Calculating effective nuclear charge

  • Thread starter escryan
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  • #1
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Homework Statement



If ionization energy is 899.4 kJ/mol for Be, what is the effective nuclear charge?

Homework Equations


Zeff = Z - S

E=RH(Z2/n2) ??
E=RH(Zeff2/n2)??

The Attempt at a Solution



My attempted solution was subbing into
Zeff = Z - S
Zeff = 4 - 2
= 2

But I suspect that is wrong... because why ionization energy is given.. so shouldn't it be used in the calculation?

And somewhere I think I read that "S" was supposed to be a "constant" of some sort, and I just subbed 2 in because I thought that it was the number of electrons in the first orbital ?
 

Answers and Replies

  • #2
Borek
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One of the equations you listed contains both ionization energy and effective nuclear charge, why don't you use it?
 
  • #3
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Ah..
So subbing in values
E=899.4 kJ/mol
RH=2.178 x 10-21 kJ
n=1

I get Z2eff= 4.129 x 10 23 mol

How does one get to the units/value of Z2eff after this?
 
  • #4
Borek
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Ionization energy was per mole, not per molecule.
 
  • #5
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Oops, just noticed that "Z2eff= 4.129 x 10 23 mol " should actually read "Z2eff= 4.129 x 10 23 mol-1 "

I think that the italicized part is what confuses me the most -- what are the units for this portion? I'm going to take a guess that it is currently molecules/mol, but if so, is this always the case whenever expressing a value and the unit mol-1?

Like for this example, what was given was in kJ/mol. When the kJs were cancelled, what resulted was just mol-1...
 

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