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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

鈭弍 Orbital in O24+ contains two unpaired electrons so paramagnetic. The鈭弍 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+: (蟽2s)2 (蟽 *2s)2 (鈭�2辫测鈭�2辫锄)4(蟽0 2px)

O24+ : (蟽2s)2 (蟽 *2s)2(蟽 2px)2 (鈭�2辫测鈭�2辫锄)2

The 蟽p orbital is lifted above the鈭弍, 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,蟽p orbital will fill before鈭弍, orbital in O2,and鈭弍, orbital inO24+ has two unpaired electrons and is paramagnetic. The鈭弍 orbital in theN22+ is completely filled and it is diamagnetic.

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Most popular questions from this chapter

Why is the concept of hybridization required in valence bond theory?

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 \({{\rm{O}}_{\rm{2}}}\).

Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two S orbitals and from two P orbitals.

Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?

What is the hybridization of the central atom in each of the following?

(a) \({\rm{Be}}{{\rm{H}}_{\rm{2}}}\)

(b) \({\rm{S}}{{\rm{F}}_{\rm{6}}}\)

(c) \({\rm{PO}}_{\rm{4}}^{{\rm{3 - }}}\)

(d) \({\rm{PC}}{{\rm{l}}_{\rm{5}}}\)
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