Chemical Bonding Shape Lab

1)One structural isomer only

i)water, H 2 O shape: angular

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

ii)methane, CH 4

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

iii)methanol, CH 3 OH

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

iv)carbon tetrachloride, CCl 4

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

v)ammonia, NH 3

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

vi)hydrazine, N 2 H 4

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

vii)hydrogen sulfide, H 2 S

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

viii)nitrogen triiodide, NI 3

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

ix)hydrogen peroxide, H 2 O 2

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

x)chlorine, Cl 2

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, CO 2

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

ii)nitrogen, N 2

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, O 2

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, C 2 H 2

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

v)hydrogen cyanide, HCN

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

vi)carbon disulfide, CS 2

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, CH 2 O

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

viii)ethene, C 2 H 4

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

3)Special Compounds

i)beryllium hydride, BeH 2

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

ii)boron trichloride, BCl 3

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

iii)phosphorus pentabromide, PBr 5

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

iv)sulfur hexachloride, SCl 6

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

v)cyclohexane, C 6 H 12

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

vi)benzene, C 6 H 6

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