Answer
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 $Pb^{2+}$.
Work Step by Step
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 $Pb^{2+}$.
