The experimental value of the dipole moment of HCl is 1.03 D. The length of the H-Cl bond is $1.275{A^0}$ . The percentage of ionic character in HCl is :
A) 43
B) 21
C) 17
D) 7
Answer
606.6k+ views
Hint: Firstly calculate dipole moment assuming 100 percent ionic character. Then find the ratio of dipole moment when we assume behaviour to be 100 percent ionic to actual experimental dipole moment. This ratio will tell us percentage of ionic character
Dipole moments can be calculated using the formula \[\mu = q \times d\] where q is the charge of particles and d is the distance between charged particles.
Complete step by step answer:
Here, we have to find out percentage ionic character, so to get that we will use the formula as:
Percent ionic character = $\dfrac{{Dipole\; moment\; observed}} {{Calculated\;Dipole\; moment}}\times 100$
Now, from the above formula we know, observed dipole moment, and we need to calculate dipole moment for 100 percent ionic.
Given: Observed Dipole moment, \[\mu = 1.03D\]
Bond length, d= $1.275{A^0}$ = \[1.275 \times {10^{ - 8}}cm\]
Assuming HCl to be 100 percent ionic, then Charge on H = +1, and Charge on Cl = -1.
Therefore, q = 1 unit charge = \[4.8 \times {10^{ - 10}}esu\]
Now, dipole moment when 100 percent ionic can be calculated using same formula as given below:
\[\mu = q \times d\]
Substitute the value of q to be 1, as charge is 1 on both cation and anion, and distance or bond length is also given.
\[\mu = 4.8 \times {10^{ - 10}} \times 1.275 \times {10^{ - 8}} \\
\mu = 6.12 \times {10^{ - 18}}esu.cm \\
\therefore \mu = 6.12D \\\]
We get dipole moment value in unit of esu.cm, so to convert it into Debye (D),
As we know \[{10^{ - 18}}esu.cm = 1D\]
Thus dipole moment = 6.12 D
Now,
Percent ionic character = $\dfrac{{Dipole\; moment\; observed}} {{Calculated\;Dipole\; moment}}\times 100$
Substitute the values, and thus we get:
Percent ionic character = $\dfrac{1.03}{6.12}\times 100$
Percent ionic character = 16.83 % (which is nearly equal to 17 %).
So, thus the correct option is (C) 17.
Additional information: Here we are taking CGS units as we know \[{10^{ - 18}}esu.cm = 1D\] .
Dipole moment is a calculated value which is a product of charge (q) separating and the distance between them (d).
After calculation of percent ionic character, we can also calculate percent covalent character, by using formula,
Percent Covalent character = 100 - percent ionic character.
Note: We should be very careful with units and unit conversion as many units are involved. We can also understand that any bond will have both ionic and covalent character. Thus, we can calculate both ionic and covalent character. This rule was also given by Fajan’s Rule.
Dipole moments can be calculated using the formula \[\mu = q \times d\] where q is the charge of particles and d is the distance between charged particles.
Complete step by step answer:
Here, we have to find out percentage ionic character, so to get that we will use the formula as:
Percent ionic character = $\dfrac{{Dipole\; moment\; observed}} {{Calculated\;Dipole\; moment}}\times 100$
Now, from the above formula we know, observed dipole moment, and we need to calculate dipole moment for 100 percent ionic.
Given: Observed Dipole moment, \[\mu = 1.03D\]
Bond length, d= $1.275{A^0}$ = \[1.275 \times {10^{ - 8}}cm\]
Assuming HCl to be 100 percent ionic, then Charge on H = +1, and Charge on Cl = -1.
Therefore, q = 1 unit charge = \[4.8 \times {10^{ - 10}}esu\]
Now, dipole moment when 100 percent ionic can be calculated using same formula as given below:
\[\mu = q \times d\]
Substitute the value of q to be 1, as charge is 1 on both cation and anion, and distance or bond length is also given.
\[\mu = 4.8 \times {10^{ - 10}} \times 1.275 \times {10^{ - 8}} \\
\mu = 6.12 \times {10^{ - 18}}esu.cm \\
\therefore \mu = 6.12D \\\]
We get dipole moment value in unit of esu.cm, so to convert it into Debye (D),
As we know \[{10^{ - 18}}esu.cm = 1D\]
Thus dipole moment = 6.12 D
Now,
Percent ionic character = $\dfrac{{Dipole\; moment\; observed}} {{Calculated\;Dipole\; moment}}\times 100$
Substitute the values, and thus we get:
Percent ionic character = $\dfrac{1.03}{6.12}\times 100$
Percent ionic character = 16.83 % (which is nearly equal to 17 %).
So, thus the correct option is (C) 17.
Additional information: Here we are taking CGS units as we know \[{10^{ - 18}}esu.cm = 1D\] .
Dipole moment is a calculated value which is a product of charge (q) separating and the distance between them (d).
After calculation of percent ionic character, we can also calculate percent covalent character, by using formula,
Percent Covalent character = 100 - percent ionic character.
Note: We should be very careful with units and unit conversion as many units are involved. We can also understand that any bond will have both ionic and covalent character. Thus, we can calculate both ionic and covalent character. This rule was also given by Fajan’s Rule.
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