ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
7th Edition
ISBN: 9781319403959
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 1, Problem 1D.24E

(a)

Interpretation Introduction

Interpretation:

Number of orbitals that can have the given quantum numbers has to be written.

    n=3,l=1

Concept Introduction:

There are four set of quantum numbers present for explaining the distribution of electron density in an atom.  They are principal quantum number, angular momentum quantum number, magnetic quantum number and spin quantum number.

    NameSymbolValuesSpecifiesPrincipaln1,2,...shellangular momentuml0,1,...n-1subshell:l=0,1,2,3,4...s,p,d,f,g...magneticmll,l-1,...,-1orbitalsofshellspinms+12,-12spinstate

For a given value of n, there will be n different values for l.  And in the case of magnetic quantum number, there will be (2l+1) different values of ml present for each values of l.

(a)

Expert Solution
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Explanation of Solution

When the principal quantum number (n) value is 3 and the angular momentum quantum number (l) value is 1, then the subshell notation is 3p.

If the angular momentum quantum number lis1, then there will be three different ml values are present, which means there will be three orbitals for the given subshell.

(b)

Interpretation Introduction

Interpretation:

Number of orbitals that can have the given quantum numbers has to be written.

    n=5,l=3,ml=1

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

When the principal quantum number (n) value is 5 and the angular momentum quantum number (l) value is 3, then the subshell notation is 5f.

If the angular momentum quantum number lis3, then there will be seven different values for  ml and they are 3,2,1,0,+1,+2,+3.  If  ml value is 1, then there will be only one orbital.

(c)

Interpretation Introduction

Interpretation:

Number of orbitals that can have the given quantum numbers has to be written.

    n=2,l=1,ml=0

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

When the principal quantum number (n) value is 2 and the angular momentum quantum number (l) value is 1, then the subshell notation is 2p.

If the angular momentum quantum number lis1, then there will be three different values for  ml and they are 1,0,+1.  If  ml value is 0, then there will be only one orbital.

(d)

Interpretation Introduction

Interpretation:

Number of orbitals that can have the given quantum numbers has to be written.

    n=7

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

If the principal quantum number (n) is 7 then the possible angular momentum quantum number (l) values are 0,1,2,3,4,5and6.

If the angular momentum quantum number lis0, then there will be only one ml value present, which means there will be only one orbital present.

If the angular momentum quantum number lis1, then there will be three different ml values are present, which means there will be three orbitals present.

If the angular momentum quantum number lis2, then there will be five different ml values are present, which means there will be five orbitals present.

If the angular momentum quantum number lis3, then there will be seven different ml values are present, which means there will be seven orbitals present.

If the angular momentum quantum number lis4, then there will be nine different ml values are present, which means there will be nine orbitals present.

If the angular momentum quantum number lis5, then there will be eleven different ml values are present, which means there will be eleven orbitals present.

If the angular momentum quantum number lis6, then there will be thirteen different ml values are present, which means there will be thirteen orbitals present.

So in total there will be forty nine orbitals.

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Chapter 1 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

