Problem 149
Question
Isostructural group of molecule is (a) \(\mathrm{XeO}_{4}, \mathrm{NH}_{4}^{+}, \mathrm{CH}_{4}\) (b) \(\mathrm{CH}_{3}^{-}, \mathrm{NH}_{3}, \mathrm{NF}_{3}\) (c) \(\mathrm{NH}_{3}, \mathrm{NF}_{3}, \mathrm{BF}_{3}\) (d) \(\mathrm{NO}_{3}, \mathrm{NO}_{2}, \mathrm{SF}_{4}\)
Step-by-Step Solution
Verified Answer
(a) \(\mathrm{XeO}_{4}, \mathrm{NH}_{4}^{+}, \mathrm{CH}_{4}\) are isostructural.
1Step 1: Understand Isostructural Molecules
Isostructural molecules have the same shape and symmetry but can contain different atoms. They typically have the same number of bonding pairs and lone electron pairs on the central atom.
2Step 2: Identify VSEPR Shapes of Molecules
Determine the VSEPR shapes for each molecule or ion in the given options.- (a) - \(\mathrm{XeO}_{4}\) is tetrahedral, - \(\mathrm{NH}_{4}^{+}\) is tetrahedral, - \(\mathrm{CH}_{4}\) is tetrahedral.(b) - \(\mathrm{CH}_{3}^{-}\) is trigonal pyramidal,- \(\mathrm{NH}_{3}\) is also trigonal pyramidal,- \(\mathrm{NF}_{3}\) is also trigonal pyramidal.(c) - \(\mathrm{NH}_{3}\) is trigonal pyramidal,- \(\mathrm{NF}_{3}\) is trigonal pyramidal,- \(\mathrm{BF}_{3}\) is trigonal planar.(d) - \(\mathrm{NO}_{3}^{-}\) is trigonal planar,- \(\mathrm{NO}_{2}\) is bent,- \(\mathrm{SF}_{4}\) is seesaw-shaped.
3Step 3: Compare Shapes and Structures
Compare the geometries of the molecules in each set of the given options:
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For option (a), all molecules are tetrahedral.
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For option (b), all molecules are trigonal pyramidal.
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For option (c), two molecules are trigonal pyramidal and one is trigonal planar.
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For option (d), all have different shapes: trigonal planar, bent, and seesaw.
4Step 4: Choose the Isostructural Group
The isostructural group is the one where all species have the same geometry. Compare the geometries from Step 3 and identify the matching group. You see that both (a) and (b) have consistent shapes across their molecules, but they have different geometries, where (a) is tetrahedral and (b) is trigonal pyramidal.
Key Concepts
VSEPR TheoryMolecular GeometryTetrahedral GeometryTrigonal Pyramidal Geometry
VSEPR Theory
VSEPR stands for Valence Shell Electron Pair Repulsion theory. It is a model used to predict the geometry of individual molecules based on the number of electron pairs surrounding their central atoms. The key idea is that electron pairs, whether bonding or non-bonding, will arrange themselves as far apart as possible around the central atom to minimize repulsion.
- Electron pairs include both bonding pairs and lone pairs.
- The geometrical arrangement of these electron pairs dictates the overall shape of the molecule.
Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. This structure is determined by the distances and angles between atoms, which are influenced by electron pair repulsion, as explained by VSEPR theory. Understanding molecular geometry helps in predicting and explaining the behavior and reactivity of molecules.
- The geometry affects properties like polarity and intermolecular forces.
- Diverse configurations include linear, bent, tetrahedral, and more.
Tetrahedral Geometry
Tetrahedral geometry is one of the most common molecular shapes and occurs when four pairs of electrons are placed around a central atom, creating an angle of about 109.5° between them. This formation is characteristic of molecules like \(\mathrm{CH}_{4}\) and \(\mathrm{NH}_{4}^{+}\).
- A tetrahedral molecule has four bonds arranged symmetrically around the central atom.
- It is highly symmetric and typically does not have lone pairs on the central atom.
Trigonal Pyramidal Geometry
Trigonal pyramidal geometry occurs when a molecule has three bonding pairs and one lone pair on the central atom. This lone pair exerts a greater repulsive force on the bonding pairs, which results in a shape different from the ideal triangular base. An example of a trigonal pyramidal molecule includes \(\mathrm{NH}_{3}\).
- The bond angles are slightly less than the ideal 109.5° found in a perfect tetrahedron.
- The central atom is located at the apex of the pyramid formed by the three atoms at the base.
Other exercises in this chapter
Problem 147
Mark out the incorrect match of shape for a given molecule/ion. (a) \(\mathrm{ICl}_{4}^{-}-\)square planar (b) \(\mathrm{NH}_{2}^{-}-\)Pyramidal (c) \(\mathrm{S
View solution Problem 148
Which of the pairs have identical values of bond order? (a) \(\mathrm{F}_{2}\) and \(\mathrm{Ne}_{2}\) (b) \(\mathrm{N}_{2}^{+}\)and \(\mathrm{O}_{2}^{+}\) (c)
View solution Problem 150
The correct statements if the following are: (a) The bond angle of hybrid bonds increases as \(\mathrm{sp}^{3}
View solution Problem 152
Which combination of the compound and their shapes are correct? (a) \(\mathrm{ClF}_{3}-\) see saw (b) \(\mathrm{ICl}_{4}^{-}-\)square planar (c) \(\mathrm{ICl}_
View solution