NAME___Average 8/10_________________

Chemistry 101

Quiz—Chapter 7 Nov. 3, 2003

NA = 6.022 x 1023                            

h = 6.63 x 10 -34 J.s/photon               c = 3.0 x 10 8 m/s

E = hn             E= hc/l           R = 0.082 L atm / K mol = 8.314 J / K mol

 


            Calculate the wavelength of the photon involved in the transition from n=2 to n=3 in the hydrogen atom.  Draw an energy diagram and state whether the photon is emitted or absorbed.

 

 

 

 

 

 

 

 


A photon is absorbed.

Note:  the value of RH = 2.18 x 10 –18 J (wrong exponent in data provided).  Using the value of RH provided you would get l = 6.52 x 10 –43 m

 

DE = 2.18 x 10 –18 J ( 1/4 – 1/9) = 3.05 x 10 –19 J

l = h c / D E = 6.63 x 10 –34 J.s x 3 x 10 8 m/s / 3.05 x 10 –19 J =

 6.52 x 10 –7 m = 652 nm

  1. Write the electronic configuration for fluorine, Z=9.

1s 2 2 s2 2p5

 

  1. How many electrons can be found in the 3p orbital?

6 electrons – 2 each in px, py, and pz.

 

  1. Explain why Chromium’s electronic configuration is: [Ar]4s13d5 and not [Ar]4s23d4.

Having a half-filled d shell has a special stability. (p 221)

 

  1. Given the following electronic configuration:

a)    

 
 
Is this paramagnetic?  Why?  Yes, because 2 electrons are unpaired. (p 215)

b)    Circle the valence electrons. 4 electrons