Question

Q1)

Use molecular orbital theory to predict whether or not each of the following molecules or ions should exist in a relatively stable form.

Drag the appropriate items to their respective bins.

C2 2+

Be2 2+

Li2

Li2 2-

*Will exist:

*Will not exist:

Q2)

Part A

What is the electron-domain (charge-cloud) geometry of ClF5?

Part B

What is the molecular geometry of ClF5?

Enter the molecular geometry of the molecule.

Part C

Ignoring lone-pair effects, what is the smallest bond angle in ClF5?

Express your answer as an integer.

Part D

Which choice best describes the polarity of ClF5?

Which choice best describes the polarity of ?

	The molecule is polar and has polar bonds.
	The molecule is nonpolar and has polar bonds.
	The molecule is polar and has nonpolar bonds.
	The molecule is nonpolar and has nonpolar bonds.

Answer 1

Concepts and reason

Molecular orbitals: Molecular orbitals are formed by linear combination of atomic orbitals. Atomic orbitals and molecular orbitals of a molecule can be shown in a molecular orbital diagram.

To predict the stable form of the molecules or ions first calculate bond orders of each molecule based on their electron configuration of molecular orbitals.

Determine the electron domain geometry and molecular geometry of the molecule based on the number of electron pairs around the central atom.

First find out number of atoms present around the central atom of a molecular geometry, then identify the possible shape of the geometry. Finally indicate possible number of different number of electron-domain geometries that are consistent with the molecular geometry.

Determine the bond angles based on molecular geometry. Each type of shape possesses specific bond angles due to the repulsions between bond pairs and lone pairs of electrons.

Fundamentals

Bond order: Half of the difference between numbers of electrons in bonding orbitals to number of electrons in antibonding orbitals is called as bond order of a molecule.

${\rm{Bond order = }}\frac{1}{2}\left[ {\left( \begin{array}{l}\\{\rm{Number of electrons}}\\\\{\rm{in bonding orbitals }}\\\end{array} \right) - \left( \begin{array}{l}\\{\rm{Number of electrons}}\\\\{\rm{in antibonding orbitals }}\\\end{array} \right)} \right]$

Electron pairs that are not involved in the bond formation are called as lone pairs. Pair of electrons that involved in the bonding are called as bond pairs. Number of atoms around the central atom is equal to number of bond pairs.

Q1

Electron configuration for molecular orbitals of ${{\rm{C}}_{\rm{2}}}^{{\rm{2 + }}}$ is as follows:

${\sigma _{1s}}^2 < {\sigma ^*}{_{1s}^2} < {\sigma _{2s}}^2 < {\sigma ^*}{_{2s}^2} < {\pi _{2{p_y}}}^1 = {\pi _{2{p_z}}}^1 < {\pi _{2{p_x}}}^0 < {\pi ^*}{_{2{p_y}}^0} = {\pi ^*}{_{2{p_z}}^0} < {\pi ^*}{_{2{p_x}}^0}$

$\begin{array}{c}\\{\rm{Bond order = }}\frac{1}{2}\left[ {{\rm{6}} - {\rm{4}}} \right]\\\\ = 1\\\end{array}$

Because the bond order is 1, the ion will exist.

Electron configuration for molecular orbitals of ${\rm{B}}{{\rm{e}}_{\rm{2}}}^{{\rm{2 + }}}$ is as follows:

${\sigma _{1s}}^2 < {\sigma ^*}{_{1s}^2} < {\sigma _{2s}}^2 < {\sigma ^*}{_{2s}^0} < {\pi _{2{p_y}}}^0 = {\pi _{2{p_z}}}^0 < {\pi _{2{p_x}}}^0 < {\pi ^*}{_{2{p_y}}^0} = {\pi ^*}{_{2{p_z}}^0} < {\pi ^*}{_{2{p_x}}^0}$

$\begin{array}{c}\\{\rm{Bond order = }}\frac{1}{2}\left[ {4 - 2} \right]\\\\ = 1\\\end{array}$

Because the bond order is 1, so the ion will exist.

Electron configuration for molecular orbitals of ${\rm{L}}{{\rm{i}}_2}$ is as follows:

${\sigma _{1s}}^2 < {\sigma ^*}{_{1s}^2} < {\sigma _{2s}}^2 < {\sigma ^*}{_{2s}^0} < {\pi _{2{p_y}}}^0 = {\pi _{2{p_z}}}^0 < {\pi _{2{p_x}}}^0 < {\pi ^*}{_{2{p_y}}^0} = {\pi ^*}{_{2{p_z}}^0} < {\pi ^*}{_{2{p_x}}^0}$

$\begin{array}{c}\\{\rm{Bond order = }}\frac{1}{2}\left[ {4 - 2} \right]\\\\ = 1\\\end{array}$

Because the bond order is 1, the ion will exist.

Electron configuration for molecular orbitals of ${\rm{L}}{{\rm{i}}_{\rm{2}}}^{2 - }$ ion is as follows:

${\sigma _{1s}}^2 < {\sigma ^*}{_{1s}^2} < {\sigma _{2s}}^2 < {\sigma ^*}{_{2s}^2} < {\pi _{2{p_y}}}^0 = {\pi _{2{p_z}}}^0 < {\pi _{2{p_x}}}^0 < {\pi ^*}{_{2{p_y}}^0} = {\pi ^*}{_{2{p_z}}^0} < {\pi ^*}{_{2{p_x}}^0}$

$\begin{array}{c}\\{\rm{Bond order = }}\frac{1}{2}\left[ {4 - 4} \right]\\\\ = 0\\\end{array}$

Because the bond order is 0, the ion will not exist.

Q2. A

Central atom of the molecule is chlorine. Chlorine is surrounded with five fluorine atoms. Electron domain geometry of ${\rm{Cl}}{{\rm{F}}_5}$ is octahedral.

Q2. B

Molecular geometry of ${\rm{Cl}}{{\rm{F}}_5}$ is square pyramidal.

Q2. C

Geometry of the molecule is square pyramidal. The smallest bond angle in the square pyramidal geometry is ${90^{\rm{o}}}$ .

Q2. D

The molecule ${\rm{Cl}}{{\rm{F}}_5}$ is polar and it has polar bonds.

Ans: Part Q1

The specie ${{\rm{C}}_{\rm{2}}}^{{\rm{2 + }}}$ , ${\rm{B}}{{\rm{e}}_{\rm{2}}}^{{\rm{2 + }}}$ and ${\rm{L}}{{\rm{i}}_2}$ are exist, and the ion ${\rm{L}}{{\rm{i}}_{\rm{2}}}^{2 - }$ does not exist.

Part Q2. A

Electron domain geometry of ${\rm{Cl}}{{\rm{F}}_5}$ is octahedral.

Part Q2. B

Molecular geometry of ${\rm{Cl}}{{\rm{F}}_5}$ is square pyramidal.

Part Q2. C

The smallest bond angle in the ${\rm{Cl}}{{\rm{F}}_5}$ is ${90^{\rm{o}}}$ .

Part Q2. D

The molecule ${\rm{Cl}}{{\rm{F}}_5}$ is polar and it has polar bonds.