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# Engineering

The quantum number of four electrons are given below:

I.n=4,l=2,m_{l}=-2,m_{s}=-1/2

I\! I.n=3,l=2,m_{l}=1,m_{s}=+1/2

I\!I \! I.n=4,l=1,m_{l}=0,m_{s}=+1/2

IV.n=3,l=1,m_{l}=1,m_{s}=-1/2

The correct order of their increasing energies will be :

 

  • Option 1)

    I<II<III<IV

  • Option 2)

    IV<III<II<I

  • Option 3)

    IV<II<III<I

     

  • Option 4)

    I<III<II<IV

 

Answers (1)
S solutionqc
  n l n+l  
I 4 2 6 4d
II 3 2 5 3d
III 4 1 5 4p
IV 3 1 4 3p

I<III<II<IV Higher the value of (n+l) energy. If two orbitals have same value of (n+l) , the orbital with value of n will have higher energy.


Option 1)

I<II<III<IV

Option 2)

IV<III<II<I

Option 3)

IV<II<III<I

 

Option 4)

I<III<II<IV

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