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Match the following: List I (Compounds) 1\. \(\mathrm{CS}_{2}\) 2\. \(\mathrm{SO}_{2}\) 3\. \(\mathrm{BF}_{3}\) 4\. \(\mathrm{NH}_{3}\) List II (Structures) (i) Bent (ii) Linear (iii) Trigonal planar (iv) Tetrahedral (v) Trigonal pyramidal

Short Answer

Expert verified
1-(ii), 2-(i), 3-(iii), 4-(v).

Step by step solution

01

Understanding the Compounds

First, we need to identify the compounds given in List I and understand their common structural characteristics. This requires basic knowledge of molecular geometry. Let's recall the shapes associated with each common configuration: - \(\mathrm{CS}_{2}\): Typically forms a linear shape due to double bonds between carbon and sulfur. - \(\mathrm{SO}_{2}\): Known for having a bent shape due to the lone pair on sulfur. - \(\mathrm{BF}_{3}\): Forms a trigonal planar shape because boron has three bonds without lone pairs. - \(\mathrm{NH}_{3}\): Typically forms a trigonal pyramidal shape, as the nitrogen atom has one lone pair.
02

Matching Compounds with Structures

Now that we have identified the common shapes associated with each compound, we can match them to the descriptions in List II:1. \(\mathrm{CS}_{2}\): Matches with (ii) Linear2. \(\mathrm{SO}_{2}\): Matches with (i) Bent3. \(\mathrm{BF}_{3}\): Matches with (iii) Trigonal planar4. \(\mathrm{NH}_{3}\): Matches with (v) Trigonal pyramidal.Each compound has a geometry based on the number of electron pairs and bonded atoms, aligning with these descriptions.

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Key Concepts

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

Linear Shape
In molecular geometry, a linear shape arises when a molecule consists of three atoms connected in a straight line. This structure occurs because the central atom has no lone pairs that could bend the configuration. One classic example of a molecule with a linear shape is carbon disulfide (\(\mathrm{CS}_{2}\)), which features an arrangement where the central carbon atom forms double bonds with two sulfur atoms.

This linear arrangement occurs due to the electron pair geometry involving two electron domains that repel according to VSEPR (Valence Shell Electron Pair Repulsion) theory, pushing the bonded atoms as far apart as possible, thus forming a straight line.

This lack of lone electron pairs means there are no additional repulsive forces to alter the molecule's shape, maintaining the atoms in a simple linear alignment.
Bent Shape
A bent shape in molecular geometry occurs when a molecule comprises a central atom bonded to two atoms, with lone electron pairs on the central atom causing a repulsive force that bends the structure. A well-known example of this configuration is sulfur dioxide (\(\mathrm{SO}_{2}\)).

In the case of \(\mathrm{SO}_{2}\), the sulfur atom forms bonds with two oxygen atoms, and also has a lone pair.

The presence of this lone pair pushes the oxygen atoms closer together, resulting in a bent shape rather than a linear one.
  • This arrangement is typical in molecules where the central atom has at least one lone pair, and it reflects how these lone pairs influence molecular geometry.


Such a structure often results in polar molecules because the charge distribution is not symmetrical, causing an imbalance. Understanding the bent shape aids in predicting physical properties, like polarity and reactivity, in molecular interactions.
Trigonal Planar Shape
A trigonal planar shape appears in molecular structures with three atoms bonded to a central atom, all lying in a single, flat plane. This structure typically emerges when there are no lone pairs on the central atom. The molecule boron trifluoride (\(\mathrm{BF}_{3}\)) is a classic example of this geometry.

In \(\mathrm{BF}_{3}\), the central boron atom forms three equal bonds with fluorine atoms.
  • Each bond angle measures precisely 120 degrees, ensuring that the molecule remains flat without any angular distortion.


This geometry results from the optimal electronic arrangement where the electrons repel each other equally, distributing around the plane.
Such a configuration is often seen in molecules where the central atom needs to form bonds without repulsion from lone pairs, leading to minimal structural distortion and a distinct planar layout.
Trigonal Pyramidal Shape
The trigonal pyramidal shape is distinct, characterized by a central atom with lone pairs and three atoms bonded to it, forming a three-sided pyramid. The notable example here is ammonia (\(\mathrm{NH}_{3}\)), where the nitrogen atom at the center bonds to three hydrogen atoms.

Unlike the trigonal planar shape, the presence of a lone pair on the nitrogen pushes the hydrogen atoms down, away from it, giving the molecule a three-dimensional pyramidal shape.
  • This crucial lone pair results in bond angles of about 107 degrees, slightly less than the ideal 109.5 degrees in a perfect tetrahedral arrangement.


This structure is significant in understanding molecular interactions because the lone pair creates asymmetrical charge distribution, potentially influencing the molecule's polarity and chemical behavior. It exemplifies how lone pairs of electrons can substantially modify a molecule's structure and properties.

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Most popular questions from this chapter

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? (a) bond length in \(\mathrm{NO}^{+}\)is greater than in \(\mathrm{NO}\) (b) bond length in \(\mathrm{NO}\) is greater than in \(\mathrm{NO}^{+}\) (c) bond length in \(\mathrm{NO}^{+}\)is equal than in NO (d) bond length is unpredictable

The correct statements if the following are: (a) The bond angle of hybrid bonds increases as \(\mathrm{sp}^{3}<\) \(s p^{2}

The electronegativity difference between \(\mathrm{N}\) and \(\mathrm{F}\) is greater than that between \(\mathrm{N}\) and \(\mathrm{H}\), yet the dipole moment of \(\mathrm{NH}_{3}(1.5 \mathrm{D})\) is larger than that of \(\mathrm{NF}_{3}\) \((0.2 \mathrm{D})\) This is because (a) in \(\mathrm{NH}_{3}\) as well as \(\mathrm{NF}_{3}\) the atomic dipole and bond dipole are in opposite directions (b) in \(\mathrm{NH}_{3}\) the atomic dipole and bond dipole are in the opposite directions whereas in \(\mathrm{NF}_{3}\) these are in the same direction (c) in \(\mathrm{NH}_{3}\) as well as in \(\mathrm{NF}_{3}\) the atomic dipole and bond dipole are in same direction (d) in \(\mathrm{NH}_{3}\) the atomic dipole and bond dipole and in the same direction whereas in \(\mathrm{NF}_{3}\) these are in opposite directions

Which of the following are non-polar? (1) \(\mathrm{SiF}_{4}\) (2) \(\mathrm{XeF}_{4}\) (3) \(\mathrm{SF}_{4}\) (4) \(\mathrm{BF}_{3}\) (5) \(\mathrm{NF}_{3}\) Select the correct answer using the code given below: (a) 1,2 and 4 (b) 3,4 and 5 (c) 2,3 and 4 (d) 1,3 and 4

Consider the given statements about the molecule \(\left(\mathrm{H}_{3} \mathrm{C}\right)_{2} \mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} .\) 1\. Three carbon atoms are sp \(^{3}\) hybridized 2\. Three carbon atoms are sp \(^{2}\) hybridized 3\. Two carbon atoms are sp hybridized Of three statements (a) 1 and 2 are correct (b) 1 and 3 are correct (c) 2 and 3 are correct (d) 1,2 and 3 are correct

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