Chapter 7 (3.1.1)

Question 1

Arrangement of elements

Answer 1

• arranged in order of increased atomic number
• groups with same outer-shell electrons with similar chemical properties
• periods give highest electron energy shell
• periodicity - a repeating trend in properties of the elements across each period of the Periodic Table

Question 2

Periodic trend in electron configuration across a period

Answer 2

• each period starts with an electron in a new highest energy shell
• for each period, the s- and p-orbitals are filled in the same way - a periodic pattern

Question 3

Periodic trend in electron configuration down a group

Answer 3

• same number of electrons in each sub-shell gives the elements in the same group similar chemistry

Question 4

Classification of blocks

Answer 4

• division into blocks based on highest energy sub-shell: s, p, d, f
• s - left 2 groups
• p - right 6 groups
• d - middle 10 groups
  f - bottom elements

Question 5

What is first ionisation energy?

Answer 5

The energy change when each atom in 1 mole of gaseous atoms loses an electron (to form 1 mole of gaseous 1+ ions).

Question 6

What will the first electron experience?

Answer 6

The first electron will be in the highest energy level and will experience the least attraction from the nucleus.

Question 7

Equations for ionisation energies

Answer 7

FIE: He (g) –> He⁺ (g) + e⁻
SIE: He (g) –> He²⁺ (g) + e⁻

Question 8

Why is SIE larger than FIE?

Answer 8

Second ionisation energy (and successive ones) are larger than the first because after the first electron is lost, the single electron is pulled closer to the atoms nucleus. Nuclear attraction on remaining electrons increase and more ionisation energy will be needed to remove the next electrons.

Question 9

What does a large increase between 2nd and 3rd ionisation energies mean?

Answer 9

That the 3rd electron must be removed from a different shell, closer to the nucleus and with less shielding, thus element is from Group 2.

Question 10

Trends in FIE down a group

Answer 10

First ionisation energy decreases down a group:
- atomic radius increases
- more inner shells so shielding increases
- nuclear attraction on outer electrons decreases
- first ionisation energy decreases
(although nuclear charge increases, increased shielding and radius outweigh this)

Question 11

Trends in FIE across a period

Answer 11

General increases across (first 3) periods:

• nuclear charge increases
• same shell = similar shielding
• nuclear attraction increases
• atomic radius decreases
• first ionisation energy increases

Question 12

Sub-shell trends in FIE

Answer 12

• FIE falls in 2 places in the same position in each period
• suggests a periodic cause - linked to sub-shells, their energies and how orbitals fill with electrons
• links to the filling of s- and p-sub-shells

Question 13

sub-shell trend: s- and p-orbital energies

Answer 13

• e.g. Be and B
• marks the start of filling the 2p sub-shell
• the 2p sub-shell in boron has a higher energy than the 2s sub-shell in beryllium
• thus 2p electrons in B is easier to remove than one of the 2s electrons in Be.
  IN BORON THE ELECTRON IS BEING REMOVED AT A HIGHER ENERGY. AN S ELECTRON IS LOST IN BERYLIUM AND A P ELECTRON IS LOST IN BORON.

Question 14

sub-shell trend: p-orbital repulsion

Answer 14

• e.g. N and O
• marks the start of electron pairing in the p-orbitals of the 2p sub-shell
• in O, the paired electrons in one of the 2p orbitals repel one another, making it easier to remove and electron from an O atom than an N atom (2 electrons in the 2px sub-shell)

Question 15

What is metallic bonding?

Answer 15

The strong electrostatic forces of attraction between cations and delocalised electrons

Question 16

Structure of metals

Answer 16

• cations = fixed maintaining the structure
• delocalised electrons = mobile
• billions of metals atoms are held together by metallic bonding in a giant metallic lattice structure.

Question 17

Properties of metals

Answer 17

• electrical conductivity: conducts as a solid and liquid . when a voltage is applied across a metal, the delocalised electrons can move through structure, carrying charge.
• high melting and boiling points - depends on the strength of the metallic bonds - large amount of energy needed to overcome strong attraction - higher ions have stronger bonds
• solubility - metals do not dissolve - interaction lead to a reaction rather than dissolving

Question 18

Giant covalent structures

Answer 18

Non-metallic elements arranged so that many billions of atoms are held together by a network of strong covalent bonds to form a giant covalent lattice.

Question 19

Carbon and silicon

Answer 19

Group 14 - use their 4 outer electrons to form covalent bonds with other C and Si atoms, resulting in a tetrahedral arrangement, 109.5° bond angle.

Question 20

Properties of giant covalent

Answer 20

• stable structures due to strong covalent bonds
• high melting and boiling points - due to strong covalent bonds - lots of E needed to break
• solubility - insoluble in most solvents due to the bonds holding the atoms together in the lattice being too strong to be broken by the interaction with solvents
• electrical conductivity - non-conductors except graphene + graphite (allotropes of carbon). in diamond = silicon etc all 4 outer electrons are involved in covalent bonding so none available for conducting electricity / graphite has 1 delocalised electron available for conductivity - only 3 involved in bonding (bond angle 120° - hexagonal planar.

Question 21

Periodic trend in melting points (2 and 3)

Answer 21

• increases from Group 1-14, then sharp decrease between 14&15, then comparatively low from 15-18.
• sharp decrease due to change from giant structures to simple molecular - weak IM forces to overcome
• giant metallic to giant covalent to simple molecular

Question 22

Strength of forces - simple molecular structures

Answer 22

The larger the molecule the stronger its van der Waals forces - sulfur has S₈ rings so stronger IM forces than P₄ and Cl₂ etc.