In an atoms ,the order of increasing energy of electrons with number quantum numbers.(i) n=4,1=1 (ii) n=4,1=0(ii) n=3,1=2 (iv)n=3,1=1 is

Options

A. `iiiltiltivltii`
B. `iiltivltiltiii`
C. `iltiiiltiiltiv`
D. `ivltiiltiiilti`

Answer

d

Explanation

`(n+l)` is used to compute energies of orbitals in atoms.Orbital with low`(n-l)` values will have low energy and vice-versa.In case of orbitals with equal values of `(n+l)`,the orbital whose `n` value is least will be of lower energy (ii) `n=4,l=0,(n+l)=4+0=4`
`n=3,l=2,(n+l)=3+2=5`
`n=3,l=2,(n+l)=3+1=4`
`therefore ` The order is ,`(iv)lt(ii)lt(ii)lt(i)`