O2 📖 has 12 valence electrons. Its MO configuration is:
\(O_2 = KK(\sigma_{2s})^ ²(\sigma_{2s}^*)^2(\sigma_{2p})^2(\Pi_{2p})^²(\Pi_{2p})^2 (\Pi_{2p}^*)(\Pi_{2p}^*) \)with one electron in each of the \(\Pi_{2p}^*\) orbitals spin aligned (Hund’s rule) and, bond order for \(O_2= \frac12(8-4) = 2.\)
I want to know how can we write MO electrons configurations for \(NO_3^-\)(Nitrate anion) and C6H6 📖(Benzene)?
What is KK in MO electrons configurations of O2 📖?
I know the principal quantum number also identifies the shell. With the increase in the value of ‘n’, the number of allowed orbital increases and are given by \(n^2\).All the orbitals of a given value of ‘n’ constitute a single shell of atom and are represented by the following letters.
n = 1 2 3 4 5 6 7 ........
Shell = K L M N O P Q ........
Molecular orbital electrons configurations of Benzene and Nitrate anion
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Dhamnekar Winod
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Molecular orbital electrons configurations of Benzene and Nitrate anion
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