Question

18. Mark each electronic configuration as “correct" or "incorrect”. If it is incorrect, say why and write the correct electronic configuration. a. P 1s 2s 2pº3s 3p! b. Ca 1s²2s22p%3s 3pº3d? c. Mg [Ne]3s d. Mn [Kr]4s 3d

Answer 1

1) P : 1 s ² , 2 s ² , 2 p ⁶, 3 s ² , 3 p ¹

INCORRECT

Atomic number of P is 15. It contain 15 electrons. In E.C of P, there are only 13 electrons shown. Hence, correct E.C of P is 1 s ² , 2 s ² , 2 p ⁶, 3 s ² , 3 p ³

2) Ca :1 s ² , 2 s ² , 2 p ⁶, 3 s ² , 3 p⁶, 3 d ²

INCORRECT

Atomic number of Ca is 20. It contain 20 electrons. In any atom, electrons are filled first lowest energy orbitals. The energy level sequence of orbitals is as shown below.

1 s < 2 s < 2 p < 3 s < 3 p < 4 s < 3 d < 4 p .........

In Ca , electrons must be filled in the 4 s orbital before 3 d.

Therefore, correct E.C of Ca is Ca :1 s ² , 2 s ² , 2 p ⁶, 3 s ² , 3 p ⁶, 4 s ²

3) Mg : [ Ne ] 3 s ²

CORRECT

Atomic number of Mg is 12. It contain 12 electrons. Its E.C is 1 s ² , 2 s ² , 2 p ⁶, 3 s ² .

The E.C of Ne is 1 s ² , 2 s ² , 2 p ⁶. Therefore, In short hand notation we can write E,C of Mg as, [Ne] 3 s ²

4) Mn : [Kr ] 4 s ² , 3 d ⁵

INCORRECT

Atomic number of Mn is 25. It contain 25 electrons. The E.C given shows it contain 43 electrons. Hence it is incorrect E.C .  

The Correct E.C of Mn is 1 s ² , 2 s ² , 2 p ⁶, 3 s ² , 3 p ⁶, 4 s ² , 3 d ⁵

E.C of noble gas element Ar is 1 s ² , 2 s ² , 2 p ⁶, 3 s ² , 3 p ⁶. Therefore, In short hand notation we can write E,C of Mn as [Ar] 4 s ² , 3 d ⁵ .