Problem 190
Question
The pair of species having identical shapes for molecules of both species is (a) \(\mathrm{CF}_{4}, \mathrm{SF}_{4}\) (b) \(\mathrm{XeF}_{2}, \mathrm{CO}_{2}\) (c) \(\mathrm{BF}_{3}, \mathrm{PCl}_{3}\) (d) \(\mathrm{PF}_{5}, \mathrm{IF}_{5}\)
Step-by-Step Solution
Verified Answer
The pair with identical shapes is (b) \(\mathrm{XeF}_2, \mathrm{CO}_2\).
1Step 1: Understanding Molecular Geometry
Molecular geometry is determined by the arrangement of electron pairs around the central atom. This is described by the VSEPR (Valence Shell Electron Pair Repulsion) theory. For a molecule to have a specific shape, this depends on the number of bonds and lone pairs around the central atom.
2Step 2: Analyze Each Pair of Species
Examine the shapes:- \(\mathrm{CF}_4\): Tetrahedral, as there are four bonding pairs and no lone pairs around carbon.\(\mathrm{SF}_4\): Seesaw shape due to four bonding pairs and one lone pair.- \(\mathrm{XeF}_2\): Linear, with two bonding pairs and three lone pairs on xenon.\(\mathrm{CO}_2\): Linear, with double bonds on carbon and no lone pairs.- \(\mathrm{BF}_3\): Trigonal planar, with three bonding pairs and no lone pairs.\(\mathrm{PCl}_3\): Trigonal pyramidal, as there are three bonding pairs and one lone pair on phosphorus.- \(\mathrm{PF}_5\): Trigonal bipyramidal, with five bonding pairs and no lone pairs.\(\mathrm{IF}_5\): Square pyramidal, with five bonding pairs and one lone pair on iodine.
3Step 3: Identify Identical Shapes
Matching the shapes analyzed above, we see that both \(\mathrm{XeF}_2\) and \(\mathrm{CO}_2\) have a linear geometry. Each molecule of the pair has the same arrangement of atoms (linear), confirming they have identical shapes.
Key Concepts
VSEPR TheoryLinear GeometryTrigonal Planar Geometry
VSEPR Theory
When we talk about molecular geometry, VSEPR Theory is essential. It's like the roadmap that helps us figure out the shape of a molecule. VSEPR stands for Valence Shell Electron Pair Repulsion. This theory asserts that electron pairs, whether they are in bonds or unshared (also known as lone pairs), will arrange themselves around a central atom to be as far apart as possible. This minimizes the repulsion between electron pairs.
Here's how it works:
Here's how it works:
- Electron pairs are negatively charged, which means they repel each other.
- The molecule adopts a shape that minimizes this repulsion.
- Bonds and lone pairs around a central atom determine the geometry.
Linear Geometry
Linear geometry is one of the simplest shapes a molecule can have. Picture it as a straight line connecting the atoms. This linear arrangement happens when there are two bonded atoms with no lone pairs on the central atom influencing the shape. This gives a bond angle of 180 degrees.
For example:
For example:
- \(\text{XeF}_2\): Xenon forms a shape with two fluorine atoms forming a line, aided by three lone pairs that stay on the sides, maintaining the line.
- \(\text{CO}_2\): The carbon is bonded to oxygen atoms on either side, with double bonds.These arrangements create a straight path, giving the molecules their linear shape.
Trigonal Planar Geometry
In the world of molecular geometry, trigonal planar is a common shape. It occurs when there are three bonds and no lone pairs on the central atom.
Imagine a triangle lying flat. This is what a trigonal planar molecule looks like:
Imagine a triangle lying flat. This is what a trigonal planar molecule looks like:
- All three atoms around the central atom are in the same plane.
- The bond angles in a trigonal planar molecule are each 120 degrees.
Other exercises in this chapter
Problem 188
Which one of the following compounds has the smallest bond angle in its molecule? \(\quad\) [2003] (a) \(\mathrm{SO}_{2}\) (b) \(\mathrm{H}_{2} \mathrm{O}\) (c)
View solution Problem 189
Which one of the following pairs of molecules will have permanent dipole moments for both members? [2003] (a) \(\mathrm{SiF}_{4}\) and \(\mathrm{NO}_{2}\) (b) \
View solution Problem 191
The correct order of bond angles (smallest first) in \(\mathrm{H}_{2} \mathrm{~S}, \mathrm{NH}_{3}, \mathrm{BF}_{3}\) and \(\mathrm{SiH}_{4}\) is [2004] (a) \(\
View solution Problem 192
The bond order in \(\mathrm{NO}\) is \(2.5\) while that in \(\mathrm{NO}^{+}\)is 3 . Which of the following statements is true for these two species? [2004] (a)
View solution