Question

1.Which two electron configurations represent elements that would have similar chemical properties? I.1s22s22p4II.1s22s22p5III.[Ar]4s23d104p3IV. [Ar]4s23d104p4A) Iand IIB) I and IIIC) Iand IVD) IIand IVE) IIand II

Answer 1

Option C= 1 and 1V

Step-by-step explanation:

The first one is the electronic configuration of oxygen.

O₈ =1s² 2s² 2p⁴

The IV is the electronic configuration of Selenium.

Se₄ =[Ar] 4s² 3d¹⁰ 4p⁴

Oxygen and selenium both are present in group 16. The elements present in same group showed similar chemical properties.

While the II is the electronic configuration of fluorine:

F₉ =1s² 2s² 2p⁵

and the III is the electronic configuration of arsenic.

Ar₃₃ = [Ar] 4s² 3d¹⁰ 4p³

Arsenic is present in group fifteen and fluorine is present in group 17. That's why their properties are not similar with each other and also with oxygen and selenium.