The electronic configuration of an atom is written first, and then electrons are removed from the outermost shell to form the ion. (i) Cr^3+ Atomic number of Chromium (Cr) is 24. Electronic configuration of Cr atom: [Ar] 3d^5 4s^1. To form the Cr^3+ ion, three electrons are removed (one from the 4s…

Q4.2: Why are Mn^2+ compounds more stable than Fe^2+ towards oxidation to their +3 state?

The stability of Mn^2+ compounds compared to Fe^2+ towards oxidation can be explained by comparing their electronic configurations. 1. Electronic configuration of Mn^2+: The atomic number of Manganese (Mn) is 25. Its electronic configuration is [Ar] 3d^5 4s^2. When it forms the Mn^2+ ion, it loses…

The D-and F-Block Elements Class 12 Chemistry NCERT Solutions

Q4.1Exercises

Write down the electronic configuration of:

(i)

$\mathrm{Cr}^{3+}$

(ii)

$\mathrm{Pm}^{3+}$

(iii)

$\mathrm{Cu}^{+}$

(iv)

$\mathrm{Ce}^{4+}$

(v)

$\mathrm{Co}^{2+}$

(vi)

$\mathrm{Lu}^{2+}$

(vii)

$\mathrm{Mn}^{2+}$

(viii)

$\mathrm{Th}^{4+}$

Solution

The electronic configuration of an atom is written first, and then electrons are removed from the outermost shell to form the ion.

(i)

$\mathrm{Cr}^{3+}$

Atomic number of Chromium (Cr) is 24.
Electronic configuration of Cr atom: $[\mathrm{Ar}] \, 3d^5 \, 4s^1$ .
To form the $\mathrm{Cr}^{3+}$ ion, three electrons are removed (one from the $4s$ orbital and two from the $3d$ orbitals).
Electronic configuration of $\mathrm{Cr}^{3+}$ : $[\mathrm{Ar}] \, 3d^3$ .

(ii)

$\mathrm{Pm}^{3+}$

Atomic number of Promethium (Pm) is 61.
Electronic configuration of Pm atom: $[\mathrm{Xe}] \, 4f^5 \, 6s^2$ .
To form the $\mathrm{Pm}^{3+}$ ion, three electrons are removed (two from the $6s$ orbital and one from the $4f$ orbital).
Electronic configuration of $\mathrm{Pm}^{3+}$ : $[\mathrm{Xe}] \, 4f^4$ .

(iii)

$\mathrm{Cu}^{+}$

Atomic number of Copper (Cu) is 29.
Electronic configuration of Cu atom: $[\mathrm{Ar}] \, 3d^{10} \, 4s^1$ .
To form the $\mathrm{Cu}^{+}$ ion, one electron is removed from the $4s$ orbital.
Electronic configuration of $\mathrm{Cu}^{+}$ : $[\mathrm{Ar}] \, 3d^{10}$ .

(iv)

$\mathrm{Ce}^{4+}$

Atomic number of Cerium (Ce) is 58.
Electronic configuration of Ce atom: $[\mathrm{Xe}] \, 4f^1 \, 5d^1 \, 6s^2$ .
To form the $\mathrm{Ce}^{4+}$ ion, four electrons are removed (two from $6s$ , one from $5d$ , and one from $4f$ ).
Electronic configuration of $\mathrm{Ce}^{4+}$ : $[\mathrm{Xe}]$ .

(v)

$\mathrm{Co}^{2+}$

Atomic number of Cobalt (Co) is 27.
Electronic configuration of Co atom: $[\mathrm{Ar}] \, 3d^7 \, 4s^2$ .
To form the $\mathrm{Co}^{2+}$ ion, two electrons are removed from the $4s$ orbital.
Electronic configuration of $\mathrm{Co}^{2+}$ : $[\mathrm{Ar}] \, 3d^7$ .

(vi)

$\mathrm{Lu}^{2+}$

Atomic number of Lutetium (Lu) is 71.
Electronic configuration of Lu atom: $[\mathrm{Xe}] \, 4f^{14} \, 5d^1 \, 6s^2$ .
To form the $\mathrm{Lu}^{2+}$ ion, two electrons are removed from the $6s$ orbital.
Electronic configuration of $\mathrm{Lu}^{2+}$ : $[\mathrm{Xe}] \, 4f^{14} \, 5d^1$ .

(vii)

$\mathrm{Mn}^{2+}$

Atomic number of Manganese (Mn) is 25.
Electronic configuration of Mn atom: $[\mathrm{Ar}] \, 3d^5 \, 4s^2$ .
To form the $\mathrm{Mn}^{2+}$ ion, two electrons are removed from the $4s$ orbital.
Electronic configuration of $\mathrm{Mn}^{2+}$ : $[\mathrm{Ar}] \, 3d^5$ .

(viii)

$\mathrm{Th}^{4+}$

Atomic number of Thorium (Th) is 90.
Electronic configuration of Th atom: $[\mathrm{Rn}] \, 6d^2 \, 7s^2$ .
To form the $\mathrm{Th}^{4+}$ ion, four electrons are removed (two from $7s$ and two from $6d$ ).
Electronic configuration of $\mathrm{Th}^{4+}$ : $[\mathrm{Rn}]$ .

Q4.2Exercises

Why are $\mathrm{Mn}^{2+}$ compounds more stable than $\mathrm{Fe}^{2+}$ towards oxidation to their +3 state?

Solution

The stability of

\mathrm{Mn}^{2+}

compounds compared to

\mathrm{Fe}^{2+}

towards oxidation can be explained by comparing their electronic configurations.

Electronic configuration of $\mathrm{Mn}^{2+}$ : The atomic number of Manganese (Mn) is 25. Its electronic configuration is $[\mathrm{Ar}] \, 3d^5 \, 4s^2$ . When it forms the $\mathrm{Mn}^{2+}$ ion, it loses the two $4s$ electrons, resulting in the configuration $[\mathrm{Ar}] \, 3d^5$ . This configuration has a half-filled $d$ -subshell, which is exceptionally stable due to symmetrical electron distribution and high exchange energy.
Oxidation of $\mathrm{Mn}^{2+}$ to $\mathrm{Mn}^{3+}$ : The oxidation process is: $\mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}^{3+} + e^{-}$ . The electronic configuration of $\mathrm{Mn}^{3+}$ is $[\mathrm{Ar}] \, 3d^4$ . This process involves removing an electron from the very stable half-filled $3d^5$ configuration, which requires a large amount of energy. This is reflected in the high third ionization enthalpy of Mn. Therefore, $\mathrm{Mn}^{2+}$ is resistant to oxidation to $\mathrm{Mn}^{3+}$ .
Electronic configuration of $\mathrm{Fe}^{2+}$ : The atomic number of Iron (Fe) is 26. Its electronic configuration is $[\mathrm{Ar}] \, 3d^6 \, 4s^2$ . When it forms the $\mathrm{Fe}^{2+}$ ion, it loses the two $4s$ electrons, resulting in the configuration $[\mathrm{Ar}] \, 3d^6$ .
Oxidation of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ : The oxidation process is: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^{-}$ . The electronic configuration of $\mathrm{Fe}^{3+}$ is $[\mathrm{Ar}] \, 3d^5$ . This process results in the formation of a stable half-filled $3d^5$ configuration. Because the product ( $\mathrm{Fe}^{3+}$ ) is more stable than the reactant ( $\mathrm{Fe}^{2+}$ ), the oxidation of $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3+}$ occurs readily.

Conclusion:

\mathrm{Mn}^{2+}

is stable and resists further oxidation because its oxidation disrupts a stable half-filled

d^5

configuration. In contrast,

\mathrm{Fe}^{2+}

is easily oxidized to

\mathrm{Fe}^{3+}

because this process leads to the formation of the stable half-filled

d^5

configuration. Thus,

\mathrm{Mn}^{2+}

compounds are more stable than

\mathrm{Fe}^{2+}

compounds towards oxidation.

Q4.3Exercises

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

Solution

In the first half of the first row of transition elements (from Scandium to Manganese), the +2 oxidation state is formed by the loss of the two electrons from the outermost

4s

orbital. The stability of this +2 state increases as we move from Sc to Mn.

The stability of the +2 oxidation state relative to the +3 state can be assessed by considering the third ionization enthalpy (IE₃), which is the energy required to remove an electron from the M²⁺ ion.

As we move across the period from Scandium (Z=21) to Manganese (Z=25), the atomic number increases, and so does the effective nuclear charge. The additional electrons are added to the inner

3d

orbitals, which do not shield the nuclear charge very effectively. This increasing nuclear charge holds the

3d

electrons more tightly.

Let's examine the trend:

Scandium (Sc): The configuration of $\mathrm{Sc}^{2+}$ is $[\mathrm{Ar}] \, 3d^1$ . It has a low third ionization enthalpy because the removal of the single $3d$ electron leads to the highly stable noble gas configuration of $\mathrm{Sc}^{3+}$ ( $[\mathrm{Ar}]$ ). Therefore, the +2 state is not stable for scandium, and it almost exclusively exhibits the +3 state.
Titanium (Ti) to Chromium (Cr): As we move from Ti to Cr, the nuclear charge increases. This makes it progressively more difficult to remove the third electron from the $3d$ orbital. Consequently, the third ionization enthalpy (IE₃) generally increases. This trend signifies that the +2 state becomes increasingly stable against oxidation to the +3 state.
Manganese (Mn): For manganese, the configuration of $\mathrm{Mn}^{2+}$ is $[\mathrm{Ar}] \, 3d^5$ . This configuration has a half-filled $d$ -subshell, which is particularly stable. The third ionization enthalpy of Mn is exceptionally high because removing the third electron disrupts this stable configuration. As a result, $\mathrm{Mn}^{2+}$ is very stable and resistant to oxidation to $\mathrm{Mn}^{3+}$ .

In summary, the stability of the +2 oxidation state increases across the first half of the first transition series because the increasing effective nuclear charge makes it progressively harder to remove the third electron. This trend culminates in the very stable

\mathrm{Mn}^{2+}

ion, which benefits from the extra stability of a half-filled

d

-subshell.

Q4.4Exercises

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.

Solution

The electronic configurations of transition elements play a crucial role in determining the stability of their various oxidation states. The tendency to attain a stable configuration, such as an empty (

d^0

), half-filled (

d^5

), or completely filled (

d^{10}

)

d

-subshell, is a major factor. However, other factors like ionization enthalpy, lattice energy, and hydration enthalpy also contribute.

Here are some examples from the first series of transition elements (Sc to Zn) that illustrate this:

Scandium (Sc, $[\mathrm{Ar}] \, 3d^1 \, 4s^2$ ): Scandium exhibits only the +3 oxidation state. When it forms the $\mathrm{Sc}^{3+}$ ion, it loses all three of its valence electrons, attaining the stable noble gas configuration of Argon ( $[\mathrm{Ar}]$ , a $d^0$ configuration). The +2 state ( $\mathrm{Sc}^{2+}$ , $[\mathrm{Ar}] \, 3d^1$ ) is highly unstable as it can easily lose one more electron to achieve the noble gas configuration.
Manganese (Mn, $[\mathrm{Ar}] \, 3d^5 \, 4s^2$ ): Manganese shows a wide range of oxidation states, from +2 to +7. Its most stable oxidation state is +2. The $\mathrm{Mn}^{2+}$ ion has the electronic configuration $[\mathrm{Ar}] \, 3d^5$ . This half-filled $d$ -subshell provides extra stability. Consequently, Mn(II) compounds are common and resist oxidation. While Mn can show a +7 state (as in $\mathrm{MnO}_4^-$ ), where it has a $d^0$ configuration, this state is strongly oxidizing, indicating its desire to return to a lower, more stable state like +2.
Iron (Fe, $[\mathrm{Ar}] \, 3d^6 \, 4s^2$ ): Iron commonly shows +2 and +3 oxidation states. The $\mathrm{Fe}^{2+}$ ion has a $3d^6$ configuration. It can be easily oxidized to the $\mathrm{Fe}^{3+}$ ion, which has a stable half-filled $3d^5$ configuration. This explains why $\mathrm{Fe}^{3+}$ compounds are generally more stable than $\mathrm{Fe}^{2+}$ compounds.
Zinc (Zn, $[\mathrm{Ar}] \, 3d^{10} \, 4s^2$ ): Zinc exclusively shows the +2 oxidation state. The $\mathrm{Zn}^{2+}$ ion has the electronic configuration $[\mathrm{Ar}] \, 3d^{10}$ . The completely filled $d$ -subshell is very stable. Removing a third electron would require a very large amount of energy (very high third ionization enthalpy) to break into this stable configuration. Therefore, zinc does not exhibit oxidation states higher than +2.
Copper (Cu, $[\mathrm{Ar}] \, 3d^{10} \, 4s^1$ ): Copper shows +1 and +2 oxidation states. The $\mathrm{Cu}^{+}$ ion has a stable $3d^{10}$ configuration. However, in aqueous solutions, the +2 state ( $\mathrm{Cu}^{2+}$ , $[\mathrm{Ar}] \, 3d^9$ ) is more stable than the +1 state. This is because the much higher hydration enthalpy of the smaller, more highly charged $\mathrm{Cu}^{2+}$ ion compensates for the higher second ionization enthalpy required to form it. This example shows that while electronic configuration is a key determinant, thermodynamic factors related to the chemical environment can sometimes override the stability suggested by configuration alone.

