The correct order of the size of `Be`, `C`, `N`, `P` and `S` is

Options

A. `N lt C lt P lt S`
B. `C lt N lt P lt S`
C. `N lt C lt S lt P`
D. `C lt N lt S lt P`

Answer

C

Explanation

`C` and `N` are of `2nd` period element. Size decreases from `C` to `N`. Therefore, size of `N lt C`.
`P` and `S` are of `3rd` period elements. But size of `P gt S` [opposite to the expected trend, i.e.,size of `S gt P`].
Because `P` has half-filled stabe configuration.
Increase of one electrons in `S` causes repulsion between paired electrons `(3s^(2) 3p_(x)^(2) 3p_(y)^(1) 3p_(z)^(1))` and thus size of `S` is slightly increases from `P` to `S`.
Moreover, size of `3rd` period elements is greater than that of `2nd` period.
Hence the correct order of sizes is
`(P gt S)/("3rd period") gt (C gt N)/("2nd period")`