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Concept explainers
Interpretation:
The Lewis structure of
Concept Introduction:
A covalent bond is a bond that is formed from the mutual sharing of electrons between atoms. Lewis structures are representations of the covalent bond. In this, Lewis symbols show how the valence electrons are present in the molecule.
Steps to write Lewis structures are as follows:
1. The skeleton structure with single bonds between all bonded atoms has to be written
2. Sum the valence electrons of the atoms in the molecule.
(a) For cations, one electron is subtracted for each positive charge.
(b) For anions, one electron is added for each negative charge.
3. Subtract two electrons from total number of valence electrons for each bond in the skeleton structure.
4. Count the number of electrons required to satisfy the octet rule for each atom in the structure. If the number of electrons needed is less than the number remaining, add one bond for every two electrons needed between atoms to attain an octet.
5. The remaining electrons are placed as lone pairs on atoms that need them to satisfy the octet rule.
The formula to calculate formal charge of atom is,
Some molecules do not have only one Lewis structure. The Lewis structures that differ only in the arrangement of multiple bonds are called resonance structures.
Resonance structure comprises of two or more Lewis Structures that describes the arrangement of bond of a single species and include fractional bonds and fractional charges.
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Chapter 9 Solutions
Chemistry Principles And Practice
- Q: Draw the molecular orbital energy level diagram for the following molecules. 1- The SF4 molecule is seesaw molecular geometry and has C2v point group. 2- The Mn(CO)s molecule with C4v point group is square pyramidal.arrow_forwardPlease correct answer and don't use hand ratingarrow_forwardwhen a 0.150 g sample of the compound was burned, it produced 0.138 g CO2 & 0.0566 g H2O. All the nitrogen in a different 0.200 g sample of the compound was converted to NH3, which was found to weigh 0.0238 g. Finally, the chlorine in a 0.125 g sample of the compound was converted to Cl- and by reacting it with AgNO3, all of the chlorine was recovered as the solid AgCl. The AgCl, when dried was found to weigh 0.251 g. What is the empirical formulaarrow_forward
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