Question

Rank these bonds from most ionic to most covalent in character.

Ca-F
K-F
F-F
Br-F
Cl-F

Thanks.

Answer 1

Concepts and reason

The concept used is to ranks the bonds from most ionic to most covalent among the given compounds.

Fundamentals

An ionic bond is formed by the transfer of electrons between the atoms. Metal donates electrons to non-metals.

A covalent bond is formed by the sharing of electrons between the atoms so as to attain stable electronic configuration.

When the electronegativity difference between the atoms is greater than 2, it is an ionic bond.

When the electronegativity difference between the atoms is $\left( {1.00{\rm{ to }}2.00} \right)$ , it is most polar covalent bond.

When the electronegativity difference between the atoms is $\left( {0.41{\rm{ to }}1.00} \right)$ , it is slightly polar covalent bond.

When the electronegativity difference between the atoms is $\left( {0.00{\rm{ to }}0.41} \right)$ , it is non-polar covalent bond.

The bond type in ${\rm{Ca}} - {\rm{F}}$ is ionic bond. It is formed by the transfer of electrons from ${\rm{Ca}}$ to F.

The bond type in ${\rm{K}} - {\rm{F}}$ is ionic bond. It is formed by the transfer of electrons from ${\rm{K}}$ to F.

The bond type in ${\rm{F}} - {\rm{F}}$ is covalent bond. It is formed by the sharing of electrons between two F atoms.

The bond type in ${\rm{Br}} - {\rm{F}}$ is ionic bond. It is formed by the sharing of electrons between ${\rm{Br}}$ and F atoms.

The bond type in ${\rm{Cl}} - {\rm{F}}$ is ionic bond. It is formed by the sharing of electrons between ${\rm{Cl}}$ and F atoms.

Based on the electronegativity difference, the order of the compounds from most ionic to most covalent is as follows:

$\begin{array}{l}\\{\rm{K}} - {\rm{F}}\\\\{\rm{Ca}} - {\rm{F}}\\\\{\rm{Br}} - {\rm{F}}\\\\{\rm{Cl}} - {\rm{F}}\\\\{\rm{F}} - {\rm{F}}\\\end{array}$

Ans:

Therefore, the order of the compounds from most ionic to most covalent is as follows:

$\begin{array}{l}\\{\rm{K}} - {\rm{F}}\\\\{\rm{Ca}} - {\rm{F}}\\\\{\rm{Br}} - {\rm{F}}\\\\{\rm{Cl}} - {\rm{F}}\\\\{\rm{F}} - {\rm{F}}\\\end{array}$