Question

A covalent molecule, X - Y, is found to have a dipole moment 1.5 x 10-29
Cm and a bond length
of 150 pm. The percent ionic character of the bond will be
bi
a) 50%
b) 62.5%
c) 75%
d) 90%​

Answer 1

Answer:

b) is the correct answer

Explanation:

there is formula friend which is dipole observed/dipole calculated( assuming 100% ionic*100

dipole observed = 1.6*10^-19 * 150*10^-12

then do the calculation correctly u will get the right answer

Answer 2

Answer:

The percentage of ionic character of the bond will be 62.5%

Explanation:

• The % age of an ionic character may be calculated by dividing the measured dipole moment by the estimated dipole moment and multiplying by 100.
• The following formula can be used to determine the percentage of an ionic character in a bond:

$%$% $Ionic character =\frac{experimental moment of dipole moment}{theoritical value of dipole moment}$ × $100$

Given,

Observed dipole moment = $1.5$ ×$10^{-29} cm$

Charge for a covalent molecule = $1.602$ ×$10^{-19}C$

Bond length = $150 pm = 150$ × $10^{-12} m$

Calculated dipole moment = $1.602$ × $10^{-19}C$ × $150$ × $10^{-12} m$

                                             =$2.4$ × $10^{-29}D$

% of ionic character = $(\frac{1.5 \\\\X 10^{-29} }{2.4\\X 10^{-29} }) X 100$ = 62.5 %

Therefore, the percentage of ionic character of the bond will be 62.5%

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