Solutions to Mini-quiz #2

1.a. Give the formula of an ionic hydride. NaH, LiH, CaH₂ etc

b. Give the formula of a covalent, acidic hydride. HCl, HI, H₂S etc

c. Give the formula of a silver salt that is insoluble in water. AgCl, AgI, Ag₂CO₃, AgNO₂, etc.

2. Complete and balance TWO of the following equations: [4]

a. CO₃^2-(aq) + 2H⁺(aq) ---> CO₂(g) + H₂O(l)

b. CaH₂(s) + 2H₂O(l) ---> H₂(g) + Ca(OH)₂(s)

c. ¹⁴₇N + ¹₀n ---> ³₁T + ¹²₆C

or ⁶₃Li + ¹₀n ---> ³₁T + ⁴₂He

3. (See question paper for diagram.)
[3]

The diagram shows the boiling points of the hydrides ofthe group 14 - 17 elements. Explain why the boiling points of first elements of groups 15, 16, 17 (N, O, F) are higher than those of the other elements in their group while methane has the lowest boiling point of its group. (Briefly explain any special terms you use - continue on back if necessary.)

This is due to hydrogen bonding. Hydrogen bonding occurs between a hydrogen atom bound by a normal covalent bond to a very electronegative element and a lone pair of electrons on another very electronegative element. It is caused by the electrons in the "regular" bond being attracted close to the electronegative element, leaving the H atom's proton exposed and able to attract a lone pair. It is stronger than other forms of intermolecular interations. F, O & N are electronegative and their hydrides have lone pairs hence we see the effect in HF, H₂O and NH₃. Even if C were electronegative enough there is no lone pair available in the methane molecule. The lower elements in groups 15 - 17 are not electronegative enough to show significant hydrogen bonding.

(C) H.J Clase 1999