QUESTION: Compare the relative stability of the following species and indicate
their magnetic properties;
- o2
- o2+
- o2-(superoxide)
- o2
- o2+
- o2-(superoxide)
Asked by SIDDHESH
| 18th Sep, 2014,
08:51: PM
Expert Answer:
O2 = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px1= π*2py1)
Bond order = (Nb-Na) /2 = (10-6)/2= 2
It is paramagnetic as it contains two unpaired electrons.
O2+ = = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px1)
Bond order = 2.5
It is paramagnetic as it contains one unpaired electron.
O2+ = = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px2= π*2py1)
Bond order = 1.5
It is paramagnetic as it contains one unpaired electron.
O2 = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px1= π*2py1)
Bond order = (Nb-Na) /2 = (10-6)/2= 2
It is paramagnetic as it contains two unpaired electrons.
O2+ = = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px1)
Bond order = 2.5
It is paramagnetic as it contains one unpaired electron.
O2+ = = (σ1s)2(σ*1s)2 (σ2s)2(σ*2s)2(σ2z)2(π2px2= π2py2) (π*2px2= π*2py1)
Bond order = 1.5
It is paramagnetic as it contains one unpaired electron.
Answered by Prachi Sawant
| 19th Sep, 2014,
12:49: PM
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