Problem 109 - Chemistry The Central Science

Question

A compound composed of \(2.1 \% \mathrm{H}, 29.8 \% \mathrm{~N},\) and \(68.1 \% \mathrm{O}\) has a molar mass of approximately \(50 \mathrm{~g} / \mathrm{mol}\). (a) What is the molecular formula of the compound? (b) What is its Lewis structure if \(\mathrm{H}\) is bonded to \(\mathrm{O} ?\) (c) What is the geometry of the molecule? (d) What is the hybridization of the orbitals around the \(\mathrm{N}\) atom? (e) How many \(\sigma\) and how many \(\pi\) bonds are there in the molecule?

Step-by-Step Solution

Answer

(a) The molecular formula is HNO2. (b) The Lewis structure is: O .. / H - O - N \ O .. (c) The molecule has a trigonal pyramidal geometry. (d) The hybridization of the orbitals around the N atom is sp3. (e) There are three σ bonds and no π bonds in the molecule.

Step 1: Calculate the moles of each element

First, assume that the compound has 100 g. This will make it so that the percentage composition can be treated as the mass of each element. Given that the molar masses are: - H: 1 g/mol - N: 14 g/mol - O: 16 g/mol Divide the mass of each element by its molar mass to obtain the number of moles: - Moles of H: \(\frac{2.1}{1} = 2.1\) - Moles of N: \(\frac{29.8}{14} \approx 2.13\) - Moles of O: \(\frac{68.1}{16} \approx 4.26\)

Step 2: Determine the empirical formula

Find the smallest ratio of moles between the elements by dividing all the values by the smallest number of moles: - H: \(\frac{2.1}{2.1} = 1\) - N: \(\frac{2.13}{2.1} \approx 1\) - O: \(\frac{4.26}{2.1} \approx 2\) So, the empirical formula is HNO2.

Step 3: Calculate molecular formula

Determine the empirical formula's molar mass: Empirical formula molar mass = H + N + 2O = 1 + 14 + 2(16) = 47 g/mol Use the compound's molar mass (50 g/mol) and the empirical formula molar mass (47 g/mol) to calculate the ratio: \(\frac{\text{Compound's molar mass}}{\text{Empirical formula molar mass}} = \frac{50}{47} \approx 1\) Since the ratio is approximately equal to 1, the molecular formula is the same as the empirical formula, HNO2.

Step 4: Draw Lewis structure with H bonded to O

To draw the Lewis structure, follow these steps: 1. Count the total number of valence electrons: H has 1, N has 5, and O has 6. Total valence electrons = 1 + 5 + 2(6) = 18. 2. Place the least electronegative atom (N) in the center and arrange the other atoms around it. 3. Draw a single bond between the central atom (N) and the surrounding atoms (keeping H bonded to O). Fill the octet of the surrounding bonded atoms with the remaining electrons. 4. Place any remaining electrons on the central atom and check its octet. So, the Lewis structure is: O .. / H - O - N \ O ..

Step 5: Determine molecule geometry

Since nitrogen (N) has three bonding groups and one lone electron pair, the molecule has a trigonal pyramidal geometry.

Step 6: Identify hybridization around N atom

N has a total of 4 electron domain groups (3 bonding groups and 1 lone pair). Therefore, the hybridization around the N atom is sp3.

Step 7: Calculate number of σ and π bonds

Because there are three single bonds (N-O, O-H, and N-O) in the molecule, there are three σ bonds. And because there are no double or triple bonds in the molecule, there are no π bonds. To answer the questions: (a) The molecular formula is HNO2. (b) The Lewis structure is provided in Step 4. (c) The molecule has a trigonal pyramidal geometry. (d) The hybridization of the orbitals around the N atom is sp3. (e) There are three σ bonds and no π bonds in the molecule.

Key Concepts

Lewis Structure

Understanding the Lewis structure of a molecule helps us predict how atoms are bonded together and how they share electrons. For our compound, the molecular formula is HNO₂, which means we have hydrogen (H), nitrogen (N), and oxygen (O) in the mix.

In HNO₂, since hydrogen is bonded to oxygen, we start by counting the total number of valence electrons available for bonding. Hydrogen has 1, nitrogen has 5, and each oxygen has 6, leading to a total of 18 valence electrons.

To draw the Lewis structure:

• Place the least electronegative atom, nitrogen, in the center.
• Connect the nitrogen to each oxygen atom with a single bond.
• Remember, one hydrogen should be bonded to one oxygen.
• Distribute the remaining electrons to satisfy the octet rule for each atom, adding lone pairs where necessary.
• Ensure all atoms have the correct valence electrons around them, and adjust for any double bonds if octets are not satisfied.

This Lewis structure helps us visualize how atoms are interconnected, giving foundational insight for understanding other characteristics of the molecule.

Molecular Geometry

The molecular geometry describes the three-dimensional shape of a molecule, which can greatly affect its properties and interactions. In the case of HNO₂, where nitrogen is the central atom and is bound to other atoms with electron pair repulsions, the geometry can be predicted by examining its arrangement of atoms and lone pairs.

For nitrogen in HNO₂:

• There are three bonding groups (N bound to two oxygens and one oxygen bound to hydrogen).
• One lone pair is also present on the nitrogen atom due to its extra valence electron.

These four regions of electron density push against each other, causing the molecule to adopt a trigonal pyramidal shape. Trigonal pyramidal geometry is a common shape for molecules with a central atom that has three bonds and one lone pair.

Orbital Hybridization

Orbital hybridization is the concept where atomic orbitals fuse to form new hybrid orbitals. These hybrid orbitals are used to form covalent bonds in molecules. For HNO₂, we focus on the nitrogen atom as it is the central atom coordinating the structure.

Nitrogen in HNO₂ has:

• Three adjacent bonding groups (to O, an O and the bonded OH group).
• One lone pair of electrons.

This amount of bonding direction and electron density means the nitrogen exhibits sp³ hybridization. The sp³ hybrid orbitals are formed from one s orbital and three p orbitals, creating four equivalent hybridized orbitals.

The use of sp³ hybridization allows the molecule to form stable 3-dimensional structures like the trigonal pyramidal shape, also aiding in determining the potential chemical reactivity and interactions of the molecule.

Sigma and Pi Bonds

Covalent bonds within a molecule can be classified into sigma (σ) and pi (π) bonds. This distinction is essential because the type of bond affects the molecule's stability and how molecules can interact with each other.

In our compound HNO₂:

• Sigma bonds ( σ bonds) are the first bonds formed between two atoms, involving head-on overlap of orbitals. They are generally quite strong.
• In HNO₂, we identify three sigma bonds: one between H and O, and two between N and each O.
• Pi bonds ( π bonds) result from the sideways overlap of p orbitals. These frequently accompany sigma bonds in double or triple bonds.

With no double or triple bonds noted in the structure of HNO₂, there are no π bonds present. This absence of π bonds means all bonding in HNO₂ can be attributed to the three σ bonds, contributing to the molecule’s stability and expected reactivity.