Which of the following is not a permissible arrangement of electrons in an atom?

  • A
    $n = 5, l = 3, m = 0, s = +1/2$
  • B
    $n = 3, l = 2, m = -3, s = -1/2$
  • C
    $n = 3, l = 2, m = -2, s = -1/2$
  • D
    $n = 4, l = 0, m = 0, s = -1/2$

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The arrangement of orbitals on the basis of energy is based upon their $(n + l)$ value. Lower the value of $(n + l)$,lower is the energy. For orbitals having same values of $(n + l)$,the orbital with lower value of $n$ will have lower energy.
$(I)$ Based upon the above information,arrange the following orbitals in the increasing order of energy.
$(a)$ $1s, 2s, 3s, 2p$
$(b)$ $4s, 3s, 3p, 4d$
$(c)$ $5p, 4d, 5d, 4f, 6s$
$(d)$ $5f, 6d, 7s, 7p$
$(II)$ Based upon the above information,solve the questions given below:
$(a)$ Which of the following orbitals has the lowest energy? $4d, 4f, 5s, 5p$
$(b)$ Which of the following orbitals has the highest energy? $5p, 5d, 5f, 6s, 6p$

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For $n = 4$,what is the total number of orbitals?

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