The electrons, identified by quantum number n and l
i. `n = 4,l=1` ii. `n = 4, l= 0` iii. `n = 3 , l= 2 ` iv. `n= 3 , l = 1`
Can be palced in the order of increasing energy from the lowest to highest,its

Options

A. `(iv)lt(ii)lt(iii)lt(i)`
B. `(ii)lt(iv)lt(i)`
C. `(i)lt(iii)lt(ii)lt(iv)`
D. `(iii)lt(i)lt(iv)lt(ii)`

Answer

`A`
Apply Aufbau priciple :
(i) The orbital with lowest value of `(n+l)` have lowest energy.
(ii) If `n+l` value for two orbitals are same, the orbital with lower value of `n` is having lower energy.
`{:("for",(i)n+l=5),(,(ii)n+l=4),(,(iii)n+l=5),(,(iv)n+l=4):}`
for `(iii)` and `(iv) n` is lower hence correct order is `(iv) lt (ii) lt (iii) lt (i)`