DEPARTMENT OF CHEMISTRY
C10J - Structure and Bonding  
Problem Paper #2  (Week of Nov. 19, 2001)

1.	Cis-H(CH3)C=CH(CH3): 	E, C2, ,  / 
	
(the molecule is flat, so  is in the molecular plane, assuming Me groups are spherical)

Trans-H(CH3)C=CH(CH3): 	E, C2, ,  i. 

C2 is at the centre of the double bond and perpendicular to plane of the molecule, which is also planar. 
 
 is therefore in the plane of the molecule. 

(b)	XeOF4 :  E, C4, C2, 2v , 2v/

2 v  lie along the F-Xe-F bonds.

2 v/  bisect the F-Xe-F bond angles 

2.	

2  =  s + pz  (bonding)

3*  =  s - pz (antibonding)

                                                                             
See Basic Inorg. Chem. 3rd Edn. pg. 104 for diagrams.

The other orbitals are nonbonding.

(b)	Lewis representation of HF is consistent with MO treatment:

that is, one bonding pair of electrons shared by both H and F, and 3 lone pairs (non-bonding) associated with F.

3.	Electron configuration of B: 1s2  2s2 2p1

Electron confign of B2 : 12 (2*)2 12
B2 is paramagnetic with 2 unpaired electrons, 
The bond order is (4 - 2) x ½  = 1 , which corresponds to a single bond.

(i)	B2-  is obtained by adding one electron to the 1  orbital;
Electron confign of B2- : 12 (2*)2 13

B2-  is still paramagnetic with 1 unpaired electron, but the bond order is now (5 - 2) x ½  = 1.5, which corresponds to a larger bond order (the bond is intermediate in strength between a single bond and a double bond).  The bond should be shorter and its energy is greater than in B2 .  

(ii)	B2+  is obtained by subtracting one electron from the 1 orbital;
Electron confign of B2+ : 12 (2*)2 11

B2+  is still paramagnetic with 1 unpaired electron, but the bond order is now (3 - 2) x  ½  = 0.5, which corresponds to a lower bond order .  The bond should be longer and its energy is smaller than in B2 .  

4.	An AB3 solid should have a cross section as shown, with the smaller cation in the centre:

From diagram above:

 CBD = 30o  

Cos  30o = r-/ (r+  +  r-) = 0.866

0.866 r+  +  0.866 r-  =  r-

0.866r+ = r-(1-0.866)

r+/r- = 0.134/0.866 = 0.155

5.	Using trigonometry (see diagram of  2-D arrangement of ions below): 

For (a):		Central cation near to 4 anions at distance r
			Central cation near to 4 cations at distance r√2

			E =                { 4 - 2√ }

For (b):		Central cation near to 4 cations at distance 2r
			Central cation near to 4 cations at distance r2√2
			Central cation near to 8 anions at distance r√5

So, E =                { -2 – 2/√2  +  8/ √5}

For (c):		Central cation near to 4 anions at distance 3r
			Central cation near to 4 cations at distance r3√2
			Central cation near to 8 anions at distance r√13
			Central cation near to 8 cations at distance r√10

E =               { 4/3 –  4/3√2   +  8/√13  -  8/ √10}

M =  (a)  +  (b)  +  (c) terms in brackets

		(a)  4 – 2.828  =  1.172
		(b)  –2 –1.414 + 3.578 = 0.164
		(c)  1.333 – 0.942  + 2.219  - 2.53 = 0.08
	
M  = 1.172 + 0.164 + 0.08 = 1.42

6.	Hf MgCl2(s) = Hatm + Hdiss + 1st and 2nd Hion  +  HEA   + U
	
U =  Hf MgCl2(s) – (Hatm + Hdiss + 1st and 2nd Hion  +  HEA)

	   =  - 644 – ( 147.7 + 243.4  + 737.7  + 1450.6  - 348.7 x 2) kJ/mol

	   =  - 2526 kJ/mol