I think A but if it’s wrong I’m sorry ❤️
Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
The answer is B as isotopes are different versions of the same chemical element containing the same amount of protons and electrons but different amounts of neutrons.
Answer:
(B) F⁻, HCOOH
Explanation:
(A) CH₄, HCOOH
(B) F⁻, HCOOH
(C) F⁻, CH₃-O-CH₃
The hydrogen bonds are formed when the hydrogen is found between two electronegative atoms such as oxygen (O), nitrogen (N) or florine (F).
O····H-O, F····H-O, O····H-N
(A) CH₄, HCOOH
- here methane CH₄ is not capable to form hydrogen bond with water
- formic acid HCOOH can form hydrogen bonds with water
H-C(=O)-O-H····OH₂
(B) F⁻, HCOOH
-both floride (F⁻) and formic acid can form hydrogen bonds with water
F····OH₂
H-C(=O)-O-H····OH₂
(C) F⁻, CH₃-O-CH₃
- dimethyl-ether CH₃-O-CH₃ is not capable to form hydrogen bond with water
- floride (F⁻) can form hydrogen bonds with water
F····OH₂
Because it's protected from air because air makes it turn brown.
