Question

Draw a Lewis structure for methyl isocyanate, \(\mathrm{CH}_{3} \mathrm{NCO}\), showing all valence electrons. Predict all bond angles in this molecule and the hybridization of each \(\mathrm{C}, \mathrm{N}\), and \(\mathrm{O}\).

Short answer

Answer: The hybridization and bond angles in the methyl isocyanate (CH3NCO) molecule are: - C1: sp3 hybridization with bond angles of approximately 109.5° - N: sp2 hybridization with a bond angle of approximately 120° - C2: sp2 hybridization with a bond angle of approximately 120°

Step-by-step solution

Determine the number of valence electrons for each atom

To draw the Lewis structure, we first need to determine the number of valence electrons for each atom in the molecule. - Carbon (C): 4 valence electrons - Hydrogen (H): 1 valence electron - Nitrogen (N): 5 valence electrons - Oxygen (O): 6 valence electrons In the \(\mathrm{CH}_3 \mathrm{NCO}\) molecule, there are 3 hydrogen atoms, one nitrogen atom, and one oxygen atom. Therefore, there are a total of 4 + (3×1) + 5 + 6 = 18 valence electrons in the molecule.

Draw the Lewis structure

Based on the number of valence electrons, we can draw the Lewis structure for methyl isocyanate: 1. Start by connecting the central atoms (C, N, and O) with single bonds. 2. Complete the octet rule (except hydrogen) for each atom with the remaining electrons. 3. Calculate formal charge and adjust the Lewis structure if necessary. The resulting Lewis structure should look like this: H | H - C - N = C = O Now we can see that all atoms have a complete octet (except hydrogens), with a total of 18 valence electrons.

Predict the hybridization

Next, we have to determine the hybridization of each carbon, nitrogen, and oxygen atom. - Carbon (C1) in \(\mathrm{CH}_3\) is bonded to 3 hydrogen atoms and 1 nitrogen atom through sigma bonds. Hence, the hybridization of C1 is \(\mathrm{sp}^3\), since it forms a tetrahedral structure with four single bonds. - Nitrogen (N) is bonded to C1 through a sigma bond and C2 through a double bond. So the hybridization of N is \(\mathrm{sp}^2\), forming a planar structure with three electron domains. - Carbon (C2) in \(\mathrm{NCO}\) has a double bond with both N and O. Thus, the hybridization of C2 is also \(\mathrm{sp}^2\).

Predict the bond angles

Now we can predict the bond angles in the molecule based on the hybridization of each atom. - The bond angle \(\angle \mathrm{H_{1}-C_{1}-H_{2}}\) is approximately 109.5°, as it forms a tetrahedral geometry around C1 due to \(\mathrm{sp}^3\) hybridization. - Similarly, \(\angle \mathrm{H-C_{1}-N}\) is also 109.5°. - The bond angle \(\angle \mathrm{C_{1}-N-C_{2}}\) is approximately 120°, due to the planar structure formed by \(\mathrm{sp}^2\) hybridization of N. - Finally, \(\angle \mathrm{N-C_{2}-O}\) is also 120°, because C2 is \(\mathrm{sp}^2\) hybridized and forms a planar structure. In summary, the Lewis structure of methyl isocyanate exhibits both \(\mathrm{sp}^3\) and \(\mathrm{sp}^2\) hybridizations, resulting in bond angles of 109.5° and 120°, respectively.