In conclusion, the stability of oxidation states is largely dictated by the drive to achieve

d^0

,

d^5

, or

d^{10}

configurations, but it is the net result of various thermodynamic factors, not just the electronic configuration in isolation.

Q4.5Exercises

What may be the stable oxidation state of the transition element with the following $d$ electron configurations in the ground state of their atoms : $3d^{3}$ , $3d^{5}$ , $3d^{8}$ and $3d^{4}$ ?

Solution

The stable oxidation state of a transition element is often related to the electronic configuration that results in an empty (

d^0

), half-filled (

d^5

), or completely filled (

d^{10}

)

d

-subshell. We identify the element from its ground state d-electron configuration and then predict its stable oxidation state(s).

$3d^3$ configuration:
- An atom with a $3d^3$ configuration in its ground state, assuming a filled $4s$ orbital, would be Vanadium (V), with the configuration $[\mathrm{Ar}] \, 3d^3 \, 4s^2$ . It has a total of 5 valence electrons.
- Vanadium can lose all 5 valence electrons to achieve the +5 oxidation state, which gives it a stable noble gas configuration ( $[\mathrm{Ar}]$ , $d^0$ ). This state is stable, as seen in compounds like $\mathrm{V}_2\mathrm{O}_5$ .
- The +4 state (in $\mathrm{VO}^{2+}$ ) and +3 state are also known.
- Stable oxidation state: +5.
$3d^5$ configuration:
- This configuration can correspond to two elements in their ground state: Chromium ( $\[\mathrm{Ar}] \, 3d^5 \, 4s^1$ ) or Manganese ( $\[\mathrm{Ar}] \, 3d^5 \, 4s^2$ ).
- For Chromium (Cr), the most stable oxidation state is +3. The $\mathrm{Cr}^{3+}$ ion has a $3d^3$ configuration, which is stable due to crystal field effects (explained in coordination chemistry). The +6 state ( $d^0$ ) is strongly oxidizing.
- For Manganese (Mn), losing the two $4s$ electrons results in the +2 oxidation state. The $\mathrm{Mn}^{2+}$ ion has a $3d^5$ configuration, which is a very stable half-filled subshell.
- Stable oxidation states: +3 for Cr, and +2 for Mn.
$3d^8$ configuration:
- An atom with a $3d^8$ configuration in its ground state is Nickel (Ni), with the configuration $[\mathrm{Ar}] \, 3d^8 \, 4s^2$ . It has 10 valence electrons.
- Nickel readily loses its two $4s$ electrons to form the $\mathrm{Ni}^{2+}$ ion. The $\mathrm{Ni}^{2+}$ ion has a $3d^8$ configuration. This is the most common and stable oxidation state for nickel. Higher oxidation states are rare and require strong oxidizing agents.
- Stable oxidation state: +2.
$3d^4$ configuration:
- No element in the first transition series has a ground state configuration of $[\mathrm{Ar}] \, 3d^4 \, 4s^2$ . The element with four $3d$ electrons is Chromium (Cr), but its actual ground state configuration is $[\mathrm{Ar}] \, 3d^5 \, 4s^1$ due to the stability of the half-filled $d$ -subshell.
- Assuming the question refers to Chromium, as discussed for the $3d^5$ case, its most stable oxidation state is +3. In the +3 state, $\mathrm{Cr}^{3+}$ has a $3d^3$ configuration, which is stable.
- If we consider the loss of electrons from a hypothetical $[\mathrm{Ar}] \, 3d^4 \, 4s^2$ atom, it could show +2 ( $3d^4$ ), +3 ( $3d^3$ ), and +6 ( $d^0$ ) states. Among these, the +3 and +6 states would be significant due to stability associated with the resulting electronic configurations.
- Stable oxidation state (for Chromium): +3.

Q4.6Exercises

Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Solution

The oxometal anions of the first series of transition metals where the metal exhibits an oxidation state equal to its group number are:

Vanadate ion ( $VO_3^-$ or $VO_4^{3-}$ ): Vanadium (V) is in Group 5. In the vanadate ion, for example $VO_4^{3-}$ , if the oxidation state of V is $x$ , then $x + 4(-2) = -3$ , which gives $x = +5$ . This oxidation state is equal to its group number.
Chromate ion ( $CrO_4^{2-}$ ): Chromium (Cr) is in Group 6. In the chromate ion, if the oxidation state of Cr is $x$ , then $x + 4(-2) = -2$ , which gives $x = +6$ . This is equal to its group number.
Dichromate ion ( $Cr_2O_7^{2-}$ ): Chromium (Cr) is in Group 6. In the dichromate ion, if the oxidation state of Cr is $x$ , then $2x + 7(-2) = -2$ , which gives $2x = 12$ or $x = +6$ . This is equal to its group number.
Permanganate ion ( $MnO_4^-$ ): Manganese (Mn) is in Group 7. In the permanganate ion, if the oxidation state of Mn is $x$ , then $x + 4(-2) = -1$ , which gives $x = +7$ . This is equal to its group number.

Q4.7Exercises

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Solution

Lanthanoid Contraction

The lanthanoid contraction is the steady and regular decrease in the atomic and ionic radii of the lanthanoid elements with increasing atomic number (from Lanthanum, La, to Lutetium, Lu).

Cause: As we move across the lanthanoid series, for each successive element, the nuclear charge increases by one unit, and one electron is added to the same inner

4f

subshell. The

4f

orbitals have a very diffuse shape and, as a result, they have a poor shielding effect. They are ineffective at shielding the outer shell electrons (in

6s

and

5d

orbitals) from the increasing positive charge of the nucleus. Consequently, the effective nuclear charge experienced by the outer electrons increases, causing the electron cloud to be pulled more strongly towards the nucleus. This results in a contraction of the atomic and ionic size.

Consequences of Lanthanoid Contraction

Similarity in Radii of 4d and 5d Series Elements: The atomic radii of the elements of the second transition series (4d) are almost identical to the corresponding elements of the third transition series (5d) that come after the lanthanoids. For example, the atomic radius of Zirconium (Zr, 4d series) is 160 pm, which is almost the same as that of Hafnium (Hf, 5d series) at 159 pm. This similarity in size leads to very similar chemical properties, making the separation of these elements extremely difficult.
Difficulty in Separation of Lanthanoids: Due to the very small change in radii between adjacent lanthanoid elements, their chemical properties are very similar. This makes it very difficult to separate them from each other in their pure form.
Basicity of Hydroxides: The basic character of the hydroxides of lanthanoids decreases from Lanthanum hydroxide, $La(OH)_3$ , to Lutetium hydroxide, $Lu(OH)_3$ . As the size of the $Ln^{3+}$ ion decreases across the series, the covalent character of the $Ln-OH$ bond increases according to Fajan's rules. This strengthens the bond, making the release of $OH^-$ ions more difficult, thus decreasing the basicity.

Q4.8Exercises

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?

Solution

Characteristics of Transition Elements

Transition elements are the d-block elements that have characteristic properties due to their partially filled d-orbitals. These include:

Variable Oxidation States: They show multiple oxidation states because of the small energy difference between the $(n-1)d$ and $ns$ orbitals. Both $(n-1)d$ and $ns$ electrons can participate in bonding.
Formation of Coloured Ions: Most of their compounds are coloured in the solid or aqueous state. This is due to the presence of unpaired d-electrons which can undergo d-d transitions by absorbing light from the visible region and radiating the complementary colour.
Formation of Complex Compounds: They have a strong tendency to form coordination complexes. This is due to their small size, high ionic charge, and the availability of vacant d-orbitals to accept lone pairs of electrons from ligands.
Catalytic Properties: They and their compounds are widely used as catalysts (e.g., Vanadium(V) oxide in the Contact process). This is attributed to their ability to exhibit variable oxidation states and form reaction intermediates.
Paramagnetic Behaviour: Most transition metal compounds are paramagnetic because of the presence of one or more unpaired electrons in their $(n-1)d$ orbitals.
Metallic Character: They are all metals, generally hard, with high melting points, boiling points, and densities. This is due to strong metallic bonding involving both $ns$ and $(n-1)d$ electrons.

Why are they called Transition Elements?

They are called transition elements because they are located in the periodic table between the s-block and p-block elements. Their properties are transitional between the highly reactive electropositive metals of the s-block and the less reactive, mostly non-metallic elements of the p-block. According to IUPAC, a transition element is defined as an element which has an incompletely filled d sub-shell in its ground state or in any one of its common oxidation states.

d-Block Elements Not Regarded as Transition Elements

The elements of Group 12—Zinc (Zn), Cadmium (Cd), and Mercury (Hg)—are considered d-block elements but not transition elements. This is because they have completely filled

d

-orbitals (

d^{10}

configuration) in their ground state as well as in their most common oxidation state (+2). For example, Zinc (atomic number 30) has the electronic configuration

[Ar] 3d^{10} 4s^2

. In its common

+2

oxidation state, the configuration is

[Ar] 3d^{10}

. Since the d-subshell is complete in both states, it does not meet the definition of a transition element.

Q4.9Exercises

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

Solution

The electronic configurations of transition elements are different from non-transition elements (s-block and p-block elements) primarily in terms of which electron shell is being filled.

Non-Transition Elements (s- and p-block): In non-transition elements, the last electron enters the outermost shell (the valence shell).

For s-block elements, the differentiating electron enters the $ns$ orbital. Their general valence shell configuration is $ns^{1-2}$ .
For p-block elements, the differentiating electron enters the $np$ orbital. Their general valence shell configuration is $ns^2 np^{1-6}$ . In both cases, all the inner shells, including the penultimate shell (the one before the outermost), are completely filled.

Transition Elements (d-block): In transition elements, the last electron enters the

d

orbital of the penultimate shell, i.e., the

(n-1)d

subshell. Their general electronic configuration is

(n-1)d^{1-10} ns^{0-2}

.

The key difference is: For non-transition elements, electrons are added to the outermost shell, while the inner shells are already complete. For transition elements, electrons are added to an inner shell (the penultimate d-subshell) after the outermost s-subshell has already been partially or completely filled. This results in transition elements having an incomplete penultimate shell, which is responsible for their characteristic properties.

Q4.10Exercises

What are the different oxidation states exhibited by the lanthanoids?

Solution

Lanthanoids exhibit different oxidation states, but one state is overwhelmingly the most common and stable.

1. Common Oxidation State (+3): The most characteristic and stable oxidation state for all the lanthanoid elements is +3. This is because the sum of the first three ionization enthalpies is low, and by losing three electrons (typically two from the

6s

orbital and one from the

5d

or

4f

orbital), the elements form the tripositive ion,

Ln^{3+}

. This

Ln^{3+}

state is energetically very stable for all lanthanoids.

2. Other Oxidation States (+2 and +4): Some lanthanoids also show +2 and +4 oxidation states. These oxidation states are considered anomalous and are generally observed when they lead to a particularly stable electronic configuration for the ion, such as:

an empty f-subshell ( $f^0$ )
a half-filled f-subshell ( $f^7$ )
a completely filled f-subshell ( $f^{14}$ )

Examples of +2 Oxidation State:

Europium ( $Eu^{2+}$ ): Europium (atomic no. 63) has the configuration $[Xe] 4f^7 6s^2$ . When it forms $Eu^{2+}$ , its configuration becomes $[Xe] 4f^7$ , which has a stable half-filled f-subshell. Thus, $Eu^{2+}$ is quite stable.
Ytterbium ( $Yb^{2+}$ ): Ytterbium (atomic no. 70) has the configuration $[Xe] 4f^{14} 6s^2$ . When it forms $Yb^{2+}$ , its configuration becomes $[Xe] 4f^{14}$ , which has a stable completely filled f-subshell.

Examples of +4 Oxidation State:

Cerium ( $Ce^{4+}$ ): Cerium (atomic no. 58) has the configuration $[Xe] 4f^1 5d^1 6s^2$ . When it forms $Ce^{4+}$ , it loses four electrons to achieve the noble gas configuration of Xenon ( $[Xe]$ or $4f^0$ ). This makes $Ce^{4+}$ very stable. It is a strong oxidizing agent as it can readily revert to the common +3 state.
Terbium ( $Tb^{4+}$ ): Terbium (atomic no. 65) has the configuration $[Xe] 4f^9 6s^2$ . When it forms $Tb^{4+}$ , its configuration becomes $[Xe] 4f^7$ , which has a stable half-filled f-subshell.

