Chapter 9, Problem 9B.2P

(a)

Interpretation Introduction

Interpretation:

The probability of the presence of an electron at nucleus A in the bonding and also in the antibonding orbitals is to be calculated.

Concept introduction:

Electronic configuration tells about the arrangement of the electrons in each subshell and each orbital of an atom.  The distribution of the electrons in the molecules is described by the molecular orbital theory.  The unoccupied molecular orbital having lowest energy is known as the LUMO and the occupied molecular orbital having highest energy is known as the HOMO.

Answer to Problem 9B.2P

The probability of the presence of an electron at nucleus A in the bonding orbital is $\underset{\_}{8.7\times 10^{-7}}$. The probability of the presence of an electron at nucleus A in the antibonding orbital is $\underset{\_}{1.9\times 10^{-6}}$.

Explanation of Solution

It is given that a small-electron probe having volume $1.00{\text{pm}}^{\text{3}}$ is inserted in ${\text{H}}_{\text{2}}^{+}$ molecule-ion in the ground state.

The value of R is $2.00a_0$.

The value of $N_{+}$ for bonding is $0.56$.

The value of $N_{-}$ for antibonding is $1.10$.

The expression that is used represent probability of finding an electron is given below.

    $P_{\pm }={\text{N}}_{\pm }^2\left(\frac{1}{\pi a_0^3}\right){\left(e^{-r_A/a_0}\pm e^{-r_B/a_0}\right)}^2\delta \tau \mathrm{.................}\left(1\right)$

Where,

$\delta \tau$ is the volume.

$r_A$ is the distance measure from point A.

$r_B$ is the distance measure from point B.

The distance between A and B is $\text{106}\text{pm}$ as shown in the Figure below.

Figure 1

The value of $a_0$ is $\text{52}\text{.9}\text{pm}$.

Substitute the values of $\delta \tau$, $a_0$ $r_A$ as $0$ and $r_B$ as $\text{106}\text{pm}$, $N_{+}$ for bonding orbital in equation (1).

    $\begin{array}{c}{P}_{+}={\left(0.56\right)}^2\left(\frac{1}{3.14\times {\left(52.9\text{pm}\right)}^3}\right){\left(e^{-0/52.9\text{pm}}+e^{-160\text{pm}/52.9{\text{pm}}_0}\right)}^2\times 1{\text{pm}}^{\text{3}}\\ =0.3136\times \left(\frac{1}{3.14\times \left(148035.889{\text{pm}}^3\right)}\right){\left(0+e^{-3.02457}\right)}^2{\text{pm}}^{\text{3}}\\ =0.3136\times \left(\frac{1}{\left(464832.6915{\text{pm}}^3\right)}\right){\left(0.048578\right)}^2{\text{pm}}^{\text{3}}\\ =\underset{\_}{8.7\times 10^{-7}}\end{array}$

Substitute the values of $\delta \tau$, $a_0$ $r_A$ as $0$ and $r_B$ as $\text{106}\text{pm}$, $N_{-}$ for antibonding orbital in equation (1).

    $\begin{array}{c}{P}_{-}={\left(1.10\right)}^2\left(\frac{1}{3.14\times {\left(52.9\text{pm}\right)}^3}\right){\left(e^{-0/52.9\text{pm}}-e^{-160\text{pm}/52.9{\text{pm}}_0}\right)}^2\times 1{\text{pm}}^{\text{3}}\\ =1.21\times \left(\frac{1}{3.14\times \left(148035.889{\text{pm}}^3\right)}\right){\left(0-e^{-3.02457}\right)}^2{\text{pm}}^{\text{3}}\\ =1.21\times \left(\frac{1}{\left(464832.6915{\text{pm}}^3\right)}\right){\left(0.048578\right)}^2{\text{pm}}^{\text{3}}\\ =\underset{\_}{1.9\times 10^{-6}}\end{array}$

Therefore, the probability of the presence of an electron at nucleus A in the bonding orbital is $\underset{\_}{8.7\times 10^{-7}}$. The probability of the presence of an electron at nucleus A in the antibonding orbital is $\underset{\_}{1.9\times 10^{-6}}$.

