Answer
The iso-electronic species are,
$NH_{4}^{+}$ and $CH_{4}$ with 10 electrons.
$C_{6}H_{6}$ and $B_{3}N_{3}H_{6}$ with 42 electrons.
$CO$ and $N_{2}$ with 14 electrons.
Work Step by Step
Atoms, compounds, or ions with the same number of electrons are called iso-electronic species.
The atomic number of N is 7.
The atomic number of H is 1.
Therefore $NH_{4}^{+}$ contains $[(1\times7)+(4\times1)- 1]$ electrons, that is 10 electrons.
The atomic number of C is 6.
The atomic number of H is 1.
Therefore $C_{6}H_{6}$ contains $[(6\times6)+(6\times1)]$ electrons, that is 42 electrons.
The atomic number of C is 6.
The atomic number of O is 8.
Therefore $CO$ contains $[(1\times6)+(1\times8)]$ electrons, that is 14 electrons.
The atomic number of C is 6.
The atomic number of H is 1.
Therefore $CH_{4}$ contains $[(1\times6)+(4\times1)]$ electrons, that is 10 electrons.
The atomic number of N is 7.
Therefore $N_{2}$ contains $(2\times7)$ electrons, that is 14 electrons.
The atomic number of B is 5.
The atomic number of N is 7.
The atomic number of H is 1.
Therefore $B_{3}N_{3}H_{6}$ contains $[(3\times5)+(3\times7)+(6\times1)]$ electrons, that is 42 electrons.
Thus iso-electronic species are,
$NH_{4}^{+}$ and $CH_{4}$ with 10 electrons.
$C_{6}H_{6}$ and $B_{3}N_{3}H_{6}$ with 42 electrons.
$CO$ and $N_{2}$ with 14 electrons.
