(a) Draw the Lewis structures for at least four species that have the general formula $$[ : x \equiv Y :]^{n}$$ where \(\mathrm{X}\) and \(\mathrm{Y}\) may be the same or different, and \(n\) may have a value from \(+1\) to \(-2 .\) (b) Which of the compounds

A Lewis structure is a representation of the valence electrons of an atom, molecule, or ion. In a Lewis structure, the chemical symbol for the element is used along with dots to represent its valence electrons. To draw a Lewis structure, we need to know the valence electrons of the elements involved and follow certain rules: 1. Count the total number of valence electrons from all the atoms in the molecule. 2. Place the least electronegative atom in the center and surround it with the remaining atoms. 3. Connect each surrounding atom to the central atom with a single bond (a line). 4. Fill the octets of the surrounding atoms with the remaining electrons. 5. If there are electrons left, place them around the central atom.

Since the given formula contains a triple bond between X and Y, the elements forming the species must be capable of forming triple bonds. Elements like nitrogen (N), carbon (C), and oxygen (O) can form triple bonds. We can choose different combinations for X and Y from these elements to create the four required species.

We will choose the following combinations for X and Y: 1. X = N, Y = N 2. X = C, Y = N 3. X = C, Y = C 4. X = N, Y = O (a) Lewis structures for the species: 1. N ≡ N: Nitrogen has 5 valence electrons. Both nitrogens share three pairs of electrons, forming a triple bond. Additionally, each nitrogen has one lone pair of electrons. The structure is: \[ [\,:\mathrm{N}\equiv\mathrm{N}:\,] \] 2. C ≡ N: Carbon has 4 valence electrons, and nitrogen has 5 valence electrons. The carbon and nitrogen atoms share three pairs of electrons, forming a triple bond. The nitrogen atom has one lone pair of electrons. The structure is: \[ [\mathrm{C}\equiv\mathrm{N}:\,] \] 3. C ≡ C: Carbon has 4 valence electrons. Both carbon atoms share three pairs of electrons, forming a triple bond. The structure is: \[ [\mathrm{C}\equiv\mathrm{C}] \] 4. N ≡ O: Nitrogen has 5 valence electrons, and oxygen has 6 valence electrons. The nitrogen and oxygen atoms share three pairs of electrons, forming a triple bond. The oxygen atom has one lone pair of electrons. The structure is: \[[:\mathrm{N}\equiv\mathrm{O}\,] \] (b) Analyzing the charges of each compound: 1. N ≡ N: There is no net charge, as both nitrogen atoms share electrons equally. 2. C ≡ N: The carbon atom has a formal charge of +1, and the nitrogen atom has a formal charge of -1. The net charge of the compound is 0. 3. C ≡ C: There is no net charge, as both carbon atoms share electrons equally. 4. N ≡ O: The nitrogen atom has a formal charge of -1, and the oxygen atom has a formal charge of +1. The net charge of the compound is 0. In conclusion, we have drawn Lewis structures for four different species that satisfy the given formula. The charges of the species are as described in part (b) of the solution.