(b)

Interpretation Introduction

Interpretation:

The probability of the presence of an electron at nucleus B in the bonding and also in the antibonding orbitals is to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 9B.2P

The probability of the presence of an electron at nucleus B in the bonding orbital is $\underset{\_}{8.7\times 10^{-7}}$. The probability of the presence of an electron at nucleus B in the antibonding orbital is $\underset{\_}{1.9\times 10^{-6}}$.

Explanation of Solution

It is given that a small-electron probe having volume $1.00{\text{pm}}^{\text{3}}$ is inserted in ${\text{H}}_{\text{2}}^{+}$ molecule-ion in the ground state.

The value of R is $2.00a_0$.

The value of $N_{+}$ for bonding is $0.56$.

The value of $N_{-}$ for antibonding is $1.10$.

The expression that is used represent probability of finding an electron is given below.

    $P_{\pm }={\text{N}}_{\pm }^2\left(\frac{1}{\pi a_0^3}\right){\left(e^{-r_A/a_0}\pm e^{-r_B/a_0}\right)}^2\delta \tau \mathrm{.................}\left(1\right)$

Where,

$\delta \tau$ is the volume.

$r_A$ is the distance measure from point A.

$r_B$ is the distance measure from point B.

The distance between A and B is $\text{106}\text{pm}$ as shown in the Figure below.

Figure 1

The value of $a_0$ is $\text{52}\text{.9}\text{pm}$.

Substitute the values of $\delta \tau$, $a_0$, $r_A$ as $\text{106}\text{pm}$ and $r_B$ as $0$,, $N_{+}$ for bonding orbital in equation (1).

    $\begin{array}{c}{P}_{+}={\left(0.56\right)}^2\left(\frac{1}{3.14\times {\left(52.9\text{pm}\right)}^3}\right){\left(e^{-160\text{pm}/52.9{\text{pm}}_0}+e^{-0/52.9\text{pm}}\right)}^2\times 1{\text{pm}}^{\text{3}}\\ =0.3136\times \left(\frac{1}{3.14\times \left(148035.889{\text{pm}}^3\right)}\right){\left(e^{-3.02457}+0\right)}^2{\text{pm}}^{\text{3}}\\ =0.3136\times \left(\frac{1}{\left(464832.6915{\text{pm}}^3\right)}\right){\left(0.048578\right)}^2{\text{pm}}^{\text{3}}\\ =\underset{\_}{8.7\times 10^{-7}}\end{array}$

Substitute the values of $\delta \tau$, $a_0$, $r_A$ as $\text{106}\text{pm}$ and $r_B$ as $0$, $N_{-}$ for antibonding orbital in equation (1).

    $\begin{array}{c}{P}_{-}={\left(1.10\right)}^2\left(\frac{1}{3.14\times {\left(52.9\text{pm}\right)}^3}\right){\left(e^{-160\text{pm}/52.9{\text{pm}}_0}-e^{-0/52.9\text{pm}}\right)}^2\times 1{\text{pm}}^{\text{3}}\\ =1.21\times \left(\frac{1}{3.14\times \left(148035.889{\text{pm}}^3\right)}\right){\left(e^{-3.02457}-0\right)}^2{\text{pm}}^{\text{3}}\\ =1.21\times \left(\frac{1}{\left(464832.6915{\text{pm}}^3\right)}\right){\left(0.048578\right)}^2{\text{pm}}^{\text{3}}\\ =\underset{\_}{1.9\times 10^{-6}}\end{array}$

Therefore, the probability of the presence of an electron at nucleus B in the bonding orbital is $\underset{\_}{8.7\times 10^{-7}}$. The probability of the presence of an electron at nucleus B in the antibonding orbital is $\underset{\_}{1.9\times 10^{-6}}$.

(c)

Interpretation Introduction

Interpretation:

The probability of the presence of an electron at half way between A and B in the bonding and also in the antibonding orbitals is to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 9B.2P

The probability of the presence of an electron at half way between A and B in the bonding orbital is $\underset{\_}{3.63\times 10^{-7}}$. The probability of the presence of an electron at half way between A and B in the antibonding orbital is $\underset{\_}{0}$.

