Atomic Structure Suggested Solutions
Topic 2 : Atomic Structure AJC/P2/Q1a 1. The first ionisation energy of aluminium will be lower than that of sulphur, as S has a higher nuclear charge than Al, but the shielding effect is similar for both atoms. Hence, the valence electrons of S experience a stronger attraction by the nucleus, and require more energy to be ionised. CJC/P2/Q1 2. (a)(i) Mg+ (g) Mg2+ (g) + e(ii)
The second ionisation energy involves removal of an electron from the 3p subshell from Si+, and the 3s subshell from Al+. Since the 3p subshell is on average further from the nucleus / at a higher energy level / experiences more shielding than the 3s subshell, the electron removed from Si+ is less tightly held and requires less energy to remove.
(a)
Neon, being already in the gaseous state, is ionised first and emits light, whereas sodium is in the solid state and takes time to vaporise before it can be ionised. OR Neon has a higher first ionisation energy than sodium. The initial power surge when the light is turned on is able to ionize neon, giving rise to the red light, after which the power drops and is only enough to ionise sodium, giving rise to the orange light.
CJC/P3/Q1c 3. (i) P3– has a larger radius than S2–. P3– has a smaller nuclear charge but the same number of electrons as S2–, hence the electrons experience less attraction to the nucleus and spread out over a larger volume. (ii)
State: PCl5 or phosphorus in oxidation state +5 Phosphorus is able to expand its octet to accommodate 5 electron pairs as it is in Period 3 and thus has empty 3d orbitals of similar energy, whereas nitrogen, being in Period 2, does not have suitable orbitals of similar energy for this purpose.
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Atomic Structure Suggested Solutions
AJC/P2/Q2d 4. The Cl- anion has the same nuclear charge and one extra electron compared to the Cl atom. Thus, there is more repulsion between the electrons in Cl-, and a decrease in effective nuclear charge due to the increased shielding in Cl-. Hence, the anionic radius of chlorine is larger than its atomic radius.
NYJC/P3/Q1a(i) 6. Element A The sharp drop in value from G to H indicates that G is from Group I where the second ionization energy involves the removal of an electron from the inner shell. Hence Element A is in Group III and is Aluminium.
RI/P2/Q1 7. (a)(i) 1s2 2s2 2p5 (a)(ii) z y x
(b)(i) 1st I.E of O = 1310 kJmol-1 ; 1st I.E of F = 1680 kJmol-1 O: 1s22s22p4 The 1st IE increases from O to F as the number of protons increases in F but the shielding effect remains effectively constant. (b)(ii) 2nd I.E of O = 3390 kJmol-1 ; 2nd I.E of F = 3370 kJmol-1 O+: 1s22s22p3 F+: 1s22s22p4 The 2nd IE decreases from O to F due to inter–electronic repulsion between the 2p electrons in F
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Atomic Structure Suggested Solutions
SRJC/P2/Q1 8(a) (i)
Across the period from Na to Cl, • • •
(ii)
(b)
nuclear charge increases but increase in shielding effect is negligible as electron is added to the same quantum shell; Hence, the effective nuclear charge increases OR the electrostatic forces of attraction between the nucleus and the outermost/ valence electrons increases across the period. Therefore, the atomic radius decreases across the period from Na to Cl. Na+ has one electron less than Na or Na+ has one quantum shell less than Na. Nuclear charge remains the same. The electrostatic forces of attraction between the nucleus and the remaining electrons are greater for Na+.
(i)
(ii)
The anionic radii of P3–, S2– and Cl– are larger than their respective atoms. This is due to electrons are being added to the same valence shell, resulting in greater inter-electronic repulsions between the valence electrons. Across the period from P3–, S2– and Cl–, the anionic radii decrease. The nuclear charge increases across the period with negligible increase in shielding effect resulting in the corresponding increase in the effective nuclear charge OR electrostatic forces of attraction between the nucleus and the valence electrons.
(c)
(i)
1s22s22p63s23p4 or [Ne] 3s23p4
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Atomic Structure Suggested Solutions
(ii)
(d)
Sulfur is in Period 3 and hence has available 3d orbitals to expand beyond its octet structure while oxygen cannot expand its octet due to the absence of d orbitals. 740,
1500,
760
7700,
6200
10500, 2800
13600,
3100
4400
18000,
21700
3700 (difference in IE)
- Element A belongs to Group II - Largest energy difference between the 2nd and 3rd IE. - Removal of the 3rd electron is from an inner quantum shell which requires more energy. Formula: ACl2 TJC/P2/Q3 9 (a) Element B belongs to Group V. There is a big jump in I.E. between the 5th and 6th electron removed which implies that the 6th electron removed lies in an inner quantum shell and is more strongly attracted by the nucleus. Hence more energy is required to remove the 6th electron. (b) Element A (Group IV) and B (Group V): The nuclear charge of B is higher than that of A, while their screening effect is approximately the same since both have the same number of occupied quantum shells. The increase in nuclear charge outweighs the increase in screening effect resulting in a higher effective nuclear charge for B. The valence electron to be removed from B is more strongly attracted by the nucleus, therefore it has a higher 1st I.E. than A. Element B (Group V) and C (Group VI): The valence electron to be removed from C is one of the paired electrons in the p orbital. It experiences inter-electron repulsion, therefore less energy is required to remove it. TPJC/P3/Q1b 10. Cr: [Ar]3d54s1 Al: [Ne]3s23p1 Atomic radius of chromium is smaller than that of aluminium as it has higher nuclear charge and 3d orbitals provide poor shielding effective nuclear charge higher hence radius smaller.
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Atomic Structure Suggested Solutions
Cr3+: [Ar]3d3 Al3+: [Ne]3s0 Cr3+ has one more quantum shell compared to Al3+, hence the bigger ionic radius.
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