Lewis Structure of the Guanidinium Ion

Question

In a problem, I was asked to find the Lewis structure of the guanidinium ion $\ce{C(NH2)3}^{+1}$. I followed the following steps which led me to an incorrect structure and I was hoping someone could help me find my mistake.

1. The total number of valence electrons is given by $4+3(5+2(1)) +1 = 26.$
2. The total number of electrons with each atom having an octet is $8+3(8+2(2)) = 44.$
3. The number of bonding electrons is thus $44-26 = 18.$
4. I drew the following skeletal structure which includes $9$ bonds (corresponding to $18$ bonding electrons):

5. I have $26 - 18 = 8$ non-bonding electrons remaining, which I assigned as lone pairs to the three nitrogens and carbon:

However, this structure is clearly incorrect because carbon rarely forms three bonds. The correct structure (including resonance) is

Answer 1

You missed a sign. When you incorporate the charge into your electron count, you subtract positive charges and add negative charges because an electron is negatively charged.

$4+[3×(5+2)]\color{red}{+}1=26$ 🚫

$4+[3×(5+2)]\color{blue}{-}1=24$ ✔️

With the correct electron count you should now be able to render a double bond from the carbon to one of the nitrogen atoms. In reality the pi bond this represents is delocalized through the whole ion, providing an extra stabilization that makes guanidine about as strongly basic as sodium hydroxide.