Most common oxidation states of Ce (Cerium) are
(a) + 3, + 4
(b) + 2, + 3
(c) + 2, + 1
(d) + 3, + 5

Q: Why is +3 the most common oxidation state for lanthanides?

The +3 state is most stable because the sum of the first three ionization enthalpies is relatively low, and the resulting hydration or lattice energy is highly negative, making the +3 state thermodynamically favorable.

Q: Why does Cerium show a +4 oxidation state?

After losing 3 electrons, Cerium has a $4f^1$ configuration. Losing this last $4f$ electron gives it the highly stable $[\text{Xe}]$ noble gas core, making the +4 state very stable.

Q: Is Ce+4 stable in aqueous solutions?

While Ce+4 is stable, it is a strong oxidizing agent in aqueous solutions because it tends to gain an electron to revert to the more common +3 state.

Q: Why don't we remove electrons from the 4f orbital first?

Electrons are removed from the outermost principal quantum shell first. So, 6s electrons are removed first, followed by 5d, and finally 4f.

Q: Does Cerium show a +2 oxidation state?

Cerium rarely shows a +2 state because the +3 state is significantly more stable thermodynamically. The +2 state is more common in lanthanides like Europium or Ytterbium where it leads to half-filled or fully-filled f-orbitals.

Question

Most common oxidation states of Ce (Cerium) are

Options:

(A) + 3, + 4

(B) + 2, + 3

(C) + 2, + 1

(D) + 3, + 5

Sigma Percentile · Accepted Answer

### The Enigma of Lanthanide Oxidation States

When we dive into the f-block of the periodic table, we encounter the lanthanides—a series of elements that share remarkably similar chemical properties. One of their defining characteristics is their overwhelming preference for the **+3 oxidation state**.

But chemistry is rarely without its exceptions and fascinating nuances. Enter Cerium (Ce), the first element of the lanthanide series, with an atomic number of 58. While it faithfully follows the lanthanide tradition of exhibiting a +3 state, it also proudly displays a **+4 oxidation state**. Let's unravel the mystery behind this dual behavior.

### Decoding the Electronic Configuration

To understand an element's chemical behavior, we must first look at its electronic configuration. Cerium sits just after Lanthanum, and its electrons begin to fill the deep-seated 4f orbitals.

Using Xenon (Xe, atomic number 54) as our noble gas core, the electronic configuration of Cerium is:
$$ 	ext{Ce}_{58} = [	ext{Xe}] 4f^1 5d^1 6s^2 $$

Notice the arrangement: we have two electrons in the outermost 6s orbital, one in the 5d orbital, and a lone electron sitting in the 4f orbital.

### The Journey to +3 and +4

When Cerium reacts, it first loses its outermost electrons. Removing the two 6s electrons and the single 5d electron requires a reasonable amount of energy, which is easily compensated by the hydration or lattice energy when it forms compounds.

This loss of three electrons gives us the Cerium(III) ion:
$$ 	ext{Ce}^{3+} = [	ext{Xe}] 4f^1 $$

This +3 state is the hallmark of all lanthanides. But look closely at the $	ext{Ce}^{3+}$ configuration. It has just one electron left in the 4f orbital. If it can somehow lose this single electron, it will achieve the electronic configuration of Xenon—a highly stable, completely filled noble gas core.

$$ 	ext{Ce}^{4+} = [	ext{Xe}] 4f^0 $$

Because achieving a noble gas configuration provides immense thermodynamic stability, Cerium readily loses that fourth electron to form the $	ext{Ce}^{4+}$ ion. This makes the +4 oxidation state exceptionally common and stable for Cerium, alongside the standard +3 state.

### The Oxidizing Power of Ce(IV)

It is worth noting that while $	ext{Ce}^{4+}$ is stable, the +3 state is still the "default" comfort zone for lanthanides. As a result, $	ext{Ce}^{4+}$ in aqueous solutions has a strong tendency to snatch an electron from other species to revert to $	ext{Ce}^{3+}$. This makes Cerium(IV) compounds, like Ceric Ammonium Nitrate (CAN), excellent and widely used oxidizing agents in organic chemistry.

In conclusion, the most common oxidation states of Cerium are **+3 and +4**, a beautiful consequence of its electronic structure and the pursuit of noble gas stability.

Animated Solution for Chemistry - d and f-Block Elements: Most common oxidation states of Ce (Cerium) are

Select Answer:

Visualized Solution

Cerium (Ce)

Electronic Configuration

Formation of $Ce^{3 +}$

Formation of $Ce^{4 +}$

Common Oxidation States

The Sigma Insight: Inner Transition Elements

The Enigma of Lanthanide Oxidation States

Decoding the Electronic Configuration

The Journey to +3 and +4

The Oxidizing Power of Ce(IV)

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