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Methyl isocyanate, \(\mathrm{CH}_{3} \mathrm{NCO},\) was made infamous in 1984 when an accidental leakage of this compound from a storage tank in Bhopal, India, resulted in the deaths of about 3800 people and severe and lasting injury to many thousands more. (a) Draw a Lewis structure for methyl isocyanate. (b) Draw a ball-and-stick model of the structure, including estimates of all the bond angles in the compound. (c) Predict all the bond distances in the molecule. (d) Do you predict that the molecule will have a dipole moment? Explain.

Short Answer

Expert verified
The Lewis structure of methyl isocyanate (CH3NCO) can be represented as H-C-N=C=O with the H-C-H bond angle being approximately 109.5° and the N-C-O bond angle being 180°. The predicted bond distances are approximately 1.09 Å for C-H, 1.47 Å for C-N, 1.24 Å for N=C, and 1.21 Å for C=O. Methyl isocyanate has a net dipole moment due to the polar C-N and C-H bonds, despite the cancellation of the N=C and C=O bond dipoles.

Step by step solution

01

Draw the Lewis structure of methyl isocyanate

First, count the total number of valence electrons for the molecule. Carbon (C) has 4, Nitrogen (N) has 5, Oxygen (O) has 6, and Hydrogen (H) has 1 valence electron each. In the CH3NCO molecule, there are 3 Hydrogen atoms, so the total number of valence electrons is 4 (from C) + 5 (from N) + 6 (from O) + 3 * 1 (from the 3 H atoms) = 16. Now, arrange the atoms and distribute the electrons: 1. Place the central atom (Carbon) in the center. 2. Place the other atoms around it (N next to C, O next to N, and the 3 H atoms around the central C). 3. Distribute the electrons by placing pairs between the atoms to form single bonds. Add remaining electrons as lone pairs. 4. Make multiple bonds if needed to satisfy the octet rule for all atoms except Hydrogen. The Lewis structure of methyl isocyanate is: H | H--C--N==C==O | H
02

Draw a ball-and-stick model and estimate the bond angles

To draw a ball-and-stick model, represent each atom as a sphere connected by sticks indicating the bonds. You can estimate the bond angles based on the molecular geometry and the bond pair repulsion theory. For methyl isocyanate: 1. The central Carbon atom is surrounded by three groups (three Hydrogen atoms) with a tetrahedral molecular geometry. The H-C-H bond angle is approximately 109.5°. 2. Nitrogen and Oxygen atoms are involved in a double bond with the Carbon atom. The N-C-O bond angle is approximately 180° since NCO has a linear geometry. The ball-and-stick model of methyl isocyanate with the bond angles will look like the following: H | H--C--N==C==O (180°) | H (109.5°)
03

Predict bond distances

To predict bond distances, you can use known bond lengths for similar chemical species: 1. C-H bond: The bond length in methyl isocyanate is expected to be the same as in other aliphatic hydrocarbons, approximately 1.09 Å. 2. C-N bond: The bond length in methyl isocyanate is expected to be similar to a C-N single bond length, which is approximately 1.47 Å. 3. N=C double bond: The bond length in methyl isocyanate is expected to be similar to a N=C double bond length, which is approximately 1.24 Å. 4. C=O double bond: The bond length in methyl isocyanate is expected to be similar to a C=O double bond length, which is approximately 1.21 Å.
04

Determine if the molecule has a dipole moment and explain

A molecule has a dipole moment when there is a difference in electronegativity and an asymmetric distribution of charge. In methyl isocyanate, the C-N bond and the N=C bond are polar due to a difference in electronegativity between Carbon and Nitrogen. The C=O bond is also polar due to a difference in electronegativity between Carbon and Oxygen. However, since the NCO group in the molecule is linear, the bond dipoles of the N=C and C=O bonds are pointing in opposite directions and therefore cancel each other out. As a result, the net dipole moment of the molecule is due to the C-N bond and the three C-H bonds creating a slight asymmetry in the electron distribution. Therefore, methyl isocyanate does have a dipole moment.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lewis Structure
Understanding molecular structures starts with grasping the **Lewis Structure**. Lewis structures represent the valence electron configuration around atoms in a molecule. They help us visualize the placement of atoms and how they bond together to achieve stable electronic arrangements.

For methyl isocyanate, \({\text{CH}}_3{\text{NCO}}\), the first step is to determine the total number of valence electrons. Carbon (C), having 4, Nitrogen (N) with 5, Oxygen (O) with 6, and each Hydrogen (H) with 1, sum up to a total of 16 valence electrons in the molecule.

