Question

The accompanying diagrams represent the three lowest energy wave functions for three "atoms." As in all truly molecular states we consider, these states are shared among the atoms. At such large atomic separation, however, the energies are practically equal, so anelectron would be just as happy occupying any combination.

(a) Identify algebraic combinations of the states (for instance, 5+11/2+11/2 ) that would place the electron in each of the three atoms.

(b) Were the atoms closer together, the energies of states 1.11, and III would spread out and an electron would occupy the lowest energy one. Rank them in order of increasing energy as the atoms draw closer together. Explain your reasoning.

Short answer

(a) The combination for first atom is l+ll-2lll , for second atom is I - II and for third atom is l+ll-2lll .

(b) The rank of the energies in increasing order is $E_l<E_{lll}<E_{ll}$.

Step-by-step solution

Given data

The diagram that represents the three lowest energy wave function for three atoms is given in figure 1 asl, ll, lll .

The energies are practically equal to each other.

Concept of the linear combination of the three states

The expression for the linear combination of the three states is given as:

l+a ll+b lll

Where, a and b are the two coefficients.

Step 3:Calculate the linear combination of three states

(a)

The states of each atom are denoted by the signs that represent the wave function.

State I is denoted as, l+1,+2,+3 >.

State II is denoted as, l+1,+2,+3 > .

State III is denoted as, l-1,+3 > .

Calculate the linear combination of three states as follows:

$$\begin{gathered} \mathrm{l}+\mathrm{all}+\mathrm{blll}=\mathrm{l}+1,+2,+3>+\mathrm{al}+1,2,+3>+\mathrm{b}\mathrm{l}-1,+3> \\ =\mathrm{l}\left(1+\mathrm{a}-\mathrm{b}\right)+1.\left(1-\mathrm{a}\right)+2.\left(1+\mathrm{a}+\mathrm{b}\right)+3> \end{gathered}$$

If the electron is kept only in first atom, the electron cannot be in second or third atom and their function is taken equal to zero.

$$\begin{gathered} \left(1-a\right)=0 \\ a=-1 \\ \left(1+a+b\right)=0 \\ b=-2 \end{gathered}$$

Similarly, if the electron is kept in second atom, the function of first and third atom is taken equal to zero.

$$\begin{gathered} \left(1+a-b\right)=0 \\ \left(1+a-b\right)=0 \\ a=-1 \\ b=0 \end{gathered}$$

Similarly, if the electron is kept in third atom, the function of first and second atom is equal to zero.

$$\begin{gathered} \left(1+a-b\right)=0 \\ \left(1-a\right)=0 \\ a=1 \\ b=2 \end{gathered}$$

Thus, the combination for first atom is l+ll-2lll , for second atom is l-ll and for third atom is l+ll+2lll .

Determine the rank of the energies in increasing order

(b)

The wave function of the energy is directly proportional to the wave number.

The atoms are closer together and their energies of each state will spread out and the electron will take the place that has lowest energy.

As the wave function's energy varies directly to its wave number.

Since the atoms are closer to each other, the state I will form one distorted anti-node, state II will have three anti-nodes and state III will have two antinodes.

Thus, the state I has the lowest energy and state II has the highest energy.

Thus, the energies in increasing order are$E_l<E_{lll}<E_{ll}$.

Therefore, the rank of the energies in increasing order is $E_l<E_{lll}<E_{ll}$.