In summary, while +3 is the dominant oxidation state for all lanthanoids, +2 and +4 states are also exhibited by some elements due to the extra stability of

f^0

,

f^7

, and

f^{14}

configurations.

Q4.11Exercises

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.

Solution

(i)

Paramagnetic behaviour: Transition metals and their compounds exhibit paramagnetic behaviour due to the presence of unpaired electrons in their d-orbitals. An atom, ion, or molecule with one or more unpaired electrons is attracted by a magnetic field, and this property is known as paramagnetism. The magnetic moment arises from the spin and orbital angular momentum of these unpaired electrons. For the first transition series, the contribution of the orbital angular momentum is effectively quenched, so the magnetic moment is determined by the number of unpaired electrons and is calculated using the 'spin-only' formula:

\mu = \sqrt{n(n+2)}

where

\mu

is the magnetic moment in Bohr magnetons (BM) and

n

is the number of unpaired electrons. For example, in

\text{Ti}^{3+}

(electronic configuration:

[\text{Ar}] 3d^1

), there is one unpaired electron, so it is paramagnetic. In contrast,

\text{Sc}^{3+}

(electronic configuration:

[\text{Ar}] 3d^0

) has no unpaired electrons and is diamagnetic.

(ii)

High enthalpies of atomisation: The enthalpy of atomisation is the energy required to break the metallic bonds in one mole of a solid metal to form gaseous atoms. Transition metals have high enthalpies of atomisation due to the presence of a large number of unpaired electrons in their

(n-1)d

and

ns

orbitals. These electrons participate in strong interatomic metallic bonding. The more unpaired electrons there are, the stronger the metallic bond and the higher the enthalpy of atomisation. This is why the values are high for elements in the middle of each series (e.g., V, Cr, Mo) and lower at the ends (e.g., Zn, Cd, Hg), where there are fewer or no unpaired d-electrons.

(iii)

Formation of coloured compounds: Most of the compounds of transition metals are coloured in the solid or solution state. This is due to the absorption of radiation from the visible light region to promote an electron from a lower energy d-orbital to a higher energy d-orbital within the same d-subshell. This phenomenon is known as a d-d transition. In the presence of ligands, the d-orbitals of the transition metal ion split into two sets of different energies. When light falls on the compound, an electron can jump from the lower energy set to the higher energy set. The energy of this transition corresponds to a specific wavelength (and thus colour) of light. The compound appears to have the complementary colour of the light absorbed. For example, a solution of

[\text{Cu(H}_2\text{O)}_6]^{2+}

is blue because it absorbs orange-red light for the d-d transition. Ions with an empty d-orbital (e.g.,

\text{Sc}^{3+}

,

d^0

) or a completely filled d-orbital (e.g.,

\text{Zn}^{2+}

,

d^{10}

) are colourless as d-d transitions are not possible.

(iv)

Catalytic activity: Transition metals and their compounds are known for their catalytic activity. This is attributed to two main reasons:

Variable Oxidation States: Transition metals can exhibit a wide range of oxidation states. This allows them to form unstable intermediate compounds with reactants, providing a new reaction path with lower activation energy. The transition metal can then easily change its oxidation state to release the products and regenerate the original catalyst. For example, iron(III) catalyses the reaction between iodide and persulphate ions.
Large Surface Area: In their finely divided state, transition metals provide a large surface area for reactants to be adsorbed. This increases the concentration of reactants on the catalyst surface, weakening the bonds in the reacting molecules and facilitating the reaction. For example, iron is used as a catalyst in the Haber process for ammonia synthesis, and nickel is used for hydrogenation reactions.

Q4.12Exercises

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

Solution

Interstitial Compounds: Interstitial compounds are non-stoichiometric compounds formed when small non-metal atoms, such as hydrogen (H), boron (B), carbon (C), or nitrogen (N), are trapped in the interstitial voids (empty spaces) of the crystal lattice of a metal, particularly transition metals. These compounds are not strictly chemical compounds as they do not follow the laws of constant proportion.

Why are such compounds well known for transition metals? Transition metals are well known for forming interstitial compounds because their crystal structures have empty spaces or voids of appropriate size to accommodate these small atoms. The close-packed structures (like cubic close-packed and hexagonal close-packed) common in transition metals have tetrahedral and octahedral voids. The small size of atoms like H, C, and N allows them to fit into these voids without causing significant distortion of the metal lattice.

Properties of Interstitial Compounds:

Hardness: They are very hard and rigid because the small atoms occupy the interstitial sites and form strong bonds with the metal atoms. For example, steel and cast iron are hard because of the carbon atoms in the iron lattice.
High Melting Points: They have higher melting points than the pure metals.
Chemical Inertness: They are chemically inert.
Conductivity: They retain metallic conductivity.

Examples of interstitial compounds include TiC, Mn₄N, Fe₃H, and VH₀.₅₆.

Q4.13Exercises

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

Solution

The variability in oxidation states of transition metals is different from that of non-transition metals (p-block elements) in two main aspects: the range of states and the difference between successive states.

Transition Metals (d-block elements): Transition metals exhibit a wide variety of oxidation states in their compounds. This is due to the very small energy difference between the

(n-1)d

and

ns

orbitals. As a result, electrons from both subshells can participate in bond formation. The oxidation states typically differ from each other by one unit.

Example: Manganese (Mn, Atomic No. 25) has the electronic configuration $[\text{Ar}] 3d^5 4s^2$ $[Ar] 3 d^{5} 4 s^{2}$ . By losing electrons from the $4s$ $4 s$ and $3d$ $3 d$ orbitals, it can show all oxidation states from +2 to +7.
- MnCl₂ (+2)
- Mn₂O₃ (+3)
- MnO₂ (+4)
- K₃[Mn(CN)₆] (+3), but also +5 is known
- MnO₄²⁻ (+6)
- MnO₄⁻ (+7)

Non-transition Metals (mainly p-block elements): Non-transition elements also show variable oxidation states, but the variation is not as extensive as in transition metals. The successive oxidation states generally differ by two units, not one. This is often explained by the inert pair effect, especially for heavier elements in the p-block. The inert pair effect is the reluctance of the

ns^2

pair of electrons to participate in bonding.

Example: Lead (Pb, Atomic No. 82) belongs to Group 14 and has the electronic configuration $[\text{Xe}] 4f^{14} 5d^{10} 6s^2 6p^2$ $[Xe] 4 f^{14} 5 d^{10} 6 s^{2} 6 p^{2}$ . It can show +2 and +4 oxidation states.
- PbO, PbCl₂ (+2, by losing two 6p electrons)
- PbO₂, PbCl₄ (+4, by losing two 6p and two 6s electrons) The +2 state is more stable than the +4 state for lead due to the inert pair effect. The difference between the two common oxidation states is $4 - 2 = 2$ . Other examples include Sn (+2, +4) and Tl (+1, +3).

Summary of Differences:

Feature	Transition Metals	Non-transition Metals
Reason	Small energy gap between $(n-1)d$ and $ns$ orbitals.	Inert pair effect; involvement of p and s electrons.
Successive States	Differ by one unit (e.g., +2, +3, +4).	Differ by two units (e.g., +2, +4 or +3, +5).
Range	Wide range of oxidation states.	Limited number of oxidation states.

Q4.14Exercises

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Solution

Preparation of Potassium Dichromate (K₂Cr₂O₇) from Iron Chromite Ore (FeCr₂O₄)

The preparation involves the following three steps:

Step 1: Conversion of chromite ore to sodium chromate. The finely powdered iron chromite ore (

FeCr_2O_4

) is fused with sodium carbonate (

Na_2CO_3

) in the presence of a free supply of air in a reverberatory furnace. This process oxidizes chromium(III) in the ore to chromium(VI) in sodium chromate.

4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2

The resulting mass is extracted with water, which dissolves the soluble sodium chromate (

Na_2CrO_4

), leaving behind the insoluble ferric oxide (

Fe_2O_3

).

Step 2: Conversion of sodium chromate to sodium dichromate. The yellow solution of sodium chromate is filtered and then acidified with concentrated sulphuric acid. This converts the chromate ions into dichromate ions.

2\text{Na}_2\text{CrO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}

On concentration, the less soluble sodium sulphate (

Na_2SO_4 \cdot 10H_2O

) crystallizes out and is removed by filtration. The solution now contains sodium dichromate.

Step 3: Conversion of sodium dichromate to potassium dichromate. A hot, concentrated solution of sodium dichromate is treated with a calculated amount of potassium chloride (KCl). Potassium dichromate is less soluble than sodium dichromate in water. Therefore, on cooling the solution, crystals of orange-red potassium dichromate (

K_2Cr_2O_7

) separate out, while the more soluble sodium chloride (NaCl) remains in the solution.

\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}

The crystals of potassium dichromate are then separated by filtration and purified by recrystallization.

Effect of increasing pH on a solution of potassium dichromate: In aqueous solution, chromate (

CrO_4^{2-}

) and dichromate (

Cr_2O_7^{2-}

) ions exist in equilibrium, which is pH-dependent. The dichromate ion (orange) is stable in acidic solutions, while the chromate ion (yellow) is stable in alkaline (basic) solutions.

If the pH of a potassium dichromate solution is increased (i.e., an alkali like NaOH or KOH is added), the hydrogen ion concentration decreases. According to Le Chatelier's principle, the equilibrium will shift to the right to consume the added hydroxide ions. This converts the orange dichromate ions into yellow chromate ions.

The ionic equation for the reaction is:

\underset{\text{(Orange)}}{\text{Cr}_2\text{O}_7^{2-}} (aq) + 2\text{OH}^- (aq) \rightleftharpoons \underset{\text{(Yellow)}}{2\text{CrO}_4^{2-}} (aq) + \text{H}_2\text{O} (l)

Therefore, increasing the pH of a potassium dichromate solution will cause the colour to change from orange to yellow.

Q4.15Exercises

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i)

iodide

(ii)

iron(II) solution and

(iii)

$\mathrm{H}_{2} \mathrm{~S}$

Solution

Potassium dichromate (

K_2Cr_2O_7

) is a powerful oxidising agent, especially in acidic medium. In the presence of an acid, the dichromate ion (

Cr_2O_7^{2-}

) takes up electrons and is reduced to the green chromium(III) ion (

Cr^{3+}

). The oxidation state of chromium changes from +6 to +3.

The half-reaction for this reduction is:

Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \quad (E^\circ = +1.33 \text{ V})

The ionic equations for its reaction with the given species are as follows:

(i) Reaction with iodide (I⁻): Potassium dichromate oxidises iodide ions to iodine (

I_2

). The solution turns brown due to the formation of iodine.

Oxidation half-reaction: $2I^- \rightarrow I_2 + 2e^-$ To balance the electrons, this equation is multiplied by 3: $6I^- \rightarrow 3I_2 + 6e^-$
Reduction half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Overall ionic equation: Combining the two half-reactions gives: $Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$

(ii) Reaction with iron(II) solution (Fe²⁺): Potassium dichromate oxidises green iron(II) ions to yellow iron(III) ions (

Fe^{3+}

).

Oxidation half-reaction: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ To balance the electrons, this equation is multiplied by 6: $6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$
Reduction half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Overall ionic equation: Combining the two half-reactions gives: $Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$

(iii) Reaction with H₂S: Potassium dichromate oxidises hydrogen sulphide (

H_2S

) to elemental sulphur (S), which appears as a yellow precipitate. The oxidation state of sulphur changes from -2 to 0.

Oxidation half-reaction: $H_2S \rightarrow S + 2H^+ + 2e^-$ To balance the electrons, this equation is multiplied by 3: $3H_2S \rightarrow 3S + 6H^+ + 6e^-$
Reduction half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Overall ionic equation: Combining the two half-reactions gives: $Cr_2O_7^{2-} + 14H^+ + 3H_2S \rightarrow 2Cr^{3+} + 3S + 6H^+ + 7H_2O$ After cancelling the common $H^+$ ions (6 $H^+$ from both sides), the final balanced equation is: $Cr_2O_7^{2-} + 8H^+ + 3H_2S \rightarrow 2Cr^{3+} + 3S + 7H_2O$

Q4.16Exercises

Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) $\mathrm{SO}_{2}$ and (iii) oxalic acid? Write the ionic equations for the reactions.

Solution

Preparation of Potassium Permanganate ( $KMnO_4$ )

Potassium permanganate is prepared commercially from pyrolusite ore (

MnO_2

). The preparation involves two main steps:

Step 1: Conversion of $MnO_2$ to potassium manganate ( $K_2MnO_4$ ) Pyrolusite ore is fused with an alkali, like potassium hydroxide (

KOH

), in the presence of an oxidizing agent like atmospheric oxygen (

O_2

) or potassium nitrate (

KNO_3

). This process yields green potassium manganate.

