Tutorial Questions 2

You have now completed several areas of the lecture course, below again are some problems to understand with answers supplied. Following consultation with Dr Tony Haynes (of the department), I decided to restrict my discussion of VSEPR to coordination numbers 2, 3 and 4, and not to consider the greater steric requirements of double bonds. This is scheduled for CHM102 later this semester. Consequently, the second part of Question 3 is not strictly relevant, and the comparison of S=O to S-Cl in part (c) has not been covered in the lectures, although they should be able to determine the overall coordination polyhedron. Benefit is certainly to be had by reading up on these topics.

Questions

1) Look up the covalent radii and first ionisation energies of the Group 2 elements [Be ==> Ba] and of the elements of the second main period (principal quantum number n=3) [Na ==> Cl], and see if they follow the same trends as those of Group 1 and the first main period.

2) Determine the Pauling electronegativity differences () for HCl, HBr and BrCl from tabulated bond dissociation energies (look in a text book for values) and check if they are in agreement with each other as a set, and whether they agree with tabulated values of the Pauling electronegativities.

3) Using a VSEPR approach, determine the likely geometries of:-

(a) H₂S (as compared with H₂O);

(b) IF₇ ;

(c) SOCl₂ (look up the bonding if you do not know it).

Answers

1)

• Covalent radii (pm) of Be(89), Mg(136), Ca(174), Sr(191), Ba(198) - progressive increase as expected.
• First ionisation energies (kJ mol^-1) of Be(900), Mg(738), Ca(590), Sr(549), Ba(502) - progressive decrease as expected.
• Covalent radii (pm) of Na(156), Mg(136), Al(125), Si(117), P(110), S(104), Cl(99) - progressive decrease as expected.
• First ionisation energies (kJ mol^-1) of Na(495), Mg(738), Al(577), Si(787), P(1060), S(1000), Cl(1255) - general increase as expected, but with dips at ionisation of first 3p electron and first from a pair in a 3p-orbital.

• Using the information that: Hdiss (kJ mol^-1) for HCl(428), HBr(362), BrCl(216), H₂(432), Cl₂(240), Br₂(190).
• One can then use the relationship below to give the desired answer.

  




• (H v Cl)=0.98; H v (Br)=0.73; (Br v Cl)=0.10 (ill-determined): (H)=2.20; (Cl)=3.16; (Br)=2.96.

3)

• (a) H₂S has two bond pairs and two lone pairs, four electron pairs in total; geometry based on a tetrahedron with H-S-H bond angle reduced by greater LP-Lp and LP-BP interactions than Bp-BP. Since S is less electronegative than O there is a higher probability of finding the BP closer in the bond towards O than the S, thereby subtending a smaller angle at S. If there is less electron density near the S then there will be less electron electron repulsions, allowing the H-S-H to be less than the H-O-H angle.
• (b) IF7 has seven bonding pairs and no lone pairs and is pentagonal bipyramidal.
• (c) SOCl₂ has two S-Cl single bonds, one S=O double bond and a sulphur lone pair, it is tetrahedral with the lone pair and S=O compressing the S-Cl together.