Hole formalism

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Hole formalism in quantum chemistry states that for many electronic properties one may consider systems with e or (n-e), the number of unoccupied sites or “holes”, to be equivalent.[1] The number of microstates (N) of a system corresponds to the total number of distinct arrangements for “e” number of electrons to be placed in “n” number of possible orbital positions:

N=n!/(e!(n-e)!)

In hole formalism, commutative property of multiplication is used. It means in the above equation (e!(n-e)!)=((n-e)!e!).

Example[edit]

For a set of p orbitals n = 6 since there are 2 positions in each orbital. Therefore, p2 (e=2 and n=6) so,

N = 6!/(2!(6-2)!) = (6*5*4*3*2*1)/((2*1) * (4*3*2*1)) = 15

Based on the hole formalism, microstates of P4 (e = 4 and n = 6) are:

N = 6!/(4!(6-4)!) = 15

so P4 (2 holes) and P2 (4 holes) give equivalent values of N. The same is true for all p, d, f, … systems such as d1/d9 or f2/f12.

See also[edit]

References[edit]

  1. ^ K Veera Reddy (1998). Symmetry And Spectroscopy Of Molecules. New Age International. ISBN 8122411428.