General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
Question
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Chapter 5, Problem 5.40P

(a)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with O2 has to be written.

Concept Introduction:

Electrons of an atom are arranged in orbitals by the order of increasing energy.  This arrangement is known as electronic configuration of atom.  This can be represented using noble-gas shorthand notation also.

Ions are formed from neutral atom either by removal or addition of electrons from the valence shell.

Each orbital has two electrons and they both are in opposite spin.  Electrons are filled up in the orbitals of a sub-shell by following Hund’s rule.  This rule tells that all the orbitals are singly filled in a sub-shell and then the pairing occurs.

Isoelectronic species are the ones that have different number of protons but same number of electrons.  No two neutral atom can be isoelectronic.  A neutral atom can be isoelectronic with an ion.

(a)

Expert Solution
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Explanation of Solution

Given ion is O2Atomic number of oxygen is 8.  Ground state electronic configuration of oxygen is shown below.

    O=1s2 2s2 2p4

The given O2 ion is formed when two electrons are added to the valence shell.  In this case the two electrons are added to 2p orbital.  Therefore, the electronic configuration of O2 ion is written as shown below.

    O2=1s2 2s2 2p6

The total number of electrons present in O2 ion is found by summing up the superscript.

    Totalnumberofelectrons=2+2+6=10

Total number of electrons that is present in O2 ion is 10.  The neutral atom that has eighteen electrons is found to be neon.  Therefore, the O2 ion is found to be isoelectronic with neon atom.

(b)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with Ca+ has to be written.

Concept Introduction:

Refer part (a).

(b)

Expert Solution
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Explanation of Solution

Given ion is Ca+.  Atomic number of calcium is 20.  Ground state electronic configuration of calcium is shown below.

    Ca=1s2 2s2 2p6 3s2 3p6 4s2

The given Ca+ ion is formed when one electron is removed from the valence shell.  In this case the one electron is removed 4s orbital.  Therefore, the electronic configuration of Ca+ ion is written as shown below.

    Ca+=1s2 2s2 2p6 3s2 3p6 4s1

The total number of electrons present in Ca+ ion is found by summing up the superscript.

    Totalnumberofelectrons=2+2+6+2+6+1=19

Total number of electrons that is present in Ca+ ion is 19.  The neutral atom that has nineteen electrons is found to be potassium.  Therefore, the Ca+ ion is found to be isoelectronic with potassium atom.

(c)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with He+ has to be written.

Concept Introduction:

Refer part (a).

(c)

Expert Solution
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Explanation of Solution

Given ion is He+.  Atomic number of helium is 2.  Ground state electronic configuration of helium is shown below.

    He=1s2

The given He+ ion is formed when one electron is removed from the valence shell.  In this case the one electrons are removed from the 1s orbital.  Therefore, the electronic configuration of He+ ion is written as shown below.

    He+=1s1

The total number of electrons present in He+ ion is found by summing up the superscript.

    Totalnumberofelectrons=1

Total number of electrons that is present in He+ ion is 1.  The neutral atom that has one electron is found to be hydrogen.  Therefore, the He+ ion is found to be isoelectronic with hydrogen atom.

(d)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with Pb2+ has to be written.

Concept Introduction:

Refer part (a).

(d)

Expert Solution
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Explanation of Solution

Given ion is Pb2+.  Atomic number of lead is 82.  Ground state electronic configuration of lead is shown below.

    Pb=1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p2

The given Pb2+ ion is formed when two electrons are removed from the valence shell.  In this case the two electrons are removed from p orbital.  Therefore, the electronic configuration of Pb2+ ion is written as shown below.

    Pb2+=1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10

The total number of electrons present in Pb2+ ion is found by summing up the superscript.

    Totalnumberofelectrons=2+2+6+2+6+2+10+6+2+10+6+2+14+10=80

Total number of electrons that is present in Pb2+ ion is 80.  The neutral atom that has eighty electrons is found to be mercury.  Therefore, the Pb2+ ion is found to be isoelectronic with mercury atom.

(e)

Interpretation Introduction

Interpretation:

Neutral atom that is isoelectronic with N3 has to be written.

Concept Introduction:

Refer part (a).

(e)

Expert Solution
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Explanation of Solution

Given ion is N3.  Atomic number of nitrogen is 7.  Ground state electronic configuration of nitrogen is shown below.

    N=1s2 2s2 2p3

The given N3 ion is formed when three electrons are added to the valence shell.  In this case the three electrons are added to p orbital.  Therefore, the electronic configuration of N3 ion is written as shown below.

    N3-=1s2 2s2 2p6

The total number of electrons present in N3 ion is found by summing up the superscript.

    Totalnumberofelectrons=2+2+6=10

Total number of electrons that is present in N3 ion is 10.  The neutral atom that has eighteen electrons is found to be neon.  Therefore, the N3 ion is found to be isoelectronic with neon atom.

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