JEE MAIN - Chemistry (2023 - 15th April Morning Shift - No. 23)
The total number of isoelectronic species from the given set is ___________.
$\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Al}, \mathrm{Mg}^{2+}, \mathrm{Na}^{+}, \mathrm{O}^{+}, \mathrm{Mg}, \mathrm{Al}^{3+}, \mathrm{F}$
$\mathrm{O}^{2-}, \mathrm{F}^{-}, \mathrm{Al}, \mathrm{Mg}^{2+}, \mathrm{Na}^{+}, \mathrm{O}^{+}, \mathrm{Mg}, \mathrm{Al}^{3+}, \mathrm{F}$
Answer
5
Explanation
Isoelectronic species are atoms, ions or molecules that have the same number of electrons. To find the isoelectronic species from the given set, we need to determine the number of electrons in each species.
The electron configuration of each species is:
- $\mathrm{O}^{2-}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{F}^{-}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{Al}$: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^1$, 13 electrons
- $\mathrm{Mg}^{2+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{Na}^{+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{O}^{+}$: $1s^2\ 2s^2\ 2p^3$, 7 electrons
- $\mathrm{Mg}$: $1s^2\ 2s^2\ 2p^6\ 3s^2$, 12 electrons
- $\mathrm{Al}^{3+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{F}$: $1s^2\ 2s^2\ 2p^5$, 9 electrons
The electron configuration of each species is:
- $\mathrm{O}^{2-}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{F}^{-}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{Al}$: $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^1$, 13 electrons
- $\mathrm{Mg}^{2+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{Na}^{+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{O}^{+}$: $1s^2\ 2s^2\ 2p^3$, 7 electrons
- $\mathrm{Mg}$: $1s^2\ 2s^2\ 2p^6\ 3s^2$, 12 electrons
- $\mathrm{Al}^{3+}$: $1s^2\ 2s^2\ 2p^6$, 10 electrons
- $\mathrm{F}$: $1s^2\ 2s^2\ 2p^5$, 9 electrons
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