Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 7, Problem 7.83QP

(a)

Interpretation Introduction

Interpretation: The numbers of orbitals in an atom for each of the given principal quantum number are to be stated.

Concept introduction: There are four quantum numbers that indicate the size, energy and shape of an atomic orbital. The quantum number n is the principal quantum number that indicates the energy level and size of the orbital and l is the azimuthal quantum number that determines the shape and angular momentum of an orbital. The magnetic quantum number is an integer that takes the values from (l to +l) .

To determine: The number of orbitals in an atom for the given principal quantum number.

(a)

Expert Solution
Check Mark

Answer to Problem 7.83QP

Solution

The number of orbital for n=1 is 1_ .

Explanation of Solution

Explanation

The given principal quantum number (n) is 1 . For n=1 , the value of l is,

l=n1l=11l=0

When l=0 then m=0 that is only one orbital is present. For l=0 indicates the presence of s orbital. Hence, there is only 1 orbital in the first shell of an atom that is for n=1 .

(b)

Interpretation Introduction

To determine: The number of orbitals in an atom for the given principal quantum number.

(b)

Expert Solution
Check Mark

Answer to Problem 7.83QP

Solution

The number of orbital for n=2 is 4_ .

Explanation of Solution

Explanation

The given principal quantum number (n) is 2 . For n=2 , the value of l is,

l=n1l=21l=1

The allowed values of l are 0and1 . For n=2 , the value of l represents 2sand2p sub shells. The possible values of ml take the values from (l to +l) which are,

For l=0

ml=0

The combination of l and ml represents one 2s orbital.

For l=1

ml=1,0,+1

The combination of l and ml represents three 2p orbitals.

Hence, the total number of orbitals for n=2 is 1+3=4 orbitals.

(c)

Interpretation Introduction

To determine: The number of orbitals in an atom for the given principal quantum number.

(c)

Expert Solution
Check Mark

Answer to Problem 7.83QP

Solution

The number of orbital for n=3 is 9_ .

Explanation of Solution

Explanation

The given principal quantum number (n) is 3 . For n=3 , the value of l is,

l=n1l=31l=2

The allowed values of l are 0,1and2 . For n=2 , the value of l represents 3s,3p and 3d sub shells. The possible values of ml take the values from (l to +l) which are,

For l=0

ml=0

The combination of l and ml represents one 3s orbital.

For l=1

ml=1,0,+1

The combination of l and ml represents three 3p orbitals.

For l=2

ml=2,1,0,+1,+2

The combination of l and ml represents five 3d orbitals.

Hence, the total number of orbitals for n=3 is 1+3+5=9 orbitals.

(d)

Interpretation Introduction

To determine: The number of orbitals in an atom for the given principal quantum number.

(d)

Expert Solution
Check Mark

Answer to Problem 7.83QP

Solution

The number of orbital for n=4 is 16_ .

Explanation of Solution

Explanation

The given principal quantum number (n) is 4 . For n=4 , the value of l is,

l=n1l=41l=3

The allowed values of l are 0,1,2and3 . For n=4 , the value of l represents 4s,4p,4d and 4f sub shells. The possible values of ml take the values from (l to +l) which are,

For l=0

ml=0

The combination of l and ml represents one 4s orbital.

For l=1

ml=1,0,+1

The combination of l and ml represents three 4p orbitals.

For l=2

ml=2,1,0,+1,+2

The combination of l and ml represents five 4d orbitals.

For l=3

ml=3,2,1,0,+1,+2,+3

The combination of l and ml represents seven 4f orbitals

Hence, the total number of orbitals for n=4 is 1+3+5+7=16 orbitals.

(e)

Interpretation Introduction

To determine: The number of orbitals in an atom for the given principal quantum number.

(e)

Expert Solution
Check Mark

Answer to Problem 7.83QP

Solution

The number of orbital for n=5 is 25_ .

Explanation of Solution

Explanation

The given principal quantum number (n) is 5 . For n=5 , the value of l is,

l=n1l=51l=4

The allowed values of l are 0,1,2,3and4 . For n=5 , the value of l represents 5s,5p,5d,5f and 5g sub shells. The possible values of ml take the values from (l to +l) which are,

For l=0

ml=0

The combination of l and ml represents one 5s orbital.

For l=1

ml=1,0,+1

The combination of l and ml represents three 5p orbitals.

For l=2

ml=2,1,0,+1,+2

The combination of l and ml represents five 5d orbitals.

For l=3

ml=3,2,1,0,+1,+2,+3

The combination of l and ml represents seven 5f orbitals.

For l=4

ml=4,3,2,1,0,+1,+2,+3+4

The combination of l and ml represents nine 5g orbitals.

Hence, the total number of orbitals for n=5 is 1+3+5+7+9=25 orbitals.

Conclusion

  1. a. The number of orbital for n=1 is 1_ .
  2. b. The number of orbital for n=2 is 4_ .
  3. c. The number of orbital for n=3 is 9_ .
  4. d. The number of orbital for n=4 is 16_ .
  5. e. The number of orbital for n=5 is 25_ .