Explanation of Solution

It is given that a small-electron probe having volume $1.00{\text{pm}}^{\text{3}}$ is inserted in ${\text{H}}_{\text{2}}^{+}$ molecule-ion in the ground state.

The value of R is $2.00a_0$.

The value of $N_{+}$ for bonding is $0.56$.

The value of $N_{-}$ for antibonding is $1.10$.

The expression that is used represent probability of finding an electron is given below.

    $P_{\pm }={\text{N}}_{\pm }^2\left(\frac{1}{\pi a_0^3}\right){\left(e^{-r_A/a_0}\pm e^{-r_B/a_0}\right)}^2\delta \tau \mathrm{.................}\left(1\right)$

Where,

$\delta \tau$ is the volume.

$r_A$ is the distance measure from point A.

$r_B$ is the distance measure from point B.

The distance between A and B is $\text{106}\text{pm}$ so half way becomes $\text{53}\text{pm}$ as shown in the Figure below.

Figure 1

The value of $a_0$ is $\text{52}\text{.9}\text{pm}$.

Substitute the values of $\delta \tau$, $a_0$, $r_A$ as $\text{53}\text{pm}$ and $r_B$ as $\text{53}\text{pm}$, $N_{+}$ for bonding orbital in equation (1).

    $\begin{array}{c}{P}_{+}={\left(0.56\right)}^2\left(\frac{1}{3.14\times {\left(52.9\text{pm}\right)}^3}\right){\left(e^{-\text{53}\text{pm}/52.9{\text{pm}}_0}+e^{-\text{53}\text{pm}/52.9\text{pm}}\right)}^2\times 1{\text{pm}}^{\text{3}}\\ =0.3136\times \left(\frac{1}{3.14\times \left(148035.889{\text{pm}}^3\right)}\right){\left(e^{-1.002}+e^{-1.002}\right)}^2{\text{pm}}^{\text{3}}\\ =0.3136\times \left(\frac{1}{\left(464832.6915{\text{pm}}^3\right)}\right){\left(0.734288\right)}^2{\text{pm}}^{\text{3}}\\ =\underset{\_}{3.63\times 10^{-7}}\end{array}$

Substitute the values of $\delta \tau$, $a_0$, $r_A$ as $\text{53}\text{pm}$ and $r_B$ as $\text{53}\text{pm}$, $N_{-}$ for antibonding orbital in equation (1).

    $\begin{array}{c}{P}_{-}={\left(1.10\right)}^2\left(\frac{1}{3.14\times {\left(52.9\text{pm}\right)}^3}\right){\left(e^{-\text{53}\text{pm}/52.9{\text{pm}}_0}-e^{-\text{53}\text{pm}/52.9\text{pm}}\right)}^2\times 1{\text{pm}}^{\text{3}}\\ =1.21\times \left(\frac{1}{3.14\times \left(148035.889{\text{pm}}^3\right)}\right){\left(e^{-1.002}-e^{-1.002}\right)}^2{\text{pm}}^{\text{3}}\\ =1.21\times \left(\frac{1}{\left(464832.6915{\text{pm}}^3\right)}\right){\left(0\right)}^2{\text{pm}}^{\text{3}}\\ =\underset{\_}{0}\end{array}$

Therefore, the probability of the presence of an electron at half way between A and B in the bonding orbital is $\underset{\_}{3.63\times 10^{-7}}$. The probability of the presence of an electron at half way between A and B in the antibonding orbital is $\underset{\_}{0}$.

(d)

Interpretation Introduction

Interpretation:

The probability of the presence of an electron at a point $\text{20}\text{pm}$ along the bond from A and $\text{10}\text{pm}$ perpendicularly in the bonding and also in the antibonding orbitals is to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 9B.2P

The probability of the presence of an electron at a point $\text{20}\text{pm}$ along the bond from A and $\text{10}\text{pm}$ perpendicularly in the bonding orbital is $\underset{\_}{4.9\times 10^{-7}}$. The probability of the presence of an electron at a point $\text{20}\text{pm}$ along the bond from A and $\text{10}\text{pm}$ perpendicularly in the antibonding orbital is $\underset{\_}{5.5\times 10^{-7}}$.

