Oxidation of the cyanide ion produces the stable cyanate ion, \(\mathrm{OCN}^{-}\) . The fulminate ion, \(\mathrm{CNO}^{-}\), on the other hand, is very unstable. Fulminate salts explode when struck; \(\mathrm{Hg}(\mathrm{CNO})_{2}\) is used in blasting caps. Write the Lewis structures and assign formal charges for the cyanate and fulminate ions. Why is the fulminate ion so unstable? (C is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{N}\) is the central atom in \(\mathrm{CNO}^{-}\) )

For both ions, we will first calculate the total number of valence electrons, then distribute these valence electrons in the structure, and finally optimize the structure by creating double or triple bonds if necessary. For the cyanate ion (OCN⁻): 1. Oxygen has 6 valence electrons, Carbon has 4, and Nitrogen has 5. 2. Additionally, there is 1 extra electron due to the negative charge, making a total of 16 valence electrons. 3. Distribute these electrons to create the structure: Oxygen is single bonded to Carbon, which is single bonded to Nitrogen. Oxygen has 6 nonbonding electrons (3 lone pairs), Carbon has 2 nonbonding electrons (1 lone pair), and Nitrogen has 4 nonbonding electrons (2 lone pairs). 4. Since Carbon has less than an octet, we can form a double bond with Oxygen and a triple bond with Nitrogen to complete the octet for all atoms. For the fulminate ion (CNO⁻): 1. Carbon has 4 valence electrons, Nitrogen has 5, and Oxygen has 6. 2. An extra electron due to the negative charge makes a total of 16 valence electrons. 3. Distribute these electrons to create the structure: Carbon is single bonded to Nitrogen, which is single bonded to Oxygen. Carbon has 4 nonbonding electrons (2 lone pairs), Nitrogen has 3 nonbonding electrons (1 lone pair and 1 single electron), and Oxygen has 6 nonbonding electrons (3 lone pairs). 4. To complete the octet for all atoms, we can form a triple bond between Carbon and Nitrogen, leaving a single bond between Nitrogen and Oxygen.

For each ion, we will calculate the formal charge for each atom using the formula: \[Formal\,Charge = Valence\,Electrons - Nonbonding\,Electrons - \frac{Bonding\,Electrons}{2}\] For cyanate ion (OCN⁻): 1. Oxygen: \(6 - 4 - \frac{4}{2} = -1\) 2. Carbon: \(4 - 0 - \frac{10}{2} = 0\) 3. Nitrogen: \(5 - 2 - \frac{6}{2} = 0\) For fulminate ion (CNO⁻): 1. Carbon: \(4 - 0 - \frac{10}{2} = -1\) 2. Nitrogen: \(5 - 1 - \frac{8}{2} = 0\) 3. Oxygen: \(6 - 6 - \frac{2}{2} = 0\)

The instability of the fulminate ion (CNO⁻) can be attributed to the presence of a negative formal charge on the central Carbon atom. Carbon normally forms bonds with atoms to achieve a formal charge of zero, which is more stable. In the case of the fulminate ion, Carbon has three bonds, yet it still carries a negative formal charge. This property makes the molecule highly reactive and unstable, leading to its explosive nature.