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 hybridizations of the C, N, and O atoms in methyl isocyanate are as follows: - C in the CH3 group: sp2 - N: sp3 - C attached to N and O: sp3 - O: sp2

Step-by-step solution

1. Determine the total number of valence electrons

To determine the total number of valence electrons for methyl isocyanate \(\mathrm{(CH}_{3} \mathrm{NCO)}\), we first find the number of valence electrons for each atom: - C (carbon) has 4 valence electrons (group 14 in the periodic table). - H (hydrogen) has 1 valence electron (group 1 in the periodic table). - N (nitrogen) has 5 valence electrons (group 15 in the periodic table). - O (oxygen) has 6 valence electrons (group 16 in the periodic table). In methyl isocyanate, we have 1 carbon, 3 hydrogen, 1 nitrogen, and 1 oxygen atom. Therefore, the total number of valence electrons is: \(4 + 3(1) + 5 + 6 = 18\).

2. Draw the Lewis structure

Now, we'll draw the Lewis structure for methyl isocyanate, remembering that the molecule's atomic arrangement is \(\mathrm{CH}_{3} \mathrm{NCO}\). Start by arranging the atoms and connecting them with single bonds, which use 2 valence electrons each. H - C - N - C (double bond) O Now, complete the octet for each atom by adding lone pairs: - Carbon in \(\mathrm{CH}_{3}\) group: already has 4 bonds (4 electrons). - Hydrogen atoms: Each has 1 bond (1 electron) and doesn't need lone pairs. - Nitrogen: has 3 bonds (6 electrons) so add 1 lone pair (2 electrons). - Carbon attached to N and O: has 4 bonds (8 electrons) with a double bond to the O. - Oxygen: has 2 bonds (4 electrons) from the double bond, add 2 lone pairs (4 electrons). Final Lewis Structure: H - C - N (lone pair) - C (double bond) O (2 lone pairs)

3. Predict the bond angles

Using the VSEPR theory, we can predict the bond angles in the molecule: - \(\angle{\mathrm{H-C-H}}\): The C atom in the \(\mathrm{CH}_{3}\) group has 3 bonding and 0 lone pair electrons (AX3). It is trigonal planar with a bond angle of 120°. - \(\angle{\mathrm{C-N-C}}\): The N atom has 3 bonding and 1 lone pair electrons (AX3E). Its structure is trigonal pyramidal with a bond angle of around 107°. - \(\angle{\mathrm{N-C-O}}\): The C atom attached to N and O has 4 bonding and 0 lone pair electrons (AX4). Its structure is tetrahedral with a bond angle of 109.5°.

4. Determine the hybridization

To find the hybridization of each C, N, and O atom, we count the number of electron groups (bonding + lone pairs) around each atom: - Carbon in the \(\mathrm{CH}_{3}\) group: 3 electron groups → sp2 hybridization. - Nitrogen: 4 electron groups → sp3 hybridization. - Carbon attached to N and O: 4 electron groups → sp3 hybridization. - Oxygen: 3 electron groups → sp2 hybridization. In summary, the hybridization of each C, N, and O atom in methyl isocyanate is as follows: - C in the \(\mathrm{CH}_{3}\) group: sp2 - N: sp3 - C attached to N and O: sp3 - O: sp2

Key concepts

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial in determining how the atom interacts with others. In methyl isocyanate (CH₃NCO), calculating the total valence electrons involves summing up the electrons from each constituent atom. Carbon has four valence electrons, hydrogen has one, nitrogen has five, and oxygen has six.
This totals to 18 valence electrons for CH₃NCO, as calculated by:

• Carbon: 4 valence electrons
• 3 Hydrogens: 3 × 1= 3 valence electrons
• Nitrogen: 5 valence electrons
• Oxygen: 6 valence electrons

Together, these electrons form the bonds that hold the atoms in the molecule together and establish the molecule's connectivity through the sharing or transferring of electrons.

VSEPR Theory

The VSEPR (Valence Shell Electron Pair Repulsion) theory is vital for predicting the geometric arrangement of electrons around each atom in a molecule, which further determines the molecular shape. This theory states that electron pairs will position themselves as far apart as possible to minimize repulsion.
In methyl isocyanate, we apply VSEPR theory to estimate the bond angles:

• CH₃ group's Carbon: With three hydrogen bonds, it's an AX3 system, implying a trigonal planar shape with an ideal bond angle of 120°.
• Nitrogen bound to C: This has three bonds and one lone pair in an AX3E configuration, creating a trigonal pyramidal shape with a bond angle slightly less than 109.5°, around 107°.
• Carbon between N and O: The conjunction of three single bonds and one double bond forms a tetrahedral shape, maintaining bond angles close to 109.5°.

Bond Angles

Bond angles are the angles between two adjacent bonds at an atom in a molecule. These angles help in determining the molecular shape and are dictated by VSEPR theory. Understanding bond angles is essential for predicting the geometry of molecules such as methyl isocyanate.
In CH₃NCO, the key angles include:

• H-C-H: The bond angles between the hydrogen atoms and carbon in the CH₃ group are approximately 120°, because of the trigonal planar geometry derived from sp² hybridization.
• C-N-C: The nitrogen atom, with its lone pair, forms around 107° due to the steric effects of the lone pair reducing the bond angle from the ideal tetrahedral angle of 109.5°.
• N-C-O: This angle approaches the ideal tetrahedral bond angle of 109.5°.

These angles are critical for understanding the molecular geometry, which influences the molecule's physical and chemical properties.

Hybridization

Hybridization is about the mixing of atomic orbitals to form new hybrid orbitals that can form sigma bonds to other atoms. It provides insight into the geometry and bonding properties of a molecule. In methyl isocyanate, each atom has its unique hybridization:

• Carbon in CH₃: Exhibits sp² hybridization as it is bonded to three other atoms and no lone pairs, adopting a trigonal planar shape.
• Nitrogen: With three sigma bonds and a lone pair, nitrogen undergoes sp³ hybridization, influencing its pyramidal shape.
• Carbon bonded to N and O: This carbon shows sp³ hybridization due to its tetrahedral environment of four single bonds/interaction sites.
• Oxygen: Oxygen in the molecule typically presents sp² hybridization due to its strong double bonding with carbon and presence of lone pairs influencing its bent shape.

Hybridization explains the bonding and molecular shape by showing how orbitals overlap and interact, leading to the formation of a stable molecular structure.