2MnO_2 + 4KOH + O_2 \xrightarrow{\Delta} 2K_2MnO_4 + 2H_2O

Step 2: Oxidation of potassium manganate to potassium permanganate The potassium manganate is then oxidized to potassium permanganate. This can be done in two ways:

(a) Electrolytic Oxidation: In this method, an aqueous solution of potassium manganate is electrolyzed. The manganate ions are oxidized to permanganate ions at the anode.

MnO_4^{2-} \rightarrow MnO_4^- + e^-

(Manganate) (Permanganate)

(b) Chemical Oxidation: The manganate solution is treated with an oxidizing agent like chlorine (

Cl_2

) or by passing carbon dioxide (

CO_2

) through the solution. The disproportionation of manganate in a neutral or acidic medium also yields permanganate.

Using chlorine:

2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KCl

Using carbon dioxide (which provides a weakly acidic medium):

3K_2MnO_4 + 4CO_2 + 2H_2O \rightarrow 2KMnO_4 + MnO_2 + 4KHCO_3

Reactions of Acidified Permanganate Solution

Acidified potassium permanganate acts as a strong oxidizing agent. The permanganate ion (

MnO_4^-

) is reduced to

Mn^{2+}

ion. The ionic half-equation for the reduction is:

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O

(i) Reaction with iron(II) ions ( $Fe^{2+}$ ): Acidified permanganate solution oxidizes green iron(II) ions to yellow iron(III) ions.

Oxidation half-reaction: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
To balance the electrons, we multiply the oxidation half-reaction by 5.
Overall ionic equation: $MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

(ii) Reaction with sulfur dioxide ( $SO_2$ ): Acidified permanganate solution oxidizes sulfur dioxide (which forms sulfurous acid in water) to sulfuric acid (or sulfate ions).

Oxidation half-reaction: $SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^-$
To balance the electrons, we multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5.
Overall ionic equation: $2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+$

(iii) Reaction with oxalic acid ( $H_2C_2O_4$ ): Acidified permanganate solution oxidizes oxalic acid to carbon dioxide.

Oxidation half-reaction: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
To balance the electrons, we multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5.
Overall ionic equation: $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$

Q4.17Exercises

For $\mathrm{M}^{2+} / \mathrm{M}$ and $\mathrm{M}^{3+} / \mathrm{M}^{2+}$ systems the $E^{\mathrm{o}}$ values for some metals are as follows: $\mathrm{Cr}^{2+} / \mathrm{Cr}$ -0.9 V $\mathrm{Cr}^{3} / \mathrm{Cr}^{2+}$ -0.4 V $\mathrm{Mn}^{2+} / \mathrm{Mn}$ -1.2 V $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$ +1.5 V $\mathrm{Fe}^{2+} / \mathrm{Fe}$ -0.4 V $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ +0.8 V Use this data to comment upon:

(i)

the stability of $\mathrm{Fe}^{3+}$ in acid solution as compared to that of $\mathrm{Cr}^{3+}$ or $\mathrm{Mn}^{3+}$ and

(ii)

the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Solution

The given standard electrode potential (

E^o

) values are:

$E^o(Cr^{2+}/Cr) = -0.9 \text{ V}$
$E^o(Cr^{3+}/Cr^{2+}) = -0.4 \text{ V}$
$E^o(Mn^{2+}/Mn) = -1.2 \text{ V}$
$E^o(Mn^{3+}/Mn^{2+}) = +1.5 \text{ V}$
$E^o(Fe^{2+}/Fe) = -0.4 \text{ V}$
$E^o(Fe^{3+}/Fe^{2+}) = +0.8 \text{ V}$

(i) Stability of $Fe^{3+}$ in acid solution as compared to that of $Cr^{3+}$ or $Mn^{3+}$

The stability of these ions in their +3 oxidation state can be compared using the

E^o(M^{3+}/M^{2+})

values. A more positive

E^o

value indicates that the

M^{3+}

ion is more easily reduced to the

M^{2+}

ion, and hence, is less stable.

For Chromium: $E^o(Cr^{3+}/Cr^{2+}) = -0.4 \text{ V}$ . The negative value indicates that $Cr^{3+}$ is very stable and is not easily reduced. This stability is due to the half-filled $t_{2g}$ level in its electronic configuration ( $[Ar]3d^3$ ).
For Manganese: $E^o(Mn^{3+}/Mn^{2+}) = +1.5 \text{ V}$ . The large positive value shows that $Mn^{3+}$ is a strong oxidizing agent and is readily reduced to $Mn^{2+}$ . This is because $Mn^{2+}$ has a very stable half-filled d-orbital configuration ( $[Ar]3d^5$ ). Therefore, $Mn^{3+}$ is highly unstable.
For Iron: $E^o(Fe^{3+}/Fe^{2+}) = +0.8 \text{ V}$ . The positive value indicates that $Fe^{3+}$ can be reduced to $Fe^{2+}$ . $Fe^{3+}$ has a stable half-filled d-orbital configuration ( $[Ar]3d^5$ ), making it more stable than $Mn^{3+}$ . However, it is less stable than $Cr^{3+}$ .

Comparing the reduction potentials for the

M^{3+}/M^{2+}

couple:

Mn^{3+}/Mn^{2+}

(+1.5 V) >

Fe^{3+}/Fe^{2+}

(+0.8 V) >

Cr^{3+}/Cr^{2+}

(-0.4 V).

Conclusion: The order of stability for the +3 ions is

Cr^{3+} > Fe^{3+} > Mn^{3+}

. Thus,

Fe^{3+}

is more stable than

Mn^{3+}

but less stable than

Cr^{3+}

in an acidic solution.

(ii) The ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

The ease of oxidation of a metal M to

M^{2+}

is determined by its standard oxidation potential,

E^o(M/M^{2+})

, which is equal to

-E^o(M^{2+}/M)

. A higher (more positive) oxidation potential indicates that the metal is more easily oxidized.

Oxidation of Chromium: $E^o(Cr/Cr^{2+}) = -E^o(Cr^{2+}/Cr) = -(-0.9 \text{ V}) = +0.9 \text{ V}$
Oxidation of Manganese: $E^o(Mn/Mn^{2+}) = -E^o(Mn^{2+}/Mn) = -(-1.2 \text{ V}) = +1.2 \text{ V}$
Oxidation of Iron: $E^o(Fe/Fe^{2+}) = -E^o(Fe^{2+}/Fe) = -(-0.4 \text{ V}) = +0.4 \text{ V}$

Comparing the oxidation potentials: Manganese (+1.2 V) > Chromium (+0.9 V) > Iron (+0.4 V).

Conclusion: The ease of oxidation follows the order:

Mn > Cr > Fe

. Therefore, iron is the most difficult to oxidize among the three metals.

Q4.18Exercises

Predict which of the following will be coloured in aqueous solution? $\mathrm{Ti}^{3+}, \mathrm{V}^{3+}$ , $\mathrm{Cu}^{+}, \mathrm{Sc}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{Co}^{2+}$ . Give reasons for each.

Solution

The color of transition metal ions in an aqueous solution is typically due to the absorption of light in the visible region, which promotes an electron from a lower energy d-orbital to a higher energy d-orbital. This phenomenon is known as a d-d transition. For a d-d transition to be possible, the d-subshell of the metal ion must be partially filled (i.e., contain 1 to 9 electrons). Ions with an empty (

d^0

) or completely filled (

d^{10}

) d-subshell are generally colorless.

Let's examine the electronic configuration of each ion:

$Ti^{3+}$ : Titanium ( $Z=22$ ) has the electronic configuration $[Ar]3d^24s^2$ . $Ti^{3+}$ has the configuration $[Ar]3d^1$ . Since it has one unpaired electron in the d-orbital, it can undergo d-d transition. Therefore, $Ti^{3+}$ will be colored (purple).
$V^{3+}$ : Vanadium ( $Z=23$ ) has the electronic configuration $[Ar]3d^34s^2$ . $V^{3+}$ has the configuration $[Ar]3d^2$ . It has two unpaired electrons in the d-orbitals, allowing for d-d transitions. Therefore, $V^{3+}$ will be colored (green).
$Cu^{+}$ : Copper ( $Z=29$ ) has the electronic configuration $[Ar]3d^{10}4s^1$ . $Cu^{+}$ has the configuration $[Ar]3d^{10}$ . Its d-subshell is completely filled. No d-d transition is possible. Therefore, $Cu^{+}$ will be colorless.
$Sc^{3+}$ : Scandium ( $Z=21$ ) has the electronic configuration $[Ar]3d^14s^2$ . $Sc^{3+}$ has the configuration $[Ar]3d^0$ . Its d-subshell is empty. No d-d transition is possible. Therefore, $Sc^{3+}$ will be colorless.
$Mn^{2+}$ : Manganese ( $Z=25$ ) has the electronic configuration $[Ar]3d^54s^2$ . $Mn^{2+}$ has the configuration $[Ar]3d^5$ . The d-subshell is half-filled with five unpaired electrons. It can undergo d-d transitions. Therefore, $Mn^{2+}$ will be colored (pale pink).
$Fe^{3+}$ : Iron ( $Z=26$ ) has the electronic configuration $[Ar]3d^64s^2$ . $Fe^{3+}$ has the configuration $[Ar]3d^5$ . The d-subshell is half-filled with five unpaired electrons, allowing for d-d transitions. Therefore, $Fe^{3+}$ will be colored (yellow).
$Co^{2+}$ : Cobalt ( $Z=27$ ) has the electronic configuration $[Ar]3d^74s^2$ . $Co^{2+}$ has the configuration $[Ar]3d^7$ . It has three unpaired electrons in its partially filled d-subshell. It can undergo d-d transitions. Therefore, $Co^{2+}$ will be colored (pink).

Summary:

Colored ions: $Ti^{3+} (3d^1)$ , $V^{3+} (3d^2)$ , $Mn^{2+} (3d^5)$ , $Fe^{3+} (3d^5)$ , $Co^{2+} (3d^7)$ .
Colorless ions: $Sc^{3+} (3d^0)$ , $Cu^{+} (3d^{10})$ .

Q4.19Exercises

Compare the stability of +2 oxidation state for the elements of the first transition series.

Solution

The stability of the +2 oxidation state for the elements of the first transition series (Scandium to Zinc) varies across the period and is influenced by several factors, including ionization enthalpy, electronic configuration, and hydration enthalpy.

General Trend: With the exception of Scandium, all elements of the first transition series exhibit a +2 oxidation state. Scandium does not form a +2 ion because achieving the +3 state by losing its three valence electrons ( $3d^14s^2$ ) results in a very stable noble gas configuration ( $[Ar]$ ).
Ionization Enthalpies: The stability of the +2 state can be related to the sum of the first and second ionization enthalpies ( $IE_1 + IE_2$ ). As we move from left to right across the series, the effective nuclear charge increases, and the atomic size decreases. This generally leads to an increase in the values of ( $IE_1 + IE_2$ ), making the formation of $M^{2+}$ ions more difficult. However, this trend is not smooth.
Stability due to Electronic Configuration: The stability of the resulting $M^{2+}$ ion's electronic configuration plays a crucial role.
- Manganese ( $Mn^{2+}$ ): Manganese shows a prominent and stable +2 oxidation state. The $Mn^{2+}$ ion has the electronic configuration $[Ar]3d^5$ , which is a half-filled and exceptionally stable configuration. This stability is reflected in its high third ionization enthalpy.
- Zinc ( $Zn^{2+}$ ): Zinc almost exclusively shows the +2 oxidation state. The $Zn^{2+}$ ion has the configuration $[Ar]3d^{10}$ , which is a completely filled and very stable configuration. Removing a third electron would require a very large amount of energy.
- Chromium ( $Cr^{2+}$ ) and Iron ( $Fe^{2+}$ ): The $Cr^{2+}$ ion ( $[Ar]3d^4$ ) is a strong reducing agent as it readily loses an electron to form $Cr^{3+}$ ( $[Ar]3d^3$ ), which has a stable half-filled $t_{2g}$ level in octahedral complexes. The $Fe^{2+}$ ion ( $[Ar]3d^6$ ) is also easily oxidized to $Fe^{3+}$ ( $[Ar]3d^5$ ), which has a more stable half-filled d-subshell.
Standard Electrode Potentials ( $E^o(M^{2+}/M)$ ): The values of the standard electrode potentials for the $M^{2+}/M$ couple give an indication of the stability of the +2 state. The values generally become less negative across the series, indicating a decrease in the tendency to form $M^{2+}$ ions from the metal. However, the values for Mn and Zn are more negative than expected from the general trend. This is due to the extra stability of the half-filled ( $d^5$ ) configuration in $Mn^{2+}$ and the completely filled ( $d^{10}$ ) configuration in $Zn^{2+}$ .
Copper ( $Cu^{2+}$ ): Although forming $Cu^{2+}$ ( $[Ar]3d^9$ ) requires more energy (higher second ionization enthalpy) than forming $Cu^{+}$ ( $[Ar]3d^{10}$ ), the $Cu^{2+}$ ion is more stable than $Cu^{+}$ in aqueous solutions. This is because the much higher hydration enthalpy of $Cu^{2+}(aq)$ compared to $Cu^{+}(aq)$ compensates for the high second ionization enthalpy of copper.

