Chemical Bonding

Chemical Bonding

Shape Lab

1) One structural isomer only

i) water, H2O

• shape: angular

O

H

H

• END = O – H = 3.5 – 2.1 = 1.4• polar covalent bond

– bond dipoles exist, – molecule is asymmetrical– dipoles do not cancel– molecule is polar

O H

ii) methane, CH4

C

H

H

H

H

C

H

H

H

H

• Shape: Tetrahedral

• END = C – H = 2.5 – 2.1 = 0.4

• polar covalent bond

– bond dipoles exist, – molecule is symmetrical – the forces cancel– molecule is non-polar

C H

iii) methanol, CH3OH

C

HO

H

H

HC

H

O

H

H

H

• Shape: Tetrahedral about C Angular about O

• END = C – H = 2.5 – 2.1 = 0.4• END = C – O = 2.5 – 3.5 = 1.0• END = O – H = 3.5 – 2.1 = 1.4• all bonds are polar covalent

– bond dipoles exist– molecule is not symmetrical because different atoms

are bonded to the C and the O is angular– the forces do not cancel– molecule is polar

C H

O H

C O

iv) carbon tetrachloride, CCl4

C

Cl

Cl Cl

Cl

C

Cl

Cl Cl

Cl

• Shape: Tetrahedral

• END = C – Cl = 2.5 – 3.0 = 0.5

• polar covalent bond– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar

C Cl

v) ammonia, NH3

NH

H

H

N

H H

H

• Shape: Trigonal pyramidal

• END = N - H = 3.0 – 2.1 = 0.9



N H

vi) hydrazine, N2H4

N

H

H

N H

H

N

H

H

N H

H

• Shape: Trigonal pyramidal about each N• END = N - H = 3.0 – 2.1 = 0.9• END = N – N = 3.0 – 3.0 = 0.0• N – H is polar covalent bond• N – N is covalent bond


N H

vii) hydrogen sulfide, H2S

SH

H

• Shape: Angular

• END = S – H = 2.5 – 2.1 = 0.4



S H

viii) nitrogen triiodide, NI3

N

I

I

I

N

I

I

I

• Shape: Trigonal pyramidal

• END = N - I = 3.0 – 2.5 = 0.4



N I

ix) hydrogen peroxide, H2O2

O

OH H

• Shape: Angular about each O

• END = O – H = 3.5 – 2.1 = 1.4



O H

x) chlorine, Cl2

Cl

Cl

• Shape: only 2 atoms (linear)• END: Cl – Cl = 3.0 – 3.0 = 0.0• covalent bond• no bond dipoles exist, so molecule is

non-polar

2) Double and triple bonds

(use the springs)

i) carbon dioxide, CO2

C

O

O

• Shape: Linear (bonded to 2 atoms with no lone pairs)

• END = C – O = 2.5 – 3.5 = 1.0• END = C – C = 2.5 – 2.5 = 0.0• C – O is polar covalent bond

– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar

C O

ii) nitrogen, N2

N

N

• Shape: only 2 atoms (linear)• END: N – N = 3.0 – 3.0 = 0.0• covalent bond

– no bond dipoles exist– molecule is non-polar

iii) oxygen, O2

O O

• Shape: only 2 atoms• END: O – O = 3.5 – 3.5 = 0.0• covalent bond

– no bond dipoles exist– molecule is non-polar

iv) ethyne, C2H2

C CH H

• Shape: Linear (each C bonded to 2 atoms with no lone pairs)

• END = C – H = 2.5 – 2.1 = 0.4• END = C – C = 2.5 – 2.5 = 0.0• C – H is polar covalent bond, • C - C is covalent


C H

v) hydrogen cyanide, HCN

C NH

• Shape: Linear (C bonded to 2 atoms with no lone pairs)

• END = C – H = 2.5 – 2.1 = 0.4• END = C – N = 2.5 – 3.0 = 0.5• both are polar covalent bonds

– bond dipoles exist– molecule is symmetrical but the C is bonded to

2 different atoms– the forces do not cancel– molecule is polar

C H

C N

vi) carbon disulfide, CS2

C SS

• Shape: Linear (bonded to 2 atoms with no lone pairs)

• END = C – S = 2.5 – 2.5 = 0.0

• END = C – C = 2.5 – 2.5 = 0.0

• both are covalent bonds– no bond dipoles exist– molecule is non-polar

vii) methanal, CH2O

CO

H

H

• Shape: Trigonal planar (bonded to 3 atoms with no lone pairs)

• END = C – O = 2.5 – 3.5 = 1.0• END = C – H = 2.5 – 2.1 = 0.4• both are polar covalent bonds

– bond dipoles exist– molecule is symmetrical but the C is bonded to 2

different atoms– the forces do not cancel– molecule is polar

C H

C O

viii) ethene, C2H4

C C

H

HH

H

• Shape: Planar trigonal (each C bonded to 3 atoms with no lone pairs)

• END = C – H = 2.5 – 2.1 = 0.4• END = C – C = 2.5 – 2.5 = 0.0• C – H is polar covalent bond


C H

3) Special Compounds

i) beryllium hydride, BeH2

Be

H

H

• Shape: Linear• END: Be – H = 1.5 – 2.1 = 0.6• polar covalent bond


Be H

ii) boron trichloride, BCl3

B

Cl

Cl

Cl

• Shape: Planar trigonal

• END: B – Cl = 2.0 – 3.0 = 1.0


B Cl

iii) phosphorus pentabromide, PBr5

P

Br

Br

Br

Br

Br

P

Br

Br

Br

Br

Br

• Shape: Trigonal bipyramidal

• END: P – Br = 2.1 – 2.8 = 0.7


P Br

iv) sulfur hexachloride, SCl6

S

Cl

Cl

Cl

Cl

Cl

Cl

S

Cl

Cl

Cl

Cl

Cl

Cl

• Shape: Octahedral

• END: S – Cl = 2.5 – 3.0 = 0.5


S Cl

v) cyclohexane, C6H12

CC

C

CC

C

H

H

H

H

H

H

H

H

H

H

H

H

• Shape: Tetrahedral about each C• END: C – H = 2.5 – 2.1 = 0.4• END: C – C = 2.5 – 2.5 = 0.0• C – H is polar covalent bond; C – C is

covalent– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar

C H

vi) benzene, C6H6

CC

C

CC

C

H

H

H

H

H

H

• Shape: Planar trigonal about each C (bonded to 3 atoms with no lone pairs)

• END: C – H = 2.5 – 2.1 = 0.4• END: C – C = 2.5 – 2.5 = 0.0• C – H is polar covalent bond; • C – C is covalent


C H

Documents

Chemical Bonding