Question

The dipole moment of $LiH$ is $1.964 \times {10^{ - 29}}cm$ and the interatomic distance between $Li$ and $H$ in this molecule is $1.596\mathop {\text{A}}\limits^ \circ $. What is the percentage ionic character in $LiH$?
A.$75$
B.$76.8$
C.$79.8$
D.$100$

Answer 1

Hint: Dipole moment: It is defined as the product of charge of an atom in a molecule to the distance between the atoms of the molecule. The unit of dipole moment is Coulomb-centi-meter, also known as Diopter. Percentage ionic character is defined as the ratio of observed dipole moment to the theoretical dipole moment multiply by $100$.

Complete step by step solution:
First of all we will talk about bonds.
Bonds: In molecules atoms are linked together by bonds. Bonds are of many types, for example:
Ionic bond: The bond in the molecules which is formed by give and take of electrons i.e. one atom gives the electron and other atoms take the electron.
Covalent bonds: The bond in the molecules in which the atoms share their electrons to each other. There are two types of covalent bonds: polar covalent bond and nonpolar covalent bond.
Polar covalent bond: The covalent bonds in which the electronegativity difference of the two atoms is more i.e. one atom can attract the electron of another atom, are known as polar covalent bonds.
Non-polar covalent bond: The covalent bonds in which the electronegativity difference of the two atoms is less i.e. one atom cannot attract the electron of another atom, are known as non-polar covalent bonds.
Now we will talk about dipole moment and dipole length.
Dipole moment: It is defined as the product of charge of an atom in a molecule to the distance between the atoms of the molecule. The unit of dipole moment is Coulomb-centi-meter, also known as Diopter.
Dipole length: It is defined as the distance between the atoms of the molecules. It is measured in Angstrom.
Percentage ionic character is defined as the ratio of observed dipole moment to the theoretical dipole moment multiply by $100$.
In the question we are given with the observed dipole moment as $1.964 \times {10^{ - 29}}cm$.
Dipole length is given as $1.596\mathop {\text{A}}\limits^ \circ $.
So theoretical dipole moment will be as one coulomb charge multiply by dipole length:
Theoretical dipole moment $ = (1.6 \times {10^{ - 19}}) \times 1.596 \times {10^{ - 10}}C - cm$
Theoretical dipole moment $ = 2.5536 \times {10^{ - 29}}C - cm$.
So percentage ionic character will be as :
$ = \dfrac{{1.964 \times {{10}^{ - 29}}}}{{2.5536 \times {{10}^{ - 29}}}} \times 100$

$ = 76.8\% $

Hence, option B is correct.

Note: Electronegativity is defined as the tendency of an atom to attract a bonding pair of electrons. The atom which has high electronegativity will take the electrons and the atom which has low electronegativity will lose the electrons.