Conclusion: The stability of the +2 oxidation state increases generally from left to right, but with significant variations.

Mn^{2+}

and

Zn^{2+}

are particularly stable due to their

d^5

and

d^{10}

configurations. The stability of

Cu^{2+}

in solution is a notable case governed by hydration enthalpy. Towards the left of the series (e.g., Ti, V), higher oxidation states are more stable, while towards the right (e.g., Ni, Cu, Zn), the +2 state becomes the most common and stable oxidation state.

Q4.20Exercises

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(i)

electronic configuration

(ii)

atomic and ionic sizes and

(iii)

oxidation state

(iv)

chemical reactivity.

Solution

The chemistry of actinoids is compared with that of lanthanoids with reference to the following points:

(i) Electronic Configuration

Lanthanoids: The general electronic configuration is $[Xe]4f^{1-14}5d^{0-1}6s^2$ . The differentiating electron enters the $4f$ subshell. The energy of the $4f$ orbital is significantly lower than that of the $5d$ orbital.
Actinoids: The general electronic configuration is $[Rn]5f^{1-14}6d^{0-1}7s^2$ . The differentiating electron enters the $5f$ subshell. In the beginning of the series (up to Uranium), the energies of the $5f$ , $6d$ , and $7s$ orbitals are very comparable. Consequently, electrons can occupy any of these orbitals, leading to more complex electronic configurations.

(ii) Atomic and Ionic Sizes

Both lanthanoids and actinoids show a gradual decrease in atomic and ionic radii across their respective series. This is known as the lanthanoid contraction and actinoid contraction, respectively.
The cause is the same in both cases: the imperfect shielding of the nuclear charge by the inner f-electrons ( $4f$ for lanthanoids, $5f$ for actinoids). As the atomic number increases, the effective nuclear charge experienced by the valence electrons increases, pulling them closer to the nucleus.
The actinoid contraction is greater from element to element than the lanthanoid contraction. This is because the shielding effect of $5f$ electrons is poorer than that of $4f$ electrons, as the $5f$ orbitals are more extended in space.

(iii) Oxidation State

Lanthanoids: The most common and stable oxidation state for all lanthanoids is +3. Some elements also show +2 (e.g., $Eu^{2+}$ , $Yb^{2+}$ ) and +4 (e.g., $Ce^{4+}$ ) states, which are attained to acquire stable empty ( $f^0$ ), half-filled ( $f^7$ ), or completely filled ( $f^{14}$ ) configurations. The range of oxidation states is limited.
Actinoids: Actinoids exhibit a much wider range of oxidation states. While +3 is a common state, the earlier actinoids (e.g., Th, Pa, U, Np, Pu) show higher oxidation states, up to +7 for Np and Pu. This is because the $5f$ , $6d$ , and $7s$ orbitals are of comparable energies, allowing all their electrons to participate in bonding. For the later actinoids, the +3 oxidation state becomes more stable, similar to the lanthanoids.

(iv) Chemical Reactivity

Lanthanoids: The earlier members of the lanthanoid series are quite reactive, similar to calcium. They react with hot water, tarnish in air, and combine with non-metals at moderate temperatures. The reactivity decreases across the series as the atomic size decreases.
Actinoids: Actinoids are highly reactive metals, especially when finely divided. They react with boiling water or dilute acids to liberate hydrogen gas. They combine with most non-metals at moderate temperatures. The reactivity of actinoids is generally higher than that of lanthanoids.
A significant difference is that all actinoids are radioactive, whereas among lanthanoids, only promethium is radioactive. The radioactivity of actinoids adds to their chemical complexity and handling difficulties.

Q4.21Exercises

How would you account for the following:

(i)

Of the $d^{4}$ species, $\mathrm{Cr}^{2+}$ is strongly reducing while manganese(III) is strongly oxidising.

(ii)

Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.

(iii)

The $d^{1}$ configuration is very unstable in ions.

Solution

(i)

Cr $^{2+}$ is reducing while Mn $^{3+}$ is oxidising: Both Cr

^{2+}

and Mn

^{3+}

have a

d^4

electronic configuration.

Cr $^{2+}$ as a reducing agent: Cr $^{2+}$ readily loses an electron to form Cr $^{3+}$ . $\text{Cr}^{2+} \rightarrow \text{Cr}^{3+} + e^-$ The electronic configuration changes from $3d^4$ to $3d^3$ . The $3d^3$ configuration in an octahedral field corresponds to a half-filled $t_{2g}$ level ( $t_{2g}^3$ ). This is a particularly stable configuration. The tendency to achieve this stable state makes Cr $^{2+}$ a strong reducing agent. The standard electrode potential $E^\circ(\text{Cr}^{3+}/\text{Cr}^{2+}) = -0.41 \text{ V}$ also supports this, as the negative value indicates that oxidation of Cr $^{2+}$ is favoured.
Mn $^{3+}$ as an oxidising agent: Mn $^{3+}$ readily gains an electron to form Mn $^{2+}$ . $\text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+}$ The electronic configuration changes from $3d^4$ to $3d^5$ . The $3d^5$ configuration is a half-filled d-subshell, which is exceptionally stable due to maximum exchange energy. The tendency to achieve this highly stable state makes Mn $^{3+}$ a strong oxidising agent. The standard electrode potential $E^\circ(\text{Mn}^{3+}/\text{Mn}^{2+}) = +1.57 \text{ V}$ is large and positive, indicating that reduction of Mn $^{3+}$ is highly favoured.

(ii)

Stability of Cobalt(II): Cobalt(II) is stable in aqueous solution but is easily oxidised in the presence of complexing agents.

In aqueous solution, Co(II) exists as the hexaaqua complex, $[\text{Co(H}_2\text{O)}_6]^{2+}$ . The standard electrode potential for the oxidation to Co(III) is very high, $E^\circ(\text{Co}^{3+}/\text{Co}^{2+}) = +1.97 \text{ V}$ . This high positive value indicates that Co(II) is thermodynamically stable with respect to oxidation in water.
In the presence of strong-field ligands (complexing reagents) such as $\text{NH}_3$ or $\text{CN}^-$ , Co(II) is easily oxidised to Co(III). For example, $[\text{Co(NH}_3)_6]^{2+}$ is readily oxidised to $[\text{Co(NH}_3)_6]^{3+}$ . The reason for this is the large crystal field stabilisation energy (CFSE) of the Co(III) complex. Co(III) has a $d^6$ configuration. In the presence of strong-field ligands, it forms a low-spin octahedral complex with the electronic configuration $t_{2g}^6 e_g^0$ . This is a very stable arrangement. The energy released due to the high CFSE of the Co(III) complex is greater than the energy required for the third ionisation of cobalt, thus making the oxidation favourable.

(iii)

Instability of the $d^1$ configuration: The

d^1

configuration is very unstable in ions for several reasons:

Tendency to form $d^0$ configuration: Ions with a $d^1$ configuration have a strong tendency to lose their single d-electron to achieve a stable $d^0$ configuration, which corresponds to the noble gas configuration of the preceding period. For example, $\text{Ti}^{3+}$ ( $d^1$ ) is a good reducing agent as it readily oxidises to the more stable $\text{Ti}^{4+}$ ( $d^0$ ). Similarly, $\text{Sc}^{2+}$ ( $d^1$ ) is extremely unstable and immediately oxidises to $\text{Sc}^{3+}$ ( $d^0$ ).
Disproportionation: Some ions with a $d^1$ configuration may disproportionate into more stable oxidation states.
Redox activity in solution: In aqueous solution, ions with a $d^1$ configuration are often strong reducing agents, reacting with water or other species to lose their electron.

Q4.22Exercises

What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Solution

Disproportionation is a specific type of redox reaction in which an element in a particular oxidation state is simultaneously oxidised and reduced. This occurs when an intermediate oxidation state is unstable relative to a lower and a higher oxidation state.

Two examples of disproportionation reactions in aqueous solution are:

Manganate(VI) ion in acidic or neutral solution: The manganate ion, $\text{MnO}_4^{2-}$ , where manganese is in the +6 oxidation state, is unstable in acidic or neutral solutions and disproportionates to permanganate ion, $\text{MnO}_4^{-}$ (Mn in +7 state), and manganese dioxide, $\text{MnO}_2$ (Mn in +4 state). The reaction is: $3\text{MnO}_4^{2-}(\text{aq}) + 4\text{H}^+(\text{aq}) \rightarrow 2\text{MnO}_4^{-}(\text{aq}) + \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l})$ Here, Mn(+6) is oxidised to Mn(+7) and reduced to Mn(+4).
Copper(I) ion in aqueous solution: The copper(I) ion, $\text{Cu}^+$ , is unstable in aqueous solution and disproportionates to copper(II) ion, $\text{Cu}^{2+}$ , and solid copper metal, $\text{Cu}(s)$ . The reaction is: $2\text{Cu}^{+}(\text{aq}) \rightarrow \text{Cu}^{2+}(\text{aq}) + \text{Cu}(\text{s})$ Here, Cu(+1) is oxidised to Cu(+2) and reduced to Cu(0). The greater stability of $\text{Cu}^{2+}(\text{aq})$ compared to $\text{Cu}^{+}(\text{aq})$ is due to its much higher hydration enthalpy, which more than compensates for the second ionisation enthalpy of copper.

Q4.23Exercises

Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Solution

The metal in the first series of transition metals (3d series) that exhibits the +1 oxidation state most frequently is Copper (Cu).

Reason: The electronic configuration of copper is

[\text{Ar}] 3d^{10} 4s^1

. It can easily lose its single

4s

electron to form the

\text{Cu}^+

ion.

\text{Cu} ([\text{Ar}] 3d^{10} 4s^1) \rightarrow \text{Cu}^+ ([\text{Ar}] 3d^{10}) + e^-

The resulting

\text{Cu}^+

ion has an electronic configuration of

[\text{Ar}] 3d^{10}

. This configuration has a completely filled d-subshell, which provides extra stability. Although the

\text{Cu}^+

ion is unstable in aqueous solution and undergoes disproportionation, it is stable in many solid compounds (e.g.,

\text{Cu}_2\text{O}

,

\text{CuCl}

) and in complex ions. No other element in the first transition series forms a +1 ion with such a stable electronic configuration, which is why this oxidation state is most common for copper.

Q4.24Exercises

Calculate the number of unpaired electrons in the following gaseous ions: $\mathrm{Mn}^{3+}$ , $\mathrm{Cr}^{3+}, \mathrm{V}^{3+}$ and $\mathrm{Ti}^{3+}$ . Which one of these is the most stable in aqueous solution?

Solution

To calculate the number of unpaired electrons, we first write the electronic configuration of each gaseous ion.

Mn $^{3+}$ : The atomic number of Mn is 25. Its electronic configuration is $[\text{Ar}] 3d^5 4s^2$ . To form $\text{Mn}^{3+}$ , it loses two $4s$ electrons and one $3d$ electron. The configuration of $\text{Mn}^{3+}$ is $[\text{Ar}] 3d^4$ . The four $3d$ electrons will occupy four different orbitals with parallel spins. Number of unpaired electrons = 4.
Cr $^{3+}$ : The atomic number of Cr is 24. Its electronic configuration is $[\text{Ar}] 3d^5 4s^1$ . To form $\text{Cr}^{3+}$ , it loses one $4s$ electron and two $3d$ electrons. The configuration of $\text{Cr}^{3+}$ is $[\text{Ar}] 3d^3$ . The three $3d$ electrons will occupy three different orbitals with parallel spins. Number of unpaired electrons = 3.
V $^{3+}$ : The atomic number of V is 23. Its electronic configuration is $[\text{Ar}] 3d^3 4s^2$ . To form $\text{V}^{3+}$ , it loses two $4s$ electrons and one $3d$ electron. The configuration of $\text{V}^{3+}$ is $[\text{Ar}] 3d^2$ . The two $3d$ electrons will occupy two different orbitals with parallel spins. Number of unpaired electrons = 2.
Ti $^{3+}$ : The atomic number of Ti is 22. Its electronic configuration is $[\text{Ar}] 3d^2 4s^2$ . To form $\text{Ti}^{3+}$ , it loses two $4s$ electrons and one $3d$ electron. The configuration of $\text{Ti}^{3+}$ is $[\text{Ar}] 3d^1$ . There is only one electron in the $3d$ subshell. Number of unpaired electrons = 1.

