If electronegativity difference between two atoms is 2 then percentage of ionic character is. please answer with solution?

Question

Answer

You can use Hanny and Smith formula for this purpose%ionic character=16 (X) - 3.5 (X)^2where X= diff. in EN of 2 elements

Answer

Answer

Calculating Percentage of Ionic Character in a Compound with Electronegativity Difference of 2

When two atoms bond together, the electronegativity difference between them determines the type of bond that is formed. If the electronegativity difference is greater than 1.7, the bond is considered ionic. If the electronegativity difference is less than 1.7 but greater than 0.5, the bond is considered polar covalent. If the electronegativity difference is less than 0.5, the bond is considered nonpolar covalent.

The electronegativity difference between two atoms can be calculated by subtracting the electronegativity value of the less electronegative atom from the electronegativity value of the more electronegative atom. For example, the electronegativity value of chlorine is 3.16 and the electronegativity value of sodium is 0.93. The electronegativity difference between chlorine and sodium is:

3.16 - 0.93 = 2.23

This means that the bond between chlorine and sodium is ionic because the electronegativity difference is greater than 1.7.

To calculate the percentage of ionic character in a compound with an electronegativity difference of 2, we can use the following equation:

% ionic character = [(electronegativity difference) / (electronegativity difference + 3.2)] x 100

In this case, the electronegativity difference is 2. Plugging in this value into the equation, we get:

% ionic character = [(2) / (2 + 3.2)] x 100
% ionic character = (0.625) x 100
% ionic character = 62.5%

This means that the bond between the two atoms in the compound has about 62.5% ionic character and 37.5% covalent character.

In conclusion, the percentage of ionic character in a compound with an electronegativity difference of 2 can be calculated using the equation % ionic character = [(electronegativity difference) / (electronegativity difference + 3.2)] x 100. In this case, the bond has about 62.5% ionic character and 37.5% covalent character.

Answer

The percentage ionic character is related to electronegativity difference between the two atoms. It is given by Hannay and Smith relationship.
%age ionic character.
= 16 (XA − XB) + 3.5 (XA − XB)2
Here, XA and XB are electronegativities of two atoms A and B
\ % ionic character = 16(2) + 3.5(2)2 = 46%

Answer

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Answer

According to me my answer is 64.9%.
Formula for %of ionic character is
( 1-e^(Xa-Xb/4))*100.
(1-e^(2/4))*100=0.64872*100=64.872=64.9.

Answer

What is hubble constant

Answer

%age ionic character=16(Xa-Xb)+3.5(Xa-Xb)2.
=(16×2.)+3.5(2)2
=46%

Answer

16(2)+3.5(2)^2=32+14=46% ans

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