Solutions 2b (Suggested Problems from Chapters 8 and 2) Chem151 [Kua]
8.20 (a) This configuration has only 8 electrons, while the F atom possesses 9 electrons.
(b) This is an excited-state configuration, because the 2s orbital is not full.
(c) This is Pauli-forbidden: s orbitals can hold no more than two electrons.
(d) This configuration uses a non-existent orbital: there is no 1p orbital.
8.24 Use the filling rules to determine the correct configuration, being aware of exceptions.
N (7 electrons): 1s2 2s2 2p3;
Ti (22 electrons): 1s2 2s2 2p6 3s2 3p6 4s2 3d2;
As (33 electrons): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 ;
Xe (54 electrons): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6.
8.26 The ground state for F is 1s2 2s2 2p5. Excited states with no electron having n > 2 can be
formed by moving electrons out of the 1s and/or 2s orbital and placing them in the 2p
orbital. There are only two ways to do this: 1s1 2s2 2p6; and 1s2 2s1 2p6.
8.56 To determine the correct configuration of a cation, first determine the configuration of the
neutral atom, then remove the appropriate number of electrons, removing valence s
electrons before valence d electrons:
Co (27 electrons): 1s2 2s2 2p6 3s2 3p6 4s2 3d7;
Co3+ (remove 3 electrons from Co): 1s2 2s2 2p6 3s2 3p6 3d6;
Quantum numbers: n = 3; l = 2; ml = -2, -1, 0, 1, 2
ms = +1/2 or –1/2
Since there are six electrons, Hund’s Rule suggests 5 up and 1 down spin.
For example:
n l ml m2
3 2 -2 +1/2
3 2 -1 +1/2
3 2 0 +1/2
3 2 +1 +1/2
3 2 +2 +1/2
3 2 0 -1/2
8.58 Total electron spin depends on the number of unpaired electrons. In partially filled
shells, each unpaired electron contributes 1/2 to the total spin.
Gd: [Xe] 6s2 5d1 4f7, each 4f orbital is half filled, so net spin = (7+1)(1/2) = 4;
Sr: [Kr] 5s2, all orbitals are filled, so net spin = 0;
Ag+: [Kr] 4d10 all orbitals are filled, so net spin = 0.
