Answer : The correct option is, (d) [tex]Mg^{+2}[/tex]
Explanation :
First we have to determine the number of electrons present in [tex]F^{-1}[/tex] ion.
Atomic number of fluorine (F) = 9
As we know that,
Atomic number = Number of electrons (for neutral atom)
Number of electrons in [tex]F^{-1}[/tex] ion = 9 + 1 = 10
Now we have to determine the number of electrons present in given option ions.
For [tex]B^{-3}[/tex] ion:
Atomic number of boron (B) = 5
Number of electrons in [tex]B^{-3}[/tex] ion = 5 + 3 = 8
For [tex]N^{+1}[/tex] ion:
Atomic number of nitrogen (N) = 7
Number of electrons in [tex]N^{+1}[/tex] ion = 7 - 1 = 6
For [tex]Na^{-1}[/tex] ion:
Atomic number of sodium (Na) = 11
Number of electrons in [tex]Na^{-1}[/tex] ion = 11 + 1 = 12
For [tex]Mg^{+2}[/tex] ion:
Atomic number of magnesium (Mg) = 12
Number of electrons in [tex]Mg^{+2}[/tex] ion = 12 - 2 = 10
For [tex]Ne^{-1}[/tex] ion:
Atomic number of neon (Ne) = 10
Number of electrons in [tex]Ne^{-1}[/tex] ion = 10 + 1 = 11
From this we conclude that magnesium ion, [tex]Mg^{+2}[/tex] has the same number of electrons as a fluoride ion, [tex]F^{-1}[/tex].
Hence, the correct option is, (d) [tex]Mg^{+2}[/tex]
