Topic 2 Atomic Structure & the Periodic Table

Question 1

Briefly describe how a mass spectrometer works.

Answer 1

1. The sample injected into the mass spectrometer and vaporized.
2. The vapor is bombarded with high energy electrons, which will collide with the atoms/molecules and knock off one or more electrons from each atom/molecule, forming positive ions. The positive ions could be positively charged atoms, positively charged molecules or positively charged fragments of molecules.
3. A magnetic field deflects the anions. The amount of deflection depends on the mass to charge ratio of the anion.

Question 2

Explain why there are three peaks which correspond to molecular masses of 70, 72 and 74 on the mass spectrum of chlorine and the ratio of their heights is roughly 9:6:1.

Hint: The ratio of chlorine-35 to chlorine-37 is usually 3:1.

Answer 2

70 (chlorine-35 & chlorine-35): 3/4 * 3/4 = 9/16
72 (chlorine-35 & chlorine-37): 3/4 * 1/4 * 2 = 6/16
74 (chlorine-37 & chlorine-37): 1/4 * 1/4 = 1/16

Question 3

How many orbitals does an s sub-shell, a p sub-shell and a d sub-shell have respectively?

Answer 3

s: 1
p: 3
d: 5

Question 4

Why is the electronic configuration of K (19) [Ar] 4s¹ instead of [Ar] 3d¹?

Answer 4

The 4s orbital has a lower energy level than the 3d orbitals. Hence electrons will fill the 4s orbital before filling in the 3d orbitals.

Question 5

Why is the electronic configuration of Cr (24) [Ar] 4s¹ 3d⁵ instead of [Ar] 4s² 3d⁴?

Answer 5

There are 5 orbitals in the 3d sub-shell. Electrons are more stable when all of the orbitals are occupied in a sub-shell. Hence, when four of the five orbitals in the 3d sub-shell are already occupied, one electron from the 4s orbital will move to occupy the last 3d orbital.

Question 6

Write an equation to represent the first ionization energy of an element A.

Answer 6

A(g) → A⁺(g) + e^–

Question 7

Give the formula for ionisation energy and explain it.

Answer 7

IE = energy of electron when removed – energy of electron when in its orbital
Ionisation requires an electron to be so far away from the nucleus that it no longer experiences an attraction force from it. The energy of the electron needs to be increased to a particular value to remove it. For any given atom, the energy value an electron has when it reaches this position will always be the same, regardless of which orbital the electron originally occupied. The IE is the amount of energy an electron needs to gain in order to be removed.

Question 8

Factors that affect the IE of an electron

Answer 8

• the orbital in which the electron exists
• distance between the electron and the nucleus
• the nuclear charge of the atom
• the repulsion (shielding) experience by the electron from all the other electrons present

Question 9

Explain why hydrogen has a high ionisation energy.

Answer 9

• single electron exists in orbital of the lowest energy level
• short distance to nucleus
• no inner electrons

→ outermost electron experiences a very strong attraction towards the nucleus
→ high IE

Question 10

Explain why He has a higher first IE than H.

Answer 10

He: 1s²
H: 1s¹

• same orbital
• both have no inner electrons

but

• He has a higher nuclear charge than H

→ He has a higher first IE than H.

Question 11

Explain why the first ionisation energies generally increase across a period.

Answer 11

• orbitals of similar energy levels
• similar number of inner electrons

but

• nuclear charge increases

→ first IE generally increases across a period.

Question 12

Explain the drop in first IE between the Group 2 and Group 3 element. Use the period 2 elements (Be & B) as an example.

Answer 12

TL;DR: higher energy level + greater shielding effect
Be: 1s² 2s²
B: 1s² 2s² 2p¹

• B has a higher nuclear charge than Be
• the outermost electron of B is in a 2p orbital while the outermost electron of Be is in the 2s orbital
• 2p orbital has a slightly higher energy level than the 2s orbital
• outermost electron of Be experiences shielding from Be’s 1s² electrons
• outermost electron of B experiences shielding from B’s 1s² electrons and 2s² electrons
• effect of B’s outermost electron occupying a higher energy orbital and the greater shielding effect it experiences > effect of B’s higher nuclear charge

→ B has a lower first IE than Be.

Question 13

Explain the drop in first IE between the Group 5 and Group 6 element. Use the period 2 elements (N & O) as an example.

Answer 13

TL;DR: electron-electron repulsion between electrons occupying the same orbital
N: 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹
O: 1s² 2s² 2p_x² 2p_y¹ 2p_z¹

• O has a higher nuclear charge than N
• outermost electron of O shares the 2p_x orbital with another electron while the outermost electron of N is the only electron in the orbital it occupies
  → the outermost electron of O experiences greater electron-electron repulsion
• effect of this repulsion outweighs the effect of O’s higher nuclear charge

→ O has a lower first IE than N

Question 14

Explain why first IE decreases down the group.

Answer 14

• higher nuclear charge

but

• greater number of inner electrons
• outermost electron occupies orbital of higher energy level
• greater distance from the nucleus

Question 15

3 types of atomic radii

Answer 15

Covalent radius: distance between two bonded atoms/2
van der Waals radius: distance between two atoms that are just touching/2
Metallic radius

Question 15

How is atomic radius calculated?

Answer 15

Distance between two nuclei/2