Question

Write Lewis structures for these compounds. Show all valence electrons. None of them contains a ring of atoms. (a) Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\) (b) Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\) (c) Methanol, \(\mathrm{CH}_{3} \mathrm{OH}\)

Short answer

Answer: The Lewis structures for the given compounds are as follows: 1. Hydrogen peroxide (H2O2): H-O-O-H, with each Oxygen atom having two lone pairs of electrons. 2. Hydrazine (N2H4): H-N-N-H, with each Nitrogen atom bonded to two Hydrogen atoms, each other, and carrying one lone pair of electrons. 3. Methanol (CH3OH): CH3O-H, with the Carbon atom bonded to three Hydrogen atoms, an Oxygen atom, and the Oxygen atom bonded to another Hydrogen atom and having two lone pairs of electrons.

Step-by-step solution

Determine the number of valence electrons for each compound

For each compound, we will add up the number of valence electrons that each type of atom contributes. (a) Hydrogen peroxide, H2O2: Hydrogen has 1 valence electron, Oxygen has 6 valence electrons. So the total is 2(1) + 2(6) = 14 valence electrons. (b) Hydrazine, N2H4: Nitrogen has 5 valence electrons, Hydrogen has 1 valence electron. So the total is 2(5) + 4(1) = 12 valence electrons. (c) Methanol, CH3OH: Carbon has 4 valence electrons, Hydrogen has 1 valence electron, and Oxygen has 6 valence electrons. So the total is 1(4) + 4(1) + 1(6) = 12 valence electrons.

Draw the Lewis structures with the given valence electrons

(a) For H2O2, arrange two Oxygen atoms in the center and two Hydrogen atoms on the sides, with the following bonds: H-O-O-H. Now, distribute the remaining 10 valence electrons (14 - 4 used in forming the single bonds), as evenly as possible to satisfy the octet rule for each Oxygen atom. The final Lewis structure is H-O-O-H with a single bond between each adjacent atom and each Oxygen atom having two lone pairs of electrons. (b) For N2H4, arrange two Nitrogen atoms in the center and two Hydrogen atoms on each side of each Nitrogen atom, with single bonds between Nitrogen and Hydrogen atoms. Distribute the remaining 4 valence electrons (12 - 8 used in forming the single bonds) to form a single bond between the two Nitrogen atoms and have each Nitrogen atom carrying one lone pair of electrons. Thus, the final Lewis structure is H-N-N-H with each Nitrogen atom bonded to two Hydrogen atoms and also to each other. (c) For CH3OH, arrange the Carbon atom in the center and three Hydrogen atoms around the Carbon atom, forming single bonds. Next, connect the Oxygen atom to the Carbon atom with a single bond and attach one Hydrogen atom to the Oxygen atom. Distribute the remaining 4 valence electrons (12 - 8 used in forming the single bonds) by providing two lone pairs of electrons to the Oxygen atom. The final Lewis structure is CH3O-H with the Carbon atom bonded to three Hydrogen atoms, an Oxygen atom, and the Oxygen atom bonded to another Hydrogen atom.