Question 1

6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Accepted Answer

The highest oxidation state of a metal is exhibited in its oxide or fluoride because these compounds contain highly electronegative elements (oxygen and fluorine) which can stabilize the high positive charge on the metal ion by strong ionic or covalent bonding. Oxygen and fluorine atoms have high electronegativity and can effectively attract electron density towards themselves, thus stabilizing the metal in its highest oxidation state. Other ligands or anions are less electronegative and cannot stabilize such high oxidation states as effectively. — The stabilization of high oxidation states requires ligands that can accommodate the high positive charge on the metal. Oxides and fluorides, being highly electronegative, stabilize these states through strong bonding, making the highest oxidation states observable only in these compounds.

Question 2

7 Which is a stronger reducing agent Cr2+ or Fe2+ and why?

Accepted Answer

Cr2+ is a stronger reducing agent than Fe2+. This is because Cr2+ has a greater tendency to lose electrons and get oxidized to Cr3+ compared to Fe2+ which oxidizes to Fe3+. The standard electrode potential for the Cr3+/Cr2+ couple is more negative than that for the Fe3+/Fe2+ couple, indicating Cr2+ is more easily oxidized and thus a stronger reducing agent. — Reducing agents lose electrons. The more negative the standard reduction potential, the stronger the reducing agent. Cr2+ has a more negative potential, so it is more readily oxidized and hence stronger as a reducing agent than Fe2+.

Question 3

Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

Accepted Answer

Atomic number 25 corresponds to manganese (Mn). The divalent ion Mn2+ has the electronic configuration [Ar] 3d5, meaning it has 5 unpaired electrons. Using the spin-only formula for magnetic moment:

μ = √(n(n+2))

where n = number of unpaired electrons = 5

μ = √(5 × (5 + 2)) = √(5 × 7) = √35 ≈ 5.92 BM (Bohr Magneton)

Therefore, the magnetic moment of Mn2+ ion in aqueous solution is approximately 5.92 BM. — The magnetic moment depends on the number of unpaired electrons. Mn2+ has 5 unpaired electrons, so applying the spin-only formula gives the value of magnetic moment.

Question 4

8 Calculate the 'spin only' magnetic moment of M2+_{isq} ion (Z = 27).

Accepted Answer

The atomic number Z = 27 corresponds to cobalt (Co). The divalent ion Co2+ has the electronic configuration [Ar] 3d7.

Number of unpaired electrons in high spin Co2+ (3d7) is 3.

Using the spin-only formula:

μ = √(n(n+2))

μ = √(3 × (3 + 2)) = √(3 × 5) = √15 ≈ 3.87 BM

Therefore, the spin-only magnetic moment of Co2+ ion is approximately 3.87 Bohr Magnetons. — Determine the number of unpaired electrons for Co2+ (3d7). For high spin, 3 unpaired electrons are present. Applying the spin-only formula yields the magnetic moment.

Question 5

1 Write down the electronic configuration of:

(i) Cr3+

(ii) Pm3+

(iii) Cu+

(iv) Ce4+

(v) Co2+

(vi) Lu2+

(vii) Mn2+

(viii) Th4+

Accepted Answer

The electronic configurations are as follows:

(i) Cr3+ : Atomic number of Cr = 24
Electronic configuration of Cr: [Ar] 3d5 4s1
Cr3+ means loss of 3 electrons, first from 4s and then from 3d:
Cr3+ : [Ar] 3d3

(ii) Pm3+ : Atomic number of Pm = 61
Electronic configuration of Pm: [Xe] 4f5 6s2
Pm3+ means loss of 3 electrons, first from 6s and then from 4f:
Pm3+ : [Xe] 4f4

(iii) Cu+ : Atomic number of Cu = 29
Electronic configuration of Cu: [Ar] 3d10 4s1
Cu+ means loss of 1 electron, from 4s:
Cu+ : [Ar] 3d10

(iv) Ce4+ : Atomic number of Ce = 58
Electronic configuration of Ce: [Xe] 4f1 5d1 6s2
Ce4+ means loss of 4 electrons, first from 6s, then 5d and 4f:
Ce4+ : [Xe]

