Following graph shows variation of ionization energies with atomis number in second period `(Li -Ne)`. Value of ionization energies of `Na(11)` will be-

Options

A. above `Ne`
B. below `Ne` but above `O`
C. below `Li`
D. between `N` and `O`.

Explanation

`Na` is `3rd` period element and is bigger than `Li`. The distance between the nucleus and outer most electron is more as compared to `Li`. Thus the outer most electron is loosely bound with nucleus and removal of electron is easier. So option `(C )` is correct.