Question

The correct order of basicity is

• A. Al(OH3) > Ca(OH)2 > Lu(OH)3 > La(OH)3
• B. Ca(OH)2 > La(OH)3 > Lu(OH)3 > Al(OH)3
• C. Lu(OH)3 > La(OH)3 > Ca(OH)2 > Al(OH)3
• D. La(OH)3 > Ca(OH)2 > Al(OH)2 > Lu(OH)3

Answer 1

Answer: (B)

B

Answer 2

To determine the correct order of basicity for the given hydroxides, we need to consider the periodic trends and the properties of the metal ions. Basicity of metal hydroxides $M(OH)_n$ depends on the ease with which the $M-O$ bond can break, releasing $OH^⁻$ ions. This ease is favored by:

1. Larger ionic size of the metal (M): A larger cation leads to a weaker electrostatic attraction for the hydroxide ion, making the $M-O$ bond more ionic and easier to break.
2. Lower charge on the metal ion (M): A lower charge on the cation results in less polarizing power, leading to a more ionic $M-O$ bond and thus higher basicity.

Let's analyze each hydroxide:

• $Al(OH)_3$: Aluminum is in Group 13, Period 3. $Al^{3+}$ has a small size (53.5 pm) and a high charge (+3). This leads to significant covalent character in the $Al-O$ bond. $Al(OH)_3$ is amphoteric, meaning it can act as both an acid and a base, but it is a very weak base.
• $Ca(OH)_2$: Calcium is an alkaline earth metal in Group 2, Period 4. $Ca^{2+}$ has a moderate size (100 pm) and a charge of +2. $Ca(OH)_2$ is a strong base, though its solubility is limited.
• $La(OH)_3$: Lanthanum is the first element of the lanthanide series (f-block). $La^{3+}$ has a relatively large size (103.2 pm) and a charge of +3. $La(OH)_3$ is a strong base.
• $Lu(OH)_3$: Lutetium is the last element of the lanthanide series. $Lu^{3+}$ has a smaller size (86.1 pm) than $La^{3+}$ due to lanthanoid contraction, and a charge of +3. $Lu(OH)_3$ is a base, but weaker than $La(OH)_3$.

Now let's compare them:

1. $Al(OH)_3$ vs. other hydroxides: $Al(OH)_3$ is amphoteric and generally considered a very weak base compared to alkaline earth metal hydroxides and lanthanide hydroxides. Therefore, $Al(OH)_3$ will be the least basic.

2. $La(OH)_3$ vs. $Lu(OH)_3$: Both are lanthanide hydroxides. Due to the lanthanoid contraction, the ionic radius decreases from $La^{3+}$ to $Lu^{3+}$. A smaller cation has a higher charge density, leading to greater polarizing power and increased covalent character in the $M-OH$ bond. This makes the hydroxide less basic.
   Thus, $La(OH)_3 > Lu(OH)_3$.

3. $Ca(OH)_2$ vs. $La(OH)_3$: This is the most crucial comparison.

   • $Ca^{2+}$: Charge +2, ionic radius ≈ 100 pm.
   • $La^{3+}$: Charge +3, ionic radius ≈ 103.2 pm.
     While $La^{3+}$ is slightly larger than $Ca^{2+}$ (which favors higher basicity for $La(OH)_3$), the higher charge of $La^{3+}$ (+3 vs +2) significantly increases its polarizing power. A higher charge on the metal ion generally leads to a more covalent $M-O$ bond, making the hydroxide less basic. The effect of charge is usually more dominant than a small difference in size.
     Therefore, $Ca(OH)_2 > La(OH)_3$.

Combining these comparisons, the order of basicity is:

• $Ca(OH)_2$ (most basic among the given)
• $La(OH)_3$ (stronger than $Lu(OH)_3$ due to larger size)
• $Lu(OH)_3$ (weaker than $La(OH)_3$ due to lanthanoid contraction)
• $Al(OH)_3$ (least basic, being amphoteric)

So, the correct order is:
$Ca(OH)_2 > La(OH)_3 > Lu(OH)_3 > Al(OH)_3$