Question

Do solve the question of this attachment : ​

Answer 1

Answer:

→ Option C → C < N < F < O.

Explanation:

After removal of one electron :

C+ = 1s² 2s² 2p¹ .

N+ = 1s² 2s² 2p² .

O+ = 1s² 2s² 2p³ .

F+ = 1s² 2s² 2p⁴ .

Removal of second electron will require maximum energy in case of O because of its half filled electronic configuration after removal of one electron.

•°• increasing order of second ionization enthalpy : C<N<F<O.

Hence, it is solved.

Answer 2

Answer:

CORRECT ORDER IS OPTION (3)

C<N<F<O

Hope it helps you

Thank u very much

Explanation:

Second ionisation potential is the amount of energy required to remove an electron from the outer shell of unipositive ion and the electronic configuration of unipositive ions is as follows

C+ (5) = 1s2,2s2 2p1

N+ (6) = 1s2, 2s2 2p2

O+ (7) = 1s2, 2s2 2p3

F+ (8) = 1s2, 2s2 2p4

In O+ half-filled p-orbital is present, so its second ionisation potential is greater than that of F+.

Thus, the correct order is O>F>N>C

PLEASE MARK MY ANSWER AS A BRAINLIEST ANSWER... I AM IN NEED OF THAT... PLEASE

FOLLOW ME