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Topic: Conceptual doubt in Molecular Orbital Theory  (Read 4356 times)

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Offline umangarora

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Conceptual doubt in Molecular Orbital Theory
« on: July 30, 2011, 12:25:12 PM »
According to my teacher, Stability of N2 , N2- and N2+ is in the order N2 > N2+ = N2-.
BUT,In hydrogen, stability is in the order H2 > H2+ > H2-.
I know its beacuse in H- and H+ have same bond order, 1/2. and that is H- an electron is present in antibonding orbital. i wanna know why it doesnt apply to Nitrogen. and what is the reason for equating stability of N2+ and N2-.
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Offline Cavillus

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Re: Conceptual doubt in Molecular Orbital Theory
« Reply #1 on: July 30, 2011, 10:06:56 PM »
Quote from: umangarora on July 30, 2011, 12:25:12 PM
According to my teacher, Stability of N2 , N2- and N2+ is in the order N2 > N2+ = N2-.
BUT,In hydrogen, stability is in the order H2 > H2+ > H2-.
I know its beacuse in H- and H+ have same bond order, 1/2. and that is H- an electron is present in antibonding orbital. i wanna know why it doesnt apply to Nitrogen. and what is the reason for equating stability of N2+ and N2-.

IMHO, the stability order should be the same, because the explanation is the same even in this case.

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Offline umangarora

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Re: Conceptual doubt in Molecular Orbital Theory
« Reply #2 on: July 31, 2011, 06:58:18 AM »
Thanks. Thats what i thought too, but my answer on the test was wrong, apparently :P
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Offline cheese (MSW)

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Re: Conceptual doubt in Molecular Orbital Theory
« Reply #3 on: August 22, 2011, 02:04:12 AM »
N2: 10 valence e-  σ1(2e-) σ2 (2e-) π1(4e-) σ3(2e-)      π2*(0e-) σ4*(0e-)
Bond Order = ½[Σ(bonding e-) - Σ(antibonding e-)]
bo = ½[Σ σ1(2e-) + π1(4e-)] = 3.0 (σ2, σ3 are considered nonbonding or their effects cancel)
[N2]^+    bo =? Take σ3 to be bonding;  N2^- bo = ?   π2*(1e-)
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