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Class 12th (Chemistry) Chapters
1. The Solid State 2. Solutions 3. Electrochemistry
4. Chemical Kinetics 5. Surface Chemistry 6. General Principles And Processes Of Isolation Of Elements
7. The P-Block Elements 8. The D-And F-Block Elements 9. Coordination Compounds
10. Haloalkanes And Haloarenes 11. Alcohols, Phenols And Ethers 12. Aldehydes, Ketones And Carboxylic Acids
13. Amines 14. Biomolecules 15. Polymers
16. Chemistry In Everyday Life



Unit 8 The D- And F- Block Elements



Position In The Periodic Table

The elements located in the middle section of the periodic table, encompassing groups 3 through 12, constitute the d-block. These elements are characterized by the progressive filling of their d orbitals across each of the four extended periods.

Situated separately at the bottom of the periodic table are the elements forming the f-block. This block includes elements where the 4f and 5f orbitals are being filled sequentially.

Commonly, elements of the d-block are referred to as transition metals, while those of the f-block are termed inner transition metals.

According to the International Union of Pure and Applied Chemistry (IUPAC), transition metals are defined as metals possessing an incomplete d subshell either in their neutral atomic state or in their common ionic states.

Elements in Group 12, namely Zinc (Zn), Cadmium (Cd), and Mercury (Hg), have a completely filled d$^{10}$ electron configuration in both their ground state and typical oxidation states. Consequently, they are not classified as transition metals based on the IUPAC definition. However, due to their position at the end of the 3d, 4d, and 5d series, their chemical properties are often studied alongside those of transition metals.



Electronic Configurations Of The D-Block Elements

The d-block elements reside in the central portion of the periodic table, positioned between the s-block and p-block elements. The filling of d-orbitals in the second to outermost energy level ((n-1)d) gives rise to four series of transition metals: the 3d, 4d, 5d, and 6d series.

The general electronic configuration for the outer orbitals of these elements is commonly represented as $(n-1)d^{1-10} ns^{1-2}$. Here, (n-1) refers to the inner d orbitals which can accommodate between one and ten electrons, while the outermost ns orbital typically holds one or two electrons.

It is important to note that this general configuration has exceptions. These exceptions arise because the energy difference between the (n-1)d and ns orbitals is very small. Additionally, electron configurations resulting in half-filled (d$^5$) or completely filled (d$^{10}$) d orbitals exhibit enhanced stability.

Notable examples in the 3d series include Chromium (Cr) and Copper (Cu). Chromium adopts the configuration 3d$^5$ 4s$^1$ instead of the expected 3d$^4$ 4s$^2$. Similarly, Copper has the configuration 3d$^{10}$ 4s$^1$ instead of 3d$^9$ 4s$^2$. These deviations maximise the number of half-filled or fully filled orbitals for greater stability.

ElementScTiVCrMnFeCoNiCuZn
Z21222324252627282930
4s2221222212
3d123556781010
ElementYZrNbMoTcRuRhPdAgCd
Z39404142434445464748
5s2211111012
4d1245678101010
ElementLaHfTaWReOsIrPtAuHg
Z57727374757677787980
6s2222222112
5d123456791010
ElementAcRfDbSgBhHsMtDsRgCn
Z89104105106107108109110111112
7s2222222212
6d123456781010

As noted earlier, elements like Zn, Cd, Hg, and Cn have a general electronic configuration of $(n-1)d^{10} ns^2$. Since their d orbitals are completely filled in both their ground state and typical ionic states, they do not fit the definition of transition elements.

A key characteristic is that the d orbitals extend more outwards compared to s and p orbitals. This makes them highly susceptible to influence from their surroundings and allows them to significantly affect surrounding atoms or molecules. The presence of partially filled d orbitals (d$^{1}$ to d$^{9}$) is responsible for many characteristic properties of transition elements, including:

Unlike non-transition elements, transition elements in a horizontal series often exhibit greater similarities in properties. However, some similarities also exist within groups.

Example 8.1. On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not?

Answer:

Scandium (Z=21) has an electronic configuration of [Ar] 3d$^1$ 4s$^2$. In its common oxidation state, Sc$^{3+}$, the configuration is [Ar] 3d$^0$. Although the ion has an empty d subshell, the neutral atom has a partially filled 3d subshell (3d$^1$). Thus, according to the IUPAC definition, scandium is considered a transition element.

Zinc (Z=30) has an electronic configuration of [Ar] 3d$^{10}$ 4s$^2$. In its common oxidation state, Zn$^{2+}$, the configuration is [Ar] 3d$^{10}$. Both the neutral atom and its typical ion have a completely filled 3d subshell (3d$^{10}$). Therefore, zinc is not classified as a transition element by the IUPAC definition.



General Properties Of The Transition Elements (D-Block)

Physical Properties

Most transition elements exhibit characteristic metallic properties. These include high tensile strength, the ability to be drawn into wires (ductility), shaped by hammering (malleability), excellent thermal and electrical conductivity, and a typical metallic lustre.

With the notable exceptions of Zinc (Zn), Cadmium (Cd), Mercury (Hg), and Manganese (Mn), transition metals typically possess one or more standard metallic crystal structures at normal temperatures.

