(a) The electronic configuration of Scandium (Sc) is [Ar] 3d¹ 4s².

We can see that the valence shell of the Sc atom contains only 3 electrons (1 in 3d and 2 in 4s).

If it loses these 3 electrons, it can attain the nearest noble gas configuration, Ar, and we know that noble gases are the most stable element on the periodic table.

As a result, it would prefer to lose 3 electrons to form Sc³⁺, which the most stable. So, Sc prefers this oxidation state only.

(b) We know that the colour of cations depend upon the number of unpaired electrons present in the d-orbital. If no ‘d’ electrons are present in a particular oxidation state of a metal, it means that every compound where that oxidation state is present, will be colourless.

In [Ti(H₂O)₆]⁴⁺, let us consider the oxidation state of Ti to be x, and we know that H₂O is a neutral ligand (0 oxidation state). Also the species has +4 charge overall. So, we represent this as:

∴ x = +4

So, the oxidation state of Ti in [Ti(H₂O)₆]⁴⁺is +4.

The electronic configuration of Ti is [Ar] 3d² 4s². Ti⁴⁺ would mean 4 electrons are removed from the valence shell (2 from d and 2 from s).

So, Ti⁴⁺ = 3d⁰4s⁰. As ‘d’ orbital is empty, having no electrons,[Ti(H₂O)₆]⁴⁺is colourless.

(c) Mn₂O₇ is the anhydride (removal of water molecule) of an acid, permanganic acid (H₂Mn₂O₈), so the anhydride is also acidic.

H₂Mn₂O₈ - H₂O = Mn₂O₇

But, MnO is the anhydride of a base, manganese (II) hydroxide, Mn(OH)₂, so the anhydride is also basic.

Mn(OH)₂ - H₂O = MnO

OR

(a) Lanthanoids primarily show three oxidation states - +2, +3 and +4. Here, amongst these oxidation states, +3 oxidation state is the most common. As the energy difference between the 4f, 5d and 6s orbitals is quite large, lanthanoids display a limited number of oxidation states.

We know that the stability of a half filled orbital (like d⁵) or a full filled orbital (like d¹⁰) is always higher than the other configurations. So, an element or an ion would always prefer the half filled or full filled configuration to gain stability, thus they always try to go somehow to that configuration. So, amongstlanthanoids, Ln (III) compounds are predominant. However, occasionally in solution or in solid compounds, +2 and +4 ions are also obtained due to the extra stability of half filled, empty or fully filled f orbitals.

But the general electronic configuration is given as: 4f ^1-145d^0-16s².

(b) Cerium is an important member of the lanthanide family having the general configuration of 4f ^1-145d^0-16s².

Its atomic number is 58, and we can easily find out its electronic configuration to be [Xe] 4f � 5d � 6s �.

The most stable oxidation state would be that of its nearest noble gas, Xe, and it is obtained by the removal of 4 electrons (1 from f, 1 from d and 2 from s).

So, +4 is the most common state.

Ce can also show +3 oxidation state, but again, it is not as stable as the +4 one.

(c) Actinoidshave the common oxidation state of +3. But, the energy difference between 5f, 6d, and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states unlike the lanthanoids where the difference between the 4f, 5d and 6s orbitals is quite large, so they display a limited number of oxidation states.