Respuesta :
The question is incomplete. The complete question is :
The table compares the number of electrons in two unknown neutral atoms.
Comparison of Electrons
Atom Number of Electrons
A 9
D 11
Use this information to determine the number of valence electrons in the atoms. Which of the following correctly compares the stability of the two atoms?
A. Both are unreactive.
B. Both are highly reactive.
C. A is unreactive and D is reactive.
D. A is reactive and D is unreactive.
Answer:
B. Both are highly reactive.
Explanation:
Atomic number of electrons given is 9 and 11, it means that atom A is fluorine (F) and atom D is Sodium (Na).
In order to gain stability 8 electrons are required in outer shell. So atoms either gain or lose electrons in its outer shell to become more reactive.
Sodium has electronic configuration (2,8,1) and easily loses one electron and become (2, 8) which is highly reactive.
Fluorine has electronic configuration (2, 7) and can gain easily an electron and becomes reactive to form (2, 8).
So, both they both attain stability by losing and gaining only one electron.
Hence, they both are highly reactive and the correct option is B.
Both atoms are highly reactive because of incomplete outermost shell.
The first atom is fluorine that has atomic number 9 and the second atom is sodium having atomic number 11. Fluorine belongs to group 7a which is highly reactive due to incomplete outermost shell while on the other hand, sodium belongs to group 1a which is also very reactive due to incomplete outermost shell.
Sodium gets stability by losing of outermost electron whereas fluorine gets stability by gaining one electron so we can conclude that both atoms are highly reactive because of incomplete outermost shell.
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