Chapter 6, Problem 67QRT

(a)

Interpretation Introduction

Interpretation:

The MO diagram, the number of unpaired electrons, the number bonds in peroxide ion has to be predicted using MO theory.

Concept Introduction:

Molecular orbital theory:

The atomic orbitals of the atoms constituted in a molecule are combined to produce new orbitals are called Molecular Orbitals.

Like atomic orbitals, a molecular orbital can accommodate maximum two electrons and the two electrons must have opposite spins (Pauli Exclusion Principle).

The numbers of MO’s are equals to the number of atomic orbitals are combined in such a way that the linear combination of similar atomic orbitals to form one bonding and one anti-bonding MO’s.

The bonding MO’s are lower in energy than the anti-bonding MO’s.

HOMO is the highest energized occupied orbital in the MO’s.

Relative energy levels of molecules are according to the energy levels of atomic orbitals.

LUMO is the lowest energized orbital in the MO’s.

Bond order can be calculated using below formula

  $\text{bond order}\text{=}\frac{\text{bonding}\text{molecularorbital}\text{-antibonding}\text{molecular}\text{orbitals}}{\text{2}}$

Explanation of Solution

The total number of valence electrons present in Peroxide ion is $14$ electrons.

The molecular orbital diagrams of the peroxide ion can be written as,

  $\begin{array}{l}{\sigma }_{2s}{\sigma }_{2s}*{\pi }_{2p}{\pi }_{2p}{\sigma }_{2p}{\pi }_{2p}*{\pi }_{2p}*{\sigma }_{2p}*\\ O_2^{2-}(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )()\end{array}$

The bond order can be calculated using bonding and anti-bonding orbitals, the bond order is $\frac{(8-6)}{2}=1$.  Hence, the bond between two oxygen atoms is single bond.  The peroxide ions have no unpaired electrons.

Therefore, the Peroxide ion has single bond and no unpaired electrons.

(b)

Interpretation Introduction

Interpretation:

The MO diagram, the number of unpaired electrons, the number bonds in ${\text{B}}_{\text{2}}^{\text{+}}\text{ion}$ has to be predicted using MO theory.

Concept Introduction:

Refer part (a).

Explanation of Solution

The total number of valence electrons present in ${\text{B}}_{\text{2}}^{\text{+}}\text{ion}$ are $5$ electrons.

The molecular orbital diagrams of the ${\text{B}}_{\text{2}}^{\text{+}}\text{ion}$ can be written as,

  $\begin{array}{l}{\sigma }_{2s}{\sigma }_{2s}*{\pi }_{2p}{\pi }_{2p}{\sigma }_{2p}{\pi }_{2p}*{\pi }_{2p}*{\sigma }_{2p}*\\ B_2^{+}(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow )()()()()()\end{array}$

The bond order can be calculated using bonding and anti-bonding orbitals, the bond order is $\frac{(3-2)}{2}=0.5$.  Hence, the bond between Boron and Boron is a half bond.  The ${\text{B}}_{\text{2}}^{\text{+}}\text{ion}$ have one unpaired electrons.

Therefore, the ${\text{B}}_{\text{2}}^{\text{+}}\text{ion}$ has a half bond and one unpaired electrons.

(c)

Interpretation Introduction

Interpretation:

The MO diagram, the number of unpaired electrons, the number bonds in ${\text{Li}}_{\text{2}}^{\text{+}}\text{ion}$ has to be predicted using MO theory.

Concept Introduction:

Refer part (a).

Explanation of Solution

The total number of valence electrons present in ${\text{Li}}_{\text{2}}^{\text{+}}\text{ion}$ is $1$ electrons.

The molecular orbital diagrams of the ${\text{Li}}_{\text{2}}^{\text{+}}\text{ion}$ can be written as,

  $\begin{array}{l}{\sigma }_{2s}{\sigma }_{2s}*{\pi }_{2p}{\pi }_{2p}{\sigma }_{2p}{\pi }_{2p}*{\pi }_{2p}*{\sigma }_{2p}*\\ L{i}_2^{+}(\uparrow )()()()()()()()\end{array}$

The bond order can be calculated using bonding and anti-bonding orbitals, the bond order is $\frac{(1-0)}{2}=0.5$.  Hence, the bond between two Lithium atoms is half bond.  The ${\text{Li}}_{\text{2}}^{\text{+}}\text{ion}$ have one unpaired electrons.

Therefore, the ${\text{Li}}_{\text{2}}^{\text{+}}\text{ion}$ has half bond and one unpaired electrons.

(d)

Interpretation Introduction

Interpretation:

The MO diagram, the number of unpaired electrons, the number bonds ${\text{O}}_{\text{2}}^{\text{+}}\text{ion}$ has to be predicted using MO theory.

Concept Introduction:

Refer part (a).

Explanation of Solution

The total number of valence electrons present in ${\text{O}}_{\text{2}}^{\text{+}}\text{ion}$ is $11$ electrons.

The molecular orbital diagrams of the ${\text{O}}_{\text{2}}^{\text{+}}\text{ion}$ can be written as,

  $\begin{array}{l}{\sigma }_{2s}{\sigma }_{2s}*{\pi }_{2p}{\pi }_{2p}{\sigma }_{2p}{\pi }_{2p}*{\pi }_{2p}*{\sigma }_{2p}*\\ O_2^{2-}(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow \downarrow )(\uparrow )()()\end{array}$

The bond order can be calculated using bonding and anti-bonding orbitals, the bond order is $\frac{(8-3)}{2}=2.5$.  Hence, the bond between two oxygens is half way between a double bond triple bond.  The ${\text{O}}_{\text{2}}^{\text{+}}\text{ion}$ have one unpaired electrons.

Therefore, the ${\text{O}}_{\text{2}}^{\text{+}}\text{ion}$ has half way between a double bond triple bond and one unpaired electrons.