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Chapter 2: Problem 168

Which of the following shows isostructural species? [A.F.M.C. 2003] (a) \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NH}_{2}^{-}\) (b) \(\mathrm{CH}_{3}^{-}\) and \(\mathrm{CH}_{3}^{+}\) [ (c) \(\mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}\) and \(\mathrm{BF}_{4}\) (d) \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NH}_{3}\)

Short Answer

Expert verified
Option (c) are isostructural: \(\mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}, \mathrm{BF}_{4}^{-}\).

Step by step solution

01

Understanding Isostructural Species

Isostructural species have the same number of atoms, the same type of hybridization, and the same shape or geometry. To determine if pairs are isostructural, we need to compare their molecular geometry.
02

Analyze Option (a): \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NH}_{2}^{-}\)

\(\mathrm{NH}_{4}^{+}\) (ammonium) has tetrahedral geometry with sp3 hybridization, while \(\mathrm{NH}_{2}^{-}\) (amide ion) has a bent shape with sp3 hybridization. Since their geometries differ, these species are not isostructural.
03

Analyze Option (b): \(\mathrm{CH}_{3}^{-}\) and \(\mathrm{CH}_{3}^{+}\)

Both \(\mathrm{CH}_{3}^{-}\) and \(\mathrm{CH}_{3}^{+}\) have the carbon atom bonded to three hydrogen atoms, but \(\mathrm{CH}_{3}^{-}\) has a trigonal pyramidal shape with sp3 hybridization, while \(\mathrm{CH}_{3}^{+}\) is planar (trigonal planar) with sp2 hybridization. They are not isostructural.
04

Analyze Option (c): \(\mathrm{SO}_{4}^{2-}, \mathrm{PO}_{4}^{3-}, \mathrm{BF}_{4}^{-}\)

All three ions - \(\mathrm{SO}_{4}^{2-}\), \(\mathrm{PO}_{4}^{3-}\), and \(\mathrm{BF}_{4}^{-}\) - have a tetrahedral shape with sp3 hybridization and four atoms bonded to a central atom. Hence, these species are isostructural.
05

Analyze Option (d): \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NH}_{3}\)

\(\mathrm{NH}_{4}^{+}\) has a tetrahedral geometry with sp3 hybridization, while \(\mathrm{NH}_{3}\) has a trigonal pyramidal shape with sp3 hybridization. The presence of one lone pair in \(\mathrm{NH}_{3}\) changes the shape, so they are not isostructural.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Molecular geometry is all about the three-dimensional arrangement of atoms within a molecule. This arrangement impacts the molecule's properties and functions. Understanding molecular geometry helps in predicting interactions and reactivity of molecules in chemistry and biology.
Here are some key points to remember:
  • **Linear** - Atoms are arranged in a straight line.
  • **Trigonal Planar** - Three atoms form a triangle around a central atom in one plane.
  • **Tetrahedral** - Four atoms are positioned symmetrically around a central atom.
  • **Bent** - Atoms form an angle due to the presence of lone pairs affecting bond angles.
The geometry affects various molecular aspects, from boiling points to polarity, making it a crucial concept in understanding chemistry.
Hybridization
Hybridization is a concept that describes the mixing of atomic orbitals to form new hybrid orbitals. These hybrids help explain the geometry and bonding in complex molecules. By understanding hybridization, students can predict molecular shapes and bond angles accurately.
Here are some common types:
  • **sp Hybridization** - Involves one s and one p orbital, forming two linear hybrid orbitals.
  • **sp2 Hybridization** - Involves one s and two p orbitals, creating three planar, 120-degree hybrid orbitals.
  • **sp3 Hybridization** - Involves one s and three p orbitals, resulting in four hybrid orbitals positioned in a tetrahedral shape.
Hybridization provides a valuable framework for understanding how molecules form and interact.
Tetrahedral Shape
The tetrahedral shape is one of the most common molecular geometries and is characterized by four atoms bonded to a central atom, forming a shape similar to a pyramid with a triangular base. It is highly symmetrical and typically involves sp3 hybridization.
Here are some crucial points about tetrahedral shape:
  • **Bond Angles** - The bond angles in a perfect tetrahedral shape are approximately 109.5 degrees.
  • **Examples** - Examples include methane ( \( \text{CH}_4 \)) and the ammonium ion ( \( \text{NH}_4^+ \)).
  • **Geometry Impact** - This geometry enables equal distribution of electrostatic forces, making these molecules relatively stable.
Understanding the tetrahedral shape is essential for appreciating the spatial arrangement in complex organic and inorganic molecules.
Trigonometry Pyramidal Shape
The trigonal pyramidal shape occurs in molecules where a central atom is bonded to three atoms and has one lone pair. This gives the appearance of a pyramid with a triangular base, connected to the central atom at the apex. The presence of a lone pair is what differentiates it from trigonal planar geometry.
Here are some key aspects:
  • **Bond Angles** - The bond angles are slightly less than 109.5 degrees due to the lone pair repulsion compressing the bond angles.
  • **Examples** - Ammonia ( \( \text{NH}_3 \)) is a classic example with a trigonal pyramidal shape.
  • **Hybridization** - Typically involves sp3 hybridization, accounting for the three bond pairs and one lone pair.
This shape is crucial for understanding molecules' behavior and interaction that contain lone pairs, affecting their chemical properties and reactions.

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Most popular questions from this chapter

\text { Number of bonds in benzene: } (a) \(6 \sigma\) and \(3 \pi\) \(\square\) (b) \(3 \sigma\) and \(12 \pi\) (c) \(12 \sigma\) and \(3 \pi\) \(\square\) (d) \(6 \sigma\) and \(6 \pi\)

In a triple bond there is sharing of : (a) 3 electrons \(\square\) (b) 4 electrons (c) several electrons \(\square\) (d) 6 electrons

Acetylene molecule contains: (a) 5 sigma bonds \(\square\) (b) 4 sigma and \(1 \pi\) -bond (c) 3 sigma and \(2 \pi\) -bonds \(\square\) (d) 2 sigma and \(3 \pi\) -bonds

The electronegativity of cesium is \(0.7\) and that of fluorine is \(4.0\). The bond formed between the two is: (a) covalent \(\square\) (b) electrovalent (c) coordinate \(\square\) (d) metallic

Four elements \(P, Q, R, S\) have atomic numbers \(Z-1, Z\), \(Z+1\) and \(Z+2\) respectively. If \(Z\) is 17, then bond between which pair of elements will be most ionic ? (a) \(\mathrm{S}\) and \(\mathrm{Q}\) \(\square\) (b) \(\mathrm{P}\) and \(\mathrm{R}\) (c) \(\mathrm{S}\) and \(\mathrm{R}\) \(\square\) (d) S and \(\mathrm{P}\)
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