## Chemistry: The Central Science (13th Edition)

There is no neutral atom isoelectronic with $Ga^{3+}$. $Kr$ is isoelectronic with $Zr^{4+}$. $Ar$ is isoelectronic with $Mn^{7+}$. $Xe$ is isoelectronic with $I^-$. $Hg$ is isoelectronic with $Pb^{2+}$.
Isoelectronic means having the same number of electrons and the same electron configurations. So, to find the neutral atom isoelectronic with the ions, we must: - Find the number of electrons each ion has. - Look at the periodic table to find the neutral atom that has the same number of electrons as the ion. - Compare their electron configurations to find whether they are similar. 1) $Ga^{3+}$ A neutral $Ga$ atom has 31 electrons. Losing 3 electrons to become $Ga^{3+}$ cation, it now has 28 electrons. The neutral atom that has 28 electrons is $Ni$. The electron configuration of $Ga$ is $[Ar]3d^{10}4s^24p^1$. Losing 3 electrons (atoms would lose the most easily removed ones, normally the outermost) means the electron configuration of $Ga^{3+}$ is $[Ar]3d^{10}$. However, the electron configuration of $Ni$ is $[Ar]3d^84s^2$. Since, the electron configurations are not similar, $Ni$ is not isoelectronic with $Ga^{3+}$. There is no neutral atom isoelectronic with $Ga^{3+}$. 2) $Zr^{4+}$ Cation $Zr^{4+}$ has 36 electrons. The neutral atom that also has 36 electrons is $Kr$. The electron configuration of $Zr$ is $[Kr]4d^25s^2$. Losing 4 electrons means the electron configuration of $Zr^{4+}$ is $[Kr]$, which is exactly the electron configuration of $Kr$. Therefore, $Kr$ is isoelectronic with $Zr^{4+}$. 3) $Mn^{7+}$ Cation $Mn^{7+}$ has 18 electrons. The neutral atom that also has 18 electrons is $Ar$. The electron configuration of $Mn$ is $[Ar]3d^54s^2$. Losing 7 electrons means the electron configuration of $Mn^{7+}$ is $[Ar]$, which is exactly the electron configuration of $Ar$. Therefore, $Ar$ is isoelectronic with $Mn^{7+}$. 4) $I^-$ Anion $I^-$ has 54 electrons. The neutral atom that also has 54 electrons is $Xe$. The electron configuration of $I$ is $[Kr]4d^{10}5s^25p^5$. Adding 1 electron means the electron configuration of $I^-$ is $[Kr]4d^{10}5s^25p^6$, which is also the electron configuration of $Xe$. Therefore, $Xe$ is isoelectronic with $I^-$. 5) $Pb^{2+}$ Cation $Pb^{2+}$ has 80 electrons. The neutral atom that also has 80 electrons is $Hg$. The electron configuration of $Pb$ is $[Xe]4f^{14}5d^{10}6s^26p^2$. Losing 2 electrons means the electron configuration of $Pb^{2+}$ is $[Xe]4f^{14}5d^{10}6s^2$, which is also the electron configuration of $Hg$. Therefore, $Hg$ is isoelectronic with $Pb^{2+}$.