EL 5&6: Covalent bonding, shapes of molecules, amount of substance

Question 1

List the names for compounds containing 1-4 atoms of the same elements.

Answer 1

• Monoatomic
• Diatomic
• Triatomic
• Tetra-atomic

Question 2

In covalent bonding, there is a balance between which forces?

Answer 2

• Repulsive forces between nuclei
• Attractive forces between nuclei + electrons

Question 3

Draw dot and cross diagrams showing the bonding in:

• ammonia
• water

Answer 3

Question 4

What is a dative covalent bond and how is it represented?

Answer 4

• Both bonding electrons come from one atom
• Arrow points away from donor atom

Question 5

Tetrahedral bond angle

Answer 5

109.5^o

Question 6

Octahedral bond angle

Answer 6

90^o

Question 7

Pyramidal bond angle

Answer 7

107^o

Question 8

By what angle does a lone pair distort bonding pair angles?

Answer 8

2.5^o

Question 9

Bent bond angle

Answer 9

104.5^o

Pyramidal - 2.5^o since lone pair

Question 10

• Trigonal bipyramidal bond angles
• Draw a generic shape

Answer 10

90^o, 120^o

Trig planar with groups vertically above + below

Question 11

Shape of methane?

Answer 11

Tetrahedral

Question 12

Shape of sulfur hexachloride?

Answer 12

Octahedral

Question 13

Shape of water?

Answer 13

Bent

Question 14

Shape of ammonium?

Answer 14

Tetrahedral

Question 15

Shape of ammonia?

Answer 15

Pyramidal

Question 16

Draw xenon tetrafluoride.

Answer 16

Question 17

Draw ethene.

Answer 17

Trigonal planar around both carbons

Question 18

Draw PCl₄^-, tetrachlorophosphanium ;)

Answer 18

Trigonal bipyramidal; one bonding pair replaced by lone pair

Question 19

Draw chlorine pentafluoride / chloride (v) fluoride

Answer 19

Octahedral; one bonding pair replaced by lone pair

Question 20

Urea has the formula CO(NH₂)₂. Draw it.

Answer 20

Trigonal planar (120)^o around C. All bond angles around Ns = 107^o

Question 21

Use electron pair repulsion theory to deduce the shape of BF₃ ← NH₃.

Answer 21

NH₃ = pyramidal

BF₃ = trigonal planar

BF₃NH₃ = tetrahedral around both B and N

Question 22

Draw a dot-and-cross diagram for NH₂^-. Use this to suggest and explain the bond angle in the ion.

Compare this bond angle with that in the ammonia molecule, giving your reasons.

Answer 22

• [Dot-and-cross]
• [Bent (2 lone pairs, bond angle 104.5^o)]
  
  • 4 electron groups repel to be as far apart as possible;
  • Lone pairs repel more strongly than bonding pairs
• Ammonia is pyramidal: bond angle = 107^o
  
  • Angle greater because only 1 lone pair, so less repulsion

Question 23

0.84 g of magnesium was burnt in 0.56 g of oxygen. What is the formula of the magnesium oxide produced?

Answer 23

Mol Mg = 0.84 / 24 = 0.035 mol

Mol O = 0.56 / 16 = 0.035

1 : 1 so MgO

Question 24

Experimental results that a hydrocarbon is 75% carbon and 25% hydrogen, by mass. Derive its empirical formula.

Answer 24

Let mass of hydrocarbon = 100 g

Mol C = 75 / 12 = 6.25 mol

Mol H = 25 / 1 = 25 mol

C : H = 1 : 4 so CH₄

Question 25

Calculate the mass of aluminium oxide formed when 135 g of aluminium is burned in air.

Answer 25

4Al + 3O₂ →2Al₂O₃ Mol Al = 135 / 27 = 5 mol Mol Al₂O₃ = 5 x 2/4 = 2.5 mol Mass Al₂O₃ = 2.5 x 102 = 255 g

Question 26

2.53 g of hydrated magnesium chloride, MgCl₂ • xH₂O, was heated to constant mass. 1.17 g of solid remained. What is the formula of the hydrated compound?

Answer 26

Mol pure solid = 1.17 / 95.3 Mass water in salt = 2.53 - 1.17 = 1.36 g Mol water = 1.36 / 18 Mol salt to water in ratio 1 : x = 1 : 6.15 Formula = MgCl₂ • 6H₂O

Question 27

What is the volume of 319.5 g of chlorine at room temperature and pressure?

Answer 27

Volume = mol x 24 = (319.5 / 35.5 x 2) x 24 = 108 dm³

Question 28

What is the formula for percentage yield?

Answer 28

100 (experimental mass / theoretical mass)

Question 29

Give 5 reasons why yields are inefficient.

Answer 29

* Impure reactants * Side reactions * Product lost during transfer between vessels * Reversible reactions * Changes in temperature / pressure

Question 30

0.84 g of magnesium was burnt in excess oxygen, producing 1.1 g of magnesium oxide. Work out the percentage yield of the reaction.

Answer 30

Mg + 1/2O₂ → MgO Mol Mg = 0.84 / 24 = 0.035 mol Mol MgO = 0.035 mol Theoretical mass MgO = 0.035 x 40 = 1.4 g % yield = 1.4 x 100 / 1.1 = 79%