(v) Co2+ : Atomic number of Co = 27
Electronic configuration of Co: [Ar] 3d7 4s2
Co2+ means loss of 2 electrons from 4s:
Co2+ : [Ar] 3d7

(vi) Lu2+ : Atomic number of Lu = 71
Electronic configuration of Lu: [Xe] 4f14 5d1 6s2
Lu2+ means loss of 2 electrons from 6s:
Lu2+ : [Xe] 4f14 5d1

(vii) Mn2+ : Atomic number of Mn = 25
Electronic configuration of Mn: [Ar] 3d5 4s2
Mn2+ means loss of 2 electrons from 4s:
Mn2+ : [Ar] 3d5

(viii) Th4+ : Atomic number of Th = 90
Electronic configuration of Th: [Rn] 6d2 7s2
Th4+ means loss of 4 electrons from 7s and 6d:
Th4+ : [Rn] — The electrons are removed first from the outermost shell (s orbital) and then from the d or f orbitals as per Aufbau principle and observed ionization behavior.

Question 6

2 Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?

Accepted Answer

Mn2+ compounds are more stable than Fe2+ compounds towards oxidation to +3 state because Mn2+ has a half-filled d5 configuration (3d5), which is particularly stable due to exchange energy and symmetry. On the other hand, Fe2+ has a 3d6 configuration which is less stable. Therefore, Mn2+ resists oxidation to Mn3+ more than Fe2+ resists oxidation to Fe3+. — The half-filled d5 configuration in Mn2+ provides extra stability due to exchange energy and symmetrical distribution of electrons, making Mn2+ less prone to oxidation compared to Fe2+.

Question 7

3 Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Accepted Answer

In the first half of the first row transition elements (Sc to Mn), the +2 oxidation state becomes more stable with increasing atomic number because the removal of two 4s electrons leads to a stable electronic configuration. As the nuclear charge increases, the 3d electrons are held more tightly, making it difficult to remove more electrons beyond +2. Also, the half-filled d5 configuration of Mn2+ is particularly stable, which contributes to the stability of the +2 state. — The stability of the +2 state increases due to the increasing effective nuclear charge and the attainment of stable or half-filled d subshell configurations, which resist further oxidation.

Question 8

4 To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Accepted Answer

Electronic configurations play a crucial role in deciding the stability of oxidation states in the first series of transition elements. The stability depends on the attainment of stable electronic arrangements such as half-filled (d5) or fully filled (d10) d subshells.

For example:
- Mn2+ (3d5) is very stable due to half-filled d subshell.
- Cr3+ (3d3) is more stable than Cr2+ (3d4) because removing one more electron leads to a half-filled d3 configuration.
- Cu+ (3d10) is more stable than Cu2+ (3d9) due to fully filled d subshell.

Thus, the electronic configuration influences the oxidation state stability by favoring configurations with extra stability from half-filled or fully filled d orbitals. — The stability of oxidation states is influenced by the electronic configuration because certain configurations (half-filled or fully filled) provide extra stability due to exchange energy and symmetrical electron distribution.

Question 9

5 What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d3, 3d5, 3d8 and 3d4?

Accepted Answer

The stable oxidation states for the given d electron configurations are:

- 3d3: The element with 3d3 configuration (e.g., V) generally shows +3 oxidation state as stable.
- 3d5: The element with 3d5 configuration (e.g., Mn) shows +2 oxidation state as stable due to half-filled d subshell.
- 3d8: The element with 3d8 configuration (e.g., Ni) commonly shows +2 oxidation state.
- 3d4: The element with 3d4 configuration (e.g., Cr2+) is less stable; +3 oxidation state (3d3) is more stable.

Thus, stable oxidation states correspond to achieving half-filled or stable d subshells. — Stability arises from achieving half-filled (d5) or stable configurations by losing electrons to reach such states.

