Learners' AlleyDesigned by Khow Willard

Atomic Structure

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From the lesson

16 marks

species	protons	neutrons	electrons
²³Na⁺	11	12	10	[1]
³²S²⁻	16	16	18	[1]
⁵⁶Fe³⁺	26	30	23	[1]

Exam tipCharge changes only the electrons: number of electrons = proton number − charge. Protons and neutrons never change in chemical reactions.

Angle of deflection = k × chargemass

For ¹H⁺: +12 = k × +11, so k = 12

(a) ²H⁺: 12 × +12 = +6°[1]

(b) ⁴He²⁺: 12 × +24 = +6°[1]

Note²H⁺ and ⁴He²⁺ have the same charge to mass ratio, so they are deflected by the same angle.

% ²⁶Mg = 100 − 79.0 − x = 21.0 − x

A_r = 24(79.0) + 25x + 26(21.0 − x)100 = 24.32[1]

1896 + 25x + 546 − 26x = 2432

x = 10.0[1]

Each atom is ³⁵Cl with probability 0.75 and ³⁷Cl with probability 0.25.

Mass 70 (³⁵Cl–³⁵Cl): 0.75 × 0.75 = 0.5625

Mass 72 (³⁵Cl–³⁷Cl or ³⁷Cl–³⁵Cl): 2 × 0.75 × 0.25 = 0.375

Mass 74 (³⁷Cl–³⁷Cl): 0.25 × 0.25 = 0.0625[1]

Relative abundances 70 : 72 : 74 = 9 : 6 : 1[1]

(a)

1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹[1]

(b)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²[1]

(c)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵[1]

The large jump is between the 3rd and 4th ionisation energies, so Y has 3 electrons in its outermost shell. Group 13, and in Period 3, Y is aluminium.[1]

Group 1: the large jump is after the 1st ionisation energy.[1]

The 2nd electron is removed from an inner principal quantum shell, nearer the nucleus and less shielded, so it is much more strongly attracted.[1]

Practice

22 marks

BThey have 10 neutrons in their nuclei.[1]

¹⁸O²⁻: 8 protons, 10 neutrons, 10 electrons. ¹⁹F⁻: 9 protons, 10 neutrons, 10 electrons.

AWrong: each has as many electrons as neutrons (10). CWrong: both have an outer configuration of 2s² 2p⁶. DWrong: the nucleon numbers are 18 and 19.

CCu⁺[1]

particle	electronic configuration	unpaired electrons
Fe³⁺	[Ar] 3d⁵	5
Ge⁺	[Ar] 3d¹⁰ 4s² 4p¹	1
Cu⁺	[Ar] 3d¹⁰	0
Ti³⁺	[Ar] 3d¹	1

NoteTransition metal ions lose their 4s electrons before their 3d electrons.

BN⁻[1]

particle	before losing an electron	after losing an electron
C⁻	1s² 2s² 2p³	1s² 2s² 2p²
N⁻	1s² 2s² 2p⁴	1s² 2s² 2p³ (half-filled)
O⁺	1s² 2s² 2p³	1s² 2s² 2p²
P⁺	1s² 2s² 2p⁶ 3s² 3p²	1s² 2s² 2p⁶ 3s² 3p¹

species	protons	neutrons	electrons	explanation (not required)
²H	1	1	1	each ²H atom has 1 electron, 1 proton and 1 neutron
¹H₂⁺	2	0	1	each ¹H atom has 1 electron, 1 proton and 0 neutrons
³H₃⁺	3	3 × 2 = 6	2	each ³H atom has 1 electron, 1 proton and 2 neutrons

[1] for any 3 boxes correct[3]

(a)

Si: 1s² 2s² 2p⁶ 3s² 3p²[1]

(b)

Similarity (one of the following):[1]

The 3s and 3p orbital both belong to the third principal quantum shell or the 3s and 3p orbital can each hold a maximum of two electrons (of opposite spins).

Difference (one of the following):[1]

The 3s orbital is spherical while the 3p orbital has a dumb-bell shape or the 3s orbital is lower in energy than the 3p orbital.

NoteVague answers ("both hold electrons", "different energy") were rejected. Compare one 3s orbital with one 3p orbital, not the number of orbitals in each subshell.

(c)

Difference (any one of the following): the 3p orbital is higher in energy than the 2p orbital or the 3p orbital is larger (in size) than the 2p orbital or the 3p orbital belongs to the third quantum shell, while the 2p orbital belongs to the second quantum shell.[1]

(a)

Let the nucleon number of X be m.

1/281/m = 5.62.0[1]

m = 78.4, so X is Br.[1]

(b)

Isotopes of bromine with a lower charge/mass ratio.[1]

(a)

Across the period from sodium to argon, nuclear charge increases. There is the same number of inner shell electrons, so the shielding effect is approximately constant.[1]

Effective nuclear charge increases, so the electrostatic forces of attraction between the nucleus and the valence electron increase. More energy is needed to remove the valence electron.[1]

(b)

Mg: 1s² 2s² 2p⁶ 3s² Al: 1s² 2s² 2p⁶ 3s² 3p¹[1]

The 3p electron removed from Al is at a higher energy level than the 3s electron removed from Mg, so less energy is required to remove it.[1]

(c)

P: 1s² 2s² 2p⁶ 3s² 3p³ S: 1s² 2s² 2p⁶ 3s² 3p⁴[1]

In S, two electrons are paired in the same 3p orbital, so there is inter-electronic repulsion. Less energy is required to remove a 3p electron from S than from P.[1]

(a)

B³⁺(g) → B⁴⁺(g) + e⁻[1]

(b)

B³⁺: 1s² C³⁺: 1s² 2s¹

C³⁺ has one more electron shell than B³⁺: the 2s electron in C³⁺ is further from the nucleus and more shielded than the 1s electron in B³⁺.[1]

Despite the greater nuclear charge of C³⁺, the electrostatic attraction between the nucleus and its 2s electron is weaker. Less energy is required to remove the 2s electron from C³⁺ than the 1s electron from B³⁺, so B has the higher fourth ionisation energy.[1]