# 22.3.4: iv. Review Exercise Solutions

## Q1

a. For non-degenerate point groups one can simply multiply the representations (since only one representation will be obtained):
$a_1 \otimes b_1 = b_1$
Constructing a "box" in this case is unnecessary since it would only contain a single row. Two unpaired electrons will result in a singlet (S=0, $$M_S$$=0), and three triplets (S=1, $$M_S$$=1, S=1, $$M_S$$=0, S=1, $$M_S=-1$$). The states will be: $$^3B_1(M_S$$=1), $$^3B_1(M_S=0$$), (^3B_1(M_S=-1\)), and $$^1B_1(M_S=0$$).

b. Remember that when coupling non-equivalent linear molecule angular momenta, one simple adds the individual Lz values and vector couples the electron spin. So, in this case $$(1\pi_u^12\pi_u^1)$$, we have $$M_L$$ values of 1+1, 1-1, -1+1, and -1-1 (2,0,0, and -2). The term symbol $$\Delta$$ is used to denote the spatially doubly degenerate level $$(M_L = \pm 2$$) and there are two distinct spatially non-degenerate levels denote by the term symbol $$\sum (M_L=0).$$ Again, two unpaired electrons will result in a singlet ($$S=0 \text{, } M_S=0$$), and three triplets $$(S=1\text{, } M_S=1\text{; } S=1 \text{, } M_S=0 \text{; } S=1 \text{, } M_S=-1$$). The states generate are then:
$^1\Delta \text{ }(M_L=2) \text{; one states } (M_S=0),$
$^1\Delta \text{ }(M_L=-2) \text{; one states } (M_S=0),$
$^3\Delta \text{ }(M_L=2) \text{; one states } (M_S=\text{1, 0, and -1),}$
$^3\Delta \text{ }(M_L=-2) \text{; one states } (M_S=\text{1, 0, and -1),}$
$^1\Delta \text{ }(M_L=0) \text{; one states } (M_S=0),$
$^1\Delta \text{ }(M_L=0) \text{; one states } (M_S=0),$
$^3\Delta \text{ }(M_L=0) \text{; one states } (M_S=\text{1, 0, and -1), and }$
$^3\Delta \text{ }(M_L=0) \text{; one states } (M_S=\text{1, 0, and -1)}$

c. Constructing the "box" for two equivalent $$\pi$$ electrons one obtains:

From this "box" one obtains six states:
$^1\Delta (M_L=2)\text{; one state }(M_S=0),$
$^1\Delta (M_L=-2)\text{; one state }(M_S=0),$
$^1\Delta (M_L=0)\text{; one state }(M_S=0),$
$^3\Delta (M_L=0)\text{; three states }(M_S=\text{1, 0, and -1}),$

d. It is not necessary to construct a "box" when coupling non-equivalent angular momenta since the vector coupling results in a range from the sum of the two individual angular momenta to the absolute value of their difference. In this case, $$3d^14d^1$$, L=4, 3, 2, 1, 0, and S=1, 0. The term symbol are: $$^3G, ^1G, ^3F, ^1F, ^3D, ^1D, ^3P, ^1P, ^3S\text{, and } ^1S.$$ THe L and S angular momenta can be vector coupled to produce further splitting into levels:
$\text{ J = L + S ... |L - S|. }$
Denoting J as a term symbol subscript one can identify all the levels and the subsequent (2J + 1) states:
$^3G_5 \text{ (11 states),}$
$^3G_4 \text{ (9 states),}$
$^3G_3 \text{ (7 states),}$
$^1G_4 \text{ (9 states),}$
$^3F_4 \text{ (9 states),}$
$^3F_3 \text{ (7 states),}$
$^3F_2 \text{ (5 states),}$
$^1F_3 \text{ (7 states),}$
$^3D_3 \text{ (7 states),}$
$^3D_2 \text{ (5 states),}$
$^3D_1 \text{ (3 states),}$
$^1D_2 \text{ (5 states),}$
$^3P_2 \text{ (5 states),}$
$^3P_1 \text{ (3 states),}$
$^3P_0 \text{ (1 states),}$
$^1P_1 \text{ (3 states),}$
$^3S_5 \text{ (3 states), and}$
$^1S_0 \text{ (1 states).}$

e. Construction of a "box" for the two equivalent d electrons generates (note the "box" has been turned side ways for convenience):

The term symbols are: $$^1G \text{, } ^3F \text{, } ^1D \text{, } ^3P \text{, and } ^1S.$$ The L and S angular momenta can be vector coupled to produce further splitting into levels:
$^1G_4 \text{ (9 states),}$
$^3F_4 \text{ (9 states),}$
$^3F_3 \text{ (7 states),}$
$^3F_2 \text{ (5 states),}$
$^1D_2 \text{ (5 states),}$
$^3P_2 \text{ (5 states),}$
$^3P_1 \text{ (3 states),}$
$^3P_0 \text{ (1 states), and}$
$^1S_0 \text{ (1 states).}$