Chapter 3 : Chemical Bonding

Example 3.4 pageno : 59

In [1]:
# Variables
e1 = -13.6;			#ionization potential of hydrogen in volts
n = 3.; 			#third orbit
m = 5.; 			#fifth orbit

# Calculations
e3 = e1/(n**2);			#Energy of third orbit in volts
e5 = e1/(m**2);			#Energy of fifth orbit in volts

# Results
print "Energy of third orbit (in volts)  =  %.2f eV"%e3
print "Energy of fifth orbit (in volts)  =  %.2f eV"%e5

Energy of third orbit (in volts)  =  -1.51 eV
Energy of fifth orbit (in volts)  =  -0.54 eV

Example 3.5 pageno : 61

In [2]:
# Variables
d = 0.9178;			#molecular distance in angstorm

# Calculations
d1 = d*10**-10;	    		#in m
e = 1.602*10**-19;			#in C
dm = e*d1;      			#dipole moment in fully ionic state in Cm
dm_m = 6.375*10**-30;			#meaured dipole moment in Cm
p = (dm_m*100)/dm;			#Percentage ionic character


# Results
print "Dipole moment of HF in fully ionic state (in Cm)  =  %.3e Cm"%dm
print "Percentage ionic character : %.1f %%"%p

Dipole moment of HF in fully ionic state (in Cm)  =  1.470e-29 Cm
Percentage ionic character : 43.4 %