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:
Option C= 1 and 1V
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.