Question

Draw resonance structures for (a) \(\mathrm{SeO}_{3}\) (b) \(\mathrm{CS}_{3}{ }^{2-}\) (c) CNO \(^{-}\)

Short answer

In summary, the resonance structures are as follows: (a) SeO3: There are three resonance structures, each having a double bond between Se and one O atom. The formal charges are 0 on Se and the bonded O atom, and -1 on the two unbonded O atoms. (b) CS32-: There are three resonance structures, each having a double bond between the C atom and one S atom. The formal charges are +1 on the C atom, 0 on the bonded S atom, and -1 on the two unbonded S atoms. (c) CNO-: There are two resonance structures, one having a triple bond between C and N and a single bond between N and O, and the other having a double bond between C and N and a double bond between N and O. The formal charges on each atom in both resonance structures are 0.

Step-by-step solution

(a) Draw Lewis structure for \(\mathrm{SeO}_{3}\)

First, we need to find the total number of valence electrons for SeO3. Selenium (Se) has 6 valence electrons, and each oxygen (O) atom has 6 valence electrons. So, the total number of valence electrons is \(6(Se) + 3 \times 6(O) = 24\). Now, let's draw the Lewis structure for SeO3: place the Se atom as the central atom and bond each O atom to the central atom, using one line (representing a single bond) between each O and Se atom. This would use 6 electrons for the bonds. Using the remaining 18 electrons, we should complete the octet for each O atoms by adding three lone pairs around each O atoms.

(a) Identify possible resonance positions in \(\mathrm{SeO}_{3}\)

The Se atom has a lone pair in the currently drawn Lewis structure, and it can form double bonds with O atoms to attain a formal charge of zero. So, there could be resonance structures with a double bond between Se and an O atom.

(a) Draw resonance structures for \(\mathrm{SeO}_{3}\)

By moving the lone pair of electrons on Se to form a double bond with an O atom, three possible resonance structures can be drawn: one with a Se=O double bond to each of the O atoms. These resonance structures will have a formal charge of 0 on Se and -1 for the unbonded O atoms. The bonded O atom will have a formal charge of 0. (b)

(b) Draw Lewis structure for \(\mathrm{CS}_{3}^{2-}\)

First, find the total number of valence electrons for CS3 with a 2- charge. Carbon (C) has 4 valence electrons, and each sulfur (S) atom has 6 valence electrons. So, the total number of valence electrons is \(4(C) + 3 \times 6(S) + 2 = 24\). Now, draw the Lewis structure by placing the C atom as the central atom and bond each S atom to the central atom, using one line (representing a single bond) between each S and C atom. This will use 6 electrons for the bonds. Using the remaining 18 electrons, complete the octet for each S atoms by adding three lone pairs around each S atom.

(b) Identify possible resonance positions in \(\mathrm{CS}_{3}^{2-}\)

Currently, the Lewis structure has a formal charge of -1 on each S atom and a formal charge of +2 on the C atom. By moving a lone pair from an S atom to form a double bond with the C atom, we can attain a formal charge of +1 on the C atom and reduce the charge on the S atom forming a double bond to zero, creating multiple resonance structures which may share negative charges distribution.

(b) Draw resonance structures for \(\mathrm{CS}_{3}^{2-}\)

There are three possible resonance structures, in which a double bond is formed between the C atom and one of the S atoms in each structure. The central C atom's formal charge is +1, and the other two S atoms have formal charges of -1. The S atom with the double bond has a formal charge of 0. (c)

(c) Draw Lewis structure for CNO\(^{-}\)

First, find the total number of valence electrons for CNO with a 1- charge. Carbon (C) has 4 valence electrons, nitrogen (N) has 5, and oxygen (O) has 6. So, the total number of valence electrons is \(4(C) + 5(N) + 6(O) + 1 = 16\). Now, let's draw the Lewis structure: connect the C, N, and O atoms in a linear fashion forming a triple bond between C and N, and a single bond between N and O. Then, add a lone pair to the O atom and one lone pair to the N atom. The formal charges on each atom are now 0 for C and O, and -1 for N.

(c) Identify possible resonance positions in CNO\(^{-}\)

There could be a resonance structure with a double bond between C and N atoms, and a double bond between N and O atoms.

(c) Draw resonance structures for CNO\(^{-}\)

We can move one bonding pair of electrons in the C≡N bond to form a lone pair on the N atom and thus, a double bond between the N and O atoms. This results in a resonance structure having a double bond between C and N, and a double bond between N and O. The formal charge on the C, N, and O atoms in this resonance structure are 0, also fulfilling the octet rule.