Lewis Structures and the Octet Rule: A Simple Method to write Lewis Structures > Simple Procedure for writing Lewis Structures – Example #2 > Lewis Structures for NO2+ and HCN - Examples #3
A simple procedure for writing Lewis structures is given in a previous article entitled “Lewis Structures and the Octet Rule”. Relevant worked examples were given in the following articles: Examples #1, #2.
A simple procedure for writing Lewis structures is given in a previous article entitled “Lewis Structures and the Octet Rule”. Relevant worked examples were given in the following articles: Examples #1, #2.
A few more examples for writing Lewis structures following the above procedure are given bellow:
Let us consider the case of NO2+:
Step 1: The central atom will be the N atom since it is the less electronegative. Connect the N with the O atoms with single bonds:
Step 2: Calculate the # of electrons in π bonds (multiple bonds) using formula (1) in the article entitled “Lewis Structures and the Octet Rule”.
Where n in this case is 3 since NO2+ consists of three atoms.
Where V = (6 + 5 + 6) - 1 = 16
Therefore, P = 6n + 2 – V = 6 * 3 + 2 – 16 = 4 \ there are 4 π electrons in NO2+ \
2 double bonds must be added to the structure of Step 1 or 1 triple bond.
Step 3 & 4: Two double bonds between N and O are added to the structure in step 1. Alternatively, 1 triple bond is added between N and O. Unshared electron pairs are added so that there is an octet of electrons around each atom. All the equivalent resonance structures are drawn by delocalizing electron pairs. Therefore, the Lewis structures for NO2+ are as follows:
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Figure 1: Lewis structures for the NO2+. The first Resonance form is of lower energy than the remaining two because of less charge separation. It will therefore make the largest contribution to the ground state. |
Let us consider the case of hydrogen cyanide, HCN:
Step 1: The central atom will be the C atom since it is the less electronegative (H is a terminal atom – it cannot be a central atom). Connect the atoms with single bonds
Step 2: Calculate the # of electrons in π bonds (multiple bonds) using formula (1) in the article entitled “Lewis Structures and the Octet Rule”.
Where n in this case is 2 since HCN consists of three atoms but one of them is a H atom.
Where V = (1 + 4 + 5) = 10
Therefore, P = 6n + 2 – V = 6 * 2 + 2 – 10 = 4 \ there are 4 π electrons in HCN \
2 double bonds must be added to the structure of Step 1 or 1 triple bond.
Step 3 & 4: Since only two electrons can be accommodated by a H atom double bonds cannot be added to the structure of Step 1. Alternatively, 1 triple bond is possible between C and N atoms. Unshared electron pair is added so that there is an octet of electrons around each atom.. Therefore, the Lewis structures for HCN are as follows:
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| Figure 2: Lewis structures for HCN |
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