MCQ
Arrange the following orbitals in decreasing order of energy ?

$(A)$ $n =3,1=0, m =0$

$(B)$ $n =4, l =0, m =0$

$(C)$ $n =3, l =1, m =0$

$(D)$ $n =3, l =2, m =1$

The correct option for the order is :

  • A
    $B > D > C > A$
  • $D > B > C > A$
  • C
    $A > C > B > D$
  • D
    $D > B > A > C$

Answer

Correct option: B.
$D > B > C > A$
b
$(A)$ $n =3 ; 1=0 ; m =0 ; 3 s$ orbital

$(B)$ $n =4 ; 1=0 ; m =0 ; 4 s$ orbital

$(C)$ $n =3 ; 1=1 ; m =0 ; 3 p$ orbital

$(D)$ $n =3 ; 1=2 ; m =0 ; 3 d$ orbital

As per Hund's rule energy is given by $( n +l)$ value.

If value of $( n +l)$ remains same then energy is given by $n$ only.

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