To determine the most probable value for the ionic radius of [tex]\( Al^{3+} \)[/tex], let's examine the periodic trends and the given data.
### Periodic Trends:
1. Ionic Radius Trend Across a Period:
- As you move from left to right across a period, the ionic radius generally decreases. This is because the increasing nuclear charge pulls the electrons closer to the nucleus, thereby reducing the size of the ion.
2. Ionic Radius Trend Down a Group:
- As you move down a group, the ionic radius increases. This is due to the addition of electron shells, which increases the overall size of the ion even though the nuclear charge also increases.
### Given Data:
1. Boron (B):
- Atomic radius: [tex]\( 88 \, \text{pm} \)[/tex]
- Ionic radius of [tex]\( B^{3+} \)[/tex]: [tex]\( 23 \, \text{pm} \)[/tex]
2. Aluminum (Al):
- Atomic radius: [tex]\( 143 \, \text{pm} \)[/tex]
- Ionic radius of [tex]\( Al^{3+} \)[/tex]: to be determined
### Analysis:
Since Aluminum (Al) is directly below Boron (B) in Group 13, we expect the ionic radius of [tex]\( Al^{3+} \)[/tex] to be larger than that of [tex]\( B^{3+} \)[/tex] due to the addition of an extra electron shell.
Given the ionic radius of [tex]\( B^{3+} \)[/tex] is [tex]\( 23 \, \text{pm} \)[/tex], let’s analyze the given multiple-choice options to find the most probable value for [tex]\( Al^{3+} \)[/tex]:
- [tex]\( 16 \, \text{pm} \)[/tex]: This is smaller than [tex]\( 23 \, \text{pm} \)[/tex], which contradicts the trend that the ionic radius increases down a group.
- [tex]\( 51 \, \text{pm} \)[/tex]: This is larger than [tex]\( 23 \, \text{pm} \)[/tex], fitting with the expectation that the ionic radius of [tex]\( Al^{3+} \)[/tex] should be larger than that of [tex]\( B^{3+} \)[/tex] but still a reasonable size increase.
- [tex]\( 95 \, \text{pm} \)[/tex]: This is significantly larger than [tex]\( 23 \, \text{pm} \)[/tex], which seems an unreasonable jump considering the periodic trend.
- [tex]\( 172 \, \text{pm} \)[/tex]: This is extraordinarily large and does not fit the trend. Such a large radius would be unprecedented for [tex]\( Al^{3+} \)[/tex] in its high oxidation state.
Given the logical and systematic analysis of the periodic trends and the data presented, the most probable value for the ionic radius of [tex]\( Al^{3+} \)[/tex] would be:
(B) [tex]\( 51 \, \text{pm} \)[/tex]
Hence, the ionic radius of [tex]\( Al^{3+} \)[/tex] is most likely to be:
[tex]\( 51 \, \text{pm} \)[/tex].
