A) \[O_{2}^{2+}>NO\]
B) \[O_{2}^{2+}<NO\]
C) \[O_{2}^{{}}<NO\]
D) \[O_{2}^{2+}=NO\]
Correct Answer: A
Solution :
[a] Idea| This problem includes conceptual mixing of ionisation potential and Molecular Orbital Electronic Configuration (MOEC) of molecule. |
| In order to solve this problem students are advised to write MOEC of molecules/ions. If molecular orbital contain unpaired electron in vacant MO, then it will be loosed easily hence have low value of ionisation energy. |
| MOEC of \[O_{2}^{--}=\sigma _{1s}^{2}\sigma _{1s}^{*\,2}\sigma _{2s}^{2}\sigma _{2s}^{*\,2}\sigma _{2{{p}_{z}}}^{2}\] |
| \[(\pi _{1\pi x}^{2}\equiv \pi _{2{{p}_{y}}}^{\,2})({{\pi }^{*}}_{2{{p}_{x}}}^{2}\equiv {{\pi }^{*}}_{2{{p}_{y}}}^{\,2})\] |
| All electrons are paired hence it is diamagnetic and will not electron easily MOEC of \[{{O}_{2}}={{\sigma }^{*}}_{1s}^{2}{{\sigma }^{*}}_{2s}^{2}\sigma _{2{{p}_{z}}}^{2}\] |
| \[(\pi _{2{{p}_{x}}}^{2}\equiv \pi _{2{{p}_{y}}}^{\,2})\,\,\,\,(\pi _{2{{p}_{x}}}^{1}\equiv \pi _{2{{p}_{y}}}^{\,1})n=2\] |
| Hence it can loose electron more easily due to presence of two electron having identical energy Ionization potential Molecule having available unpaired valence electron in molecular orbital looses electron easily electron show less value of ionization potential |
| MOEC of \[O_{2}^{2+}=\sigma _{1s}^{2}{{\sigma }^{*}}_{1s}^{2}\sigma _{2s}^{2}{{\sigma }^{*}}_{2s}^{2}\sigma _{2{{p}_{z}}}^{2}\pi _{2{{p}_{x}}}^{2}\equiv \pi _{2py}^{2}\] |
| All electrons are paired in molecular orbitals hence is diamagnetic |
| MOEC of \[NO=\sigma _{1s}^{2}\sigma _{1s}^{*2}\sigma _{2s}^{*2}\sigma _{2{{p}_{z}}}^{2}\pi _{2{{p}_{x}}}^{2}\equiv \pi _{2{{s}_{y}}}^{2}\]\[\pi _{2{{p}_{x}}}^{*1}=\pi _{2{{p}_{y}}}^{*}\] |
| Since, NO has unpaired electron present in antibonding molecular orbital \[\pi _{2{{p}_{x}}}^{*},\] hence it will loose the electron easily. |
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