SeriesScTiVCrMnFeCoNiCuZn
1st (3d)hcphcpbccbccX (bcc)bccccpccpccpX (hcp)
SeriesYZrNbMoTcRuRhPdAgCd
2nd (4d)hcphcpbccbcchcphcpccpccpccpX (hcp)
SeriesLaHfTaWReOsIrPtAuHg
3rd (5d)hcphcpbccbcchcphcpccpccpccpX (bcc, ccp)
SeriesAcRfDbSgBhHsMtDsRgCn
4th (6d)hcphcpbccbcchcphcpccpccpccpX (bcc)

(bcc = body centred cubic; hcp = hexagonal close packed; ccp = cubic close packed; X = a typical metal structure, variations indicated in parenthesis).

Most transition metals (again, excluding Zn, Cd, and Hg) are characterized by their significant hardness and low tendency to vaporize (low volatility). They generally exhibit high melting and boiling points.

The exceptionally high melting points are attributed to the involvement of a large number of electrons from the inner $(n-1)d$ orbitals, in addition to the outermost $ns$ electrons, in forming strong interatomic metallic bonds.

Across a horizontal series of transition metals, the melting points tend to increase, reaching a maximum around the middle of the series (elements with d$^5$ configuration), before declining as the atomic number increases. Exceptions are observed for Mn and Tc, which show anomalous values.

Transition metals also possess high enthalpies of atomisation. The peak values around the middle of each series suggest that having one unpaired electron per d orbital is particularly favourable for strong interatomic bonding interactions. Generally, a larger number of valence electrons available for bonding correlates with stronger resulting bonds and higher enthalpy of atomisation.

Since the enthalpy of atomisation is a factor influencing the standard electrode potential, metals with very high enthalpies of atomisation (meaning very high boiling points) tend to be less reactive or 'noble'.

When comparing different series, the elements of the second (4d) and third (5d) series generally have higher enthalpies of atomisation than their corresponding elements in the first (3d) series. This difference contributes to the more frequent occurrence of metal-metal bonding in compounds of heavier transition metals.

Variation In Atomic And Ionic Sizes Of Transition Metals

In a given series, ions with the same charge show a gradual decrease in radius as the atomic number increases from left to right. This trend is explained by the increasing nuclear charge (+1 for each increment in atomic number) accompanied by the addition of electrons into the inner d orbitals. The shielding effect provided by a d electron for other electrons in the same shell is not as effective as the shielding by electrons in inner shells. Consequently, the effective nuclear attraction experienced by the outermost electrons increases, leading to a contraction in ionic size.

A similar decreasing trend is observed in the atomic radii within a series, although the variation is relatively small.

An interesting comparison arises between the atomic sizes of elements in different series. Atomic radii generally increase from the first (3d) series to the second (4d) series. However, the radii of elements in the third (5d) series are remarkably similar to those of their corresponding elements in the second (4d) series.

This unexpected similarity is a direct consequence of the Lanthanoid contraction. This phenomenon occurs because the 4f orbitals are filled *before* the 5d series begins (from atomic number 58, Cerium, to 71, Lutetium). The filling of the 4f orbitals results in a steady decrease in atomic radii across the lanthanoid series. This contraction effectively offsets the expected increase in size that would normally occur when moving from the 4d to the 5d series.

The significant outcome of the lanthanoid contraction is that elements of the second and third d series have nearly identical radii (e.g., Zirconium (Zr) with 160 pm and Hafnium (Hf) with 159 pm). This similarity in size leads to very similar physical and chemical properties between corresponding elements in the 4d and 5d series, making their separation challenging.

The cause of the lanthanoid contraction is analogous to the contraction observed within a regular transition series: the imperfect shielding of valence electrons by inner electrons. Specifically, the shielding provided by one 4f electron for another is less effective than that provided by a d electron. As the nuclear charge increases along the 4f series, the effective nuclear attraction on the entire 4f$^n$ electron cloud increases, causing a regular decrease in size.

The combined effect of decreasing metallic radius (due to increasing atomic number and contraction) and increasing atomic mass results in a general increase in the density of transition elements across a series, particularly noticeable from Titanium (Z=22) to Copper (Z=29).

ElementScTiVCrMnFeCoNiCuZn
Metallic/ionic radii/pm M164147135129137126125125128137
Metallic/ionic radii/pm M$^{2+}$7982827774707375
Metallic/ionic radii/pm M$^{3+}$7367646265656160
Density/g cm$^{-3}$3.434.16.077.197.217.88.78.98.97.1

Ionisation Enthalpies

The ionisation enthalpy is the energy required to remove an electron from a gaseous atom or ion. There is a general increase in ionisation enthalpy across each series of transition elements from left to right. This increase is primarily due to the rise in nuclear charge as atomic number increases, while electrons are added to the inner (n-1)d orbitals.

However, the increase in successive ionisation enthalpies (first, second, third, etc.) for transition elements is less steep compared to that observed in non-transition elements across a period. This implies that the energy cost of removing varying numbers of electrons from the $(n-1)d$ orbitals is relatively less prohibitive, contributing to their ability to show multiple oxidation states.

The variation in ionisation enthalpy along a transition series is also much smaller than the variation observed along a period for non-transition elements.

While the first ionisation enthalpy generally increases, the magnitude of increase for the second and third ionisation enthalpies tends to be higher across the series for successive elements.

