Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \\ \text { Cyanamide } \end{aligned} $$ Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: Dicyandiamide not shown) a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\), dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the \(\mathrm{C}\) and \(\mathrm{N}\) atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(\mathrm{C}-\mathrm{N}\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\), and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

Step by step solution

01

NCN^2-

To draw Lewis structure of NCN^2-, let's count the valence electrons. N has 5, C has 4, and N has 5 with a charge of -2. Adding them all together, we have a total of 16 valence electrons. After drawing the structure, we get: \[N=C=N^{2-}\] There are no resonance structures for NCN^2-.

02

H2NCN

Counting the valence electrons for H2NCN, we have 2 from hydrogen atoms, 5 from N, 4 from C, and 5 from N for a total of 16 valence electrons. After drawing the structure, we get: \[H-N=C=N-H\] There are no resonance structures for H2NCN.

03

Dicyandiamide

Dicyandiamide has the formula H4N2C2, and we can count the valence electrons: 4 from hydrogen atoms, 10 from two N atoms, and 8 from two C atoms for a total of 22 valence electrons. After drawing the structure, we get: \[H_2N-C=N-NH_2\] There are no resonance structures for dicyandiamide.

04

Melamine

For melamine, C3H6N6, count the valence electrons: 6 from hydrogen atoms, 15 from three C atoms, and 30 from six N atoms for a total of 51 valence electrons. After drawing the structure, we get: \[\begin{array}{c} H_{2} N-C=N-H \\ || \\ -N-H \end{array}\] Please note that there is a resonance structure where the double bond is between the carbon and the nitrogen on the ring, which will be treated equally. b. Determine hybridization

05

NCN^2- hybridization

The central carbon in NCN^2- is sp hybridized, and both nitrogen atoms are also sp hybridized.

06

H2NCN hybridization

The carbon atom in H2NCN is sp hybridized, and both nitrogen atoms are sp^3 hybridized.

07

Dicyandiamide hybridization

The two carbon atoms and two nitrogen atoms in dicyandiamide are all sp hybridized.

08

Melamine hybridization

In melamine, three carbon atoms are sp^2 hybridized, and all six nitrogen atoms are also sp^2 hybridized. c. Count sigma and pi bonds

09

NCN^2- bonds

In NCN^2-, there are 2 sigma bonds and 1 pi bond.

10

H2NCN bonds

In H2NCN, there are 4 sigma bonds and 1 pi bond.

11

Dicyandiamide bonds

In dicyandiamide, there are 6 sigma bonds and 2 pi bonds.

12

Melamine bonds

In melamine, there are 12 sigma bonds and 3 pi bonds. d. Melamine ring planarity

13

Melamine ring

All the atoms in the melamine ring are sp^2 hybridized, which means they have a trigonal planar geometry. Thus, the ring in melamine is planar. e. Resonance structure importance

14

Most important resonance structure

In the case of dicyandiamide, there are two resonance structures with equivalent contributions, and their hybrid structure will show an equal distribution of bond lengths. Therefore, no single resonance structure can be deemed more important.

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