Problem 40
Question
Write the Lewis dot structural formula for each of the following. Give, also, the formula of a neutral molecule, which has the same geometry and the same arrangement of the bonding electrons as in each of the following. An example is given below in the case of \(\mathrm{H}_{3} \mathrm{O}^{+}\)
Step-by-Step Solution
Verified Answer
\(\mathrm{NH}_3\) (Ammonia) has the same geometry and electron arrangement as \(\mathrm{H}_3\mathrm{O}^+\).
1Step 1: Understand the Exercise
We need to write the Lewis dot structure for each ion given and identify a neutral molecule with the same geometry and electron arrangement.
2Step 2: Identify Valence Electrons for \(\mathrm{H}_3\mathrm{O}^+\)
Determine the valence electrons for each atom in \(\mathrm{H}_3\mathrm{O}^+\): Oxygen has 6 valence electrons, each Hydrogen has 1. Because of the positive charge, one electron is removed from the sum of valence electrons.
3Step 3: Draw the Lewis Structure for \(\mathrm{H}_3\mathrm{O}^+\)
Arrange the three Hydrogens around Oxygen. Distribute the electrons to form a single bond (2 electrons) between Oxygen and each Hydrogen, ensuring Oxygen has an octet. There will be one lone pair on Oxygen.
4Step 4: Geometry Consideration
\(\mathrm{H}_3\mathrm{O}^+\) is arranged in a trigonal pyramidal shape. This is due to the presence of three bonds and one lone pair on the Oxygen.
5Step 5: Find a Neutral Molecule with Similar Geometry
Identify a neutral molecule with the same geometry as \(\mathrm{H}_3\mathrm{O}^+\). An example is \(\mathrm{NH}_3\) (Ammonia), which also has a trigonal pyramidal geometry due to the presence of three bonds and a lone pair on Nitrogen.
Key Concepts
Valence ElectronsMolecular GeometryChemical BondingTrigonal Pyramidal Structure
Valence Electrons
Valence electrons are the outermost electrons of an atom and play a crucial role in chemical bonding and molecular formation. These electrons determine how an atom will interact with others to form molecules. In the case of a water molecule modified with an extra proton, like \( \mathrm{H}_3\mathrm{O}^+ \), understanding valence electrons helps in drawing its Lewis structure.
Let's review how valence electrons work:
Let's review how valence electrons work:
- An element's group number in the periodic table gives a clue about the number of valence electrons. For instance, Oxygen, in group 16, has 6 valence electrons.
- Hydrogen, in group 1, has 1 valence electron.
- When forming ions, like \( \mathrm{H}_3\mathrm{O}^+ \), remember to adjust the count. The positive charge indicates a loss of one electron compared to the neutral molecule.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms in a molecule. This geometric shape greatly influences the physical and chemical properties of the substance.
Why is molecular geometry important?
Why is molecular geometry important?
- It affects molecular polarity, determining solubility and boiling point.
- It influences reaction behaviors and mechanisms.
- It assists in predicting molecular vibrations useful in IR spectroscopy.
Chemical Bonding
Chemical bonding explains how atoms are held together in molecules and compounds. It primarily involves the sharing or transferring of valence electrons between atoms to achieve stable electron configurations.
Key aspects of chemical bonding involve:
Key aspects of chemical bonding involve:
- **Covalent Bonds:** Sharing of electrons between atoms. In \( \mathrm{H}_3\mathrm{O}^+ \), covalent bonds form between Oxygen and Hydrogen atoms.
- **Ionic Bonds:** Transfer of electrons from one atom to another, resulting in positively and negatively charged ions. Although \( \mathrm{H}_3\mathrm{O}^+ \) has a net positive charge, it maintains its structure through covalent bonding.
- Chemical bonds define molecular geometry and play a major role in determining molecular properties.
Trigonal Pyramidal Structure
The trigonal pyramidal structure is a common molecular shape, especially in chemistry concerning elements like nitrogen and phosphorus. It describes a geometric configuration where three atoms are positioned around a central atom, forming a base, with one lone pair creating a fourth apex point.
In \( \mathrm{H}_3\mathrm{O}^+ \):
In \( \mathrm{H}_3\mathrm{O}^+ \):
- The central atom is Oxygen with a lone pair, causing the molecule to adopt a trigonal pyramidal structure instead of the planar trigonal configuration.
- This results in bond angles that are slightly less than the 109.5° found in a perfect tetrahedral geometry.
- An example of a neutral molecule with similar geometry is \( \mathrm{NH}_3 \) (Ammonia), where Nitrogen is central, too, with a lone pair affecting its shape.
Other exercises in this chapter
Problem 39
The compound(s) with TWO lone pairs of electrons on the central atom is(are (a) \(\mathrm{BrF}_{5}\) (b) \(\mathrm{ClF}_{3}\) (c) \(\mathrm{XeF}_{4}\) (d) \(\ma
View solution Problem 39
The molecule that has linear structure is (a) \(\mathrm{CO}_{2}\) (b) \(\mathrm{NO}_{2}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{SiO}_{2}\)
View solution Problem 40
The linear structure is assumed by : (a) \(\mathrm{SnCl}_{2}\) (b) \(\mathrm{NCO}^{-}\) (c) \(\mathrm{CS}_{2}\) (d) \(\mathrm{NO}_{2}^{+}\)
View solution Problem 40
Hydrogen bonding is maximum in (a) Ethanol (b) Diethyl ether (c) Ethyl chloride (d) Triethylamine
View solution