Chemical Bonding

Question 1

Definite metallic bonding

Answer 1

Metallic bonding is the electrostatic attraction between positively charged cations and the sea of delocalised negatively charged electrons

Question 2

Factors affecting strength of metallic bonding

Answer 2

1. number of valence electrons contributed to sea of delocalised electrons (by higher charged cations)
2. size of cation (smaller size of cation, stronger metallic bond)

Question 3

Define ionic bonding

Answer 3

Ionic bond is the electrostatic forces of attraction between oppositely-charged ions

Question 4

What factors affect coordination number?

Answer 4

Coordination number is the number of ions that surround another ion of opposite charge.
Factors:
1. Cationic and anionic radius. (larger cationic radius, more anions can surround, higher coordination number)
2. Cation to Anion ratio

Question 5

Factors affecting strength of ionic bond

Answer 5

lattice energy= q+q-/r+ + r-

Question 6

Factors determining covalent character in ionic bonds

Answer 6

1. higher charge density of cation = charge/cationic radius.
   - higher charge density, higher polarising power
2. larger electron cloud of anion (more electrons or larger anionic radius), the anion is more easily polarised

-AlCl3, AlBr3 and many Be compounds are covalent

Question 7

Define covalent bonding

Answer 7

Covalent bonding is the electrostatic forces of attraction between a shared pair of electrons and positively-charged nucleus.

Question 8

Draw bonding in Al3Cl6 and ammonium ion

Answer 8

-

Question 9

Criteria for dative bond

Answer 9

Donor atom has lone pair of electrons
Acceptor atom is electron deficient

Question 10

State the shapes and bond angles for species

Answer 10

2EP:
2 bp –> Linear, 180

3EP:
3 bp –> trigonal planar, 120
2 bp, 1 lp –> bent, 110-120

4EP:
4 bp –> tetrahedral , 109.5
3 bp, 1 lp –> trigonal pyramidal, 107
2 bp, 2 lp –> bent, 105

5EP:
5 bp –> trigonal bipyramidal, 90&120
4 bp, 1 lp –> see-saw, 90&120
3 bp, 2 lp–> T-shaped, 90&120

6EP:
6 bp–> octahedral, 90
5 bp, 1 lp–> square pyramidal, 90
4 bp, 2 lp –> square planar, 90

Question 11

State the VSEPR theory

Answer 11

-Electrons are negatively-charged and repel each other. Hence electron pairs around the central atom of a molecule are arranged as far apart as possible to minimise electrostatic repulsion and maximise stability
- Lone pair of electrons are closer to central atom and repel more than bonding electrons. Hence lplp repulsion > bplp repulsion > bpbp repulsion

Question 12

Explain the effect of electronegativity on bond angle. Taking H2O and H2S as an example.

Answer 12

• Both H2O and H2S are covalent molecules with 2 bond pairs of and 2 lone pairs of electrons.
• In H2O, O has a greater electronegativity than S in H2S.
• Hence, the bond pair of electrons
  are attracted closer to the central atom O, resulting in greater repulsion between thebond pairs of electrons of O in H2O.
• Hence, H2O has a larger bond angle than H2S.

Question 12

Explain the difference in C-C bond length

Answer 12

A sp3 hybrid orbital has more p character than a sp2 hybrid orbital and it is longer/more diffuse. Hence, a sp3 hybrid orbital undergoes less effective overlap, giving rise to longer
bond length and weaker bond.

Question 13

Explain how idid attractions arise

Answer 13

• As electrons are constantly moving, there could be instantaneous dipole formed when there are more electrons formed on one side of the molecule. Hence there could be a small partial positive charge on one atom and small partial negative charge on the other atom.
• When a neighbouring molecule approaches, this instantaneous dipole induces a temporary dipole on the neighbouring molecule by attracting or repelling its electrons.
• This intermolecular forces of attraction between 2 molecules of opposite partial charged are called idid.

Question 13

Explain factors affecting covalent bond strength. (covalent bond between atoms !)

Answer 13

1. effectiveness of overlap of orbitals
   (size of atomic orbital) smaller atoms can approach each other more closely, more effective overlap, stronger bond and shorter bond length
2. bond order
   triple bond>double bond>single bond
3. bond polarity
   greater EN diff, more polar bond, stronger the bond

Question 14

Factors affecting idid attractions

Answer 14

1. Number of electrons
   more electrons, more polarisable electron cloud, stronger idid
2. Larger surface area, stronger idid

Question 15

Explain how pdpd arise

Answer 15

When two different atoms of different EN come together, the more EN atom pulls the bonding electron of the covalent bond towards itself and acquire a small positive charge while the other acquires a small negative charge. This results in charge separation and bond is polar.

Question 16

criteria for hydrogen bonding

Answer 16

• HO, HF or HN in one molecule
• lone pair of electrons on F, O, N molecule

Question 17

factors affecting hydrogen bond

Answer 17

1. Extensiveness of hydrogen bonds (number of hydrogen bonds formed per molecule)
2. EN diff between H and F,O or N
   (F more EN than O and N, forming more polar H-F bond than H-O and
   H-N, hence stronger H bond)

Question 18

Intramolecular hydrogen bond arising from close proximity of substituents, leading to less extensive intermolecular hydrogen bonding hence lesser energy to…

Answer 18

-

Question 19

Characteristics, structure, shape, bond angle of diamond

Answer 19

• Each carbon atom is covalently bonded to 4 other carbon atoms (4 σ bonds) arranged tetrahedrally around it, with a bond angle of 109.5
• Diamond does not conduct electricity as it does not possess delocalised electrons.
• Diamond has a melting point of 3550 due to the strong covalent bonds between all the carbon atoms.
• Diamond does not dissolve in water or any other solvent as a lot of energy is required to overcome the strong covalent bonds between the carbon atoms.

Question 20

Explain the structure and properties of ice

Answer 20

• When temperature is low at 0oC, the kinetic energy of the H2O molecules is low and do not move around that readily. Each O atom in H2O is tetrahedrally bonded to four hydrogen atoms, two by covalent bonds and two by intermolecular hydrogen bonds, giving rise to a highly ordered 3–D structure in ice.
• This highly ordered 3–D arrangement of H2O molecules in ice creates a very open
  structure that occupies a larger volume for the same mass of liquid water.
• Hence ice is less dense than liquid water.

Question 21

Why graphite can conduct electricity

Answer 21

• Each carbon atom is covalently bonded to 3 other carbon atoms (3 σ bonds) arranged in a trigonal planar manner, with a bond angle of 120.
• Each carbon is sp2 hybridised and has an unhybridised p-orbital containing 1 electron. The p orbital of each adjacent carbon atom will overlap sideways, resulting in a pi-
  electron cloud above and below the plane containing the carbon atoms.
• These pi-electrons are delocalised within each layer and hence, graphite only conduct electricity in a direction parallel to the layers. As the delocalised electrons cannot jump
  across the layers, graphite is an insulator in a direction perpendicular to the layers.