To determine which ion is isoelectronic with $$Be$^{2+},$ we need to understand the concept of isoelectronic species.$\\$Isoelectronic species are atoms or ions that have the same number of electrons.$\\$First, let's find the number of electrons in $$Be$^{2+}: \\$• Beryllium (Be) has an atomic number of 4, which means it has 4 electrons in its neutral state.$\\$• The ion $$Be$^{2+}$ has lost 2 electrons, so it has $4 - 2 = 2$ electrons.$\\$Now, let's analyze each option to find the number of electrons in each ion:$\\$Option 1: $$H$^+ \\$• Hydrogen (H) has an atomic number of 1, so it has 1 electron in its neutral state.$\\$• The ion $$H$^+$ has lost 1 electron, so it has $1 - 1 = 0$ electrons.$\\$Option 2: $$Li$^+ \\$• Lithium (Li) has an atomic number of 3, so it has 3 electrons in its neutral state.$\\$• The ion $$Li$^+$ has lost 1 electron, so it has $3 - 1 = 2$ electrons.$\\$Option 3: $$Na$^+ \\$• Sodium (Na) has an atomic number of 11, so it has 11 electrons in its neutral state.$\\$• The ion $$Na$^+$ has lost 1 electron, so it has $11 - 1 = 10$ electrons.$\\$Option 4: $$Mg$^{2+} \\$• Magnesium (Mg) has an atomic number of 12, so it has 12 electrons in its neutral state.$\\$• The ion $$Mg$^{2+}$ has lost 2 electrons, so it has $12 - 2 = 10$ electrons.$\\$Comparing the number of electrons:$\\$• $$Be$^{2+}$ has 2 electrons.$\\$• $$Li$^+$ also has 2 electrons.$\\$Therefore, $$Be$^{2+}$ is isoelectronic with $$Li$^+. \\$The correct option is Option 2.