A particular metal complex has an electronic configuration of np1 nf1. The number of microstates possible in an isolated gaseous state for this metal complex is :
84
A microstate is one distinct way of distributing the electrons among the available spin-orbitals. The configuration here has the two electrons in different subshells, which makes the counting simpler than for equivalent electrons.
Count the spin-orbitals in each subshell. A subshell with orbital quantum number \(l\) has \(2l+1\) orbitals, each holding two spin states, so it offers \(2(2l+1)\) spin-orbitals.
For the p subshell, \(l = 1\), giving \(2(2 \times 1 + 1) = 6\) spin-orbitals.
For the f subshell, \(l = 3\), giving \(2(2 \times 3 + 1) = 14\) spin-orbitals.
Combine them. Because the two electrons are non-equivalent — they are in different subshells — the Pauli principle imposes no extra restriction between them, and any choice for the p electron can pair with any choice for the f electron. The count is therefore a simple product:
\(6 \times 14 = 84\) microstates.
It is worth seeing why the answer is not a combination formula. Had both electrons been in the same subshell, say f2, they would be indistinguishable and we would need \(\binom{14}{2} = 91\) instead. The separate-subshell case avoids that correction entirely.
The values 6 and 14 are the individual subshell counts taken alone.
Hence the number of microstates is 84.
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