Question 10

8 What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Accepted Answer

Characteristics of transition elements:
- They have partially filled d orbitals in their atoms or common oxidation states.
- Exhibit variable oxidation states.
- Form coloured compounds.
- Show paramagnetism.
- Have high melting and boiling points.
- Act as good catalysts.

They are called transition elements because they represent the transition of properties from the highly electropositive s-block elements to the more electronegative p-block elements.

Sc and Zn are d-block elements but are not regarded as transition elements because Sc3+ has no d electrons and Zn2+ has fully filled d10 configuration, so they do not show typical transition element properties. — Transition elements are defined by their partially filled d orbitals and characteristic properties. Elements like Sc and Zn do not fulfill this criterion.

Question 11

9 In what way is the electronic configuration of the transition elements different from that of the non transition elements?

Accepted Answer

The electronic configuration of transition elements is characterized by the filling of inner d orbitals (n-1)d along with the outer s orbital (ns). In contrast, non-transition elements do not have partially filled d orbitals in their atoms or common oxidation states. Transition elements have (n-1)d1-10 ns0-2 configurations, whereas non-transition elements have s and p orbitals being filled without involvement of d orbitals. — Transition elements have partially filled d orbitals which contribute to their unique properties, unlike non-transition elements.

Question 12

10 What are the different oxidation states exhibited by the lanthanoids?

Accepted Answer

The lanthanoids predominantly exhibit the +3 oxidation state. Some lanthanoids also show +2 and +4 oxidation states in certain compounds. For example, Eu and Yb commonly show +2 state, while Ce and Tb can show +4 state. However, +3 is the most stable and common oxidation state for lanthanoids. — The stability of +3 oxidation state is due to the removal of three electrons (two 6s and one 4f or 5d) leading to stable electronic configurations.

Question 13

11 Explain giving reasons:

(i) Transition metals and many of their compounds show paramagnetic behaviour.

(ii) The enthalpies of atomisation of the transition metals are high.

(iii) The transition metals generally form coloured compounds.

(iv) Transition metals and their many compounds act as good catalyst.

Accepted Answer

(i) Transition metals and many of their compounds show paramagnetic behaviour because they have unpaired d electrons which contribute to magnetic moments.

(ii) The enthalpies of atomisation of transition metals are high due to strong metallic bonding arising from the involvement of (n-1)d electrons in bonding.

(iii) Transition metals generally form coloured compounds because of d-d electronic transitions within the partially filled d orbitals when they absorb visible light.

(iv) Transition metals and their compounds act as good catalysts because they can adopt multiple oxidation states and form intermediate complexes with reactants, facilitating reaction pathways. — The properties arise from the presence of unpaired d electrons, variable oxidation states, and ability to form coordination complexes.

Question 14

12 What are interstitial compounds? Why are such compounds well known for transition metals?

Accepted Answer

Interstitial compounds are compounds formed when small atoms like hydrogen, carbon, or nitrogen occupy the interstitial spaces (holes) in the metal lattice without significantly changing the metal's structure. Transition metals form such compounds because they have small atomic sizes and large number of vacant d orbitals, allowing small atoms to fit into the interstices and form strong metal-nonmetal bonds. — The ability of transition metals to form interstitial compounds is due to their metallic lattice structure and availability of vacant orbitals.

Question 15

13 How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.

Accepted Answer

Transition metals exhibit a wide range of oxidation states because of the involvement of both (n-1)d and ns electrons in bonding. They can lose different numbers of d and s electrons leading to multiple oxidation states.

Non-transition metals usually show fixed oxidation states because their valence electrons are only in the outermost s and p orbitals.

For example, iron shows +2 and +3 oxidation states, manganese shows +2 to +7, whereas non-transition metals like sodium show only +1 oxidation state. — The availability of d electrons for bonding in transition metals leads to variable oxidation states unlike non-transition metals.

f-block elements in the periodic

f-block elements in the periodic — Study Notes

The d- and f-Block Elements

4.1 Position in the Periodic Table

4.2 Electronic Configurations of the d-Block Elements

Practice Questions — f-block elements in the periodic

All 5 Chapters in Chemistry-I