The irregular trend observed in the first ionisation enthalpy values for the 3d series has minor chemical significance. It can be understood by considering how removing an electron alters the relative energies of the 4s and 3d orbitals. When d-block elements form ions, electrons are removed from the $ns$ orbitals before the $(n-1)d$ orbitals.

Moving across the 3d series, the nuclear charge increases, but the additional electrons occupy the inner 3d orbitals. These 3d electrons provide some shielding to the outer 4s electrons from the increasing nuclear charge. This shielding is somewhat more effective than the shielding among electrons within the same shell. Consequently, the atomic radii decrease less significantly, and the ionisation energies show only a slight increase across the series.

For doubly or more highly charged ions, the configuration is $d^n$ without any 4s electrons. A steady increase in second and third ionisation enthalpies is generally expected as the effective nuclear charge increases, and one d electron is not very effective at shielding another d electron due to the shape and spatial distribution of d orbitals.

However, the trend of steadily increasing second and third ionisation enthalpies is disrupted for the formation of Mn$^{2+}$ and Fe$^{3+}$ ions, respectively. Both these ions have a stable d$^5$ configuration (Mn$^{2+}$ is [Ar] 3d$^5$, Fe$^{3+}$ is [Ar] 3d$^5$). Similar disruptions occur at corresponding elements in the later transition series.

The variation in ionisation enthalpy for a configuration $d^n$ is influenced by three main factors:

  1. Attraction of electrons towards the nucleus.
  2. Repulsion between electrons.
  3. Exchange energy.

Exchange energy provides stability to an electronic state. It is roughly proportional to the number of possible pairs of electrons with parallel spins in degenerate orbitals. According to Hund's rule, the lowest energy state occurs when maximum orbitals are singly occupied with parallel spins. The loss of exchange energy upon ionisation increases stability, making ionisation more difficult.

There is no loss of exchange energy when forming a d$^6$ configuration. Mn$^+$ has 3d$^5$4s$^1$, and Cr$^+$ has 3d$^5$. Removing an electron from Mn$^+$ forms Mn$^{2+}$ (3d$^5$) which is stable. Removing an electron from Cr$^+$ forms Cr$^{2+}$ (3d$^4$) which loses the d$^5$ stability. Hence, the second ionisation enthalpy (IE2) of Mn (going from Mn$^+$ to Mn$^{2+}$) is lower than that of Cr (going from Cr$^+$ to Cr$^{2+}$).

Similarly, Fe$^{2+}$ has a d$^6$ configuration, and Mn$^{2+}$ has a d$^5$ configuration. The third ionisation enthalpy (IE3) involves removing an electron from Fe$^{2+}$ (d$^6$ $\rightarrow$ d$^5$) and Mn$^{2+}$ (d$^5$ $\rightarrow$ d$^4$). Removing an electron from the stable d$^5$ of Mn$^{2+}$ requires more energy than removing one from d$^6$ of Fe$^{2+}$ (which gains d$^5$ stability). Therefore, the third ionisation enthalpy of Fe is lower than that of Mn.

The lowest common oxidation state is +2. The energy required to form M$^{2+}$ ions from the solid metal involves the enthalpy of atomisation plus the sum of the first and second ionisation enthalpies. The second ionisation enthalpy is a significant factor. Notably high values for the sum of IE1 + IE2 are seen for Cr and Cu because forming M$^+$ ions involves achieving stable d$^5$ and d$^{10}$ configurations, respectively (Cr: 3d$^5$4s$^1$, Cu: 3d$^{10}$4s$^1$). The value for Zn is low because forming Zn$^+$ (3d$^{10}$4s$^1$) followed by Zn$^{2+}$ (3d$^{10}$) involves removing a 4s electron first, leading to a stable d$^{10}$ configuration.

The trend in third ionisation enthalpies (IE3) is simpler as it doesn't involve the 4s orbital factor. High IE3 values are observed for removing an electron from stable d$^5$ (Mn$^{2+}$) and d$^{10}$ (Zn$^{2+}$) ions. The high IE3 values for Cu, Ni, and Zn explain why obtaining oxidation states greater than +2 for these elements is difficult.

While ionisation enthalpies provide some insight into the relative stability of oxidation states, predicting stability is complex and not easily generalised.

ElementScTiVCrMnFeCoNiCuZn
Ionisation enthalpy/DiHV/kJ mol–1 I631656650653717762758736745906
Ionisation enthalpy/DiHV/kJ mol–1 II1235130914141592150915611644175219581734
Ionisation enthalpy/DiHV/kJ mol–1 III2393265728332990326029623243340235563837

Example 8.2. Why do the transition elements exhibit higher enthalpies of atomisation?

Answer:

Transition elements have a relatively large number of unpaired electrons in their inner (n-1)d orbitals. These electrons, along with the outer ns electrons, participate in interatomic metallic bonding. The involvement of more electrons in bonding leads to stronger metallic bonds between atoms. Therefore, more energy is required to break these bonds and convert the solid metal into individual gaseous atoms, resulting in higher enthalpies of atomisation compared to s-block and p-block elements.

Oxidation States

One of the most characteristic features of transition elements is their ability to display a variety of oxidation states in their compounds. This arises from the participation of both the $(n-1)d$ and $ns$ electrons in bonding, as the energies of these orbitals are quite similar.

