Answer:
Percent ionic character of HI bond is 4.91%.
Explanation:
<h3>Given Data:</h3>Measured Dipole = 0.380D
bond distance = d = 161pm = 1.61*10^-8 cm
<h3>Calculation:</h3>% ionic character is determined by following equation:
% ionic= (dipole measured/dipole calculated)*100
Now,
(In above step 3*10^8 is multiplied to convert coulomb into esu)
As,
So,
Now we can % ionic character using above equation:
%ionic=(0.380D/7.728D)*100
% ionic character=4.91%
A molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
The bonding orbital, which would be more stable and encourages the bonding of the two H atoms into , is the orbital that is located in a less energetic state than just the electron shells of the separate atoms. The antibonding orbital, which has higher energy but is less stable, resists bonding when it is occupied.
An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as * orbitals.
Therefore, molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>
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Hence, the correct answer will be option (b)
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