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The molecule/ion which is not planar is (1) \(\mathrm{BF}_{3}\) (2) \(\mathrm{NO}_{3}\) (3) \(\mathrm{CO}_{3}^{2}\) (4) \(\mathrm{NF}_{3}\)

Short Answer

Expert verified
The non-planar molecule is \(\mathrm{NF}_{3}\).

Step by step solution

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01

Determine the Molecular Geometry

Identify the molecular geometry of each given molecule/ion. For this, use VSEPR (Valence Shell Electron Pair Repulsion) theory to predict the shapes.
02

Analyze \(\text{BF}_3\)

\(\mathrm{BF}_{3}\) has a central boron atom with three fluoride atoms. Boron has three valence electrons, which form bonds with three fluorine atoms, resulting in no lone pairs on boron. Therefore, the geometry is trigonal planar.
03

Analyze \(\text{NO}_3^-\)

\(\mathrm{NO}_{3}^{-}\) has a central nitrogen atom with three oxygen atoms. Nitrogen has five valence electrons, and there is an extra electron due to the negative charge, making a total of 6 electrons. Three of these form single bonds, and one forms a double bond with oxygen atoms. The geometry is trigonal planar.
04

Analyze \(\text{CO}_3^{2-}\)

\(\mathrm{CO}_{3}^{2-}\) has a central carbon atom with three oxygen atoms. Carbon has four valence electrons, plus two additional electrons due to the 2- charge, giving a total of 6 electrons. The ions form two single bonds and one double bond with the oxygen atoms, resulting in a trigonal planar geometry.
05

Analyze \(\text{NF}_3\)

\(\mathrm{NF}_{3}\) has a central nitrogen atom with three fluorine atoms. Nitrogen has five valence electrons, forming three bonds with the fluorine atoms and leaving two lone pairs around the nitrogen. This results in a trigonal pyramidal geometry, which is not planar.
06

Identify the Non-Planar Molecule

From the steps above, it is clear that the geometries of \(\mathrm{BF}_{3}\), \(\mathrm{NO}_{3}^{-}\), and \(\mathrm{CO}_{3}^{2-}\) are all planar (trigonal planar). The geometry of \(\mathrm{NF}_{3}\) is trigonal pyramidal, which is not planar.

Key Concepts

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

VSEPR theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory is essential for predicting molecular shapes. This theory states that electron pairs around a central atom will arrange themselves to minimize repulsion.
This means bonds and lone pairs will move as far away from each other as possible to achieve the most stable geometry.
Whether repulsions come from bonding pairs or lone pairs, their influence shapes the three-dimensional structure of the molecule.
Understanding VSEPR theory helps us figure out if a molecule is planar (flat) or has a more complex 3D shape.
Trigonal planar geometry
In trigonal planar geometry, a central atom is surrounded by three atoms at the corners of an equilateral triangle. Important characteristics include:
  • Bond angles of 120 degrees.
  • All atoms lie in a single plane, which makes the structure flat.

This geometry happens when the central atom has no lone pairs, such as in \(\text{BF}_3\), \(\text{NO}_3^-\), and \(\text{CO}_3^{2-}\). For \(\text{BF}_3\), boron has three valence electrons, each bonding to a fluorine atom without any lone pairs.
Similarly, in \(\text{NO}_3^-\) and \(\text{CO}_3^{2-}\), the central atoms share their electrons to create a flat, planar shape.
Trigonal pyramidal geometry
Trigonal pyramidal geometry features a central atom connected to three other atoms but also has a lone pair of electrons.
Key points include:
  • Bond angles slightly less than 109.5 degrees due to the lone pair.
  • The presence of a lone pair makes it a three-dimensional shape.

For example, in \(\text{NF}_3\), nitrogen forms three bonds with fluorine atoms but has one lone pair.
Unlike trigonal planar, this lone pair pushes the bonds slightly closer, creating a 3D structure that isn't flat.
Electron pair repulsion
Electron pair repulsion is a fundamental concept in VSEPR theory, determining how electron pairs shape the molecule.
Lone pairs exert more repulsion than bonding pairs. This is because lone pairs are closer to the nucleus and occupy more space.
For instance:
  • Bonding pairs are shared between atoms, reducing their repulsive effect.
  • Lone pairs push bonding pairs closer together, altering bond angles and shaping the molecule.

Considering \(\text{NF}_3\), nitrogen's lone pair pushes the fluorine atoms together, resulting in a trigonal pyramidal geometry.
Thus, understanding these repulsions helps to predict whether a molecule will be planar or not.

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Most popular questions from this chapter

\(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are converted into monoanions \(\mathrm{N}_{2}^{-}\) and \(\mathrm{O}_{2}^{-}\), respectively. Which of the following statements is wrong? (1) In \(\mathrm{N}_{2}, \mathrm{~N}-\mathrm{N}\) bond weakens. (2) In \(\mathrm{O}_{2}^{-}, \mathrm{O}-\mathrm{O}\) bond order increascs. (3) In \(\mathrm{O}_{2}^{-}, \mathrm{O}-\mathrm{O}\) bond order decreases. (4) \(\mathrm{N}_{2}^{-}\) becomes paramagnetic.

Which of the following molecules do not exist? (1) \(\mathrm{II}_{2}\) (2) \(\mathrm{Bc}_{2}\) (3) \(\mathrm{C}_{2}\) (4) \(\mathrm{N}_{2}\)

Which of the following conditions is not correct for resonating structures? (1) The contributing structures must have the same number of unpaired electrons. (2) The contributing structures should have similar energies. (3) The contributing structures should be so written that unlike charges reside on atoms that are far apart. (4) The positive charge should be present on the electropositive clement and the negative charge on the electroncgative clement.

The correct statement about resonating structure is (1) They have no real existence. (2) These are hypothetical structures. (3) These structures can never be separated. (4) All are correct.

Which is not characteristic of a \(\pi\) bond? (1) A \(\pi\) -bond is formed when a sigma bond is already formed. (2) \(\pi\) -bonds are formed by hybrid orbitals. (3) \(\pi\) -bonds may be formed by the overlapping of atomic orbitals. (4) \(\pi\) -bond results from lateral overlap of atomic orbitals.

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