ElementScTiVCrMnFeCoNiCuZn
Common Oxidation States+3+2, +3, +4+2, +3, +4, +5+2, +3, +4, +5, +6+2, +3, +4, +5, +6, +7+2, +3, +4, +6+2, +3, +4+2, +3, +4+1, +2+2

(Most common oxidation states are in bold type).

The elements exhibiting the largest number of oxidation states are typically found in or near the middle of the series. Manganese (Mn), for instance, shows oxidation states ranging from +2 to +7.

At the beginning and end of the series, the number of accessible oxidation states is limited. Elements like Scandium (Sc) and Titanium (Ti) at the start have only a few d electrons to lose or share. Scandium primarily exhibits +3, and Ti(IV) is more stable than Ti(III) or Ti(II).

At the end of the series, elements like Copper (Cu) and Zinc (Zn) have almost or completely filled d subshells. Zinc exhibits only the +2 oxidation state, as no d electrons are involved in achieving this state. Copper commonly shows +1 and +2.

The maximum oxidation states with reasonable stability often correspond to the sum of the $s$ and $d$ electrons, particularly up to Manganese (e.g., Ti(IV) in TiO$_2$, V(V) in VO$_2^+$, Cr(VI) in CrO$_4^{2-}$, Mn(VII) in MnO$_4^-$). Beyond Manganese, there is a rapid decrease in the stability of higher oxidation states, with more typical states being Fe(II, III), Co(II, III), Ni(II), Cu(I, II), and Zn(II).

A key characteristic of transition elements' variable oxidation states is that they often differ by a unit of one (e.g., V(II), V(III), V(IV), V(V)). This contrasts with non-transition elements, where oxidation states typically differ by two (e.g., P(III), P(V) or S(II), S(IV), S(VI)).

Within groups (specifically groups 4 to 10), an interesting trend is observed, which is opposite to the inert pair effect seen in p-block elements. In d-block groups, the higher oxidation states are favoured by heavier members. For example, in Group 6, Molybdenum(VI) and Tungsten(VI) are more stable than Chromium(VI). This explains why acidified dichromate (Cr(VI)) is a strong oxidising agent, while MoO$_3$ and WO$_3$ are not.

Low oxidation states (like 0 or +1) are typically observed in complex compounds containing ligands that can accept electron density from the metal via $\pi$ bonding ($\pi$-acceptor ligands), in addition to forming $\sigma$ bonds. For instance, in metal carbonyls like Ni(CO)$_4$ and Fe(CO)$_5$, the oxidation state of the metal is zero.

Example 8.3. Name a transition element which does not exhibit variable oxidation states.

Answer:

Scandium (Z = 21) is considered a transition element that primarily exhibits only a single oxidation state of +3.

Trends In The M2+/M Standard Electrode Potentials

The standard electrode potential (E°) for the reduction of a divalent metal ion to the solid metal (M$^{2+}$/M) is influenced by the energy changes involved in transforming the solid metal into gaseous atoms (enthalpy of atomisation, $\Delta_{a}H^{\circ}$), ionising the gaseous atoms to gaseous ions (first and second ionisation enthalpies, $\Delta_{i}H^{\circ}_1 + \Delta_{i}H^{\circ}_2$), and hydrating the gaseous ions to form aqueous ions (enthalpy of hydration, $\Delta_{hyd}H^{\circ}$).

The overall energy change relates to the standard electrode potential. A more negative E° indicates a greater tendency for the metal to be oxidised (or the ion to be reduced with more difficulty).

Element (M)$\Delta_{a}H^{\circ}$ (M)$\Delta_{i}H^{\circ}_1$$\Delta_{i}H^{\circ}_2$$\Delta_{hyd}H^{\circ}$ (M$^{2+}$)E$^{\circ}$/V M$^{2+}$/M
Ti4696561309-1866-1.63
V5156501414-1895-1.18
Cr3986531592-1925-0.90
Mn2797171509-1862-1.18
Fe4187621561-1998-0.44
Co4277581644-2079-0.28
Ni4317361752-2121-0.25
Cu3397451958-2121+0.34
Zn1309061734-2059-0.76

Copper (Cu) exhibits unique behaviour with a positive E° value (+0.34 V) for the Cu$^{2+}$/Cu couple. This positive value means that the reduction of Cu$^{2+}$ to Cu is thermodynamically favoured over the oxidation of Cu to Cu$^{2+}$ under standard conditions compared to other transition metals. This is why copper is unable to displace hydrogen from non-oxidising acids like dilute HCl or H$_2$SO$_4$. Copper only reacts with oxidising acids (like nitric acid or hot concentrated sulphuric acid), where the acid itself is reduced.

The high energy required to convert solid copper to aqueous Cu$^{2+}$ ions is primarily due to the relatively high enthalpy of atomisation of Cu and the sum of its first and second ionisation enthalpies ($\Delta_aH^{\circ} + \Delta_iH^{\circ}_1 + \Delta_iH^{\circ}_2$). This energy input is not fully offset by the hydration enthalpy of the Cu$^{2+}$ ion ($\Delta_{hyd}H^{\circ}$).

Across the first transition series from left to right, there is a general trend towards less negative E° values for the M$^{2+}$/M couple. This trend is largely associated with the general increase in the sum of the first and second ionisation enthalpies across the series.

