Structure 2. Models of bonding and structure

$NaCl$

$CH_4$

$SiO_2$

$C_{diamond}$

$BaSO_4$

$Ba_2SO_4$

$Ba(SO_4)_2$

$BaS$

It is composed of discrete molecules

It is a regular repeating three-dimensional arrangement of ions

It is held together by intermolecular forces

It is only found in liquid salts

Sodium atom

Chlorine atom

Sodium ion

Chloride ion

It is formed by the loss of an electron from an ammonia molecule.

It contains four ionic bonds between N and H.

It contains three covalent bonds and one dative covalent bond.

It is a planar molecule.

$HCl$

$LiF$

$SiCl_{4}$

$Cl_{2}$

Radii increase from $N^{3-}$ to $Mg^{2+}$.

Radii decrease from $N^{3-}$ to $Mg^{2+}$.

Radii remain constant as they have the same number of electrons.

Radii decrease and then increase.

It assumes ions are point charges with no specific volume.

It assumes the attraction between ions is directional.

It describes the lattice as a result of electrostatic attraction between oppositely charged ions.

It only applies to compounds containing transition metals.

2

3

5

6

$Cl^{-}$

$Br^{-}$

$I^{-}$

$F^{-}$

There are three single and three double bonds

All C-C bonds are of equal length and strength

The bonds are longer than $C-C$ single bonds

The bonds are shorter than $C=C$ double bonds

2

4

6

8

$SO_2$

$BeCl_2$

$H_2O$

$OF_2$

Trigonal pyramidal

Tetrahedral

Trigonal planar

Square planar

Phosphorus has a higher melting point because it has a larger molecular mass.

Silicon has a lower melting point because its atoms are held together by weak intermolecular forces.

Silicon has a higher melting point because melting involves breaking strong covalent bonds in a lattice.

Phosphorus has a higher melting point because it forms a giant molecular structure.

The molecule is non-polar because the bond dipoles cancel out in a symmetrical shape.

The molecule is polar because it has an asymmetrical distribution of charge.

The molecule is non-polar because it has a trigonal planar electron domain geometry.

The molecule is polar because there are two lone pairs of electrons on the central carbon atom.

The bonds in $O_{3}$ are longer and weaker than the bonds in $O_{2}$.

The bonds in $O_{3}$ are longer and stronger than the bonds in $O_{2}$.

The bonds in $O_{3}$ are shorter and stronger than the bonds in $O_{2}$.

The bonds in $O_{3}$ are shorter and weaker than the bonds in $O_{2}$.

Linear

Trigonal planar

Tetrahedral

Trigonal pyramidal

$109.5^{\circ}$

$120^{\circ}$

$107^{\circ}$

$180^{\circ}$

$C_2H_6$

$CH_3OCH_3$

$CH_3CH_2OH$

$CH_3CHO$

A layer of electrons on the surface of the metal

Electrons that are localized between two ions

Valence electrons that are free to move throughout the lattice

Inner shell electrons that are shared by all atoms

Free to move like the electrons

Fixed in a regular lattice structure

Arranged in discrete molecules

Moving randomly in a gas-like state

Larger ions lead to higher melting points

Smaller ions lead to higher melting points

Ion size has no effect on melting point

Larger ions increase the density only

Brass

Magnesium ribbon

Graphite

Gold ring

The positive cations become mobile charge carriers only once the metal has been melted.

The delocalized 'sea of electrons' remains mobile and free to move even when the rigid lattice of cations is disrupted by melting.

Melting the metal breaks the electrostatic attractions, allowing the valence electrons to be released as a gas.

The metallic bond is converted into a covalent bond in the liquid state, allowing for easier electron flow.

$K^+$ has a smaller radius than $Rb^+$, leading to stronger attraction to delocalized electrons

$Rb$ has more delocalized electrons than $K$

$K$ has more occupied electron shells than $Rb$

$Rb$ is more metallic than $K$

Higher charge density leads to stronger metallic bonds

Higher charge density leads to weaker metallic bonds

Charge density does not affect bond strength

Lower charge density leads to higher melting points

Breaking metallic bonds requires a small amount of energy.

The electrostatic attraction between cations and delocalized electrons is very strong and acts throughout the lattice.

The covalent bonds between metal atoms are very difficult to break.

The intermolecular forces in metals are stronger than in ionic compounds.

Conductivity decreases because the ions become smaller.

Conductivity increases because the number of delocalized electrons per atom increases.

Conductivity remains constant as it depends only on the state of the matter.

Conductivity decreases because the electrons are more tightly held by the nucleus.

Group 2 metals have larger atomic radii.

Group 2 metals have a higher ratio of delocalized electrons to cations.

Group 2 metals have a lower nuclear charge.

Group 2 metals form covalent networks.

Decrease in molar mass

Increase in the number of electrons

Decrease in boiling point

Increase in electronegativity difference

Ionic

Covalent

Metallic

Hydrogen

$CH_{4}$

$NH_{3}$

$BF_{3}$

$CO_{2}$

Strong London dispersion forces

Strong hydrogen bonding

A giant network of strong covalent bonds

Strong electrostatic attraction between ions

Ethanol > Fluoroethane > Ethane

Ethane > Fluoroethane > Ethanol

Fluoroethane > Ethanol > Ethane

Ethanol > Ethane > Fluoroethane

$SiO_{2}$

$C_{10}H_{8}$

$CaCl_{2}$

$Fe$

London dispersion < Hydrogen bonding < Dipole-dipole

Hydrogen bonding < Dipole-dipole < London dispersion

London dispersion < Dipole-dipole < Hydrogen bonding

Dipole-dipole < London dispersion < Hydrogen bonding

1

2

1.5

1.33

The ionic radius of the metal cation increases.

The number of delocalized electrons decreases.

The charge on the metal cation increases and the radius decreases.

The electronegativity of the metal increases.

$PCl_{3}$

$CO_{3}^{2-}$

$H_{3}O^{+}$

$NH_{3}$