Question

1. list the following ions in order of increasing radius: li⁺, mg²⁺, br⁻, te²⁻.

Explanation:

Step1: Analyze ion types (cation/anion)

Cations: \( \text{Li}^+ \) (group 1, +1 charge), \( \text{Mg}^{2+} \) (group 2, +2 charge). Anions: \( \text{Br}^- \) (group 17, -1 charge), \( \text{Te}^{2-} \) (group 16, -2 charge). Cations are smaller than neutral atoms; anions are larger.

Step2: Compare cation radii

For cations, higher positive charge → smaller radius (same period trend, or ionic radius: more protons pull electrons tighter). \( \text{Mg}^{2+} \) (12 protons, 10 electrons) vs \( \text{Li}^+ \) (3 protons, 2 electrons). Wait, \( \text{Li}^+ \) has electron configuration \( 1s^2 \) (noble gas He), \( \text{Mg}^{2+} \) has \( 2s^2 2p^6 \) (noble gas Ne). The principal quantum number \( n \) for \( \text{Li}^+ \) is 1, for \( \text{Mg}^{2+} \) is 2. Wait, no—\( \text{Li}^+ \) is \( 1s^2 \) (n=1), \( \text{Mg}^{2+} \) is \( [\text{Ne}] \) (n=2). Wait, but ionic radius: when comparing cations, if they are not isoelectronic, we look at n and charge. Wait, \( \text{Li}^+ \) (n=1) vs \( \text{Mg}^{2+} \) (n=2). Wait, no—\( \text{Li}^+ \) has 2 electrons, \( \text{Mg}^{2+} \) has 10. Wait, actually, \( \text{Li}^+ \) is in the 1st shell, \( \text{Mg}^{2+} \) in the 2nd. But wait, \( \text{Li}^+ \) radius: ~0.76 Å, \( \text{Mg}^{2+} \) ~0.72 Å? Wait, no, maybe I messed up. Wait, isoelectronic? No. Wait, let's check electron configurations:

\( \text{Li}^+ \): \( 1s^2 \) (n=1)

\( \text{Mg}^{2+} \): \( 1s^2 2s^2 2p^6 \) (n=2)

Wait, but \( \text{Mg}^{2+} \) has more protons (12) than \( \text{Li}^+ \) (3), but higher n. Wait, actually, the radius of \( \text{Li}^+ \) is about 0.76 pm, \( \text{Mg}^{2+} \) is about 0.72 pm? Wait, maybe because \( \text{Mg}^{2+} \) has a higher charge (2+ vs 1+), so even with n=2, the higher charge pulls electrons tighter. So \( \text{Mg}^{2+} < \text{Li}^+ \)? Wait, no, wait: \( \text{Li}^+ \) is in the first period, \( \text{Mg}^{2+} \) in the third? Wait, Li is period 2? No, Li is period 2? Wait, Li is atomic number 3: period 2, group 1. Mg is atomic number 12: period 3, group 2. So \( \text{Li}^+ \) is formed by losing 1 electron from Li (electron config \( 1s^2 2s^1 \) → \( 1s^2 \)). \( \text{Mg}^{2+} \) is losing 2 electrons from Mg (\( [\text{Ne}] 3s^2 \) → \( [\text{Ne}] \)). So \( \text{Li}^+ \) has n=1, \( \text{Mg}^{2+} \) has n=2. But the radius of \( \text{Li}^+ \) is ~0.76 Å, \( \text{Mg}^{2+} \) ~0.72 Å. So \( \text{Mg}^{2+} < \text{Li}^+ \) because higher charge (2+ vs 1+) and even though n is higher, the charge effect dominates? Wait, maybe I should compare anions first.

Step3: Compare anion radii

Anions: \( \text{Br}^- \) (atomic number 35, gains 1 electron: \( [\text{Kr}] \)), \( \text{Te}^{2-} \) (atomic number 52, gains 2 electrons: \( [\text{Xe}] \)). For anions, more negative charge → larger radius (more electrons, same n or higher n). Also, \( \text{Te}^{2-} \) is in period 5, \( \text{Br}^- \) in period 4. So \( \text{Br}^- < \text{Te}^{2-} \) (since Te is lower in the periodic table, larger n, more electrons).

