Question

On the basis of electronic configuration, if the ion $X^{3-}$ has 14 electrons, then the number of electrons present in $X^{2+}$ is

Answer 1

9

Answer 2

Let Z be the atomic number of the element X. The atomic number represents the number of protons in the nucleus of an atom of element X.

For a neutral atom of X, the number of electrons is equal to the number of protons, which is Z.

When an atom forms an ion by gaining or losing electrons, the number of protons (and thus the atomic number Z) remains unchanged. The charge of the ion is determined by the difference between the number of protons and the number of electrons.

For an ion with charge $q$, the number of electrons is given by: Number of electrons = Number of protons - charge = Z - q.

The ion $X^{3-}$ has a charge of -3. The number of electrons in $X^{3-}$ is given as 14. Using the formula: Number of electrons in $X^{3-}$ = Z - (-3) = Z + 3. Given that the number of electrons in $X^{3-}$ is 14, we have: 14 = Z + 3. Solving for Z: Z = 14 - 3 Z = 11. So, the atomic number of element X is 11. This means a neutral atom of X has 11 protons and 11 electrons.

Now we need to find the number of electrons in the ion $X^{2+}$. The ion $X^{2+}$ has a charge of +2. Using the formula: Number of electrons in $X^{2+}$ = Z - (+2) = Z - 2. We know that Z = 11. Substituting the value of Z: Number of electrons in $X^{2+}$ = 11 - 2 Number of electrons in $X^{2+}$ = 9.

Therefore, the number of electrons present in the ion $X^{2+}$ is 9.