Question at position 11 Place the following ions in order of increasing radius: Al3+, F–, Mg2+, and N3–. F–

Question

Question at position 11 Place the following ions in order of increasing radius: Al3+, F–, Mg2+, and N3–. F–< N3– < Al3+< Mg2+F–< N3–< Mg2+ < Al3+Al3+< Mg2+< F–< N3–Al3+< N3– < F–< Mg2+Mg2+ < F–< N3–< Al3+

BlackTom AI · Accepted Answer

The question asks you to place Al3+, F–, Mg2+, and N3– in order of increasing ionic radius, i.e., from smallest to largest.
First, recall a key principle: among ions that share the same number of electrons (isoelectronic species), the one with the higher nuclear charge (more protons) is smaller because it pulls the electron cloud in more tightly. Here, all ions have 10 electrons except for the relative proton counts: Al3+ (13 protons), Mg2+ (12 protons), F– (9 protons), N3– (7 protons). Therefore, the smallest ion will be the one with the largest positive charge (Al3+), followed by Mg2+, then F–, and finally N3– as the largest due to the lowest effective nuclear charge.
Option 1: "F–< N3– < Al3+< Mg2+" places F– as smallest and Mg2+ as largest, with Al3+ larger than F– and Mg2+ not in the expected order. This contradicts the principle that Al3+ (highest charge) should be smallest, so this arrangement is incorrect.
Option 2: "F–< N3–< Mg2+ < Al3+" starts with F–, then N3–, then Mg2+, and ends with Al3+ as largest. This completely reverses the expected trend, since Al3+ should be the smallest, not the largest, so this option is incorrect.
Option 3: "Al3+< Mg2+< F–< N3–" aligns with the isoelectronic trend described: Al3+ (most protons) smallest, then Mg2+, then F–, and N3– largest. This sequence matches the increasing radius pattern expected from effective nuclear charge, so this option is correct.
Option 4: "Al3+< N3– < F–< Mg2+" places N3– before F– and Mg2+, implying N3– is smaller than F–, which contradicts the isoelectronic trend (N3– has the fewest protons among the four) and is therefore incorrect.
Option 5: "Mg2+ < F–< N3–< Al3+" suggests Al3+ is the largest, which is opposite to the actual order where Al3+ should be smallest due to its high charge. This makes the option incorrect.
In summary, only option 3 follows the correct logic for ionic radii among isoelectronic species, reflecting the increasing radius from the ion with the greatest effective nuclear attraction to the least.

Question at position 11 Place the following ions in order of increasing radius: Al3+, F–, Mg2+, and N3–. F–< N3– < Al3+< Mg2+F–< N3–< Mg2+ < Al3+Al3+< Mg2+< F–< N3–Al3+< N3– < F–< Mg2+Mg2+ < F–< N3–< Al3+

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