# Solutions 19

- Page ID
- 47404

## S19.1

\(CO_2\) is experimentally determined to be diamagnetic. If a student proposed the following structure to explain. Is a valid Lewis diagram? Does this Lewis diagram predict a diamagnetic species and if not, why?

*This is not a correct Lewis structure. The formal charge is not minimized, the right oxygen does not have a full octet, and the carbon is only making two bonds. However, this Lewis structure does predict a diamagnetic species since all the electrons are paired off. *

## S19.2

Does \(\ce{Br_2C=CHBr}\) have a dipole moment?

*Yes. Because the molecule is asymmetrical, the molecule will have a net dipole moment.*

## S19.3

What is the value of \(\psi^2_g\) (The probability density of finding the electron) at either nucleus in \(Li^+_2\) using the below equation for \(\psi_g\) ? What is the value of \(\psi^2_g\) at the midpoint of the bond? Use \(R_e =106\; pm \) and \(S = 0.86\)

\[\psi_g = \dfrac{1}{\sqrt{2(1+S)}}(1s_a + 1s_b) \]

The hydrogenic radial function of a 1s orbital is

\[R_{1,0} (r) = 2 \left(\dfrac{Z}{a_o}\right)^{3/2} e^{-\frac{Zr}{a_o}} \]

with the angular wave function (in this case \(Y_{0,0}\))

\[Y_{0,0} = \dfrac{1}{\sqrt{4\pi}} \]

**A.** First we have to figure out what \(1s_a\) and \(1s_b\) are. These are the 1s orbitals of hydrogen, or the hydrogenic radial functions. The difference between a and b is which of the Li molecules is being referred to but it should not matter in this case. The hydrogenic radial function of a 1s orbital is

\[R_{1,0} (r) = 2 \left(\dfrac{Z}{a_o}\right)^{3/2} e^{-\frac{Zr}{a_o}} \]

That, multiplied with the angular wave function (in this case \(Y_{0,0}\)) gives us the equation of the 1s orbitals of hydrogen.

\[Y_{0,0} = \dfrac{1}{\sqrt{4\pi}} \]

\[R_{1,0}(r) * Y_{0,0} = (2 \left(\dfrac{3}{a_o}\right)^{3/2} e^{-\frac{3r}{a_o}})(\dfrac{1}{\sqrt{4\pi}}) \]

Here we have \(R_e=106 \; pm\), convert to meters \(R_e=106 \times 10^{-12}\). Z is the atomic number of the element, in this case \(Z=3\) for Lithium. \(a_o\) is the Bohr radius and is a constant.

\[\psi_g = \dfrac{1}{\sqrt{2(1+S)}}(2(2 \left(\dfrac{3}{a_o} \right)^{3/2} \times e^{-\dfrac{(318 \times 10^{-12})}{a_o}}\dfrac{1}{\sqrt{4\pi}})) \]

plug in \(S= 0.86\) into the equation

\[\psi_g = \dfrac{1}{\sqrt{2(1+.86)}}(2( 2 \left(\dfrac{3}{a_o}\right)^{3/2} \times e^{-\dfrac{(318 \times 10^{-12})}{a_o}})\dfrac{1}{\sqrt{4\pi}}) \]

Square \(\psi_g\) to get your final answer

\[\psi^2_g = (\dfrac{1}{\sqrt{3.72}}( 4 \left(\dfrac{3}{a_o}\right)^{3/2} \times e^{-\dfrac{(318 \times 10^{-12})}{a_o}})\dfrac{1}{\sqrt{4\pi}}))^2 \]

\[\psi^2_g = \dfrac{9.24}{a_o^2} \times e^{-\dfrac{636 \times 10^{-12}}{a_o}} \]

**B.** For the midpoint, use \(R_e=53\; pm \)

\[\psi^2_g = (\dfrac{1}{\sqrt{3.72}}(4(\dfrac{3}{a_o})^{3/2} \times e^{-\dfrac{(159 \times 10^{-12})}{a_o}})\dfrac{1}{\sqrt{4\pi}}))^2 \]

\[\psi^2_g = \dfrac{9.24}{a_o^2} \times e^{-\dfrac{318 \times 10^{-12}}{a_o}} \]

## S19.4

The first excited states of \(H_2^-\) are formed by exciting an electron from the antibonding \(1\sigma_u\) molecular orbital to the bonding \(2\sigma_g\) orbital. Write the electron configuration of the ground state and excited-state \(H_2^-\) molecules. What is the bond order of \(H_2^-\) in both states? Is \(H_2^-\) diamagnetic or paramagentic in either state?

*\(H_2^-\) is a three electron system with the electron configuration for the first excited state is \( (1\sigma_{1s})^2 (2\sigma_{1s})^1 \). The bond order is \(\dfrac{3}{2}\). The ground state electron configuration for \(H_2^-\) is \( (1\sigma_{1s})^2 (1\sigma_{1s}^*)^1 \). The bond order is \(\frac{1}{2}\). In both states there is an unpaired electron so the ion is paramagnetic. *

## S19.5

Which two second row homonuclear diatomic molecules do not follow conventional Molecular Orbital energy levels of the other second row homonuclear diatomic? Which are they and what is the difference?

*Both \(O_2\) and \(F_2\) have molecular orbital diagrams that do not follow to others in the row with the 2p \(\sigma\) and \(\pi\) bonding orbitals switched in terms of which is lower in energy.*

## S19.6

A excited state of \(O_2^*\) is formed by exciting a single electron in \(O_2\) from the molecular orbital \(1\pi_g\) to the orbital \(3\sigma_u\). Note that \(^*\) is used to represent an excited state species, not an antibonding molecular orbital.

- Write the electron configuration.
- What is the bond order of \(O_2^*\)?

(a)

Ground state electron configuration for \(O_2\):

$$(1\sigma_g)^2(1\sigma_u)^2(2\sigma_g)^2(2\sigma_u)^2(1\pi_u)^4(3\sigma_g)^2(1\pi_g)^2\]

Therefore, the electron configuration for \(O_2^*\):

$$(1\sigma_g)^2(1\sigma_u)^2(2\sigma_g)^2(2\sigma_u)^2(1\pi_u)^4(3\sigma_g)^2(1\pi_g)^1(3\sigma_u)^1\]

b

$$\mbox{Bond Order}=\dfrac{\mbox{# of electrons in bonding MO - # of electrons in antibonding MO}}{2}\]

Bonding MO:

$$(2\sigma_g)^2(1\pi_u)^4(3\sigma_g)^2\]

Antibonding MO: $$(2\sigma_u)^2(1\pi_g)^1(3\sigma_u)^1\]

$$\mbox{Bond Order}=\dfrac{8-4}{2}=2\]