Chemical Bonding Shape Lab. 1)One structural isomer only.
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Transcript of Chemical Bonding Shape Lab. 1)One structural isomer only.
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
• polar covalent bond
– bond dipoles exist, – molecule is asymmetrical– dipoles do not cancel– molecule is polar
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
– bond dipoles exist, – molecule is asymmetrical– dipoles do not cancel– molecule is polar
N H
vii) hydrogen sulfide, H2S
SH
H
• Shape: Angular
• END = S – H = 2.5 – 2.1 = 0.4
• polar covalent bond
– bond dipoles exist, – molecule is asymmetrical– dipoles do not cancel– molecule is polar
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
• polar covalent bond
– bond dipoles exist, – molecule is asymmetrical– dipoles do not cancel– molecule is polar
N I
ix) hydrogen peroxide, H2O2
O
OH H
• Shape: Angular about each O
• 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
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
– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar
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
– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar
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
– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar
Be H
ii) boron trichloride, BCl3
B
Cl
Cl
Cl
• Shape: Planar trigonal
• END: B – Cl = 2.0 – 3.0 = 1.0
• polar covalent bond– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar
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
• polar covalent bond– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar
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
• polar covalent bond– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar
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
– bond dipoles exist– molecule is symmetrical– the forces cancel– molecule is non-polar
C H