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Which of the following molecule is planar? (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (d) \(\mathrm{SiCl}_{4}\)

Short Answer

Expert verified
The planar molecule is \(\mathrm{C}_{2}\mathrm{H}_{4}\).

Step by step solution

01

Determine Molecular Geometry of CH4

The molecule \(\mathrm{CH}_{4}\) has carbon as the central atom bonded to four hydrogen atoms. According to the VSEPR theory, it adopts a tetrahedral geometry with bond angles of approximately \(109.5^\circ\), making it non-planar.
02

Determine Molecular Geometry of NH3

The molecule \(\mathrm{NH}_{3}\) has nitrogen as the central atom bonded to three hydrogen atoms and a lone pair. This results in a trigonal pyramidal shape due to lone pair repulsion, making the molecule non-planar.
03

Determine Molecular Geometry of C2H4

The molecule \(\mathrm{C}_{2}\mathrm{H}_{4}\) consists of a carbon-carbon double bond. Each carbon atom is bonded to two hydrogen atoms. This arrangement provides a trigonal planar geometry around each carbon atom, resulting in an overall planar molecule.
04

Determine Molecular Geometry of SiCl4

The molecule \(\mathrm{SiCl}_{4}\) has silicon as the central atom bonded to four chlorine atoms. Similar to \(\mathrm{CH}_{4}\), it exhibits a tetrahedral geometry due to the four substituents, making it non-planar.
05

Identify the Planar Molecule

From the analysis, \(\mathrm{C}_{2}\mathrm{H}_{4}\) has a planar geometry due to its trigonal planar shape, while \(\mathrm{CH}_{4}\), \(\mathrm{NH}_{3}\), and \(\mathrm{SiCl}_{4}\) are non-planar due to their tetrahedral and trigonal pyramidal geometries, respectively.

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Key Concepts

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

Planar Molecules
Planar molecules are those where all atoms lie in the same plane. This means the molecule is flat. The structure of a molecule plays a crucial role in its chemical properties and reactions. Planarity can affect how molecules interact with each other. A classic example is ethene (\(\mathrm{C}_{2}\mathrm{H}_{4}\)), which is planar due to the nature of the carbon-carbon double bond and the trigonal planar arrangement around each carbon atom.
  • Planar molecules typically have bond angles of around \(120^\circ\) , promoting a flat structure.
  • This planarity is crucial in determining how molecules stack and interact with each other.
  • Sometimes, rigidity in planar molecules affects the molecular flexibility under physical forces.
Understanding which molecules are planar helps in predicting their behavior, especially in organic reactions where the plane can dictate stereochemistry and molecular interactions. In molecular studies and synthesis, recognizing planar configurations is vital.
VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) Theory is a simple and useful way to predict the geometry of molecules. It is based on the principle that electron pairs around a central atom will arrange themselves to be as far apart as possible to minimize repulsion.
  • VSEPR requires knowing the number of bonding and non-bonding electron pairs around the central atom.
  • Molecular geometry is determined by the positions of atomic nuclei, which adjust to ensure maximum spacing of the electron pairs.
  • This geometrical arrangement directly influences molecular shape and symmetry. For instance, methane (\(\mathrm{CH}_4\)) adopts a tetrahedral shape, while ammonia (\(\mathrm{NH}_3\)) adopts a trigonal pyramidal geometry.
Understanding VSEPR can take the mystery out of why molecules have particular shapes. For chemistry students, mastering this theory aids in predicting the basic shape of nearly any molecule based on the number and type of bonded and lone electron pairs.
Trigonal Planar Geometry
Trigonal planar geometry is one such structure predicted by VSEPR theory. It arises when there are three outward-spreading areas of electron density around a central atom, with no lone pairs obstructing the symmetry.
  • In a trigonal planar molecule, the bond angles are approximately \(120^\circ\), which facilitates the planar shape.
  • An excellent example is \(\mathrm{C}_{2}\mathrm{H}_{4}\), where double bonding constrains the molecule to lie flat.
  • This geometry often appears in molecules with double bonds or resonance structures, which lock atoms into a planar configuration.
Recognizing trigonal planar geometry is vital in advanced chemistry as it influences molecular polarity and reactivity. It significantly affects how molecules participate in chemical reactions, especially in polymers and other larger organic compounds.

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Most popular questions from this chapter

Identify the correct sequence of increasing number of \(\pi\) bonds in the structures of the following molecules 1\. \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{6}\) 2\. \(\mathrm{H}_{2} \mathrm{SO}_{3}\) 3\. \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{5}\) (a) \(1,2,3\) (b) \(2,3,1\) (c) \(2,1,3\) (d) \(1,3,2\)

Find the total number of pr-d \(\pi\) bonds present in \(\mathrm{XeO}_{4}\).

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

Arrange the following compounds in order of increasing dipole moment 1\. Toluene, 2\. m-dichlorobenzene 3\. o-dichlorobenzene, 4\. p-dichlorobenzene (a) \(1<4<2<3\) (b) \(4<1<2<3\) (c) \(4<1<3<2\) (d) \(4<2 \leq 1<3\)

Which one of the following has zero dipole moment? (a) ClF (b) \(\mathrm{PCl}_{3}\) (c) \(\mathrm{SiF}_{4}\) (d) \(\mathrm{CFCl}_{3}\)

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