}$, $Ca^{2+}$, $Cl^{-}$, $Ar$, $S^{2-}$.

(a) Which species has the smallest ionic/atomic radius? Explain your reasoning. [3 marks]

(b) Arrange the five species in order of increasing first ionisation energy. Explain. [4 marks]

(c) $K^{+}$ and $Ar$ are isoelectronic. Explain why $K^{+}$ has a higher ionisation energy than $Ar$ despite having the same number of electrons. [2 marks]

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Worked Solution

(a) All five species are isoelectronic with the electron configuration $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}$ (18 electrons each).

The ionic/atomic radius depends on the nuclear charge:

• $S^{2-}$: $Z = 16$ -- largest radius (weakest attraction)
• $Cl^{-}$: $Z = 17$
• $Ar$: $Z = 18$
• $K^{+}$: $Z = 19$
• $Ca^{2+}$: $Z = 20$ -- smallest radius (strongest attraction)

$Ca^{2+}$ has the smallest radius.

(b) Increasing first ionisation energy:

$S^{2-} \lt Cl^{-} \lt Ar \lt K^{+} \lt Ca^{2+}$

All are isoelectronic, so the nuclear charge determines how tightly the outermost electron is held. Higher nuclear charge = higher ionisation energy.

(c) $K^{+}$ has $Z = 19$ and $Ar$ has $Z = 18$. Both have 18 electrons. Since $K^{+}$ has a greater nuclear charge with the same number of electrons and identical shielding, the effective nuclear charge is greater. The outermost electron in $K^{+}$ is held more tightly, so its ionisation energy is higher.

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Integration Test 3: Multi-Concept Analysis

Question

The graph below (described verbally) shows the first ionisation energies of elements 11--18 (Na to Ar).

• Na: 496 | Mg: 738 | Al: 578 | Si: 786 | P: 1012 | S: 1000 | Cl: 1251 | Ar: 1521

(all values in kJ/mol)

(a) The general trend is increasing ionisation energy, but there are two notable dips. Identify both dips and explain each in terms of electron configuration. [4 marks]

(b) The 2nd ionisation energy of magnesium is 1451 kJ/mol, while the 2nd ionisation energy of aluminium is 1817 kJ/mol. Explain why Mg's 2nd IE is lower than Al's 2nd IE, even though Al has a greater nuclear charge. [3 marks]

(c) A scientist proposes that the 3rd ionisation energy of silicon should be approximately the same as the 1st ionisation energy of phosphorus. Evaluate this proposal quantitatively. [3 marks]

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Worked Solution

(a) Dip 1: Al (578) vs Mg (738): Mg has configuration $[Ne]\,3s^{2}$ (both $3s$ electrons paired). Al has $[Ne]\,3s^{2}3p^{1}$. The $3p$ electron in Al is at a higher energy (further from nucleus on average) and is shielded by the $3s^{2}$ electrons, so it is easier to remove.

Dip 2: S (1000) vs P (1012): P has $3p^{3}$ with three unpaired electrons in separate orbitals (minimum repulsion). S has $3p^{4}$ where the fourth electron pairs in an occupied orbital, increasing inter-electronic repulsion and making removal slightly easier.

(b) Mg's 2nd IE removes the second $3s$ electron from $Mg^{+}$ (configuration $1s^{2}2s^{2}2p^{6}3s^{1}$). This produces $Mg^{2+}$ with a noble gas configuration.

Al's 2nd IE removes a $3p$ electron from $Al^{+}$ (configuration $1s^{2}2s^{2}2p^{6}3s^{2}$).

Although $Mg^{+}$ ($1s^{2}2s^{2}2p^{6}3s^{1}$) has only one valence electron (no pairing repulsion), $Al^{+}$ ($1s^{2}2s^{2}2p^{6}3s^{2}$) has $Z = 13$ and removing one $3s$ electron leaves $Al^{2+}$ with noble gas configuration. The greater nuclear charge of Al more than compensates for the pairing repulsion in the $3s$ orbital, making Al's 2nd IE higher.

(c) Si 3rd IE: removes the third valence electron from $Si^{2+}$ ($1s^{2}2s^{2}2p^{6}3s^{1}$). This requires breaking into the noble gas core, producing $Si^{3+}$ ($1s^{2}2s^{2}2p^{5}$).

P 1st IE: removes one $3p$ electron from P ($[Ne]\,3s^{2}3p^{3}$), producing $P^{+}$ ($[Ne]\,3s^{2}3p^{2}$).

P 1st IE = 1012 kJ/mol. Si 3rd IE $\approx 16091$ kJ/mol.

The proposal is wrong. Si's 3rd IE is enormously higher because it removes an electron from the $n = 3$ shell of a $2+$ ion where the remaining electron already experiences a very high effective nuclear charge (10 core electrons shielding 14 protons). P's 1st IE is from a neutral atom with only 10 protons shielded by 10 core electrons. The difference is approximately 15,000 kJ/mol -- nowhere near "approximately the same."