However, the E° values for Manganese (Mn), Nickel (Ni), and Zinc (Zn) are more negative than the general trend would suggest. The more negative values for Mn and Zn are related to the relative stability of the d$^5$ configuration in Mn$^{2+}$ and the d$^{10}$ configuration in Zn$^{2+}$. For Nickel, the relatively more negative E° value is associated with its particularly high (most negative) enthalpy of hydration for Ni$^{2+}$.

Trends In The M3+/M2+ Standard Electrode Potentials

Examining the E° values for the redox couple M$^{3+}$/M$^{2+}$ provides insight into the relative stability of the +3 and +2 oxidation states. A more positive E° value for M$^{3+}$/M$^{2+}$ means that the M$^{3+}$ ion is more easily reduced to M$^{2+}$, implying M$^{3+}$ is a stronger oxidising agent and M$^{2+}$ is relatively stable.

Referring to Table 8.2 (reproduced below for convenience):

ElementScTiVCrMnFeCoNiCuZn
Standard electrode potential E$^{\circ}$/V M$^{3+}$/M$^{2+}$–0.37–0.26–0.41+1.57+0.77+1.97

The very low E° value for the Sc$^{3+}$/Sc$^{2+}$ couple (not listed, but Sc only stable as Sc$^{3+}$) reflects the exceptionally high stability of the Sc$^{3+}$ ion, which has a noble gas electronic configuration ([Ar] 3d$^0$). It does not readily accept an electron to form Sc$^{2+}$.

Zinc has no stable +3 state, so the Zn$^{3+}$/Zn$^{2+}$ potential is not relevant. The hypothetical removal of an electron from Zn$^{2+}$ (3d$^{10}$) is very difficult, hence a very high positive potential if we considered Zn$^{3+}$/Zn$^{2+}$.

The comparatively high positive value (+1.57 V) for the Mn$^{3+}$/Mn$^{2+}$ couple indicates that Mn$^{3+}$ is easily reduced to Mn$^{2+}$. This is because Mn$^{2+}$ has a stable d$^5$ configuration. Thus, Mn$^{3+}$ is a strong oxidising agent.

The comparatively low positive value (+0.77 V) for the Fe$^{3+}$/Fe$^{2+}$ couple suggests that Fe$^{3+}$ is also relatively stable (due to its d$^5$ configuration), but less so than Mn$^{2+}$. Fe$^{3+}$ can be reduced to Fe$^{2+}$, but it is a weaker oxidising agent compared to Mn$^{3+}$.

The very high positive value (+1.97 V) for the Co$^{3+}$/Co$^{2+}$ couple shows that Co$^{3+}$ is a very strong oxidising agent, readily reduced to Co$^{2+}$. Co$^{2+}$ (d$^7$) is significantly more stable in aqueous solution than Co$^{3+}$ (d$^6$).

The low value for the V$^{3+}$/V$^{2+}$ couple (–0.26 V) suggests that V$^{2+}$ is quite stable. This stability is related to the half-filled t$_{2g}$ level in the octahedral crystal field (a concept from Unit 9).

Example 8.4. Why is Cr$^{2+}$ reducing and Mn$^{3+}$ oxidising when both have d$^4$ configuration?

Answer:

Both Cr$^{2+}$ and Mn$^{3+}$ ions have a d$^4$ electronic configuration. However, their behaviour as reducing or oxidising agents is determined by the stability of the oxidation state they transition to.

Cr$^{2+}$ (d$^4$) acts as a reducing agent because it readily loses one electron to form Cr$^{3+}$ (d$^3$). In an octahedral complex (common in aqueous solutions), the d$^3$ configuration leads to the $t_{2g}$ level being half-filled ($t_{2g}^3 e_g^0$). This $t_{2g}^3$ configuration has significant crystal field stabilisation energy (CFSE), making Cr$^{3+}$ particularly stable in aqueous solutions compared to Cr$^{2+}$. The reaction is: Cr$^{2+}$ (d$^4$) $\rightarrow$ Cr$^{3+}$ (d$^3$) + e$^-$

Mn$^{3+}$ (d$^4$) acts as an oxidising agent because it readily gains one electron to form Mn$^{2+}$ (d$^5$). The d$^5$ configuration results in a half-filled d subshell, which confers extra stability due to exchange energy. The reaction is: Mn$^{3+}$ (d$^4$) + e$^-$ $\rightarrow$ Mn$^{2+}$ (d$^5$). The gain in stability from d$^4$ to d$^5$ is significant, driving the reduction of Mn$^{3+}$.

Trends In Stability Of Higher Oxidation States

The highest oxidation states for transition metals are most frequently observed when they combine with highly electronegative elements like Fluorine (F) and Oxygen (O). This is because fluorine and oxygen can effectively oxidise the metal to its maximum possible state.

