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The hybridization of atomic orbitals of nitrogen \(\mathrm{inNO}_{2}^{+}, \mathrm{NO}_{3}^{-}\) and \(\mathrm{NH}_{4}^{+}\) are: (a) \(\mathrm{sp}^{2}, \mathrm{sp}^{3}\) and \(\mathrm{sp}^{2}\) respectively (b) sp, sp and sp respectively (c) \(\mathrm{sp}^{2}\), sp and sp \(^{3}\) respectively (d) \(\mathrm{sp}^{2}, \mathrm{sp}^{3}\) and sp respectively

Short Answer

Expert verified
The correct option is (c) \(\mathrm{sp}, \mathrm{sp}^{2}, \mathrm{sp}^{3}\) respectively.

Step by step solution

01

Determine hybridization of NO2+

First, consider the ion \(\mathrm{NO}_{2}^{+}\). The nitrogen atom is bonded to two oxygen atoms and has no lone pairs, given the positive charge. The steric number, which is the sum of bonded atoms and lone pairs around the nitrogen, is 2. According to VSEPR theory, a steric number of 2 corresponds to \(\mathrm{sp}\) hybridization. Thus, the hybridization of nitrogen in \(\mathrm{NO}_{2}^{+}\) is \(\mathrm{sp}\) hybridization.
02

Determine hybridization of NO3-

Next, consider the ion \(\mathrm{NO}_{3}^{-}\). The nitrogen atom is bonded to three oxygen atoms and has no lone pairs due to the negative charge balancing with the bonded oxygen atoms. The steric number is 3. According to VSEPR theory, a steric number of 3 corresponds to \(\mathrm{sp}^{2}\) hybridization. Therefore, the nitrogen in \(\mathrm{NO}_{3}^{-}\) is \(\mathrm{sp}^{2}\) hybridized.
03

Determine hybridization of NH4+

Finally, consider the ion \(\mathrm{NH}_{4}^{+}\). The nitrogen atom is bonded to four hydrogen atoms and has no lone pairs. The steric number is 4. This corresponds to \(\mathrm{sp}^{3}\) hybridization according to VSEPR theory. Thus, the nitrogen in \(\mathrm{NH}_{4}^{+}\) is \(\mathrm{sp}^{3}\) hybridized.

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Key Concepts

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

Nitrogen Compounds
Nitrogen is found in various chemical compounds, exhibiting different bonding characteristics. These are known as nitrogen compounds, and they often include well-known ions like
  • NO2+
  • NO3-
  • NH4+
In these ions, nitrogen forms bonds with other atoms such as oxygen or hydrogen. The bonding and shape of these nitrogen compounds are influenced by the hybridization of nitrogen's atomic orbitals. Hybridization is a key concept here. It determines how nitrogen's atomic orbitals, usually in energy levels 2s and 2p, mix and form hybrid orbitals. These hybrid orbitals help in understanding the molecular geometry and stability of the compound.
VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory is essential for predicting the shapes of molecules. According to VSEPR theory, electron pairs around a central atom arrange themselves to minimize repulsion, which molds the shape of the molecule. Because nitrogen commonly has multiple bonds, VSEPR theory helps predict the hybridization required for bonding. For example:
  • If nitrogen bonds with only two other atoms, such as in NO2+, the steric number is 2, matching with sp hybridization.
  • In NO3-, nitrogen is bonded to three oxygen atoms, leading to a steric number of 3 and thus sp2 hybridization.
  • NH4+, with nitrogen bonded to four hydrogens, has a steric number of 4, resulting in sp3 hybridization.
The theory helps ensure that the molecule's electron cloud is distributed optimally around the center, allowing us to predict and understand the 3D shapes of molecules.
Molecular Geometry
Molecular geometry defines the three-dimensional shape of a molecule, influenced by the arrangement of electron pairs around the central atom. For nitrogen compounds, this geometry is critical. The shape of a molecule can affect everything from physical properties to reactivity. In nitrogen compounds:
  • In NO2+, the linear shape results from sp hybridization, with electrons spread 180 degrees apart.
  • The trigonal planar shape of NO3- comes from sp2 hybridization, with electrons at 120-degree angles spaced evenly.
  • NH4+ forms a tetrahedral shape due to sp3 hybridization, which positions electrons 109.5 degrees apart.
Knowing the molecular geometry aids in predicting the behavior of molecules during chemical reactions. It also impacts properties like polarity and intermolecular forces.
Steric Number
The steric number is an essential index for understanding hybridization and molecular geometry. It represents the total number of regions of electron density (bonds and lone pairs) surrounding a central atom. The steric number allows chemists to predict how atoms will hybridize. Here's how it works in context:
  • A steric number of 2 correlates with sp hybridization. This is seen in NO2+.
  • A steric number of 3 corresponds to sp2 hybridization and applies to NO3-.
  • For NH4+, with a steric number of 4, the hybridization is sp3.
Understanding the steric number helps determine not just the hybridization but also the 3D structural features of the compound. This makes it a fundamental concept in chemistry, critical for predicting and explaining molecular shapes and behaviors.

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Most popular questions from this chapter

The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}\), \(\mathrm{Mg}^{2+}, \mathrm{Be}^{2+} ?\) (a) \(\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2}+<\mathrm{Mg}^{2+}\) (b) \(\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}\) (c) \(\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}\) (d) \(\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}\)

Which one of the following pairs of molecules will have permanent dipole moments for both members? (a) \(\mathrm{SiF}_{4}\) and \(\mathrm{NO}_{2}\) (b) \(\mathrm{NO}_{2}\) and \(\mathrm{CO}_{2}\) (c) \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{3}\) (d) \(\mathrm{SiF}_{4}\) and \(\mathrm{CO}_{2}\)

Among the following compound which one has maximum number of lone pairs of electrons on central atom: (a) \(\left[\mathrm{ClO}_{3}\right]^{-}\) (b) \(\mathrm{XeF}_{A}\) (c) \(\mathrm{SF}_{4}\) (d) \(\left[\mathrm{I}_{3}\right]\)

Which contains both polar and non-polar bonds? (a) \(\mathrm{NH}_{4} \mathrm{Cl}\) (b) HCN (c) \(\mathrm{H}_{2} \mathrm{O}_{2}\) (d) \(\mathrm{CH}_{4}\)

The correct order of increasing \(\mathrm{C}-\mathrm{O}\) bond length of \(\mathrm{CO}, \mathrm{CO}_{3}^{2-}, \mathrm{CO}_{2}\) is (a) \(\mathrm{CO}_{3}^{2-}<\mathrm{CO}_{2}<\mathrm{CO}\) (b) \(\mathrm{CO}_{2}<\mathrm{CO}_{3}^{2}<\mathrm{CO}\) (c) \(\mathrm{CO}<\mathrm{CO}_{3}^{2-}<\mathrm{CO}_{2}\) (d) \(\mathrm{CO}<\mathrm{CO}_{2}<\mathrm{CO}_{3}^{2}\)

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