Chemistry - PERIODIC TABLE Question with Solution | TestHub

I.E order in second Period $\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}<\mathrm{Ne}$ Option (1) $\mathrm{T}\to \text{ Nitrogen }\mathrm{V}\to \text{ Oxygen }$ Option (2) $\mathrm{Q}\to \mathrm{Be}\mathrm{R}\to \text{ Boron }$ Option (3) Successive I.E is always greater irrespective of electronic configuration.