Most stable ion in aqueous solution: Among these ions, Cr $^{3+}$ is the most stable in aqueous solution.

Reason: In an aqueous solution, the metal ions are surrounded by water molecules in an octahedral geometry. The d-orbitals split into two sets: a lower energy

t_{2g}

set and a higher energy

e_g

set. The electronic configuration of

\text{Cr}^{3+}

is

3d^3

. These three electrons occupy the lower energy

t_{2g}

orbitals (

t_{2g}^3

), resulting in a half-filled

t_{2g}

level. This configuration gives high crystal field stabilisation energy (CFSE) and is therefore particularly stable.

Q4.25Exercises

Give examples and suggest reasons for the following features of the transition metal chemistry:

(i)

The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.

(ii)

A transition metal exhibits highest oxidation state in oxides and fluorides.

(iii)

The highest oxidation state is exhibited in oxoanions of a metal.

Solution

(i)

The lowest oxide of a transition metal is basic, while the highest is amphoteric or acidic. This trend is explained by the oxidation state of the metal.

Lowest Oxide (Basic): In the lowest oxidation state (e.g., +2), the metal atom has a larger size and low positive charge density. The bond formed with oxygen is predominantly ionic. The oxide ion ( $O^{2-}$ ) is readily available to act as a base by accepting protons. Example: Manganese(II) oxide, $\text{MnO}$ (Mn is +2), is basic. It reacts with acid: $\text{MnO} + 2\text{HCl} \rightarrow \text{MnCl}_2 + \text{H}_2\text{O}$ .
Highest Oxide (Acidic): In the highest oxidation state (e.g., +7), the metal atom is small and has a very high positive charge density. It has high polarizing power, and the bond with oxygen becomes more covalent. The metal atom strongly attracts electrons, making the oxygen atoms electron-deficient. This allows the oxide to act as a Lewis acid or react with bases. Example: Manganese(VII) oxide, $\text{Mn}_2\text{O}_7$ (Mn is +7), is acidic. It reacts with water to form permanganic acid ( $\{HMnO}_4$ ).
Intermediate Oxides (Amphoteric): Oxides with intermediate oxidation states exhibit both acidic and basic properties. Example: Chromium(III) oxide, $\text{Cr}_2\text{O}_3$ (Cr is +3), is amphoteric.

(ii)

A transition metal exhibits its highest oxidation state in oxides and fluorides. This is because oxygen and fluorine are the two most electronegative elements and are also small in size.

High Electronegativity: Their strong ability to attract electrons helps to oxidise the metal to its highest oxidation state. They can effectively remove electrons from the metal atom.
Small Size: Their small atomic size allows several oxygen or fluorine atoms to be packed around the central metal atom without causing significant steric hindrance. This allows for the formation of stable compounds even with high coordination numbers.
Multiple Bonding: Oxygen has the additional ability to form multiple bonds (p $\pi$ -d $\pi$ bonds) with the metal atom. This delocalizes the high positive charge on the metal in its high oxidation state, further stabilizing the oxide.

Examples: Vanadium forms

\text{VF}_5

and

\text{V}_2\text{O}_5

(V is +5). Manganese shows its highest oxidation state of +7 in

\text{Mn}_2\text{O}_7

. Chromium shows +6 in

\text{CrF}_6

and

\text{CrO}_3

.

(iii)

The highest oxidation state is exhibited in oxoanions of a metal. An oxoanion is a polyatomic ion containing a central metal atom bonded to one or more oxygen atoms. The ability of oxygen to stabilize high oxidation states is the reason for this phenomenon.

Reason: Oxygen is highly electronegative and can form multiple p $\pi$ -d $\pi$ bonds with the transition metal. In an oxoanion like permanganate ( $\text{MnO}_4^-$ ) or chromate ( $\text{CrO}_4^{2-}$ ), the metal is in a very high oxidation state (+7 for Mn, +6 for Cr). The high positive charge on the metal is delocalized over the oxygen atoms through the formation of multiple bonds. This delocalization stabilizes the oxoanion, making it possible for the metal to exist in such a high oxidation state.

Examples:

In permanganate ion, $\text{MnO}_4^-$ , Mn is in the +7 oxidation state.
In dichromate ion, $\text{Cr}_2\text{O}_7^{2-}$ , Cr is in the +6 oxidation state.
In vanadate ion, $\text{VO}_3^-$ , V is in the +5 oxidation state.

Q4.26Exercises

Indicate the steps in the preparation of:

(i)

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ from chromite ore.

(ii)

$\mathrm{KMnO}_{4}$ from pyrolusite ore.

Solution

(i)

Preparation of Potassium Dichromate ( $K_2Cr_2O_7$ ) from Chromite Ore ( $FeCr_2O_4$ )

The preparation involves the following three steps:

Step 1: Conversion of chromite ore to sodium chromate. Chromite ore is finely powdered and fused with molten sodium carbonate (

Na_2CO_3

) in the presence of excess air in a reverberatory furnace.

4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2

The yellow mass of sodium chromate (

Na_2CrO_4

) is extracted with water and filtered to remove the insoluble iron(III) oxide (

Fe_2O_3

).

Step 2: Conversion of sodium chromate to sodium dichromate. The yellow solution of sodium chromate is acidified with concentrated sulphuric acid, which converts it into orange sodium dichromate.

2Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O

Sodium sulphate (

Na_2SO_4

), being less soluble, is removed by filtration.

Step 3: Conversion of sodium dichromate to potassium dichromate. The resulting solution of sodium dichromate is treated with a calculated amount of potassium chloride (

KCl

). Potassium dichromate, being less soluble than sodium dichromate, crystallises out on cooling.

Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl

The orange crystals of potassium dichromate are then separated by filtration.

(ii)

Preparation of Potassium Permanganate ( $KMnO_4$ ) from Pyrolusite Ore ( $MnO_2$ )

The preparation involves the following two steps:

Step 1: Conversion of pyrolusite ore to potassium manganate. Finely powdered pyrolusite ore (

MnO_2

) is fused with an alkali, such as potassium hydroxide (

KOH

), in the presence of an oxidising agent like air or potassium nitrate (

KNO_3

). This results in the formation of dark green potassium manganate (

K_2MnO_4

).

2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O

Step 2: Oxidation of potassium manganate to potassium permanganate. The green potassium manganate is extracted with water and then converted into purple potassium permanganate by oxidation. This can be done in two ways:

(a) Chemical Oxidation: Bubbling chlorine gas or ozone through the solution of potassium manganate oxidises it to potassium permanganate.

2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KCl

(b) Electrolytic Oxidation: In this process, an aqueous solution of potassium manganate is electrolysed. The manganate ions (

MnO_4^{2-}

) are oxidised to permanganate ions (

MnO_4^-

) at the anode.

MnO_4^{2-} \rightarrow MnO_4^- + e^-

(Green) (Purple)

On evaporation, the solution yields deep purple-black crystals of potassium permanganate.

Q4.27Exercises

What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Solution

Alloys are homogeneous solid solutions of two or more metals, or a metal and one or more non-metals. They are typically prepared by mixing the molten components and then allowing them to cool. Alloys often have properties that are different from those of their constituent elements, such as increased hardness, strength, or resistance to corrosion.

An important alloy that contains some of the lanthanoid metals is Mischmetal.

Composition of Mischmetal: It is an alloy which consists of:

Lanthanoid metals: ~95% (mainly Cerium, Lanthanum, and Neodymium)
Iron: ~5%
Traces of S, C, Ca, and Al.

Uses of Mischmetal:

It is used in magnesium-based alloys to produce bullets, shells, and lighter flints.
It is used as a deoxidizer in metallurgy to remove oxygen from molten metals.
It is added to steels to improve their strength and workability.

Q4.28Exercises

What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104.

Solution

Inner transition elements (or f-block elements) are the elements in which the last electron enters the f-orbital of the antepenultimate energy level, i.e., the (n-2)f subshell. These elements are placed separately at the bottom of the periodic table in two series:

The Lanthanoid series: This series follows Lanthanum (Z=57) and includes elements from Cerium (Z=58) to Lutetium (Z=71). Here, the 4f subshell is progressively filled.
The Actinoid series: This series follows Actinium (Z=89) and includes elements from Thorium (Z=90) to Lawrencium (Z=103). Here, the 5f subshell is progressively filled.

To decide which of the given atomic numbers belong to inner transition elements, we check if they fall within the ranges for lanthanoids or actinoids.

The given atomic numbers are: 29, 59, 74, 95, 102, 104.

Atomic number 29 (Copper, Cu): This is a d-block element (Group 11). It is not an inner transition element.
Atomic number 59 (Praseodymium, Pr): This falls in the range of lanthanoids (58-71). Therefore, it is an inner transition element.
Atomic number 74 (Tungsten, W): This is a d-block element (Group 6). It is not an inner transition element.
Atomic number 95 (Americium, Am): This falls in the range of actinoids (90-103). Therefore, it is an inner transition element.
Atomic number 102 (Nobelium, No): This falls in the range of actinoids (90-103). Therefore, it is an inner transition element.
Atomic number 104 (Rutherfordium, Rf): This element comes after the actinoid series and is a d-block element (transactinide). It is not an inner transition element.

Thus, the atomic numbers of the inner transition elements from the given list are 59, 95, and 102.

Q4.29Exercises

The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.

Solution

The statement that the chemistry of the actinoid elements is not so smooth as that of the lanthanoids is justified primarily by the greater range and variability of oxidation states exhibited by the actinoids.

Lanthanoid Chemistry:

The most common and stable oxidation state for all lanthanoids is +3.
A few elements show +2 (e.g., Eu, Yb) or +4 (e.g., Ce, Tb) oxidation states, but these are exceptions and are often achieved when they lead to stable empty, half-filled, or completely filled f-orbitals.
The overall chemistry is quite uniform across the series, dominated by the properties of the Ln³⁺ ion.
This is because the 4f electrons are well-shielded and deep inside the atom, making them less available for bonding. The energy gap between the 4f, 5d, and 6s orbitals is large.

Actinoid Chemistry:

Actinoids show a much wider range of oxidation states. For example, the oxidation states vary from +3 for Actinium to a maximum of +7 for Neptunium and Plutonium.
While +3 is a common oxidation state, it is not always the most stable one, especially for the earlier actinoids. For instance, the most stable state for Thorium is +4, for Protactinium is +5, and for Uranium is +6.
The first half of the actinoid series (up to Americium) shows a variety of oxidation states. For example:
- Uranium (U): +3, +4, +5, +6
- Neptunium (Np): +3, +4, +5, +6, +7
- Plutonium (Pu): +3, +4, +5, +6, +7
This variability arises because the 5f, 6d, and 7s subshells are of comparable energies. Consequently, electrons from all these subshells can participate in chemical bonding, leading to multiple oxidation states.
The 5f orbitals are less deeply buried than the 4f orbitals and extend further in space, making them more available for bonding.

Conclusion: The wide range of oxidation states in actinoids leads to a much more complex and less predictable chemistry compared to the relatively simple and smooth chemistry of lanthanoids, which is largely based on a single, stable +3 oxidation state. This complexity is a direct consequence of the comparable energies of the valence orbitals in actinoids.

Q4.30Exercises

Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Solution

The last element in the series of the actinoids is Lawrencium.

Symbol: Lr
Atomic Number: 103

Electronic Configuration: The atomic number of Lawrencium is 103. The noble gas preceding it is Radon (Rn, Z=86). The electronic configuration of Lawrencium (Lr) is:

[Rn] 5f^{14} 6d^1 7s^2

This configuration shows that it has a completely filled 5f subshell, one electron in the 6d subshell, and two electrons in the 7s subshell.

Comment on Possible Oxidation State: The most common and stable oxidation state for Lawrencium is expected to be +3.

This is because it can readily lose its three outermost valence electrons (two from the 7s orbital and one from the 6d orbital) to achieve a stable electronic configuration. The resulting ion,

Lr^{3+}

, would have the electronic configuration:

[Rn] 5f^{14}

This configuration is highly stable due to the completely filled 5f subshell. The large energy gap required to remove an electron from the stable, filled

5f^{14}

shell makes higher oxidation states highly unlikely. Therefore, Lawrencium chemistry is expected to be dominated by the +3 oxidation state, similar to its lanthanoid homologue, Lutetium.

Q4.31Exercises

Use Hund's rule to derive the electronic configuration of $\mathrm{Ce}^{3+}$ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.

Solution

The problem asks for the electronic configuration of the

\mathrm{Ce}^{3+}

ion and its magnetic moment based on the 'spin-only' formula.