Step4: Compare cations vs anions

Cations are smaller than anions (since cations lose electrons, anions gain; cations have higher Zeff). So cations (small) < anions (large). Now, order cations: \( \text{Mg}^{2+} < \text{Li}^+ \) (from step2), order anions: \( \text{Br}^- < \text{Te}^{2-} \) (step3). Now combine: \( \text{Mg}^{2+} < \text{Li}^+ < \text{Br}^- < \text{Te}^{2-} \)? Wait, wait, no—wait \( \text{Li}^+ \) is \( 1s^2 \), \( \text{Mg}^{2+} \) is \( [\text{Ne}] \) (n=2). Wait, \( \text{Li}^+ \) radius is ~0.76 Å, \( \text{Mg}^{2+} \) ~0.72 Å. So \( \text{Mg}^{2+} \) is smaller than \( \text{Li}^+ \). Then \( \text{Br}^- \) is \( [\text{Kr}] \) (n=4), radius ~1.96 Å, \( \text{Te}^{2-} \) is \( [\text{Xe}] \) (n=5), radius ~2.21 Å. So anions are larger than cations. So putting it all together: \( \text{Mg}^{2+} < \text{Li}^+ < \text{Br}^- < \text{Te}^{2-} \)? Wait, but wait, \( \text{Li}^+ \) is in period 2, \( \text{Br}^- \) in period 4. So the order should be: smallest (most positive, least electrons) to largest (most negative, most electrons).

Wait, let's use the periodic table trends:

- Ionic radius: for isoelectronic species, more positive charge → smaller radius. But here, not isoelectronic.

- Cations: \( \text{Mg}^{2+} \) (group 2, +2) and \( \text{Li}^+ \) (group 1, +1). \( \text{Mg}^{2+} \) has higher charge, smaller radius than \( \text{Li}^+ \) (even though \( \text{Li}^+ \) is in n=1, \( \text{Mg}^{2+} \) in n=2, the charge difference makes \( \text{Mg}^{2+} \) smaller? Wait, no, actually, \( \text{Li}^+ \) is smaller than \( \text{Mg}^{2+} \)? Wait, no, let's check actual ionic radii:

- \( \text{Li}^+ \): ~0.76 Å

- \( \text{Mg}^{2+} \): ~0.72 Å

So \( \text{Mg}^{2+} \) is slightly smaller than \( \text{Li}^+ \). Then anions:

- \( \text{Br}^- \): ~1.96 Å

- \( \text{Te}^{2-} \): ~2.21 Å

So the order from smallest to largest: \( \text{Mg}^{2+} < \text{Li}^+ < \text{Br}^- < \text{Te}^{2-} \)

Wait, but let's confirm with periodic trends:

- Cations: smaller than neutral atoms; anions larger.

- For cations, higher charge (more protons) → smaller radius (same electron configuration). But here, \( \text{Mg}^{2+} \) and \( \text{Li}^+ \) have different electron configurations. \( \text{Mg}^{2+} \) has \( [\text{Ne}] \) (10 electrons), \( \text{Li}^+ \) has \( [\text{He}] \) (2 electrons). The principal quantum number for \( \text{Li}^+ \) is 1, for \( \text{Mg}^{2+} \) is 2. But the radius of \( \text{Li}^+ \) is ~0.76, \( \text{Mg}^{2+} \) ~0.72. So \( \text{Mg}^{2+} \) is smaller because even though n is higher, the charge is +2 (vs +1 for \( \text{Li}^+ \)), so the nucleus pulls electrons tighter.

- For anions, \( \text{Br}^- \) is \( [\text{Kr}] \) (period 4), \( \text{Te}^{2-} \) is \( [\text{Xe}] \) (period 5). Lower period (Br) → smaller radius than higher period (Te). Also, \( \text{Te}^{2-} \) has 2- charge, \( \text{Br}^- \) has 1- charge, so \( \text{Te}^{2-} \) is larger.

So combining: cations (small) < anions (large). Among cations, \( \text{Mg}^{2+} < \text{Li}^+ \) (due to higher charge). Among anions, \( \text{Br}^- < \text{Te}^{2-} \) (due to lower period and lower charge). So overall order: \( \text{Mg}^{2+} < \text{Li}^+ < \text{Br}^- < \text{Te}^{2-} \)

Answer:

\( \text{Mg}^{2+} < \text{Li}^+ < \text{Br}^- < \text{Te}^{2-} \)