Chemistry MCQs for NEET — Practice Questions with Answers

The stability of an oxidation state in lanthanide elements depends on the combined effect of hydration energy and ionization energy. These factors determine how easily an ion can form and remain stable in a solution.

The element with the electronic configuration $[Ar] 3d^8$ for its $M^{2+}$ ion is Nickel (Ni). The atomic number of Nickel is 28. When it loses two electrons to form $Ni^{2+}$, the configuration becomes $[Ar] 3d^8$.

Among the lanthanides, Europium (Eu) is known to show both +2 and +3 oxidation states. This is because the +2 oxidation state of Eu has a stable half-filled f-orbital configuration \(4f^7\\).

Lanthanide contraction is observed due to an increase in effective nuclear charge. As we move across the lanthanide series, electrons are added to the 4f orbitals, which do not shield the nuclear charge effectively, leading to a stronger attraction between the nucleus and the outer electrons, thus causing a contraction in atomic and ionic sizes.

In the f-block elements, also known as the lanthanides and actinides, the atomic volume decreases with an increase in atomic number. This phenomenon is due to the lanthanide contraction, where the addition of electrons to the 4f orbitals results in poor shielding, leading to an increase in the effective nuclear charge and a consequent decrease in atomic and ionic radii.

Aqueous solutions of transition metal salts are often colored due to d-d electron transitions. $ Co (NO_3)_2 $ contains cobalt, a transition metal, which typically forms colored solutions. In contrast, $ Zn (NO_3)_2 $ and $ LiNO_3 $ contain zinc and lithium, respectively, which do not have partially filled d-orbitals and thus form colorless solutions. Potash Alum, while sometimes appearing colored due to impurities, is generally colorless in its pure form.