Arrange these electrons to fulfill the octet rule (or duplet for hydrogen):
  • Carbon is often the central atom around which others are organized because of its ability to form four bonds.
  • The structure places Carbon bonded to Nitrogen, which in turn is bonded to Oxygen, and Hydrogen atoms surround Carbon.
  • Single and multiple bonds (e.g., N=C=O with double bonds) complete the electron allocation.
This representation assists in predicting the behavior and properties of the molecule.
Ball-and-Stick Model
The **Ball-and-Stick Model** provides a more tangible visualization of the molecular structure by representing atoms as spheres and bonds as connecting sticks. This model helps in appreciating the three-dimensional geometry of a molecule.

In the case of methyl isocyanate, this involves:
  • The central Carbon atom (represented as a sphere) linked via sticks to three Hydrogen atoms, which complete a tetrahedral shape around it.
  • This gives rise to an angle of approximately 109.5° between the H-C-H bonds.
  • Nitrogen and Oxygen also connect to the Carbon in a linear configuration, maintaining an angle of approximately 180° for the N=C=O bond.
By using such models, scientists and students can predict how molecules might interact with others or fit into complex structures.
Bond Angles
Bond angles define the shape and geometry of a molecule, influencing its reactivity and interaction with other molecules.

For methyl isocyanate, understanding bond angles is crucial. The angle information includes:
  • The tetrahedral geometry of the central C atom makes H-C-H bond angles approximately 109.5°.
  • The linear configuration of the N=C=O section, with the bond angle close to 180°.
These geometric configurations affect the compound's stability and participating role in chemical reactions. Such comprehension helps predict molecular behavior based on their structural geometry.
Dipole Moment Prediction
A **dipole moment** is essentially the shift of electron density across a molecule due to differences in electronegativity. It leads to having a partial positive charge and a partial negative charge in a molecule’s structure.

With methyl isocyanate, the existence of a dipole moment primarily arises from:
  • The difference in electronegativity between bonded atoms like Carbon, Nitrogen, and Oxygen.
  • The linear N=C=O group implies that individual bond dipoles along N=C and C=O cancel out since they are opposite.
  • However, the presence of the C-N bond and the asymmetrical arrangement of C-H bonds introduce a slight dipole in the molecule.
Thus, despite some cancelling out, methylene isocyanate does possess a dipole moment, influencing its intermolecular interactions and solubility.

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Most popular questions from this chapter

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2}^{*}\), orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \({ }^{4} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of \(\mathrm{M}\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(\mathrm{M}\). The CO bond axis should be on the \(x\) -axis. (c) Draw the \(\mathrm{CO} \pi_{2 p}^{*}\) orbital, with phases (see the "Closer Look" box on phases) in the plane of the paper. Two lobes should be pointing toward M. (d) Now draw the \(d_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2 p}^{*}\) orbital of \(\mathrm{CO} ?\) (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?\) (f) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.

What is the hybridization of the central atom in (a) \(\mathrm{PBr}_{5}\), (b) \(\mathrm{CH}_{2} \mathrm{O},\) (c) \(\mathrm{O}_{3},(\mathbf{d}) \mathrm{NO}_{2} ?\)

The oxygen atoms in \(\mathrm{O}_{2}\) participate in multiple bonding, whereas those in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) do not. (a) Draw Lewis structures for both molecules. (b) What is the hybridization of the oxygen atoms in each molecule? (c) Which molecule has the stronger \(\mathrm{O}-\mathrm{O}\) bond?

For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. \((\mathbf{c})\) To create a hybrid orbital, you could use the \(s\) orbital For each statement, indicate whether it is true or false. (a) The greater the orbital overlap in a bond, the weaker the bond. (b) The greater the orbital overlap in a bond, the shorter the bond. \((\mathbf{c})\) To create a hybrid orbital, you could use the \(s\) orbital

The energy-level diagram in Figure 9.36 shows that the sideways overlap of a pair of \(p\) orbitals produces two molecular orbitals, one bonding and one antibonding. In ethylene there is a pair of electrons in the bonding \(\pi\) orbital between the two carbons. Absorption of a photon of the appropriate wavelength can result in promotion of one of the bonding electrons from the \(\pi_{2 p}\) to the \(\pi_{2 p}^{*}\) molecular orbital. (a) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the HOMO in ethylene? (b) Assuming this electronic transition corresponds to the HOMO-LUMO transition, what is the LUMO in ethylene? (c) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene stronger or weaker in the excited state than in the ground state? Why? (d) Is the \(\mathrm{C}-\mathrm{C}\) bond in ethylene easier to twist in the ground state or in the excited state?

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