Explanation of Solution

It is given that a small-electron probe having volume $1.00{\text{pm}}^{\text{3}}$ is inserted in ${\text{H}}_{\text{2}}^{+}$ molecule-ion in the ground state.

The value of R is $2.00a_0$.

The value of $N_{+}$ for bonding is $0.56$.

The value of $N_{-}$ for antibonding is $1.10$.

The expression that is used represent probability of finding an electron is given below.

    $P_{\pm }={\text{N}}_{\pm }^2\left(\frac{1}{\pi a_0^3}\right){\left(e^{-r_A/a_0}\pm e^{-r_B/a_0}\right)}^2\delta \tau \mathrm{...................}\left(1\right)$

Where,

$\delta \tau$ is the volume.

$r_A$ is the distance measure from point A.

$r_B$ is the distance measure from point B.

The distance between A and B is $\text{106}\text{pm}$ so half way becomes $\text{53}\text{pm}$ as shown in the Figure below.

Figure 1

The value of $a_0$ is $\text{52}\text{.9}\text{pm}$.

Substitute the values of $\delta \tau$, $a_0$, $r_A$ as $\text{22}\text{.4}\text{pm}$ and $r_B$ as $\text{86}\text{.6}\text{pm}$, $N_{+}$ for bonding orbital in equation (1).

    $\begin{array}{c}{P}_{+}={\left(0.56\right)}^2\left(\frac{1}{3.14\times {\left(52.9\text{pm}\right)}^3}\right){\left(e^{-\text{22}\text{.4}\text{pm}/52.9\text{pm}}+e^{-\text{86}\text{.6}\text{pm}/52.9\text{pm}}\right)}^2\times 1{\text{pm}}^{\text{3}}\\ =0.3136\times \left(\frac{1}{3.14\times \left(148035.889{\text{pm}}^3\right)}\right){\left(e^{-0.4234}+e^{-1.637}\right)}^2{\text{pm}}^{\text{3}}\\ =0.3136\times \left(\frac{1}{\left(464832.6915{\text{pm}}^3\right)}\right){\left(0.849382\right)}^2{\text{pm}}^{\text{3}}\\ =\underset{\_}{4.9\times 10^{-7}}\end{array}$

Substitute the values of $\delta \tau$, $a_0$, $r_A$ as $\text{22}\text{.4}\text{pm}$ and $r_B$ as $\text{86}\text{.6}\text{pm}$, $N_{-}$ for antibonding orbital in equation (1).

    $\begin{array}{c}{P}_{-}={\left(1.10\right)}^2\left(\frac{1}{3.14\times {\left(52.9\text{pm}\right)}^3}\right){\left(e^{-\text{22}\text{.4}\text{pm}/52.9\text{pm}}-e^{-\text{86}\text{.6}\text{pm}/52.9\text{pm}}\right)}^2\times 1{\text{pm}}^{\text{3}}\\ =1.21\times \left(\frac{1}{3.14\times \left(148035.889{\text{pm}}^3\right)}\right){\left(e^{-0.4234}-e^{-1.637}\right)}^2{\text{pm}}^{\text{3}}\\ =1.21\times \left(\frac{1}{\left(464832.6915{\text{pm}}^3\right)}\right){\left(0.460258\right)}^2{\text{pm}}^{\text{3}}\\ =\underset{\_}{5.5\times 10^{-7}}\end{array}$

Therefore, the probability of the presence of an electron at a point $\text{20}\text{pm}$ along the bond from A and $\text{10}\text{pm}$ perpendicularly in the bonding orbital is $\underset{\_}{4.9\times 10^{-7}}$. The probability of the presence of an electron at a point $\text{20}\text{pm}$ along the bond from A and $\text{10}\text{pm}$ perpendicularly in the antibonding orbital is $\underset{\_}{5.5\times 10^{-7}}$.