Groups3456789101112
Oxidation NumberScTiVCrMnFeCoNiCuZn
+6CrF$_6$
+5VF$_5$CrF$_5$
+4TiX$_4$VX$^{\text{I}}_4$CrX$_4$MnF$_4$
+3ScX$_3$TiX$_3$VX$_3$CrX$_3$MnF$_3$FeX$^{\text{I}}_3$CoF$_3$
+2TiX$_2$VX$_2$CrX$_2$MnX$_2$FeX$_2$CoX$_2$NiX$_2$CuX$_2$ZnX$_2$
+1CuX$^{\text{III}}$

Key: X = F $\rightarrow$ I; X$^I$ = F $\rightarrow$ Br; X$^{II}$ = F, Cl; X$^{III}$ = Cl $\rightarrow$ I

The highest oxidation states achieved in simple halides are Ti(IV) in TiX$_4$, V(V) in VF$_5$, and Cr(VI) in CrF$_6$. Manganese's +7 state is not seen in simple halides, though MnO$_3$F exists. Beyond Manganese, achieving a +3 state in halides becomes less common, primarily limited to FeX$_3$ and CoF$_3$.

Fluorine's ability to stabilise high oxidation states is linked to either high lattice energies in ionic compounds (like CoF$_3$) or strong bond enthalpies in covalent compounds (like VF$_5$, CrF$_6$).

Fluorides can also exhibit instability in low oxidation states, such as VX$_2$ (X = Cl, Br, I) and CuX. Conversely, almost all Copper(II) halides are known, except CuI$_2$. This is because Cu$^{2+}$ ions oxidise iodide ions (I$^-$) to iodine (I$_2$) in solution:

$2\text{Cu}^{2+}\text{(aq)} + 4\text{I}^-\text{(aq)} \rightarrow 2\text{CuI(s)} + \text{I}_2\text{(aq)}$

Many Copper(I) compounds are unstable in aqueous solution and undergo disproportionation, where the ion is simultaneously oxidised and reduced:

$2\text{Cu}^+\text{(aq)} \rightarrow \text{Cu}^{2+}\text{(aq)} + \text{Cu(s)}$

The reason Cu$^{2+}$(aq) is more stable than Cu$^+$(aq) is that the hydration enthalpy ($\Delta_{hyd}H^{\circ}$) of Cu$^{2+}$ is significantly more negative than that of Cu$^+$. This large negative hydration enthalpy for Cu$^{2+}$ provides enough energy to more than compensate for the relatively high second ionisation enthalpy needed to form Cu$^{2+}$ from Cu$^+$.

Oxygen is even more effective than fluorine at stabilising the highest oxidation states. In oxides, the highest oxidation number matches the group number up to Manganese (from Sc$_2$O$_3$ for Group 3 to Mn$_2$O$_7$ for Group 7). Beyond Group 7, stable oxides do not show the maximum possible oxidation state; for instance, the highest stable oxide of Iron is Fe$_2$O$_3$ (+3 state), although unstable ferrates(VI) (FeO$_4^{2-}$) exist in alkaline media.

Oxocations also play a role in stabilising high oxidation states, such as V(V) in VO$_2^+$, V(IV) in VO$^{2+}$, and Ti(IV) in TiO$^{2+}$.

Oxygen's superiority in stabilising high oxidation states is likely due to its ability to form multiple bonds with the metal atom. In the covalent oxide Mn$_2$O$_7$, each Mn atom is linked by a Mn-O-Mn bridge, and the molecule involves multiple bonds between Mn and terminal O atoms.

Tetrahedral oxoanions [MO$_4$]$^{n-}$ are known for several metals in high oxidation states, including V(V), Cr(VI), Mn(V), Mn(VI), and Mn(VII).

Groups3456789101112
Oxidation NumberScTiVCrMnFeCoNiCuZn
+7Mn$_2$O$_7$
+6CrO$_3$
+5V$_2$O$_5$
+4TiO$_2$V$_2$O$_4$CrO$_2$MnO$_2$
+3Sc$_2$O$_3$Ti$_2$O$_3$V$_2$O$_3$Cr$_2$O$_3$Mn$_2$O$_3$Fe$_2$O$_3$
Mixed OxidesMn$_3$O$_4$*Fe$_3$O$_4$*Co$_3$O$_4$*
+2TiOVO(CrO)MnOFeOCoONiOCuOZnO
+1Cu$_2$O

* mixed oxides

Example 8.9. What is meant by ‘disproportionation’ of an oxidation state? Give an example.

Answer:

Disproportionation is a type of redox reaction where a single element in a specific oxidation state simultaneously undergoes oxidation (increase in oxidation state) and reduction (decrease in oxidation state) to form two or more different products containing the element in different oxidation states.

This occurs when a particular oxidation state is relatively unstable compared to both a lower and a higher oxidation state of the same element.

For example, the permanganate ion (MnO$_4^-$) in neutral or slightly alkaline solution contains Manganese in the +7 oxidation state. The manganate ion (MnO$_4^{2-}$) contains Manganese in the +6 state. The manganate ion is unstable in acidic solution and undergoes disproportionation:

Example: Disproportionation of Manganese(VI) in acidic solution.

Here, Manganese in the +6 oxidation state (in MnO$_4^{2-}$) is converted to Manganese(VII) (in MnO$_4^-$) and Manganese(IV) (in MnO$_2$).

$3\text{MnO}_4^{2-}\text{(aq)} + 4\text{H}^+\text{(aq)} \rightarrow 2\text{MnO}_4^-\text{(aq)} + \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$

(Manganese goes from +6 to +7 and +4).

Chemical Reactivity And EV Values

Transition metals exhibit a wide range of chemical reactivity. Many of them are electropositive enough to dissolve in mineral acids, while others are considered 'noble' because they are unreactive towards single acids.

