Chapter 4. Group III
Answers to Questions

1.  boric acid = B(OH)3   borax = Na2B4O5(OH)4.8H2O
    gypsum = CaSO4.2H2O  alumina = Al2O3

2.          (a)           (c)              (d)
           

 (b) because it has 6 valence electrons;

3.   borax, Na2B4O5(OH)4.8H2O = Na2B4O7.10 H2O
   kernite, Na2B4O5(OH)4.2H2O = Na2B4O7.4 H2O

4. bauxite, AlO(OH)

5. The Bayer process:

AlO(OH)(s) + NaOH(aq) + H2O(l)  NaAl(OH)4(aq)
SiO2(s) + 2 NaOH(aq)  Na2SiO3(aq) + H2O(l)
NaAl(OH)4(aq)  Al(OH)3(s) + NaOH(aq)
Al(OH)3(s)   Al2O3(s) + H2O
2 AlO(OH)   Al2O3 + H2O
2 Al2O3   4 Al(l) + 3 O2(g)


6.  6 HF + Al(OH)3 + 3 NaOH   Na3AlF6 + 6 H2O

7. It is the electrochemical process by which aluminum is obtained from alumina (Al2O3).  The Hall process involves the use of an artificial cryolite ore, Na3AlF6. which serves to lower the melting point of alimina:
 

Al2O3/Na3AlF6 + 3 e-   Al(l)


8. Boron is the only non-metal and there are no metalloids; Al, Ga, In, and Tl are all metals.

9. carbon, silicon

10.

 
 

11.  (a) Td (b) Oh (c) D3h (d) D2h

12.  Na = +1; B = +3; O = -2; H = +1

13. ability to conduct electricity, malleability, ductility, luster, etc.

14.
               Fe2O3(s) + 2 Al(s)   2 Fe(s) +  Al2O3
DHf         -822                0                         0         -1676
 

 DHrxn = -854 kJ


15.

 (a) 2 Ga(s) + 6 HCl(aq)   3 H2(g) + 2 GaCl3
 (b) 6 H2O + 2 In(s) + 2 OH-(aq)   3 H2(g) + 2 [In(OH)4]-


16.
 

Compound:
Uses (a - h):
Formula  (1 - 8):
alum
f
2
alumina
a
1
borax
a, b, c, d
6
boric acid
g
3
boric oxide
c
5
diborane
e
4
sodium borohydride
 h
7
sodium perborate
 b
8

17. see a periodic table