percentage of ionic character in NaCl
Answers
Explanation:
We generally calculate percentage of ionic character when observed dipole moment and actual dipole moment are given. But here, they are not mentioned. So, let's go for the generality by using the concept of Electronegativity.
- Electronegativity of Na is (x1) 0.93 (approx)
- Electronegativity of Cl is (x2) 3.36(approx)
We calculate the percentage of ionic character using this data as follows !!
Percentage of ionic character = 16| x1 - x2|+ 3.5|x1 - x2 |^2
So, 16(2.43)+3.5(2.43)^2
= 38.88+20.66715
= 59.55%
Therefore the percentage of ionic character of NaCl is 59.55%.
Answer:
The Correct Answer is 59.55%
Explanation:
The electronegativity of the two atoms involved is used to determine the ionic character (or polarity) of a connection. The larger the difference, the stronger the bond's ionic nature. The arrangement of polar bonds through the 3D structure of the molecule can be used to determine the polarity of the entire compound.
When the observed dipole moment and the true dipole moment are known, we usually calculate the proportion of ionic character. However, they are not addressed in this question. So, let's use the concept of Electronegativity to achieve universality.
Na has an electronegativity of (x1) 0.93.
Cl has an electronegativity of (x2) 3.36.
Percentage of ionic character = 16| x1 - x2|+ 3.5|x1 - x2 |^2
So, 16(2.43)+3.5(2.43)^2
= 38.88+20.66715
= 59.55%
As a result, NaCl's ionic character percentage is 59.55 percent.