The correct set of possible quantum number number (n,1,m,s,) for last electron entering entering in Lanthanum `(Z=57)` is

Options

A. `5,0,0,+(1)/(2)`
B. `4,3,0,+(1)/(2)`
C. `5,2,0,+(1)/(2)`
D. `6.0,0,+(1)/(2)`

Explanation

`La:[Xe]5d^(1)4s^(2) therefore` last electron is entering 'd' orbital
`therefore n=5,1=2 &m` could be `-2,-1,+1,+2`
`s=+(1)/(2)` or `-(1)/(2)`