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Answers to Problem Sheet 6 in CHEM1901/3: Molecular Structures and Bonding, Schemes and Mind Maps of Engineering Chemistry

Answers to problem sheet 6 in the chem1901/3 course, focusing on molecular structures and bonding. It covers various organic and inorganic compounds, including their lewis structures, bond orders, and resonance structures. Additionally, it introduces the concept of vsepr theory and its application in predicting molecular shapes.

What you will learn

  • What is the shape of the methylsulfate anion (CH3OSO3-)?
  • What is the shape of the nitrate ion (NO3-)?
  • What is the shape of the sulfur pentafluoride molecule (SF6)?
  • Which resonance structure of phosphoric acid (H3PO4) shows all atoms obeying the octet rule?
  • What is the bond order of hydrogen cyanide (HCN)?
  • Using VSEPR theory, predict the shape of the nitrate ion (NO3-).

Typology: Schemes and Mind Maps

2021/2022

Uploaded on 09/12/2022

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CHEM1901/3 Answers to Problem Sheet 6
1.
(a)
hydrogen cyanide,
HCN
C NH
Bond order = 3 (triple bond)
(b)
ethanol, CH3CH2OH
H C C O
H
H
H
H
H
Two lone pairs on oxygen.
(c)
pyridine, C5H5N
Pyridine has delocalized ‘aromatic’ electrons like benzene with a lone pair on
nitrogen instead of a C-H bond. Note that this lone pair is pointing away from the
ring and is not delocalized.
(d)
acetylene, C2H2
H C C H
(e)
SOCl2
SCl Cl
O
SCl Cl
O
The resonance structure on the left shows the (allowed) expanded valence shell
around S with 5 electron pairs (4 bonds for the S(IV) atom and 1 lone pair). The
resonance structure on the right shows all atoms obeying the octet rule leading to
charges having to be placed on S and O.
(f)
N2O4
Each resonance form is equivalent – the N-O bonds are intermediate between single
and double bonds in character. There is no multiple bond between the nitrogen atoms.
N N
O O
OO
N N
O O
OO
N N
O
O
O
O
N N
O
O
O
O
pf3
pf4

Partial preview of the text

Download Answers to Problem Sheet 6 in CHEM1901/3: Molecular Structures and Bonding and more Schemes and Mind Maps Engineering Chemistry in PDF only on Docsity!

CHEM19 01 /3 Answers to Problem Sheet 6

  1. (a) hydrogen cyanide,

HCN

H C N Bond order = 3 (triple bond)

(b) ethanol, CH

3

CH

2

OH

H C C O

H

H

H

H

H

Two lone pairs on oxygen.

(c) pyridine, C 5

H

5

N

C

C

N

C

C

C

C

C

C

N

C

C

H

H

H H

H

H

H

H H

H

Pyridine has delocalized ‘aromatic’ electrons like benzene with a lone pair on

nitrogen instead of a C-H bond. Note that this lone pair is pointing away from the

ring and is not delocalized.

(d) acetylene, C

2

H

2

H C C H

(e) SOCl 2

Cl S Cl

O

Cl S Cl

O

The resonance structure on the left shows the (allowed) expanded valence shell

around S with 5 electron pairs (4 bonds for the S(IV) atom and 1 lone pair). The

resonance structure on the right shows all atoms obeying the octet rule leading to

charges having to be placed on S and O.

(f) N 2

O

4

Each resonance form is equivalent – the N-O bonds are intermediate between single

and double bonds in character. There is no multiple bond between the nitrogen atoms.

N N

O O

O O

N N

O O

O O

N N

O

O

O

O

N N

O

O

O

O

(g) phosphoric acid,

H

3

PO

4

[=PO(OH)

3

]

P

O

O

O

H

H

H O P

O

O

O

H

H

H O

The resonance structure on the left shows the bonding with an expanded valence

shell around P (five bonds for the P(V) atom). The resonance structure on the right

shows all atoms obeying the octet rule but with charges.

(h) phosphate anion,

PO

4

3 –

P

O

O

O O

O P O

O

O

P

O

O

O O

O P O

O

O

The resonance structures all feature one double P=O and three single P-O bonds.

Overall, each P-O bond is equivalent with:

bond order = ¼ (3 × 1 (single)+ 1 × 2 (double)) = 1.

(i) methylsulfate anion,

CH

3

OSO

3

S

O

O

C O O

H

H

H

S

O

O

C O O

H

H

H

S

O

O

C O O

H

H

H

(f) SF 4

F S F

F

F

trigonal

bipyramid

see-saw

(g) ClF 3

F Cl F

F

trigonal

bipyramid

T-shaped

(h) XeF 2

F Xe F

trigonal

bipyramid

linear

(a) (b) (c)

H C

3

C H

O

H

3

C C OH

H

H

H C

3

C OH

O

(d) (e) (f)

H C

3

C O

O

CH

3

C

H

3

C

C

CH

3

H

3

C

H

H C

3

C C

H

C

C

C C

C

C

C C N CH

3

O H

H

H O

H H

H

O

H

H

H H