Which molecules have an overall dipole moment? 1 carbon monoxide, CO 2 phosphine, PH3 3 carbon dioxide, CO2

Question 31 [Chemical Bonding > Covalent bonding and co-ordinate (dative covalent) bonding including shapes of simple molecules]

Which molecules have an overall dipole moment?

1 carbon monoxide, CO

2 phosphine, PH3

3 carbon dioxide, CO2

Reference: Past Exam Paper – Chemistry (9701) March 2016 Paper 12 Q31

Solution:

  Answer: B

   Firstly, to begin elimination, remove CO₂. The molecule is linear AND symmetrical. Therefore the dipoles of the otherwise polar C=O cancel out.

Coming to Carbon Monoxide, CO, it contains only one polar carbon-oxygen bond and therefore is a polar molecule.

PH₃ was the one molecule that confused the majority of the candidates. DO NOT make the mistake of assuming that all the three bonds are equally spaced, at 120°. If that were the case then the molecule would have been symmetrical and ultimately non polar. However the first protocol in dealing with an unknown compound is, draw it. You will notice that there is still a lone pair of electrons on the central P atom; this would distort the symmetrical shape and thus, PH₃ has an overall dipole.

Solid carbon dioxide, CO2, is similar to solid iodine, I2, in its structure and properties. Carbon is in Group 14. Silica, SiO2, is a Group 14 compound.

Question 8 [Chemical Bonding >3.2 Covalent bonding and co-ordinate (dative covalent) bonding including shapes of simple molecules]

Solid carbon dioxide, CO2, is similar to solid iodine, I2, in its structure and properties. Carbon is in Group 14. Silica, SiO2, is a Group 14 compound.

Which statement about solid CO2 and solid SiO2 is correct?

A Both solids exist in a lattice structure.

B Both solids have a simple molecular structure.

C Both solids have atoms joined by single covalent bonds.

D Both solids change spontaneously to gas at s.t.p.

Reference: Past Exam Paper – Chemistry (9701) March 2016 Paper 12 Q8

Solution:

     Answer: A

     Option A:  “It should be known that solid I₂ has a simple molecular lattice, therefore it can be deduced from the information in the question that solid CO₂ does too”

     Option B can be discredited since we know that SiO₂ has a similar structure to diamond, i.e, a giant covalent structure, not simple molecular.

    Option C can be eliminated. While SiO₂ has four single bonds, CO₂ has two double bonds and is linear rather than tetrahedral like SiO₂.

    Option D can be eliminated since at stp, SiO₂ is a solid and will not change spontaneously (naturally) into a gas.

Complete the diagrams to show the energies of the electrons in a carbon atom, a C+ ion and a C– ion.

Question 1 [Chemical Bonding > Hybridisation and covalent bonding]

(a) Complete the diagrams to show the energies of the electrons in a carbon atom, a C+ ion and a C– ion.

 

Reference: Past Exam Paper – Chemistry (9701) March 2016 Paper 42 Q1

Solution:

a)     C (6 electrons): 1s² 2s² 2p² [2/3 subshells  filled with 1 electron each]

        C+ (5 electrons): 1s² 2s² 2p¹  [1/3 subshells filled with 1 electron each]

        C- (7 electrons): 1s² 2s² 2p³ [all 3 subshells filled with one unpaired electron each]

b)

(i)Since 3 bonds are seen to originate from each carbon atom, hybridisation is sp² (1+2=3)

(number of bonds due to a type of hybridisation is equal to sum of powers of s and p orbitals; eg: sp³ (1+3=4) therefore 4 bonds; tetrahedral shape. You can also work backwards, CH₄ = four bonds= 1+3= sp³ hybridisation)

(ii) in an sp² hybridisation, we have one unhybridised p orbital. Upon reaction with excess hydrogen, this p orbital takes part in bonding with the H atom. Per carbon, 1 unhybridised p orbital = 1 added H atom/ Carbon atom=> final formula= c₆₀H₆₀

c) [Substitution reactions in Arenes]

(i)

   1: Cl₂  and UV light (since substitution is made in side chain and not on benzene ring)

   2: AlCl₃ and Cl₂ (since substitution is made in benzene ring and not on side chain)

(ii) every time we use UV light, the answer is FREE RADICAL SUBSTITUTION.

(iii) Ideally, if carried out in excess, all hydrogen atoms on side chain are substituted to give benzene-CCl₃ {this is the most appropriate answer}

Solutions provided by Kashish Varshney, India