1. Electronic Configuration of Cerium (Ce): Cerium (Ce) has an atomic number (Z) of 58. Its electronic configuration is: Ce (Z=58): [Xe]

4f^1 5d^1 6s^2

2. Electronic Configuration of $\mathrm{Ce}^{3+}$ ion: To form the

\mathrm{Ce}^{3+}

ion, three electrons are removed from the neutral Ce atom. The electrons are removed from the outermost shells first. Thus, two electrons are removed from the

6s

orbital and one electron from the

5d

orbital.

Ce

\rightarrow

\mathrm{Ce}^{3+} + 3e^-

The electronic configuration of

\mathrm{Ce}^{3+}

is:

\mathrm{Ce}^{3+}

: [Xe]

4f^1

According to Hund's rule, this single electron in the

4f

subshell will occupy one of the seven

f

-orbitals, and it will be an unpaired electron. Number of unpaired electrons,

n = 1

.

3. Calculation of Magnetic Moment ( $\μ$ ): The magnetic moment is calculated using the 'spin-only' formula:

\mu = \sqrt{n(n+2)}

where

n

is the number of unpaired electrons.

Given: Number of unpaired electrons,

n = 1

Calculation: Substituting the value of

n

into the formula:

\mu = \sqrt{1(1+2)}

\mu = \sqrt{3}

\mu \approx 1.732 \text{ BM (Bohr Magnetons)}

Final Answer: The electronic configuration of

\mathrm{Ce}^{3+}

is [Xe] $4f^1$ . Its magnetic moment is $1.732$ BM.

Q4.32Exercises

Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

Solution

The most common and stable oxidation state for all lanthanoids is +3. However, some elements exhibit +2 or +4 oxidation states, which are stabilized by attaining empty (

f^0

), half-filled (

f^7

), or completely filled (

f^{14}

)

f

-orbital configurations.

Lanthanoids exhibiting +4 oxidation state: The members of the lanthanoid series that exhibit the +4 oxidation state are:

Cerium (Ce)
Praseodymium (Pr)
Neodymium (Nd)
Terbium (Tb)
Dysprosium (Dy)

Correlation with electronic configuration: The +4 state is particularly stable for Cerium and Terbium.

Cerium (Ce, Z=58): Electronic configuration is [Xe] $4f^1 5d^1 6s^2$ . When it forms the $\mathrm{Ce}^{4+}$ ion, it loses four electrons to achieve the configuration [Xe] or [Xe] $4f^0$ . This is the stable configuration of the noble gas Xenon, with an empty $f$ -orbital, making $\mathrm{Ce}^{4+}$ a strong oxidizing agent.
Terbium (Tb, Z=65): Electronic configuration is [Xe] $4f^9 6s^2$ . When it forms the $\mathrm{Tb}^{4+}$ ion, it loses four electrons to achieve the configuration [Xe] $4f^7$ . This is a very stable half-filled $f$ -subshell configuration.

Lanthanoids exhibiting +2 oxidation state: The members of the lanthanoid series that exhibit the +2 oxidation state are:

Neodymium (Nd)
Samarium (Sm)
Europium (Eu)
Thulium (Tm)
Ytterbium (Yb)

Correlation with electronic configuration: The +2 state is particularly stable for Europium and Ytterbium.

Europium (Eu, Z=63): Electronic configuration is [Xe] $4f^7 6s^2$ . When it forms the $\mathrm{Eu}^{2+}$ ion, it loses the two $6s$ electrons to achieve the configuration [Xe] $4f^7$ . This is a very stable half-filled $f$ -subshell configuration. $\mathrm{Eu}^{2+}$ is a strong reducing agent.
Ytterbium (Yb, Z=70): Electronic configuration is [Xe] $4f^{14} 6s^2$ . When it forms the $\mathrm{Yb}^{2+}$ ion, it loses the two $6s$ electrons to achieve the configuration [Xe] $4f^{14}$ . This is a very stable completely filled $f$ -subshell configuration. $\mathrm{Yb}^{2+}$ is also a reducing agent.

Q4.33Exercises

Compare the chemistry of the actinoids with that of lanthanoids with reference to:

(i)

electronic configuration

(ii)

oxidation states and

(iii)

chemical reactivity.

Solution

The chemistry of actinoids is compared with that of lanthanoids with reference to the following points:

(i) Electronic Configuration

Lanthanoids: The general electronic configuration is [Xe] $4f^{1-14} 5d^{0-1} 6s^2$ . The incoming electron enters the $4f$ orbital, which is well-shielded and deep inside the atom. The energy difference between $4f$ and $5d$ orbitals is relatively large.
Actinoids: The general electronic configuration is [Rn] $5f^{1-14} 6d^{0-1} 7s^2$ . The incoming electron enters the $5f$ orbital. The $5f$ , $6d$ , and $7s$ orbitals are of comparable energies. Because of the small energy gap, electrons can occupy any of these orbitals, leading to more irregularities in their electronic configurations compared to lanthanoids.

(ii) Oxidation States

Lanthanoids: The most common and stable oxidation state is +3. A few elements also show +2 and +4 oxidation states, which are favored if they lead to a stable $f^0$ , $f^7$ , or $f^{14}$ configuration. The range of oxidation states is limited.
Actinoids: They exhibit a much wider range of oxidation states. Although +3 is a common state, it is not always the most stable. The earlier actinoids (e.g., Th, Pa, U, Np) show higher oxidation states like +4, +5, +6, and +7. This is because the $5f$ , $6d$ , and $7s$ orbitals have comparable energies, allowing all their electrons to participate in bonding. For later actinoids, the +3 oxidation state becomes more stable.

(iii) Chemical Reactivity

Lanthanoids: The earlier members of the series are quite reactive, similar to alkaline earth metals like calcium. Their reactivity decreases as we move across the series. They are generally less reactive than actinoids.
Actinoids: Actinoids are highly reactive metals, especially when finely divided. They are more reactive than lanthanoids. This is attributed to their larger atomic size and lower ionization energies. They react with boiling water, non-metals, and most acids. The reactivity is complex due to the variety of oxidation states.

Other key differences:

Radioactivity: All actinoids are radioactive, whereas among lanthanoids, only promethium (Pm) is radioactive.
Complex Formation: Actinoids have a greater tendency to form complexes than lanthanoids due to their higher charge and smaller ion size for a given charge, leading to higher charge density.
Magnetic Properties: The magnetic properties of actinoids are more complex than those of lanthanoids and are difficult to interpret.

Q4.34Exercises

Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Solution

The electronic configurations of the elements with the given atomic numbers are determined by following the Aufbau principle and identifying the preceding noble gas.

Atomic Number 61 (Promethium, Pm): This element is a member of the lanthanoid series. The preceding noble gas is Xenon (Xe, Z=54). The electronic configuration is: [Xe] $4f^5 6s^2$
Atomic Number 91 (Protactinium, Pa): This element is a member of the actinoid series. The preceding noble gas is Radon (Rn, Z=86). The electronic configuration is: [Rn] $5f^2 6d^1 7s^2$
Atomic Number 101 (Mendelevium, Md): This element is a member of the actinoid series. The preceding noble gas is Radon (Rn, Z=86). The electronic configuration is: [Rn] $5f^{13} 7s^2$
Atomic Number 109 (Meitnerium, Mt): This element is a transactinide element, belonging to the d-block (Group 9, Period 7). The preceding noble gas is Radon (Rn, Z=86). The electronic configuration is: [Rn] $5f^{14} 6d^7 7s^2$

Q4.35Exercises

Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(i)

electronic configurations

(ii)

oxidation states

(iii)

ionisation enthalpies and (iv) atomic sizes.

Solution

A comparison of the general characteristics of the first (3d), second (4d), and third (5d) series of transition metals in their respective vertical columns is given below.

(i) Electronic Configurations

First Series (3d): The general configuration is [Ar] $3d^{1-10} 4s^{1-2}$ . There are two exceptions to the regular filling pattern: Cr ( $3d^5 4s^1$ ) and Cu ( $3d^{10} 4s^1$ ).
Second (4d) & Third (5d) Series: The general configurations are [Kr] $4d^{1-10} 5s^{0-2}$ and [Xe] $4f^{14} 5d^{1-10} 6s^{1-2}$ respectively. These series show a greater number of irregularities in electronic configurations. This is due to the very small energy difference between the $(n-1)d$ and $ns$ orbitals, which facilitates the shifting of electrons between them.

(ii) Oxidation States

First Series (3d): The elements of the 3d series show a variety of oxidation states. Lower oxidation states are generally more stable for the earlier elements, while higher oxidation states become more important for elements in the middle and are strongly oxidizing (e.g., $\mathrm{MnO_4^-}$ , $\mathrm{Cr_2O_7^{2-}}$ ).
Second (4d) & Third (5d) Series: The elements of the 4d and 5d series tend to exhibit higher oxidation states more readily and these higher states are more stable than those of the 3d series. For example, Ru and Os can show a +8 oxidation state, which is not seen in Fe. The compounds of 4d and 5d elements in higher oxidation states are less oxidizing than their 3d counterparts (e.g., $\mathrm{WO_3}$ is more stable than $\mathrm{CrO_3}$ ). This is because the $4d$ and $5d$ orbitals are more diffuse and extended, making them more available for bonding.

(iii) Ionisation Enthalpies

Trend down a group: Generally, ionisation enthalpy (IE) decreases down a group. However, in transition metals, the IE values for the 5d series are significantly higher than those for the 4d and 3d series. This is a direct consequence of the lanthanoid contraction, which leads to a higher effective nuclear charge for the 5d elements. The IE values generally follow the order: 3d < 4d < 5d.

(iv) Atomic Sizes

Trend down a group: Atomic radii increase from the 3d series to the 4d series, as expected due to the addition of an electron shell. However, the atomic radii of the elements of the 5d series are almost identical to those of the corresponding elements in the 4d series. This phenomenon is known as lanthanoid contraction. It is caused by the poor shielding effect of the electrons in the intervening $4f$ orbitals in the 5d series. The poor shielding results in a stronger attraction of the outermost electrons by the nucleus, leading to a contraction in size that counteracts the expected increase down the group. This similarity in size leads to the 4d and 5d series elements having very similar chemical properties.

Q4.36Exercises

Write down the number of 3d electrons in each of the following ions: $\mathrm{Ti}^{2+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Fe}^{3+}, \mathrm{Co}^{2+}, \mathrm{Ni}^{2+}$ and $\mathrm{Cu}^{2+}$ . Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Solution

The number of 3d electrons and the occupancy of 3d orbitals for the given hydrated ions (in an octahedral field) are determined as follows. First, we write the electronic configuration of the neutral atom, then the ion, to find the number of 3d electrons. For hydrated ions, water (

\mathrm{H_2O}

) acts as a weak field ligand, leading to high-spin complexes where pairing of electrons occurs only after each of the five d-orbitals is singly occupied. In an octahedral field, the d-orbitals split into two sets: the lower energy

t_{2g}

set (3 orbitals) and the higher energy

e_g

set (2 orbitals).

Ion	Atomic Number (Z)	Electronic Configuration of Ion	Number of 3d electrons	Occupancy of 3d orbitals ( $t_{2g}^x e_g^y$ )
$\mathrm{Ti}^{2+}$	22	$[\mathrm{Ar}]\,3d^2$	2	$t_{2g}^2 e_g^0$
$\mathrm{V}^{2+}$	23	$[\mathrm{Ar}]\,3d^3$	3	$t_{2g}^3 e_g^0$
$\mathrm{Cr}^{3+}$	24	$[\mathrm{Ar}]\,3d^3$	3	$t_{2g}^3 e_g^0$
$\mathrm{Mn}^{2+}$	25	$[\mathrm{Ar}]\,3d^5$	5	$t_{2g}^3 e_g^2$
$\mathrm{Fe}^{2+}$	26	$[\mathrm{Ar}]\,3d^6$	6	$t_{2g}^4 e_g^2$
$\mathrm{Fe}^{3+}$	26	$[\mathrm{Ar}]\,3d^5$	5	$t_{2g}^3 e_g^2$
$\mathrm{Co}^{2+}$	27	$[\mathrm{Ar}]\,3d^7$	7	$t_{2g}^5 e_g^2$
$\mathrm{Ni}^{2+}$	28	$[\mathrm{Ar}]\,3d^8$	8	$t_{2g}^6 e_g^2$
$\mathrm{Cu}^{2+}$	29	$[\mathrm{Ar}]\,3d^9$	9	$t_{2g}^6 e_g^3$

Explanation of Orbital Occupancy:

