Chapter 7, Problem 70SCQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# The ionization of the hydrogen atom can be calculated from Bohr's equation for the electron energy.E = −(NARhc)(Z2/n2)where NARhc =1312 kJ/mol and Z is the atomic number. Let us use this approach to calculate a possible ionization energy for helium. First, assume the electrons of the He experience the full 2+ nuclear charge. This gives us the upper limit for the ionization energy. Next, assume one electron of He completely screens the nuclear charge from the other electrons, so Z = 1. This gives us a lower limit to the ionization energy*. Compare these calculated values for the upper and lower limits to the experimental value of 2372.3 kJ/mol. What does this tell us about the ability of one electron to screen the nuclear charge?

Interpretation Introduction

Interpretation:

The upper and lower ionization energies should be calculate given the Bohr’s equations.

Concept Introduction:

Nuclear charge (Z*): The effective nuclear charge generally denoted by (Zeff or Z*) it is the net positive charge experienced by an electron in a multi-electron atom. This word “effective” is used because the shielding effect of negatively charged electron prevents higher orbital electrons experience the full nuclear charge.

Ionization energy (IE): The ionization energy is the required to remove an electron from an atom in the gas phase.

General formula of ionization energy= Atom in the ground state(g)Atom+(g)+eΔU=Ionizationenergy(IE)

As predicted by coulomb’s law, energy must be supplied to overcome the attraction between an electron and the nucleus and to separate the electron from the atom. Thus ionization energies always have positive values. An electron father from the nucleus generally has smaller ionization energy and electron closer to the nucleus has larger ionization energy.

Increase and decrease electro negativity: The less vacancy electrons an atoms has the least it will gain of electrons. Moreover the electron affinity decrease down the groups and from right to left across the periods on the periodic table, the reason is electrons are placed in a higher energy level far from the nucleus thus a decrease from its pull.

Explanation

First we calculate the upper limit ionization energy

â€‚Â LetÂ usÂ considerÂ theÂ Bohr'sÂ equationE=â€‰-(NARhc)â€‰(Z2/n2)NARhcâ€‰=1312â€‰kJ/molZ=Atomicâ€‰numbern=â€‰no.of.elctronsHereÂ E=â€‰-Câ€‰(Z2/n2)â€‰â€‰â€‰â€‰[sinceÂ C=NARhc]Z=â€‰2â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰[atomicâ€‰numberâ€‰ofâ€‰helium]En=âˆž-En=1â€‰=â€‰-Câ€‰(Z2nâˆž2)â€‰+â€‰Câ€‰(Z2n12)Substituteâ€‰1312â€‰kJ/molÂ forÂ C,Â 2Â forÂ Z,Â Â âˆžâ€‰forâ€‰nâˆžÂ andÂ 1Â forÂ n1.En=âˆž-En=1â€‰=â€‰â€‰1312â€‰kJ/molÃ—â€‰(2212)â€‰â€‰â€‰â€‰â€‰â€‰[since-Câ€‰(Z2(âˆž)2)=0]En=âˆž-En=1â€‰=â€‰5248â€‰kJ/mol

Next calculate the lower limit lionization energy.

â€‚Â LetÂ usÂ considerÂ theÂ Bohr'sÂ equationE=â€‰-(NARhc)â€‰(Z2/n2)NARhcâ€‰=1312â€‰kJ/molZ=Atomicâ€‰numbern=â€‰no

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