Problem 4.67

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Madeline_Foo_3J
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Joined: Wed Sep 21, 2016 3:00 pm

Problem 4.67

Postby Madeline_Foo_3J » Sun Nov 06, 2016 2:04 pm

Based on their valence-shell electron configurations, which of the following species would you expect to have the lowest ionization energy: a) C2+ b) C2 c) C2-

Im confused as to why the manual it put down the (pi 2 p) before the (sig 2 p) for c2 and C2- when they have more than 8 valence electrons ?

Hao 1I
Posts: 26
Joined: Wed Sep 21, 2016 3:00 pm
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Re: Problem 4.67

Postby Hao 1I » Sun Nov 06, 2016 7:14 pm

The molecular orbital diagram is based on the atom's atomic number, not the number of their valence electrons. Therefore for atoms Z<8, the order should be pi then sigma, and for atoms Z greater than or equal 8, the order would be sigma the pi.


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