Question

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in \(: \mathrm{N} \equiv \mathrm{N}-\mathrm{O}:\) be described in terms of orbitals?

Short answer

The linear and polar N2O molecule has the arrangement NNO. The Lewis structure is :N≡N-O:, and the formal charge on each atom is 0. The central nitrogen atom is sp hybridized. The triple bond between the nitrogen atoms consists of one σ bond from the overlap of sp-hybrid orbitals and two π bonds from the overlap of unhybridized p-orbitals. The single bond between the central nitrogen and the oxygen is formed due to the overlap of the other sp-hybrid orbital on the nitrogen and the hybrid orbital on the oxygen atom.

Step-by-step solution

(Determine the arrangement of N2O molecule)

We are given that the \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. The possible arrangements are NNO or NON. Since the molecule is polar, the more electronegative atom must be on one side, while the less electronegative atoms should be on the other side to maintain the polarity. Since oxygen is more electronegative than nitrogen, the correct arrangement is NNO.

(Drawing the Lewis structure of N2O)

To draw the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\), we need to count the total number of valence electrons available for bonding. Nitrogen has 5 valence electrons and oxygen has 6. There are two nitrogen atoms, so the total number of electrons is (2 x 5) + 6 = 16. The Lewis structure is: :N≡N-O:

(Determine the formal charges and hybridization of the central atom)

To find the formal charges, we use the formula: Formal charge = (valence electrons) - (non-bonding electrons) - 0.5 x (bonding electrons) For the leftmost nitrogen: Formal charge = 5 - 2 - 0.5 x (8) = 0 For the central nitrogen: Formal charge = 5 - 0 - 0.5 x (10) = 0 For the oxygen: Formal charge = 6 - 4 - 0.5 x (4) = 0 To determine the hybridization of the central nitrogen atom, we look at the electron regions around it. Since the central nitrogen is triple bonded to one nitrogen and single bonded to one oxygen, there are two regions of electrons around it. Thus, it is sp hybridized.

(Describe the multiple bonding in terms of orbitals)

The multiple bonding between the two nitrogen atoms (the triple bond) can be described in terms of orbitals as follows: 1. One \(\sigma\) bond is formed due to the overlap of the two sp-hybrid orbitals on the nitrogen atoms. 2. Two \(\pi\) bonds are formed due to the overlap of two pairs of unhybridized p-orbitals on the nitrogen atoms. The single bond between the central nitrogen and the oxygen atom is formed due to the overlap of the other sp-hybrid orbital on the nitrogen and the hybrid orbital on the oxygen atom.

Key concepts

Molecular Geometry

Molecular geometry describes the three-dimensional arrangement of atoms in a molecule. For the nitrogen monoxide (\(\mathrm{N}_{2} \mathrm{O}\)) molecule, we observe a linear geometry. The atoms in \(\mathrm{N}_{2} \mathrm{O}\) align in a straight line, typically following the arrangement of one nitrogen bonded to another nitrogen, which is then bonded to an oxygen atom (NNO configuration). This geometry affects how the molecule interacts with other molecules and can play a critical role in its chemical behavior and physical properties. Linear molecular geometry is typically characterized by a bond angle of 180 degrees.

Polarity

Polarity of a molecule is determined by the distribution of electron density across its atoms. A molecule is considered polar if it has a net dipole moment, which occurs when there is an uneven distribution of electrons. In \(\mathrm{N}_{2} \mathrm{O}\), though the geometry is linear, the polarity arises mainly due to the difference in electronegativity between nitrogen and oxygen. Oxygen, being more electronegative, attracts the bonded electron pair towards itself, generating a dipole. This leads to the molecule being polar as there is a separation of charge along the linear axis. This property is crucial in determining how the molecule interacts with other polar or nonpolar substances.

Lewis Structure

The Lewis structure is a diagrammatic representation that shows the bonding between atoms within a molecule and the lone pairs of electrons that may exist. For \(\mathrm{N}_{2} \mathrm{O}\), the Lewis structure can be drawn by arranging 16 valence electrons among the atoms, ensuring that the octet rule is satisfied where applicable. Typically, the configuration follows:

• :N≡N-O:

In this structure:

• Two nitrogen atoms form a triple bond (one sigma and two pi bonds).
• A single bond connects the central nitrogen to the oxygen atom.
• This arrangement minimizes formal charges and satisfies the valence requirements of nitrogen and oxygen.

Formal Charge

Formal charge helps us determine the most stable Lewis structure for a molecule by assigning electrons to atoms differently than their actual occupancy. It is calculated with the formula:\[\text{Formal charge} = \text{(valence electrons)} - \text{(non-bonding electrons)} - 0.5 \text{(bonding electrons)}\]Using this formula for the \(\mathrm{N}_{2} \mathrm{O}\) structure:

• Leftmost nitrogen has a formal charge of 0, with 5 valence electrons, 2 non-bonding electrons, and 8 bonding electrons.
• Central nitrogen also has a formal charge of 0, with 5 valence electrons and 10 bonding electrons.
• Oxygen also holds a formal charge of 0, with 6 valence electrons, 4 non-bonding electrons, and 4 bonding electrons.

This balances the charges across the molecule, indicating stability in the drawn Lewis structure.

Orbital Hybridization

Orbital hybridization describes the mixing of atomic orbitals to form new hybrid orbitals that influence molecular geometry and bonding properties. In \(\mathrm{N}_{2} \mathrm{O}\), the central nitrogen atom exhibits sp hybridization due to the two regions of electron density (one triple bond and one single bond).

• The formation of the triple bond involves one sigma bond, resulting from the overlap of sp-hybridized orbitals of the nitrogen atoms.
• Two pi bonds come from the unhybridized p-orbitals overlapping side by side.
• The oxygen and the central nitrogen bond involves an sp-hybridized orbital from nitrogen overlapping with an oxygen orbital.

This hybridization and the resulting bonding ensure structural integrity and chemical reactivity of the molecule.