To determine which grouping represents a collection of isoelectronic species, let's analyze the given options and compute the electron counts for each species in each group:
Isoelectronic species are atoms or ions that have the same number of electrons.
### 1) [tex]\( N^{3-}, F^-, Na^+ \)[/tex]
- [tex]\( N^{3-} \)[/tex]:
- Atomic number [tex]\( N = 7 \)[/tex] (7 protons)
- Gaining 3 electrons: [tex]\( 7 + 3 = 10 \)[/tex] electrons
- [tex]\( F^- \)[/tex]:
- Atomic number [tex]\( F = 9 \)[/tex] (9 protons)
- Gaining 1 electron: [tex]\( 9 + 1 = 10 \)[/tex] electrons
- [tex]\( Na^+ \)[/tex]:
- Atomic number [tex]\( Na = 11 \)[/tex] (11 protons)
- Losing 1 electron: [tex]\( 11 - 1 = 10 \)[/tex] electrons
So, [tex]\( N^{3-} \)[/tex], [tex]\( F^- \)[/tex], and [tex]\( Na^+ \)[/tex] each have 10 electrons. Therefore, this group is isoelectronic.
### 2) [tex]\( Ca^{2+}, Mg^{2+} \)[/tex]
- [tex]\( Ca^{2+} \)[/tex]:
- Atomic number [tex]\( Ca = 20 \)[/tex] (20 protons)
- Losing 2 electrons: [tex]\( 20 - 2 = 18 \)[/tex] electrons
- [tex]\( Mg^{2+} \)[/tex]:
- Atomic number [tex]\( Mg = 12 \)[/tex] (12 protons)
- Losing 2 electrons: [tex]\( 12 - 2 = 10 \)[/tex] electrons
Here [tex]\( Ca^{2+} \)[/tex] has 18 electrons and [tex]\( Mg^{2+} \)[/tex] has 10 electrons. Therefore, this group is not isoelectronic.
### 3) [tex]\( Cs^+, Br^- \)[/tex]
- [tex]\( Cs^+ \)[/tex]:
- Atomic number [tex]\( Cs = 55 \)[/tex] (55 protons)
- Losing 1 electron: [tex]\( 55 - 1 = 54 \)[/tex] electrons
- [tex]\( Br^- \)[/tex]:
- Atomic number [tex]\( Br = 35 \)[/tex] (35 protons)
- Gaining 1 electron: [tex]\( 35 + 1 = 36 \)[/tex] electrons
Here [tex]\( Cs^+ \)[/tex] has 54 electrons and [tex]\( Br^- \)[/tex] has 36 electrons. Therefore, this group is not isoelectronic.
### 4) [tex]\( Be, Al^{3+}, Cl^- \)[/tex]
- [tex]\( Be \)[/tex]:
- Atomic number [tex]\( Be = 4 \)[/tex] (4 protons)
- Neutral atom: [tex]\( 4 \)[/tex] electrons
- [tex]\( Al^{3+} \)[/tex]:
- Atomic number [tex]\( Al = 13 \)[/tex] (13 protons)
- Losing 3 electrons: [tex]\( 13 - 3 = 10 \)[/tex] electrons
- [tex]\( Cl^- \)[/tex]:
- Atomic number [tex]\( Cl = 17 \)[/tex] (17 protons)
- Gaining 1 electron: [tex]\( 17 + 1 = 18 \)[/tex] electrons
Here [tex]\( Be \)[/tex] has 4 electrons, [tex]\( Al^{3+} \)[/tex] has 10 electrons, and [tex]\( Cl^- \)[/tex] has 18 electrons. Therefore, this group is not isoelectronic.
### Conclusion:
Among the given groupings, the first group [tex]\( N^{3-}, F^-, Na^+ \)[/tex] is found to be isoelectronic, as all species have 10 electrons. Therefore, the correct answer is:
Option 1: [tex]\( N^{3-}, F^-, Na^+ \)[/tex]
