The Lanthanoids

The first and second ionization enthalpies of lanthanoids (elements from the lanthanide series) are quite high due to the strong effective nuclear charge experienced by the electrons as the atomic size decreases across the series. The trend of ionization enthalpy in lanthanoids is somewhat comparable to calcium because:]
- Calcium (B): Calcium belongs to the alkaline earth metals group and has relatively higher ionization enthalpies, similar to the trend seen in lanthanoids. Both show a similar increase in ionization energy across their respective series due to the increasing nuclear charge.
Thus, the correct answer is (B) Calcium because its ionization enthalpies are comparable to that of the lanthanoids.

Concept: Lanthanoids show characteristic electronic configurations with partially filled orbitals.

Step 1: Identify configuration.


Step 2: Recognize element.
Gadolinium (Gd) has this configuration due to extra stability of half-filled .

Concept: Lanthanoids show +3 oxidation state.

Step 1: Combine.

Step 1: Understanding the Concept:
The general electronic configuration of lanthanoids is . Exceptions occur for atoms with half-filled or fully-filled f-orbitals.

Step 2: Detailed Explanation:
Europium (Eu) has atomic number 63. Its expected configuration is . However, to achieve the extra stability of a half-filled f-orbital (f ), one electron is promoted, resulting in the configuration . This is similar to the chromium exception.

Step 3: Final Answer:
Europium (Eu) has the outer electronic configuration .

Lanthanoids (or lanthanides) are typically metals, and their compounds with sulfur form sulfides. The most common lanthanide sulfides have the general formula , where lanthanum is in the +3 oxidation state.

1. Step 1: Oxidation state of lanthanum. Lanthanum typically forms a +3 oxidation state, and sulfur typically forms a -2 oxidation state.

2. Step 2: Balance the charges. For a neutral compound, two ions are needed to balance three ions, giving the formula . Thus, the correct formula for lanthanum sulfide is .

Step 1: Understanding the Concept:
The color of lanthanoid ions, whether in solid state or in aqueous solution, is primarily attributed to transitions. These transitions absorb light in the visible region. For an transition to occur, the ion must have partially filled -orbitals (i.e., an to configuration).
Step 2: Key Formula or Approach:
The approach requires writing the ground-state electronic configuration of each lanthanoid ion and identifying if it has unpaired electrons in its subshell. Ions with or configurations will be colorless.
Step 3: Detailed Explanation:
Let's determine the electronic configuration of each given ion:
- A) : Lutetium ( ). Configuration of Lu is . For , remove 3 electrons (two from , one from ): . The -subshell is completely filled, so no transitions can occur. It is colorless.
- B) : Samarium ( ). Configuration of Sm is . For , remove 2 electrons from : . It has 6 unpaired electrons in the partially filled -subshell, allowing for transitions. It will be colored (typically blood-red).
- C) : Ytterbium ( ). Configuration of Yb is . For , remove 2 electrons from : . The -subshell is completely filled, so it is colorless.
- D) : Lanthanum ( ). Configuration of La is . For , remove 3 electrons: . The -subshell is completely empty, so it is colorless.
Step 4: Final Answer:
has a partially filled f-orbital ( ) and is therefore the colored ion.

Step 1: Understanding the Concept
Atomic radius usually increases as we move down a group in the periodic table due to the addition of new electron shells. However, between the second ( ) and third ( ) transition series, this trend is interrupted by a phenomenon known as Lanthanoid contraction.

Step 2: Key Formula or Approach
Lanthanoid contraction occurs due to the poor shielding effect of electrons. As we move from Lanthanum ( ) to Lutetium ( ), the shell is filled. These electrons do not shield the nucleus effectively, causing the outer electrons to be pulled closer.

Step 3: Detailed Explanation
1. Zirconium (Zr) is in the series, and Hafnium (Hf) is in the series.
2. Before Hf, the 14 lanthanoid elements are filled, where electrons enter the subshell.
3. The poor shielding of the electrons causes an increase in the effective nuclear charge acting on the outer electrons of Hf.
4. This results in a contraction of the atomic size of Hf, making it nearly identical to the size of Zr (Zr 160 pm, Hf 159 pm).

Step 4: Final Answer
The similarity in radii is due to Lanthanoid contraction.