# 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 Option 2) $IV Option 3) $IV   Option 4) $I

 $n$ $l$ $n+l$ I 4 2 6 4d II 3 2 5 3d III 4 1 5 4p IV 3 1 4 3p

$I Higher the value of ($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

Option 2)

$IV

Option 3)

$IV

Option 4)

$I

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