The carbon atom has the electronic structure 2s 2 2p 2 in its ground state. The

Question

The carbon atom has the electronic structure 2s22p2 in its ground state. The ground state and various possible excited states of C are shown in the figure. The following energies are known for the states a to e in the figure, not in any particular order, 0, 7.3eV, 4.1eV, 7.9eV and 1.2eV. Using reasonable arguments matches these energies to the states a to e. Use Hund's rule to establish the ground state with 0eV. If you have to flip a spin to go from the ground to another configuration, that would cost energy. If you have to move an electron from a lower x to p or from p to a higher s, that would cost a lot of energy. Two electrons in the same orbital(obviously with paired electrons) would have substantial Coulombic repulsion energy.

Explanation / Answer

Following the hunds rule we will get The configuration 'a'

So a has energy 0eV

and in b one electrons spin is flipped it takes less enery when compared to others So energy for b = 1.2eV

for c , One electron is excited and is coupled for the transition it takes alot energy than b but less than d since coupling the electrons is exothermic so the resultant becomes less.

energy for c is less than e therefore for c energy = 4.1eV

for d , one electron is excited but is not coupled so d>c but d<e since e has one more higher state transition

energy for d =7.3eV

For e the electron makes to third state which takes much much more than d since transition takes large amount .

So energy for d = 7.9eV

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