The correct set of four quantum numbers for the valence electrons of rubidium atom (Z=37) is :

  • Option 1)

    5,0,0,+\frac{1}{2}

  • Option 2)

    5,1,0,+\frac{1}{2}

  • Option 3)

    5,1,1,+\frac{1}{2}

  • Option 4)

    5,0,1,+\frac{1}{2}

 

Answers (2)
N neha
S solutionqc

As discussed in following concepts:

 

Principal Quantum Number (n) -

 It is a positive integer with value of n = 1,2,3.......

-

 

 

Azimuthal Quantum Number(l) -

For a given value of n, l can have n values ranging from 0 to n – 1, that is, for a given value of n, the possible value of l are : l = 0, 1, 2, ....( n –1)

-

 

 

Magnetic Quantum Number (m) -

 For any sub-shell (defined by ‘l ’value) 2\: l+1values of m are possible and these values are given by :

m = – l , – ( l –1), – ( l – 2)... 0,1... ( l – 2), ( l –1),l

-

 

 

Spin Quantum Number (s) -

It has two values +1/2 and -1/2

-

 

 _{37}^{Rb}=>1S^{2},2S^{2},2p^{6},3S^{2},3p^{6},4S^{2},3d^{10},4p^{6},5S^{1}

the value of n, l, m & s for last electrons are 

5,0,0,+\frac{1}{2}

Correct option is 1

 

 

 


Option 1)

5,0,0,+\frac{1}{2}

Correct option 

Option 2)

5,1,0,+\frac{1}{2}

Incorrect option

Option 3)

5,1,1,+\frac{1}{2}

Incorrect option

Option 4)

5,0,1,+\frac{1}{2}

Incorrect option

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