The answer is the first option (a)
Well, when an atom attains a stable valence electron, it means that the outer electrons are complete and so cannot attain any more electrons. For the first shell, it is complete when it has 2 electrons, the second shell is complete when it has 8 electrons, all the other shells also have a particular number when complete. Anyway, i believe the answer is HYDROGEN because when HYDROGEN combines with another atom of HYDROGEN, the outer shell is completed. This is because HYDROGEN has only 1 electron. If the two HYDROGENS, which both have 1 electron combine, they make the electrons 2, which is complete for the first shell, HYDROGEN ends in the first shell. Since the electrons become 2, the shell is at stable valence. In all the other options, this happens;
NEON- It has 10 electrons, 2 in the first shell and 8 in the second. So the the shells are already complete, so it can't bond with any thing, which is completely against the question.
RADON- Radon has 86 electrons.
HELIUM- Helium has 2 electrons, so the shell is already full, and cannot bond, so it goes against the question. The question says BY BONDING.
So the answer is definitely 4) HYDROGEN
Hope i helped. Have a nice day, by the way, i'm very sure it's hydrogen.
We know that each millimeter contains 10⁻³ meters. Writing this as a ratio:
1 mm : 10⁻³ m
We require a conversion from m³ to mm³, so we must take the cube of the ratio we have made:
1 mm³ = (10⁻³)³ m³
Therefore, the conversion used will be:
(1 mm / 10⁻³ m)³
When we multiply by this conversion, we will get:
32 m³ = 32 x 10⁹ mm³
The element which has the electronic configuration is CHLORINE.
The atomic number of chlorine is 17 and it has 7 valence electrons in its outermost shell. Because it needs only one more electrons to have a stable octet, it usually react with metals from group one of the periodic table who are normally willing to donate the single electrons in their outermost shells. The ground state electronic configuration of chlorine atom is 1S^2 2S^2 2P^6 3S^2 3P^5.