DEPARTMENT OF CHEMISTRY
C10J - Structure and Bonding
Paper #1 (Week of Oct. 29, 2001)

1.	Use:    O - Si  = 0.1021/2   

	3.5 – 1.7  = (0.102) (466 – (146 x Si-Si BE) 12)12	

(1.8/0.102)2  = 466 – (146 x Si-Si BE) 12

	(Si-Si BE x 146) 12  = 466 - (1.8/0.102)2	
	
Si-Si BE  =  [466 - (1.8/0.102)2 ]2/146 kJ/mol
	
    =  164 kJ/mol

2.	(a)  Electron configuration of K: 1s2 2s2 2p6 3s2 3p6 4s1 and use Z*  = Z - 
(where Z*  = effective nuclear charge; Z = atomic number and  = screening constant)
	
 can be calculated according to the following:
	
All electrons in energy levels higher than that of the electron under consideration (i.e. with n = 3 and 4) contribute zero to , so the 3s23p6  4s1 electrons are not considered.

	Consider only electrons where n = 2 and 1.
Electrons in the same principal quantum number level contribute 0.35 to .
So 7 electrons (with n = 2) contribute to : 7 x 0.35 =  2.15

Electrons in the next underlying shell contribute 0.85 to .
So 2 electrons (with n = 1) contribute to :  2 x 0.85 = 1.7

Total screening,  = 1.7 + 2.45 = 4.15
therefore, Z*  = (Z - ) = 19 – 4.15 = 14.85

(b)  Na: 1s2 2s2 2p6 3s1  and Z = 11.
      Cl: 1s2 2s2 2p6 3s2 3p5  and Z = 17.
	
	Na:	for n = 3 ,  = 0 x 0.35 = 0
		      n = 2,  =  8 x 0.85 = 6.8
		      n = 1,  = 2 x 1 =  2.0	(electrons in deeper shells contribute 1 to )
		total  = 0 + 6.8 + 2.0  =  8.8
So, Z*  = (Z - ) = 11 – 8.8 = 2.20
		
Similarly, for
	Cl:	for n = 3 ,  = 6 x 0.35 =  2.1
		      n = 2,  =  8 x 0.85 = 6.8
		      n = 1,  = 2 x 1 = 2.0
		total  = 2.1 + 6.8 + 2.0  = 10.9
So, Z*  = (Z - ) = 17 – 10.9 =  6.1

(In reality, note that the shielding constant is different for s and p electrons because they have different radial distributions.  The s electron experiences less shielding than a p electron and consequently experiences a slightly higher Z* than a p electron in the same valence shell)

3.	Allred and Rochow eqn:	 = 0.359 Z* / r2 + 0.744    (r is the covalent radius in Å)
	
Need: Si = 1.74 and covalent radius of Si = 1.17 Å 
          S = 2.44 and covalent radius of S = 1.03 Å 

(NB – use covalent radius as written; do not express as 1.17 x 10–10 m, etc)

For Si; upon rearranging, equation becomes:
	
Z*    = [(Si - 0.744) x r2 ]/0.359       
                                = [(1.74 – 0.744) x (1.172]/0.359			
		        = 3.80	

For S;  Z*   = [(S - 0.744) x (1.03)2]/0.359
       = [(2.44 - 0.744) x (1.03)2]/0.359       
       = 5.01  

4.	Lewis structures (lone pairs on the peripheral atoms have been omitted)

5.	Geometries

Zn(NH3)4 2+ - tetrahedral

BrF5 – square pyramidal
AsF5 – trigonal bipyramidal

6.   (a)	The presence of the lone pair in AX5E molecules (square pyramid) causes the bond angles to be less than 90o.  The larger space requirement of lone pair domain causes the four equatorial bonds to move up toward the apex of the square pyramid, thus decreasing the XaxAXeq bond angle.

     (b)  The ligand becomes increasingly more electronegative from I to F. 
Bonding electron density is then drawn away from P so that the space occupied by the bonding domain in the valence shell of P decreases with increasing difference of electronegativity between it and the ligand.  Consequently, the larger space occupied by the lone pair on P results in the bonding domain being pushed more closely together the greater the electronegativity of the ligand.

7. (a) E,  & i generate one operation
    
    (b)	C3 :	C3; C32
	C4 :	C4; C42; C43
	C5 :	C5; C52 ; C53 ; C54.

(c)	ICl4-: (square planar; same as XeF4) E, C4, C2, 2C2 ∕ , 2C2 ∕ ∕  , h , 2v , 2 v ∕, i , S4 (h is in the molecular plane).

	

CHCl3 : (pyramidal)  E, C3 , 3v

	H2CO : (Flat planar; C=O group means it is similar to H2O)
		 E, C2, ,  ∕ .