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+单项选择题

类似问题

Which of the following has the smallest radius?

Rank the following in order of increasing ionic radius: Br-, Kr, Rb+

Which of the following has the greatest distance between Ions?

Which one of these has the largest radius?

The higher the positive charge the [ Select ] smaller larger the ion/element. The more negative a charge is, the [ Select ] bigger smaller the ion/element

Which of the following best helps to account for the fact that the F- ion is smaller than the O2- ion?

In which security measure will you find anti-virus software?

What are user permissions?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单