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Chapter 8: Q49 E (page 451)

Explain why N22+ is diamagnetic, while O24+, which has the same number of valence electrons, is paramagnetic.

Short Answer

Expert verified

∏p Orbital in O24+ contains two unpaired electrons so paramagnetic. The∏p orbital inN22+ is fully filled so diamagnetic

Step by step solution

01

Define valence electron

A valence electron is the most probable electron to be involved in a chemical process. The electrons having the largest value of the primary quantum number, n, are usually those.

02

Explanation

This is because,

N22+: (σ2s)2 (σ *2s)2 (∏2py∏2pz)4(σ0 2px)

O24+ : (σ2s)2 (σ *2s)2(σ 2px)2 (∏2py∏2pz)2

The σp orbital is lifted above the∏p, set due to s-p mixing in N2, whereas there is no s-p mixing in O2 and it will follow the regular pattern.

As a result,σp orbital will fill before∏p, orbital in O2,and∏p, orbital inO24+ has two unpaired electrons and is paramagnetic. The∏p orbital in theN22+ is completely filled and it is diamagnetic.

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Most popular questions from this chapter

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.

For the carbonate ion, \({\rm{C}}{{\rm{O}}_{\rm{3}}}^{{\rm{2 - }}}\), draw all of the resonance structures. Identify which orbitals overlap to create each bond.

Use valence bond theory to explain the bonding in\({{\rm{F}}_2},{\rm{HF}}\), and\({\rm{ClBr}}\). Sketch the overlap of the atomic orbitals involved in the bonds.

Strike-anywhere matches contain a layer of \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) and a layer of \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\). The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. \({\rm{KCl}}{{\rm{O}}_{\rm{3}}}\) contains the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion. \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) is an unusual molecule with the skeletal structure.

  1. Write Lewis structures for \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) and the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion.
  2. Describe the geometry about the \({\rm{P}}\) atoms, the \({\rm{S}}\) atom, and the \({\rm{Cl}}\) atom in these species.
  3. Assign a hybridization to the \({\rm{P}}\) atoms, the \({\rm{S}}\)atom, and the \({\rm{Cl}}\) atom in these species.
  4. Determine the oxidation states and formal charge of the atoms in \({{\rm{P}}_{\rm{4}}}{{\rm{S}}_{\rm{3}}}\) and the \({\rm{Cl}}{{\rm{O}}_{\rm{3}}}^{\rm{ - }}\) ion.

True or false: Boron contains \({\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{1}}}\)valence electrons, so only one \({\rm{p}}\)- orbital is needed to form molecular orbitals.
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