Problem 5
Question
Sketch a picture of the following molecules based on your models: \(\mathrm{CH}_{4} ; \mathrm{NH}_{3}\); \(\mathrm{H}_{2} \mathrm{O} .\)
Step-by-Step Solution
Verified Answer
Based on VSEPR theory, Methane (CH4) has a tetrahedral shape, Ammonia (NH3) has a trigonal pyramidal shape, and Water (H2O) has a bent or V-shaped structure.
1Step 1: Preparation
Know the formulas of the compounds: CH4 is Methane, NH3 is Ammonia, and H2O is Water.
2Step 2: Determine Central Atom and Count Valence Electrons
For CH4, C is the central atom, which has 4 valence electrons. H has 1 valence electron, and since there are 4 H atoms in the molecule, that adds 4 valence electrons. Thus, CH4 has a total of 8 valence electrons. Similarly, for NH3, N is the central atom having 5 valence electrons, and H adds 3 electrons (1 electron x 3 atoms), giving a total of 8 valence electrons. For H2O, O is the central atom with 6 valence electrons, and H adds 2 electrons (1 electron x 2 atoms) for a total of 8 valence electrons.
3Step 3: Determine Number of Bonding Atoms and Number of Lone Pairs
CH4 has 4 bonding atoms (hydrogens) and no lone pairs on the central carbon atom. NH3 has 3 bonding atoms (hydrogens) and 1 lone pair on the nitrogen. H2O has 2 bonding atoms (hydrogens) and 2 lone pairs on the oxygen.
4Step 4: Apply VSEPR Theory and Predict Molecular Shape
Applying the VSEPR theory, CH4 forms a tetrahedral structure, where all 4 hydrogens are bonded to the central carbon atom. For NH3, 3 hydrogens are bonded to the central nitrogen with one lone pair, forming a trigonal pyramidal structure. In water, two hydrogens are bonded to the central oxygen with two lone pairs, forming a bent or V-shaped structure.