Question

A fluoride ion (F1-) has the same number of electrons as which of the following?

Answer 1

Answer : The correct option is, (d)
`Mg^(+2)`

Explanation :

First we have to determine the number of electrons present in
`F^(-1)` ion.

Atomic number of fluorine (F) = 9

As we know that,

Atomic number = Number of electrons (for neutral atom)

Number of electrons in
`F^(-1)` ion = 9 + 1 = 10

Now we have to determine the number of electrons present in given option ions.

For
`B^(-3)` ion:

Atomic number of boron (B) = 5

Number of electrons in
`B^(-3)` ion = 5 + 3 = 8

For
`N^(+1)` ion:

Atomic number of nitrogen (N) = 7

Number of electrons in
`N^(+1)` ion = 7 - 1 = 6

For
`Na^(-1)` ion:

Atomic number of sodium (Na) = 11

Number of electrons in
`Na^(-1)` ion = 11 + 1 = 12

For
`Mg^(+2)` ion:

Atomic number of magnesium (Mg) = 12

Number of electrons in
`Mg^(+2)` ion = 12 - 2 = 10

For
`Ne^(-1)` ion:

Atomic number of neon (Ne) = 10

Number of electrons in
`Ne^(-1)` ion = 10 + 1 = 11

From this we conclude that magnesium ion,
`Mg^(+2)` has the same number of electrons as a fluoride ion,
`F^(-1)`.

Hence, the correct option is, (d)
`Mg^(+2)`