Chemistry modelling | | DJS

Chemistry modelling

1.  Carson tells me that the bond angle of methane, CH4, is 109.5º.

i      Model this molecule as a regular tetrahedron with its base downwards, putting the carbon, C, at the centre and labelling the four hydrogen atoms EFG&H, H at the top. Let the projection of HC meet the base EFG at B and let EB projected meet FG at M. Draw the appropriate labelled diagram.                                                                                                         [2]

ii       Let all the sides of the tetrahedron be of length a and all the lines from C to the vertices be length r.  By considering triangle EFG, express length EM in terms of r.                   [2]                                       

iii      Point B is the intersection of the medians of EFG; show that EB=2BM =a/√3 [3]        

iv       Draw triangle CFG separately and label angle FCM as θ, theta.  Write the values of sin θ and CM2 in terms of a and r.                                                                                      [2]                             

v       By looking at triangle EBH, find another expression for CM2, and, by equating these two, find the non-zero value of a/r. Hence find the value of 2θ, and improve on Carson’s value for the bond angle by adding one significant figure.                                                             [5]

vi       Consider each bond length r as a vector. Any three, such as CE, CF, CG must match CH in the ‘vertical’ component, along direction CH. So three dot products, r.r equal vector CH. Symmetry says that this is the case for any direction along a bond and, indeed, any direction. Call the angle between the bonds ß, beta. By recognising that r.r  = |r|2 cosß, find the value of beta exactly and check it against your value for theta.                                             [3]

2. Ammonia, NH3, also has a tetrahedral structure; replace C with N in your diagram. At point H there is a lone pair of electrons. The symmetry is now rotational of order three along the axis NH only. The bond lengths of EN, FN, GN remain at r, while NH is a bit shorter than in Q1, at about 0.892r. This moves the nitrogen atom ‘up’ in the frame of the diagram we’re using, so the bond angle between the hydrogen atoms at EF&G becomes a little smaller.

Find the new angle, to 4 sig,fig. by solving the vector equation ( along the axis NH,      

   3r.r + 0.892r = 0, treating the vector r as a unit vector.                                           [4]

3. The situation for water, H2O, is slightly different because there are two lone pairs in the outermost shell. This reduces the symmetry further and you can represent these by drawing an encompassing cuboid, putting the hydrogen atoms on the ends of the diagonal of the base and putting the two electron pairs on the skew diagonal on the opposite face. It is not clear whether the electrons belong at the vertices or a little nearer the vertical axis.

We assume that the hydrogen bond lengths are still r; I assume that the lone pair bond lengths are shorter still than for ammonia. In turn this means that the angle between the two hydrogen bonds become smaller still than in ammonia.

We have now three repulsions: between the electron pairs, between the hydrogen atoms and between any cross-pair (electron pair and hydrogen). The nucleus attracting them to hold them close but the repulsions define the relative positions.

Writing the vector equation that assumes neutral balance along the vertical axis of the diagram (that you drew and I didn’t), we have

2r.r + 2 λr.λr = 0, which becomes 2r2cosθ + 2 λ2r2 cosβ = 0 where θ is the angle between the hydrogen bonds and β is the postulated angle between the two lone pairs.

(i)  Suggest values for  λ and β, explaining your choices.                                            [4]

Bond length varies with bond strength, temperature and pressure. Bond strength depends upon temperature, pressure, bond angle and the local dielectric constant. For water, a typical value for r is 197pm (197x10-12m). The two lone pairs on the oxygen atom in the water molecule mean that more hydrogen bonds can be formed with other water molecules. This makes bonds between water molecules strong, which explains the high values for melting and boiling (in comparison with, say, hydrogen fluoride).

(ii)  Draw a table comparing the melting point, boiling point (and, if you can, viscosity) of water, and three other simple substances of a similar structure, choosing from FON&H. Use your understanding of the molecule structure to comment on these values.                 [6,4]

4. The situation for sulphur dioxide, H2S, is slightly different to that of water because the centre atom is bigger by a shell. There are again two lone pairs in the outermost shell.

Suggest a reason for the bond angle being 92º                                                      [3]

5.  Compare the boiling points of NH3, H2O and HF with PH3, H2S and HCl respectively. Explain this by referring to the hydrogen bonding.                                                              [4]

6.  Explain the difference between a covalent bond and a hydrogen bond.             [4]

7.  Put these bonds & forces in order of strength, weakest first: 
(a) dipole-dipole; (b) van der Waals force, (c) ion-dipole, (d) hydrogen bond, (e) ionic lattice energy bond, (f) covalent bond.                                                                                [4]
Also, class these as intra-molecular and inter-molecular bonds, which you might call weak and strong but you must decide which term is appropriate.                                           [2]

DJS 20140201

tabbing doesn’t transfer from Word - very irritating

intra-something means inside the thing; e.g. intramural = within the walls
inter-somethings  means between them, e.g. international is between nations
extra-something means outside it, e.g. extra-curricular is outside the curriculum, the teaching plan.

Historically, DJS father, HIS, intended his doctorate (of 1946-51 or so) to establish a theoretical value for the bond angle of water. What I offer here is based only on my own modelling, though I looked up the information quoted (bond angles and lengths particularly. Anyone who can explain to me how to calculate a bond length will be quoted extensively.)

Thanks to CJS for fixing my grosser errors - and for being so nice about what I’d written.

1   v 109.47º     vi   cos-1(1/3) (very easy method)

2.  107.3º

3.  cos-1(-0.25) = 104..48º.  λ2 cosβ = ¼ , λ<1.

4.  The angle between the two hydrogen bonds becomes smaller still than in H20, because the change of size of the centre atom.

6  Hydrogen bond is a strong dipole-to-dipole attraction, not a true bond. A covalent bond is where two atoms share a pair of electrons, such as the bond that makes simple gaseous molecules such as H2, O2, N2.

7.  Typical values of disassociation energy in kcals/mol, thanks to Wikipedia

   a  0.5-2; b  (London) 1-15, so much the same as d;  c>d; d  1-12; e 250-4000; f 30-260;

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