For $d^1, d^2, d^3$ configurations ( $\mathrm{Ti}^{2+}, \mathrm{V}^{2+}, \mathrm{Cr}^{3+}$ ), the electrons occupy the lower energy $t_{2g}$ orbitals singly.
For $d^4$ $d^{4}$ to $d^7$ $d^{7}$ configurations, since $\mathrm{H_2O}$ $H_{2} O$ is a weak field ligand, the energy gap between $t_{2g}$ $t_{2 g}$ and $e_g$ $e_{g}$ orbitals ( $\Delta_o$ $Δ_{o}$ ) is small. Electrons will occupy the $e_g$ $e_{g}$ orbitals before pairing up in the $t_{2g}$ $t_{2 g}$ orbitals (Hund's rule). This results in high-spin complexes.
- $\mathrm{Mn}^{2+}$ and $\mathrm{Fe}^{3+}$ ( $d^5$ ): $t_{2g}^3 e_g^2$
- $\mathrm{Fe}^{2+}$ ( $d^6$ ): $t_{2g}^4 e_g^2$ (one electron pairs in $t_{2g}$ )
- $\mathrm{Co}^{2+}$ ( $d^7$ ): $t_{2g}^5 e_g^2$ (two electrons pair in $t_{2g}$ )
For $d^8$ $d^{8}$ and $d^9$ $d^{9}$ configurations ( $\mathrm{Ni}^{2+}, \mathrm{Cu}^{2+}$ $Ni^{2 +}, Cu^{2 +}$ ), the distribution is the same for both high-spin and low-spin cases.
- $\mathrm{Ni}^{2+}$ ( $d^8$ ): $t_{2g}^6 e_g^2$
- $\mathrm{Cu}^{2+}$ ( $d^9$ ): $t_{2g}^6 e_g^3$

Q4.37Exercises

Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Solution

The elements of the first transition series (3d series) exhibit properties that are significantly different from those of the heavier transition elements (4d and 5d series). The key differences are:

Atomic Radii: The atomic radii of the 4d series elements are larger than those of the 3d series. However, due to the lanthanoid contraction, the atomic radii of the 5d series elements are nearly identical to those of the 4d series elements. For example, the radii are Zr (160 pm) and Hf (159 pm).
Oxidation States: The heavier transition elements (4d and 5d series) show a greater tendency to exhibit higher oxidation states, and these higher states are more stable than for the 3d series elements. For example, in Group 6, chromium (3d) shows a maximum oxidation state of +6 (in $\mathrm{CrO_4^{2-}}$ ), but molybdenum (4d) and tungsten (5d) form stable compounds in the +6 state (e.g., $\mathrm{MoF_6}$ , $\mathrm{WF_6}$ ). Ruthenium and Osmium (Group 8) can show a +8 oxidation state, which is not seen in iron.
Magnetic Properties: For the 3d series elements, the magnetic moments can usually be calculated using the spin-only formula ( $\mu = \sqrt{n(n+2)}$ BM). For the 4d and 5d series, the orbital contribution to the magnetic moment is significant, and the spin-only formula is often inadequate.
Complex Formation: The 3d series elements form both high-spin and low-spin complexes depending on the ligand strength. The heavier 4d and 5d elements have larger d-orbitals and experience a greater crystal field splitting energy ( $\Delta_o$ ). Consequently, they predominantly form low-spin complexes, even with weak field ligands.
Metal-Metal Bonding: Heavier transition elements have a much greater tendency to form compounds with metal-metal bonds than the 3d elements. This is due to the more diffuse nature of their d-orbitals, allowing for better overlap.

Q4.38Exercises

What can be inferred from the magnetic moment values of the following complex species ? Example Magnetic Moment (BM) $\mathrm{K}_{4}[\mathrm{Mn}(\mathrm{CN})_{6})$ 2.2 $[\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$ 5.3 $\mathrm{K}_{2}[\mathrm{MnCl}_{4}]$ 5.9

Solution

The magnetic moment (

\mu

) of a complex is related to the number of unpaired electrons (n) by the spin-only formula:

\mu = \sqrt{n(n+2)} \text{ BM}

By comparing the experimental magnetic moment with the values calculated for different 'n', we can infer the number of unpaired electrons and thus the electronic configuration and geometry of the complex.

1. $\mathrm{K}_{4}[\mathrm{Mn}(\mathrm{CN})_{6}]$

Oxidation state of Mn: Let the oxidation state be $x$ . $4(+1) + x + 6(-1) = 0 \implies x = +2$ . So, we have $\mathrm{Mn}^{2+}$ .
Electronic configuration of $\mathrm{Mn}^{2+}$ : $[\mathrm{Ar}]\,3d^5$ .
Magnetic moment: Given $\mu = 2.2$ BM.
Inference:
- If $n=1$ , $\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.73$ BM.
- If $n=2$ , $\mu = \sqrt{2(2+2)} = \sqrt{8} = 2.82$ BM. The experimental value of 2.2 BM is closer to 1.73 BM, indicating the presence of one unpaired electron ( $n=1$ ).
Conclusion: The ligand $\mathrm{CN}^{-}$ is a strong field ligand, causing the pairing of electrons in the 3d orbitals. For a $d^5$ ion to have one unpaired electron, the electronic configuration in the octahedral field must be $t_{2g}^5 e_g^0$ . This is a low-spin inner orbital complex with $d^2sp^3$ hybridization.

2. $[\mathrm{Fe}(\mathrm{H}_{2}\mathrm{O})_{6}]^{2+}$

Oxidation state of Fe: Let the oxidation state be $x$ . $x + 6(0) = +2 \implies x = +2$ . So, we have $\mathrm{Fe}^{2+}$ .
Electronic configuration of $\mathrm{Fe}^{2+}$ : $[\mathrm{Ar}]\,3d^6$ .
Magnetic moment: Given $\mu = 5.3$ BM.
Inference:
- If $n=4$ , $\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90$ BM.
- If $n=5$ , $\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92$ BM. The experimental value of 5.3 BM is consistent with the presence of four unpaired electrons ( $n=4$ ). The value is slightly higher than the spin-only value due to some orbital contribution.
Conclusion: The ligand $\mathrm{H_2O}$ is a weak field ligand and does not cause electron pairing. The complex is a high-spin complex. The electronic configuration of the $d^6$ ion in the octahedral field is $t_{2g}^4 e_g^2$ , which has 4 unpaired electrons. It is an outer orbital complex with $sp^3d^2$ hybridization.

3. $\mathrm{K}_{2}[\mathrm{MnCl}_{4}]$

Oxidation state of Mn: Let the oxidation state be $x$ . $2(+1) + x + 4(-1) = 0 \implies x = +2$ . So, we have $\mathrm{Mn}^{2+}$ .
Electronic configuration of $\mathrm{Mn}^{2+}$ : $[\mathrm{Ar}]\,3d^5$ .
Magnetic moment: Given $\mu = 5.9$ BM.
Inference:
- If $n=5$ , $\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92$ BM. The experimental value of 5.9 BM is very close to the calculated value for five unpaired electrons ( $n=5$ ).
Conclusion: The complex ion $[\mathrm{MnCl}_{4}]^{2-}$ has a tetrahedral geometry. The ligand $\mathrm{Cl}^{-}$ is a weak field ligand, and tetrahedral complexes are generally high-spin. The five 3d electrons occupy the d-orbitals singly, resulting in 5 unpaired electrons. The electronic configuration in the tetrahedral field is $e^2 t_2^3$ . The hybridization is $sp^3$ .

The D-and F-Block ElementsClass 12 Chemistry NCERT Solutions

Write down the electronic configuration of:

Cr3+\mathrm{Cr}^{3+}Cr3+

Pm3+\mathrm{Pm}^{3+}Pm3+

Cu+\mathrm{Cu}^{+}Cu+

Ce4+\mathrm{Ce}^{4+}Ce4+

Co2+\mathrm{Co}^{2+}Co2+

Lu2+\mathrm{Lu}^{2+}Lu2+

Mn2+\mathrm{Mn}^{2+}Mn2+

Th4+\mathrm{Th}^{4+}Th4+

Solution

Why are Mn2+\mathrm{Mn}^{2+}Mn2+ compounds more stable than Fe2+\mathrm{Fe}^{2+}Fe2+ towards oxidation to their +3 state?

Solution

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

Solution

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.

Solution

What may be the stable oxidation state of the transition element with the following ddd electron configurations in the ground state of their atoms : 3d33d^{3}3d3, 3d53d^{5}3d5, 3d83d^{8}3d8 and 3d43d^{4}3d4 ?

Solution

Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

Solution

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Solution

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

Solution

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

Solution

What are the different oxidation states exhibited by the lanthanoids?

Solution

Explain giving reasons:

Transition metals and many of their compounds show paramagnetic behaviour.

The enthalpies of atomisation of the transition metals are high.

The transition metals generally form coloured compounds.

Transition metals and their many compounds act as good catalyst.

Solution

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

Solution

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

Solution

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Solution

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

iodide

iron(II) solution and

H2 S\mathrm{H}_{2} \mathrm{~S}H2​ S

Solution

Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2\mathrm{SO}_{2}SO2​ and (iii) oxalic acid? Write the ionic equations for the reactions.

Solution

Preparation of Potassium Permanganate (KMnO4KMnO_4KMnO4​)

Reactions of Acidified Permanganate Solution

the stability of Fe3+\mathrm{Fe}^{3+}Fe3+ in acid solution as compared to that of Cr3+\mathrm{Cr}^{3+}Cr3+ or Mn3+\mathrm{Mn}^{3+}Mn3+ and

the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Solution

Predict which of the following will be coloured in aqueous solution? Ti3+,V3+\mathrm{Ti}^{3+}, \mathrm{V}^{3+}Ti3+,V3+, Cu+,Sc3+,Mn2+,Fe3+\mathrm{Cu}^{+}, \mathrm{Sc}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}Cu+,Sc3+,Mn2+,Fe3+ and Co2+\mathrm{Co}^{2+}Co2+. Give reasons for each.

Solution

Compare the stability of +2 oxidation state for the elements of the first transition series.

Solution

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

electronic configuration

atomic and ionic sizes and

oxidation state

chemical reactivity.

Solution

How would you account for the following:

Of the d4d^{4}d4 species, Cr2+\mathrm{Cr}^{2+}Cr2+ is strongly reducing while manganese(III) is strongly oxidising.

Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.

The d1d^{1}d1 configuration is very unstable in ions.

Solution

What is meant by 'disproportionation'? Give two examples of disproportionation reaction in aqueous solution.

Solution

Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Solution

Calculate the number of unpaired electrons in the following gaseous ions: Mn3+\mathrm{Mn}^{3+}Mn3+, Cr3+,V3+\mathrm{Cr}^{3+}, \mathrm{V}^{3+}Cr3+,V3+ and Ti3+\mathrm{Ti}^{3+}Ti3+. Which one of these is the most stable in aqueous solution?

Solution

Give examples and suggest reasons for the following features of the transition metal chemistry:

The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.

A transition metal exhibits highest oxidation state in oxides and fluorides.

The highest oxidation state is exhibited in oxoanions of a metal.

Solution

Indicate the steps in the preparation of:

$\mathrm{Cr}^{3+}$

$\mathrm{Pm}^{3+}$

$\mathrm{Cu}^{+}$

$\mathrm{Ce}^{4+}$

$\mathrm{Co}^{2+}$

$\mathrm{Lu}^{2+}$

$\mathrm{Mn}^{2+}$

$\mathrm{Th}^{4+}$

Why are $\mathrm{Mn}^{2+}$ compounds more stable than $\mathrm{Fe}^{2+}$ towards oxidation to their +3 state?

What may be the stable oxidation state of the transition element with the following $d$ electron configurations in the ground state of their atoms : $3d^{3}$ , $3d^{5}$ , $3d^{8}$ and $3d^{4}$ ?

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?

$\mathrm{H}_{2} \mathrm{~S}$

Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) $\mathrm{SO}_{2}$ and (iii) oxalic acid? Write the ionic equations for the reactions.

Preparation of Potassium Permanganate ( $KMnO_4$ )

the stability of $\mathrm{Fe}^{3+}$ in acid solution as compared to that of $\mathrm{Cr}^{3+}$ or $\mathrm{Mn}^{3+}$ and

Predict which of the following will be coloured in aqueous solution? $\mathrm{Ti}^{3+}, \mathrm{V}^{3+}$ , $\mathrm{Cu}^{+}, \mathrm{Sc}^{3+}, \mathrm{Mn}^{2+}, \mathrm{Fe}^{3+}$ and $\mathrm{Co}^{2+}$ . Give reasons for each.

Of the $d^{4}$ species, $\mathrm{Cr}^{2+}$ is strongly reducing while manganese(III) is strongly oxidising.

The $d^{1}$ configuration is very unstable in ions.

Calculate the number of unpaired electrons in the following gaseous ions: $\mathrm{Mn}^{3+}$ , $\mathrm{Cr}^{3+}, \mathrm{V}^{3+}$ and $\mathrm{Ti}^{3+}$ . Which one of these is the most stable in aqueous solution?

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ from chromite ore.

$\mathrm{KMnO}_{4}$ from pyrolusite ore.

Use Hund's rule to derive the electronic configuration of $\mathrm{Ce}^{3+}$ ion, and calculate its magnetic moment on the basis of 'spin-only' formula.