Draw the Lewis structure for HCN. Indicate the hybrid orbitals, and draw a picture showing all the bonds between the atoms, labeling each bond as \(\sigma\) or \(\pi .\)

Step 1 is to determine the total number of valence electrons in the molecule: - Hydrogen (H) has 1 valence electron - Carbon (C) has 4 valence electrons - Nitrogen (N) has 5 valence electrons So, HCN has a total of 10 valence electrons. Next, we will place the atoms with the least electronegative element in the center. In HCN, Carbon has the least electronegativity, so it serves as the central atom. Now, we will connect the central atom with outer atoms (hydrogen and nitrogen) using a single bond (2 electrons for each bond). This will consume 4 of the 10 valence electrons. Then, we will distribute the remaining 6 electrons as lone pairs to achieve an octet for each atom, starting with outer atoms first (as H already satisfied with 2 electrons). Nitrogen will have one lone pair of electrons (2 electrons), and Carbon will have one lone pair of electrons (2 electrons). This completes the distribution of all valence electrons.

To identify hybrid orbitals on Carbon and Nitrogen, we will count the number of electron groups around each atom (lone pairs, single bonds, double bonds, or triple bonds). - Carbon has 3 electron groups: 1 single bond to H, 1 lone pair, and 1 triple bond to N. This corresponds to sp2 hybridization. - Nitrogen has 3 electron groups: 1 triple bond to C and 1 lone pair. This corresponds to sp hybridization.

Now, let's draw the complete structure of HCN by showing all the bonds and labeling them as σ or π. - There is a single bond between Hydrogen and Carbon, which is a σ bond. - The triple bond between Carbon and Nitrogen consists of 1 σ bond and 2 π bonds. The complete structure will look like this: H-σ-C≡N with a lone pair of electrons on C and a lone pair on N. The triple bond between C and N shows that there are 1 σ bond and 2 π bonds. The hybridization of C is sp2, and N is sp.