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Write Lewis structures for \({\rm{N}}{{\rm{F}}_{\rm{3}}}\) and \({\rm{P}}{{\rm{F}}_{\rm{5}}}\). On the basis of hybrid orbitals, explain the fact that \({\rm{N}}{{\rm{F}}_{\rm{3}}}\), \({\rm{P}}{{\rm{F}}_{\rm{3}}}\), and \({\rm{P}}{{\rm{F}}_{\rm{5}}}\) are stable molecules, but \({\rm{N}}{{\rm{F}}_{\rm{5}}}\) does not exist.

Short Answer

Expert verified

Nitrogen has four electron density zones in \({\rm{N}}{{\rm{F}}_{\rm{3}}}\) and phosphorus has four electron density regions in \({\rm{P}}{{\rm{F}}_{\rm{3}}}\) and is \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridised. Nitrogen and phosphorus will have five areas of electron density in \({\rm{N}}{{\rm{F}}_{\rm{5}}}\) and \({\rm{P}}{{\rm{F}}_{\rm{5}}}\), and will need to be \({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}\) hybridised, but nitrogen lacks a valence d orbital to accommodate the additional pair of electrons, whereas phosphorus does. So, whereas \({\rm{N}}{{\rm{F}}_{\rm{3}}}\), \({\rm{P}}{{\rm{F}}_{\rm{3}}}\), and \({\rm{P}}{{\rm{F}}_{\rm{5}}}\) are stable molecules, \({\rm{N}}{{\rm{F}}_{\rm{5}}}\) does not.

Step by step solution

01

Define molecule

A molecule is the smallest unit of a substance that holds the compound's chemical characteristics. Molecules are made up of atoms arranged in groups.

02

Explanation

\({\rm{N}}{{\rm{F}}_{\rm{3}}}\)and\({\rm{P}}{{\rm{F}}_{\rm{5}}}\)Lewis structures,


Phosphorus in \({\rm{P}}{{\rm{F}}_{\rm{3}}}\) and nitrogen in \({\rm{N}}{{\rm{F}}_{\rm{3}}}\) contain four electron density areas and are \({\rm{s}}{{\rm{p}}^{\rm{3}}}\) hybridised. Nitrogen and phosphorus will have five areas of electron density in \({\rm{N}}{{\rm{F}}_{\rm{5}}}\) and \({\rm{P}}{{\rm{F}}_{\rm{5}}}\), and will need to be \({\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}\) hybridised, but nitrogen lacks a valence d orbital to accommodate the additional pair of electrons, whereas phosphorus does. So \({\rm{N}}{{\rm{F}}_{\rm{3}}}\), \({\rm{P}}{{\rm{F}}_{\rm{3}}}\), and \({\rm{P}}{{\rm{F}}_{\rm{5}}}\) are all stable compounds, but \({\rm{N}}{{\rm{F}}_{\rm{5}}}\) isn't.

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Most popular questions from this chapter

Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds: (a) \({\rm{ClNO}}\) (\({\rm{N}}\) is the central atom) (b) \({\rm{C}}{{\rm{S}}_{\rm{2}}}\) (c) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{CO}}\) (\({\rm{C}}\) is the central atom) (d) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{SO}}\) (\({\rm{S}}\) is the central atom) (e) \({\rm{S}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\) (\({\rm{S}}\) is the central atom) (f) \({\rm{Xe}}{{\rm{O}}_{\rm{2}}}{{\rm{F}}_{\rm{2}}}\) (\({\rm{Xe}}\) is the central atom) (g) \({\rm{ClO}}{{\rm{F}}_{\rm{2}}}^{\rm{ + }}\) (\({\rm{Cl}}\) is the central atom).

A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, \({{\rm{H}}_{\rm{3}}}{\rm{CCN}}\). It is present in paint strippers. (a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule. (b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form \({\rm{\sigma }}\) bonds. (c) Describe the atomic orbitals that form the \({\rm{\pi }}\) bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.

The main component of air is N2. From the molecular orbital diagram of N2, predict its bond order and whether it is diamagnetic or paramagnetic.

Label the molecular orbital shown as σ or Π, bonding or antibonding, and indicate where the node occurs.


Use valence bond theory to explain the bonding in\({{\rm{O}}_{\rm{2}}}\). Sketch the overlap of the atomic orbitals involved in the bonds in \({{\rm{O}}_{\rm{2}}}\).

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