Ch. 1 - Prob. 1A.8ECh. 1 - Prob. 1A.9ECh. 1 - Prob. 1A.10ECh. 1 - Prob. 1A.11ECh. 1 - Prob. 1A.12ECh. 1 - Prob. 1A.13ECh. 1 - Prob. 1A.14ECh. 1 - Prob. 1A.15ECh. 1 - Prob. 1A.16ECh. 1 - Prob. 1A.17ECh. 1 - Prob. 1B.1ASTCh. 1 - Prob. 1B.1BSTCh. 1 - Prob. 1B.2ASTCh. 1 - Prob. 1B.2BSTCh. 1 - Prob. 1B.3ASTCh. 1 - Prob. 1B.3BSTCh. 1 - Prob. 1B.4ASTCh. 1 - Prob. 1B.4BSTCh. 1 - Prob. 1B.5ASTCh. 1 - Prob. 1B.5BSTCh. 1 - Prob. 1B.1ECh. 1 - Prob. 1B.2ECh. 1 - Prob. 1B.3ECh. 1 - Prob. 1B.4ECh. 1 - Prob. 1B.5ECh. 1 - Prob. 1B.6ECh. 1 - Prob. 1B.7ECh. 1 - Prob. 1B.8ECh. 1 - Prob. 1B.9ECh. 1 - Prob. 1B.10ECh. 1 - Prob. 1B.11ECh. 1 - Prob. 1B.12ECh. 1 - Prob. 1B.13ECh. 1 - Prob. 1B.14ECh. 1 - Prob. 1B.15ECh. 1 - Prob. 1B.16ECh. 1 - Prob. 1B.17ECh. 1 - Prob. 1B.18ECh. 1 - Prob. 1B.19ECh. 1 - Prob. 1B.21ECh. 1 - Prob. 1B.22ECh. 1 - Prob. 1B.23ECh. 1 - Prob. 1B.24ECh. 1 - Prob. 1B.25ECh. 1 - Prob. 1B.26ECh. 1 - Prob. 1B.27ECh. 1 - Prob. 1B.28ECh. 1 - Prob. 1C.1ASTCh. 1 - Prob. 1C.1BSTCh. 1 - Prob. 1C.1ECh. 1 - Prob. 1C.2ECh. 1 - Prob. 1C.3ECh. 1 - Prob. 1C.7ECh. 1 - Prob. 1D.1ASTCh. 1 - Prob. 1D.1BSTCh. 1 - Prob. 1D.2ASTCh. 1 - Prob. 1D.2BSTCh. 1 - Prob. 1D.1ECh. 1 - Prob. 1D.2ECh. 1 - Prob. 1D.3ECh. 1 - Prob. 1D.4ECh. 1 - Prob. 1D.5ECh. 1 - Prob. 1D.6ECh. 1 - Prob. 1D.7ECh. 1 - Prob. 1D.9ECh. 1 - Prob. 1D.10ECh. 1 - Prob. 1D.11ECh. 1 - Prob. 1D.12ECh. 1 - Prob. 1D.13ECh. 1 - Prob. 1D.14ECh. 1 - Prob. 1D.15ECh. 1 - Prob. 1D.16ECh. 1 - Prob. 1D.17ECh. 1 - Prob. 1D.18ECh. 1 - Prob. 1D.19ECh. 1 - Prob. 1D.20ECh. 1 - Prob. 1D.21ECh. 1 - Prob. 1D.22ECh. 1 - Prob. 1D.23ECh. 1 - Prob. 1D.24ECh. 1 - Prob. 1D.25ECh. 1 - Prob. 1D.26ECh. 1 - Prob. 1E.1ASTCh. 1 - Prob. 1E.1BSTCh. 1 - Prob. 1E.2ASTCh. 1 - Prob. 1E.2BSTCh. 1 - Prob. 1E.1ECh. 1 - Prob. 1E.2ECh. 1 - Prob. 1E.3ECh. 1 - Prob. 1E.4ECh. 1 - Prob. 1E.5ECh. 1 - Prob. 1E.7ECh. 1 - Prob. 1E.8ECh. 1 - Prob. 1E.9ECh. 1 - Prob. 1E.10ECh. 1 - Prob. 1E.11ECh. 1 - Prob. 1E.12ECh. 1 - Prob. 1E.13ECh. 1 - Prob. 1E.14ECh. 1 - Prob. 1E.15ECh. 1 - Prob. 1E.16ECh. 1 - Prob. 1E.17ECh. 1 - Prob. 1E.18ECh. 1 - Prob. 1E.19ECh. 1 - Prob. 1E.20ECh. 1 - Prob. 1E.21ECh. 1 - Prob. 1E.22ECh. 1 - Prob. 1E.23ECh. 1 - Prob. 1E.24ECh. 1 - Prob. 1E.25ECh. 1 - Prob. 1E.26ECh. 1 - Prob. 1F.1ASTCh. 1 - Prob. 1F.1BSTCh. 1 - Prob. 1F.2ASTCh. 1 - Prob. 1F.2BSTCh. 1 - Prob. 1F.3BSTCh. 1 - Prob. 1F.1ECh. 1 - Prob. 1F.2ECh. 1 - Prob. 1F.3ECh. 1 - Prob. 1F.4ECh. 1 - Prob. 1F.5ECh. 1 - Prob. 1F.6ECh. 1 - Prob. 1F.7ECh. 1 - Prob. 1F.8ECh. 1 - Prob. 1F.10ECh. 1 - Prob. 1F.11ECh. 1 - Prob. 1F.12ECh. 1 - Prob. 1F.13ECh. 1 - Prob. 1F.14ECh. 1 - Prob. 1F.15ECh. 1 - Prob. 1F.17ECh. 1 - Prob. 1F.18ECh. 1 - Prob. 1F.19ECh. 1 - Prob. 1F.22ECh. 1 - Prob. 1.1ECh. 1 - Prob. 1.2ECh. 1 - Prob. 1.3ECh. 1 - Prob. 1.9ECh. 1 - Prob. 1.10ECh. 1 - Prob. 1.11ECh. 1 - Prob. 1.12ECh. 1 - Prob. 1.13ECh. 1 - Prob. 1.14ECh. 1 - Prob. 1.15ECh. 1 - Prob. 1.17ECh. 1 - Prob. 1.19ECh. 1 - Prob. 1.21ECh. 1 - Prob. 1.22ECh. 1 - Prob. 1.23ECh. 1 - Prob. 1.24ECh. 1 - Prob. 1.25ECh. 1 - Prob. 1.27ECh. 1 - Prob. 1.28ECh. 1 - Prob. 1.31E
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