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Chapter 9: Problem 5

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 \(\pi\).

Short Answer

Expert verified
The Lewis structure for HCN is H-C鈮, with Carbon and Nitrogen having sp hybrid orbitals. There is one 蟽 bond between Hydrogen and Carbon, and one 蟽 bond and two 蟺 bonds between Carbon and Nitrogen.

Step by step solution

01

Determine the valence electrons

First, we need to find the total number of valence electrons for HCN. Hydrogen (H) has 1 valence electron, Carbon (C) has 4 valence electrons, and Nitrogen (N) has 5 valence electrons. Thus, the total number of valence electrons in HCN is: \(1 + 4 + 5 = 10\)
02

Arrange the atoms and form bonds

Now, arrange the atoms, placing the less electronegative atom (Carbon) in the center, and form a single bond between each pair of atoms: H-C-N
03

Complete the octets for the terminal atoms

After forming single bonds, we'll complete the octet for the terminal atoms - Hydrogen and Nitrogen. Since Hydrogen only needs 2 electrons to have a full shell, it's already satisfied. For Nitrogen, it currently has 2 electrons; we need to add 6 more electrons (3 lone pairs) to complete its octet: H-C鈮
04

Identify the hybrid orbitals for Carbon and Nitrogen atoms

Now, we need to identify the hybrid orbitals for Carbon and Nitrogen. Each of these atoms is connected to the other atoms through a triple bond, which means they are sp hybridized. In an sp hybridization, two orbitals mix to form two new hybrid orbitals, so both Carbon and Nitrogen will form two sp hybrid orbitals.
05

Draw the bonds and label them as 蟽 or 蟺

Finally, we need to draw the bonds between the atoms and label them as 蟽 (sigma) or 蟺 (pi). There will be one 蟽 bond and two 蟺 bonds between Carbon and Nitrogen, and one 蟽 bond between Hydrogen and Carbon. The final structure is: H-C鈮 (蟽 bond between H and C, one 蟽 bond and two 蟺 bonds between C and N) So, the Lewis structure for HCN is H-C鈮, with each of Carbon and Nitrogen atom having sp hybrid orbitals, one 蟽 bond between Hydrogen and Carbon, and one 蟽 bond and two 蟺 bonds between Carbon and Nitrogen.

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Most popular questions from this chapter

Use the MO model to determine which of the following has the smallest ionization energy: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2}^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.

As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, \(\mathrm{B}_{2}\). a. What is the bond order in \(\mathrm{Li}_{2}\) and \(\mathrm{B}_{2}\) ? b. How many electrons must be removed from \(\mathrm{B}_{2}\) to make it isoelectronic with \(\mathrm{Li}_{2}\) so that it might be used in the warp drive? c. The reaction to make \(\mathrm{B}_{2}\) isoelectronic with \(\mathrm{Li}_{2}\) is generalized (where \(n=\) number of electrons determined in part b) as follows: $$ \mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta H=6455 \mathrm{~kJ} / \mathrm{mol} $$ How much energy is needed to ionize \(1.5 \mathrm{~kg} \mathrm{~B}_{2}\) to the desired isoelectronic species?

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