The f-block elements are those in which the differentiating electrons enters the (n-2)f orbitals. There are two series of f-Block elements corresponding to filling of 4f and 5f-orbitals. The series of 4f-orbitals is called lanthanides. Lanthanides show different oxidation states depending upon stability of f⁰, f⁷ and f¹⁴ configurations, though the most common oxidation states is +3. There is a regular decrease in the size of lanthanides ions with increase in atomic number which is known as lanthanides contraction.

Out of the following which member of the lanthanide series which is well known to exhibit +4 oxidation state?

Cerium (Z=58)

The correct answer is option 1. Cerium (Z=58).

Among the given options, the lanthanide that is well known to exhibit the +4 oxidation state is Cerium (Z=58).

1. Cerium (Ce, Z=58): Cerium is the most well-known lanthanide to exhibit a +4 oxidation state. It can lose all four valence electrons (one from the 5d orbital and three from the 4f orbital) to form Ce^4+. This is particularly stable due to the resulting empty 4f orbital.

2. Europium (Eu, Z=63): Europium typically exhibits +2 and +3 oxidation states. The +2 state is relatively stable due to the half-filled 4f^7 configuration.

3. Lanthanum (La, Z=57): Lanthanum primarily exhibits a +3 oxidation state. It does not typically exhibit a +4 oxidation state.

4. Gadolinium (Gd, Z=64): Gadolinium usually exhibits a +3 oxidation state. It has a stable half-filled 4f^7 configuration in the +3 state and does not commonly exhibit a +4 oxidation state.

Summary: Cerium (Ce) is the lanthanide that is well known to exhibit the +4 oxidation state, making it unique among the lanthanides in this regard.