Most metals of the first (3d) series, with the exception of copper, are relatively reactive. They can be oxidised by 1 M H$^+$ ions, although their reaction rate with oxidising agents like hydrogen ions can sometimes be slow. For instance, Titanium and Vanadium are practically passive to dilute non-oxidising acids at room temperature.

The E° values for the M$^{2+}$/M couple (listed in Table 8.2) generally indicate a decreasing tendency for metals to form stable divalent cations across the series from left to right. This general trend towards less negative E° values is related to the increasing sum of the first and second ionisation enthalpies.

As discussed previously, the E° values for Mn, Ni, and Zn are more negative than expected based on this general trend. The increased stability of d$^5$ in Mn$^{2+}$ and d$^{10}$ in Zn$^{2+}$ contributes to their more negative potentials, while for Nickel, it is related to the highly negative hydration enthalpy of Ni$^{2+}$.

Looking at the E° values for the M$^{3+}$/M$^{2+}$ redox couple (Table 8.2), it is evident that Mn$^{3+}$ and Co$^{3+}$ ions are the strongest oxidising agents in aqueous solutions because they are easily reduced to the more stable Mn$^{2+}$ and Co$^{2+}$ ions respectively.

Conversely, ions like Ti$^{2+}$, V$^{2+}$, and Cr$^{2+}$ are strong reducing agents. They readily lose electrons and are strong enough reductants to liberate hydrogen gas from dilute acids. For example:

$2\text{Cr}^{2+}\text{(aq)} + 2\text{H}^+\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + \text{H}_2\text{(g)}$

Example 8.6. For the first row transition metals the E$^{\circ}$ values are:

E$^{\circ}$ V Cr Mn Fe Co Ni Cu

(M$^{2+}$/M) –1.18 – 0.91 –1.18 – 0.44 – 0.28 – 0.25 +0.34

Explain the irregularity in the above values.

Answer:

The standard electrode potentials (E$^{\circ}$) for the M$^{2+}$/M system across the first transition series are not smooth or regular. This irregularity is primarily a consequence of the irregular variations in the sum of the first and second ionisation enthalpies ($\Delta_iH^{\circ}_1 + \Delta_iH^{\circ}_2$) and, to a lesser extent, the sublimation/atomisation enthalpies of these metals.

These thermodynamic factors, coupled with the hydration enthalpies of the M$^{2+}$ ions, collectively determine the E$^{\circ}$ values. The irregularities in IE sum and atomisation enthalpy, particularly noted for elements like Manganese (due to d$^5$ stability) and Vanadium, contribute significantly to the deviations from a smooth trend in E$^{\circ}$.

Example 8.7. Why is the E$^{\circ}$ value for the Mn$^{3+}$/Mn$^{2+}$ couple much more positive than that for Cr$^{3+}$/Cr$^{2+}$ or Fe$^{3+}$/Fe$^{2+}$? Explain.

Answer:

A highly positive E$^{\circ}$ value for the M$^{3+}$/M$^{2+}$ couple indicates that the reduction of M$^{3+}$ to M$^{2+}$ is thermodynamically favourable, meaning M$^{3+}$ is a strong oxidising agent and M$^{2+}$ is relatively stable.

For the Mn$^{3+}$/Mn$^{2+}$ couple, the reaction is Mn$^{3+}$ (d$^4$) + e$^-$ $\rightarrow$ Mn$^{2+}$ (d$^5$). The high positive E$^{\circ}$ (+1.57 V) is largely due to the significant stability gained by achieving the half-filled d$^5$ configuration in Mn$^{2+}$. Removing an electron from Mn$^{2+}$ to form Mn$^{3+}$ requires a very high third ionisation energy for Mn, which makes Mn$^{3+}$ easily reduced back to Mn$^{2+}$.

For the Cr$^{3+}$/Cr$^{2+}$ couple, the reaction is Cr$^{3+}$ (d$^3$) + e$^-$ $\rightarrow$ Cr$^{2+}$ (d$^4$). Cr$^{3+}$ (d$^3$) is relatively stable (half-filled t$_{2g}$ in octahedral field), and gaining an electron to form Cr$^{2+}$ (d$^4$) is not as strongly favoured as gaining an electron to form a d$^5$ configuration. Hence, the E$^{\circ}$ (–0.41 V) is negative.

For the Fe$^{3+}$/Fe$^{2+}$ couple, the reaction is Fe$^{3+}$ (d$^5$) + e$^-$ $\rightarrow$ Fe$^{2+}$ (d$^6$). Fe$^{3+}$ is stable due to the d$^5$ configuration. While it accepts an electron to form Fe$^{2+}$, the stability gain is not as pronounced as forming Mn$^{2+}$ from Mn$^{3+}$. The E$^{\circ}$ (+0.77 V) is positive, indicating Fe$^{3+}$ is an oxidising agent, but weaker than Mn$^{3+}$.

Thus, the large positive E$^{\circ}$ for Mn$^{3+}$/Mn$^{2+}$ is primarily driven by the strong tendency to form the very stable Mn$^{2+}$ (d$^5$) ion, which is reflected in the unusually high third ionisation enthalpy of Mn compared to Cr and Fe.

Magnetic Properties

When exposed to a magnetic field, substances exhibit different behaviours. The main types are diamagnetism (repelled by the field) and paramagnetism (attracted by the field). Substances that are very strongly attracted are called ferromagnetic (an extreme form of paramagnetism, like Fe, Co, Ni).