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Chapter 7 Solutions

Chemistry

Ch. 7.8 - Prob. 11PECh. 7.9 - Prob. 12PECh. 7.9 - Prob. 13PECh. 7.10 - Prob. 14PECh. 7 - Prob. 7.1VPCh. 7 - Prob. 7.2VPCh. 7 - Prob. 7.3VPCh. 7 - Prob. 7.4VPCh. 7 - Prob. 7.5VPCh. 7 - Prob. 7.6VPCh. 7 - Prob. 7.7VPCh. 7 - Prob. 7.8VPCh. 7 - Prob. 7.9VPCh. 7 - Prob. 7.10VPCh. 7 - Prob. 7.11VPCh. 7 - Prob. 7.12VPCh. 7 - Prob. 7.13VPCh. 7 - Prob. 7.14VPCh. 7 - Prob. 7.15VPCh. 7 - Prob. 7.16VPCh. 7 - Prob. 7.17VPCh. 7 - Prob. 7.18VPCh. 7 - Prob. 7.19VPCh. 7 - Prob. 7.20VPCh. 7 - Prob. 7.21QPCh. 7 - Prob. 7.22QPCh. 7 - Prob. 7.23QPCh. 7 - Prob. 7.24QPCh. 7 - Prob. 7.25QPCh. 7 - Prob. 7.26QPCh. 7 - Prob. 7.27QPCh. 7 - Prob. 7.28QPCh. 7 - Prob. 7.29QPCh. 7 - Prob. 7.30QPCh. 7 - Prob. 7.31QPCh. 7 - Prob. 7.32QPCh. 7 - Prob. 7.33QPCh. 7 - Prob. 7.34QPCh. 7 - Prob. 7.35QPCh. 7 - Prob. 7.36QPCh. 7 - Prob. 7.37QPCh. 7 - Prob. 7.38QPCh. 7 - Prob. 7.39QPCh. 7 - Prob. 7.40QPCh. 7 - Prob. 7.41QPCh. 7 - Prob. 7.42QPCh. 7 - Prob. 7.43QPCh. 7 - Prob. 7.44QPCh. 7 - Prob. 7.45QPCh. 7 - Prob. 7.46QPCh. 7 - Prob. 7.47QPCh. 7 - Prob. 7.48QPCh. 7 - Prob. 7.49QPCh. 7 - Prob. 7.50QPCh. 7 - Prob. 7.51QPCh. 7 - Prob. 7.52QPCh. 7 - Prob. 7.53QPCh. 7 - Prob. 7.54QPCh. 7 - Prob. 7.55QPCh. 7 - Prob. 7.56QPCh. 7 - Prob. 7.57QPCh. 7 - Prob. 7.58QPCh. 7 - Prob. 7.59QPCh. 7 - Prob. 7.60QPCh. 7 - Prob. 7.61QPCh. 7 - Prob. 7.62QPCh. 7 - Prob. 7.63QPCh. 7 - Prob. 7.64QPCh. 7 - Prob. 7.65QPCh. 7 - Prob. 7.66QPCh. 7 - Prob. 7.67QPCh. 7 - Prob. 7.68QPCh. 7 - Prob. 7.69QPCh. 7 - Prob. 7.70QPCh. 7 - Prob. 7.71QPCh. 7 - Prob. 7.72QPCh. 7 - Prob. 7.73QPCh. 7 - Prob. 7.74QPCh. 7 - Prob. 7.75QPCh. 7 - Prob. 7.76QPCh. 7 - Prob. 7.77QPCh. 7 - Prob. 7.78QPCh. 7 - Prob. 7.79QPCh. 7 - Prob. 7.80QPCh. 7 - Prob. 7.81QPCh. 7 - Prob. 7.82QPCh. 7 - Prob. 7.83QPCh. 7 - Prob. 7.84QPCh. 7 - Prob. 7.85QPCh. 7 - Prob. 7.86QPCh. 7 - Prob. 7.87QPCh. 7 - Prob. 7.88QPCh. 7 - Prob. 7.89QPCh. 7 - Prob. 7.90QPCh. 7 - Prob. 7.91QPCh. 7 - Prob. 7.92QPCh. 7 - Prob. 7.93QPCh. 7 - Prob. 7.94QPCh. 7 - Prob. 7.95QPCh. 7 - Prob. 7.96QPCh. 7 - Prob. 7.97QPCh. 7 - Prob. 7.98QPCh. 7 - Prob. 7.99QPCh. 7 - Prob. 7.100QPCh. 7 - Prob. 7.101QPCh. 7 - Prob. 7.102QPCh. 7 - Prob. 7.103QPCh. 7 - Prob. 7.104QPCh. 7 - Prob. 7.105QPCh. 7 - Prob. 7.106QPCh. 7 - Prob. 7.107QPCh. 7 - Prob. 7.108QPCh. 7 - Prob. 7.109QPCh. 7 - Prob. 7.110QPCh. 7 - Prob. 7.111QPCh. 7 - Prob. 7.112QPCh. 7 - Prob. 7.113QPCh. 7 - Prob. 7.114QPCh. 7 - Prob. 7.115QPCh. 7 - Prob. 7.116QPCh. 7 - Prob. 7.117QPCh. 7 - Prob. 7.118QPCh. 7 - Prob. 7.119QPCh. 7 - Prob. 7.120QPCh. 7 - Prob. 7.121QPCh. 7 - Prob. 7.122QPCh. 7 - Prob. 7.123QPCh. 7 - Prob. 7.124QPCh. 7 - Prob. 7.125QPCh. 7 - Prob. 7.126QPCh. 7 - Prob. 7.127QPCh. 7 - Prob. 7.128QPCh. 7 - Prob. 7.129APCh. 7 - Prob. 7.130APCh. 7 - Prob. 7.131APCh. 7 - Prob. 7.132APCh. 7 - Prob. 7.133APCh. 7 - Prob. 7.134APCh. 7 - Prob. 7.135APCh. 7 - Prob. 7.136APCh. 7 - Prob. 7.137APCh. 7 - Prob. 7.138APCh. 7 - Prob. 7.139APCh. 7 - Prob. 7.140APCh. 7 - Prob. 7.141APCh. 7 - Prob. 7.142APCh. 7 - Prob. 7.143APCh. 7 - Prob. 7.144APCh. 7 - Prob. 7.145APCh. 7 - Prob. 7.146APCh. 7 - Prob. 7.147APCh. 7 - Prob. 7.148AP
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