Count one exchange for each pair of same-spin electrons among the valence ( $3 d$ and $4 s$ ) electrons. By how many exchanges does the actual ground state of Cr ( $3 d^{5} 4 s^{1}$ ) exceed the hypothetical $3 d^{4} 4 s^{2}$ state? (In $3 d^{4} 4 s^{2}$ , the four $d$ electrons and one $4 s$ electron are taken as spin-up and the second $4 s$ electron as spin-down.)

Answer: D

Actual Cr: all six valence electrons are parallel, giving

(2 6) = 15

exchanges. In

3 d^{4} 4 s^{2}

: five spin-up electrons give

(2 5) = 10

, and the lone spin-down electron gives 0. Difference

= 15 - 10 = 5