Topic

Ag has a completely filled 4d orbital (4d¹⁰ 5s¹) in its ground state. Now, silver displays two oxidation states ( +1 and +2). In the +1 oxidation state, an electron is removed from the s-orboital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d⁹). Hence, it is a transition element.

The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomization. In all transition metals (except Zn, electronic configuration: 3d¹⁰4s²), there are some unpaired electrons that account for their stronger metallic bonding. Due to the absence of these unpaired electrons, the inter-atomic electronic bonding is the weakest in Zn and as a result, it has the least enthalpy of atomization.

Mn (Z = 25) = 3d⁵ 4s²
Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2 to +7.

The E⁰ (M²⁺ /M) value of a metal depends on the energy changes involved in the following:
1. Sublimation:The energy required for converting one mole of an atom from the solid state to the gaseous state.
M_(s) &$8594; M_(g) ? _sH(sublimation energy)
2. Ionization: The energy required to take out electrons from one mole of atoms in the gaseous state.
M_(g) ? M²⁺_(g) ?_iH (Ionization energy)
3. Hydration: The energy released when one mole of ions are hydrated.
M²⁺_{(g) ?M²⁺(aq) ?hydH (Hydration energy)
Now, copper has a high energy of atomization and low hydration energy. Hence, the E⁰ (M²⁺ /M) value for copper is positive.}

Ionization enthalpies are found to increase in the given series due to a continuous filling of the inner d-orbitals. The irregular variations of ionization enthalpies can be attributed to the extra stability of configurations such as d⁰, d⁵, d¹⁰. Since these states are exceptionally stable, their ionization enthalpies are very high.
In case of first ionization energy, Cr has low ionization energy. This is because after losing one electron, it attains the stable configuration (3d⁵). On the other hand, Zn has exceptionally high first ionization energy as an electron has to be removed from stable and fully-filled orbitals (3d¹⁰ 4s²). Second ionization energies are higher than the first since it becomes difficult to remove an electron when an electron has already been taken out. Also, elements like Cr and Cu have exceptionally high second ionization energies as after losing the first electron, they have attained the stable configuration (Cr⁺ : 3d⁵ and Cu⁺: 3d¹⁰). Hence, taking out one electron more from this stable configuration will require a lot of energy.

Both oxide and fluoride ions are highly electronegative and have a very small size. Due to these properties, they are able to oxidize the metal to its highest oxidation state.

The following reactions are involved when Cr²⁺ and Fe²⁺ act as reducing agents.
Cr²⁺ →Cr³⁺
Fe²⁺ → Fe³⁺ The E^o_{Cr³⁺/Cr²⁺} value is ?0.41 V and E^o_{Fe³⁺/Fe²⁺} is 0.77 V. This means that Cr²⁺ can be easily oxidized to Cr³⁺ , but Fe²⁺ does not get oxidized to Fe³⁺ easily. Therefore, Cr²⁺ is a better reducing agent that Fe³⁺ .

In an aqueous medium, Cu²⁺ is more stable than Cu⁺. This is because although energy is required to remove one electron from Cu⁺ to Cu²⁺ , high hydration energy of Cu²⁺ compensates for it. Therefore, Cu⁺ ion in an aqueous solution is unstable. It disproportionates to give Cu²⁺ and Cu.
2 Cu⁺_(aq) → Cu²⁺_(aq) + Cu _(s)

In actinoids, 5f orbitals are filled. These 5f orbitals have a poorer shielding effect than 4f orbitals (in lanthanoids). Thus, the effective nuclear charge experienced by electrons in valence shells in case of actinoids is much more that that experienced by lanthanoids. Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

(i) Cr³⁺: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ Or, [Ar] ¹⁸ 3d³
(ii) Pm³⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f⁴ Or, [Xe] ⁵⁴ 3d³
(iii) Cu⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ Or, [Ar] ¹⁸ 3d¹⁰
(iv) Ce⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ Or, [Xe] ⁵⁴
(v) Co²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁷ Or, [Ar] ¹⁸ 3d⁷
(vi) Lu²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹ Or, [Xe] ⁵⁴ 4f¹⁴ 3d³
(vii) Mn²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ Or, [Ar] ¹⁸ 3d⁵
(viii) Th⁴⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6p⁶ Or, [Rn] ⁸⁶

Electronic configuration of Mn²⁺ is [Ar] ¹⁸ 3d⁵
. Electronic configuration of Fe²⁺ is [Ar] ¹⁸ 3d⁶
. It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in ( +2) state has a stable d⁵ configuration. This is the reason Mn²⁺ shows resistance to oxidation to Mn³⁺ . Also, Fe²⁺ has 3d⁶ configuration and by losing one electron, its configuration changes to a more stable 3d⁵ configuration. Therefore, Fe²⁺ easily gets oxidized to Fe³⁺ oxidation state.

The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states ( +2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the d-orbital. However, Sc does not show +2 oxidation state. Its electronic configuration is 4s² 3d¹. It loses all the three electrons to form Sc³ . +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar]. Ti ( +4) and V( +5) are very stable for the same reason. For Mn, 2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar] 3d⁵.

As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
Consequences of lanthanoid contraction
(i) There is similarity in the properties of second and third transition series. Separation of lanthanoids is possible due to lanthanide contraction.
(iii) It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from La(OH)₃ to Lu(OH)₃.)

(i) Vanadate, VO^-₃ Oxidation state of V is +5.
(ii) Chromate, Cr^2-₄ Oxidation state of Cr is +6.
(iii) Permanganate, MnO^-₄ Oxidation state of Mn is +7.

Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements lie in the d-block and show a transition of properties between s-block and p-block. Therefore, these are called transition elements.
Elements such as Zn, Cd, and Hg cannot be classified as transition elements because these have completely filled d-subshell.