Math  /  Geometry

QuestionPredict the shape of each molecule. (a) CF4\mathrm{CF}_{4} linear trigonal planar bent tetrahedral trigonal pyramidal (a) NCl3\quad \mathrm{NCl}_{3} linear trigonal planar bent tetrahedral trigonal pyramidal (c) H2O\mathrm{H}_{2} \mathrm{O} linear trigonal planar bent tetrahedral trigonal pyramidal

Studdy Solution

STEP 1

1. The molecular geometry is determined by the VSEPR (Valence Shell Electron Pair Repulsion) theory.
2. The central atom's electron pairs, both bonding and non-bonding, will influence the shape of the molecule.

STEP 2

1. Identify the central atom and count its valence electrons.
2. Determine the number of bonding pairs and lone pairs around the central atom.
3. Use VSEPR theory to predict the molecular shape.

STEP 3

Identify the central atom and count its valence electrons for each molecule:
(a) CF4\mathrm{CF}_{4}: The central atom is carbon (C), which has 4 valence electrons. (b) NCl3\mathrm{NCl}_{3}: The central atom is nitrogen (N), which has 5 valence electrons. (c) H2O\mathrm{H}_{2}\mathrm{O}: The central atom is oxygen (O), which has 6 valence electrons.

STEP 4

Determine the number of bonding pairs and lone pairs around the central atom:
(a) CF4\mathrm{CF}_{4}: Carbon forms 4 single bonds with fluorine atoms, resulting in 4 bonding pairs and 0 lone pairs. (b) NCl3\mathrm{NCl}_{3}: Nitrogen forms 3 single bonds with chlorine atoms, resulting in 3 bonding pairs and 1 lone pair. (c) H2O\mathrm{H}_{2}\mathrm{O}: Oxygen forms 2 single bonds with hydrogen atoms, resulting in 2 bonding pairs and 2 lone pairs.

STEP 5

Use VSEPR theory to predict the molecular shape:
(a) CF4\mathrm{CF}_{4}: With 4 bonding pairs and 0 lone pairs, the shape is tetrahedral. (b) NCl3\mathrm{NCl}_{3}: With 3 bonding pairs and 1 lone pair, the shape is trigonal pyramidal. (c) H2O\mathrm{H}_{2}\mathrm{O}: With 2 bonding pairs and 2 lone pairs, the shape is bent.
The predicted shapes are: (a) CF4\mathrm{CF}_{4}: Tetrahedral (b) NCl3\mathrm{NCl}_{3}: Trigonal Pyramidal (c) H2O\mathrm{H}_{2}\mathrm{O}: Bent

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