Question

Among the following the isoelectronic species are
I. $Mg^{2+}, Al^{3+}, F^-$
II. $O^{2-}, F^-, N^{3-}$
III. $K^+, Ca^{2+}, Sc^{3+}$
IV. $Mn^{2+}, Fe^{3+}, V^{3+}$

• A. II, III only
• B. I, II, III
• C. II, III, IV only
• D. I, III, IV

Hint:

For isoelectronic ions, the ionic radius decreases as the nuclear charge ($Z$) increases. Order of size: $N^{3-}>O^{2-}>F^->Na^+>Mg^{2+}>Al^{3+}$.

Answer 1

The Correct Option is B

Step 1: Define Isoelectronic Species:
Isoelectronic species are atoms or ions that contain the same number of electrons. Formula: Number of Electrons = Atomic Number ($Z$) - Charge.
Step 2: Analyze Set I ($Mg^{2+}, Al^{3+}, F^-$):
$Mg^{2+}$ ($Z=12$): $12 - 2 = 10$ electrons.
$Al^{3+}$ ($Z=13$): $13 - 3 = 10$ electrons.
$F^-$ ($Z=9$): $9 - (-1) = 10$ electrons.
Conclusion: Isoelectronic (10 electrons, Neon configuration).

Step 3: Analyze Set II ($O^{2-}, F^-, N^{3-}$):
$O^{2-}$ ($Z=8$): $8 - (-2) = 10$ electrons.
$F^-$ ($Z=9$): $9 - (-1) = 10$ electrons.
$N^{3-}$ ($Z=7$): $7 - (-3) = 10$ electrons.
Conclusion: Isoelectronic (10 electrons).

Step 4: Analyze Set III ($K^+, Ca^{2+}, Sc^{3+}$):
$K^+$ ($Z=19$): $19 - 1 = 18$ electrons.
$Ca^{2+}$ ($Z=20$): $20 - 2 = 18$ electrons.
$Sc^{3+}$ ($Z=21$): $21 - 3 = 18$ electrons.
Conclusion: Isoelectronic (18 electrons, Argon configuration).

Step 5: Analyze Set IV ($Mn^{2+}, Fe^{3+}, V^{3+}$):
$Mn^{2+}$ ($Z=25$): $25 - 2 = 23$ electrons.
$Fe^{3+}$ ($Z=26$): $26 - 3 = 23$ electrons.
$V^{3+}$ ($Z=23$): $23 - 3 = 20$ electrons.
Conclusion: NOT isoelectronic (23 vs 20).

Step 6: Final Selection:
Sets I, II, and III are correct. This matches Option (B).