Many transition metal ions are paramagnetic. This property arises from the presence of unpaired electrons in their d orbitals. Each unpaired electron possesses a magnetic moment due to its spin and orbital motion.

For compounds of the first transition series (3d elements), the contribution of orbital angular momentum to the magnetic moment is often negligible or "quenched" due to interaction with the surrounding environment (ligand field). Therefore, the magnetic moment is primarily determined by the number of unpaired electrons and can be calculated using the 'spin-only' formula:

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

where:

A single unpaired electron has a spin-only magnetic moment of $\sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM.

The magnetic moment value increases with the number of unpaired electrons. Therefore, experimentally measured magnetic moments are valuable indicators of the number of unpaired electrons present in a transition metal species.

IonConfigurationUnpaired electron(s) (n)Magnetic moment Calculated (BM)Magnetic moment Observed (BM)
Sc$^{3+}$3d$^0$000
Ti$^{3+}$3d$^1$11.731.75
Tl$^{2+}$3d$^2$22.842.76
V$^{2+}$3d$^3$33.873.86
Cr$^{2+}$3d$^4$44.904.80
Mn$^{2+}$3d$^5$55.925.96
Fe$^{2+}$3d$^6$44.905.3 – 5.5
Co$^{2+}$3d$^7$33.874.4 – 5.2
Ni$^{2+}$3d$^8$22.842.9 – 3.4
Cu$^{2+}$3d$^9$11.731.8 – 2.2
Zn$^{2+}$3d$^{10}$000

(Observed values are typically for hydrated ions in solution or solid state).

Example 8.8. Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

Answer:

The element with atomic number 25 is Manganese (Mn).

The electronic configuration of a neutral Manganese atom is [Ar] 3d$^5$ 4s$^2$.

A divalent ion (M$^{2+}$) is formed by the loss of the two outermost electrons from the 4s orbital.

The electronic configuration of the divalent ion, Mn$^{2+}$, is [Ar] 3d$^5$.

In the 3d$^5$ configuration, according to Hund's rule, there are 5 unpaired electrons in the 3d orbitals ($n=5$).

Using the spin-only formula for magnetic moment ($\mu$):

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

Substitute $n=5$:

$\mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35}$ BM

Calculating the value: $\mu \approx 5.916$ BM

Therefore, the magnetic moment of the divalent ion with atomic number 25 (Mn$^{2+}$) in aqueous solution is approximately 5.92 BM.

Formation Of Coloured Ions

Many transition metal ions in solution or in solid compounds are coloured. This is a characteristic property linked to the presence of partially filled d orbitals.

The colour arises from the absorption of light energy that causes an electron to transition or be excited from a lower energy d orbital to a higher energy d orbital within the same subshell. This process is called a d-d transition.

In the presence of ligands or counterions, the d orbitals of the metal ion are split into different energy levels (as explained by Crystal Field Theory in Unit 9). The energy difference between these levels corresponds to specific frequencies of electromagnetic radiation.

When white light passes through a transition metal compound, the ion absorbs light of certain frequencies corresponding to the energy needed for d-d transitions. The colours that are *not* absorbed are transmitted or reflected, and these are the colours that we perceive.

The energy required for d-d transitions typically falls within the visible region of the electromagnetic spectrum. The specific colour observed is the complementary colour of the light that was absorbed. For instance, if a substance absorbs red light, it will appear green.

The frequency of light absorbed (and thus the colour observed) depends on the energy splitting of the d orbitals, which in turn is influenced by factors like the nature of the metal ion, its oxidation state, and the nature of the ligands surrounding it.

Ions with completely empty (d$^0$) or completely filled (d$^{10}$) d subshells cannot undergo d-d transitions because there are no electrons to excite or no vacant higher-energy d orbitals to receive them. Consequently, such ions are typically colourless.

ConfigurationExampleColour
3d$^0$Sc$^{3+}$colourless
3d$^0$Ti$^{4+}$colourless
3d$^1$Ti$^{3+}$purple
3d$^1$V$^{4+}$blue
3d$^2$V$^{3+}$green
3d$^3$V$^{2+}$violet
3d$^3$Cr$^{3+}$violet
3d$^4$Mn$^{3+}$violet
3d$^4$Cr$^{2+}$blue
3d$^5$Mn$^{2+}$pink
3d$^5$Fe$^{3+}$yellow
3d$^6$Fe$^{2+}$green
3d$^6$Co$^{3+}$blue
3d$^7$Co$^{2+}$pink
3d$^8$Ni$^{2+}$green
3d$^9$Cu$^{2+}$blue
3d$^{10}$Zn$^{2+}$colourless

(Colours listed are for aquated ions, where water molecules act as ligands).

Various coloured solutions of transition metal ions

(The image would visually show examples like aqueous solutions of V$^{4+}$ (blue), V$^{3+}$ (green), Mn$^{2+}$ (pink), Fe$^{3+}$ (yellow/brown), Co$^{2+}$ (pink), Ni$^{2+}$ (green), and Cu$^{2+}$ (blue)).

Formation Of Complex Compounds

Transition metals are known for forming a vast number of complex compounds (also called coordination compounds). In these compounds, a central metal ion (usually a transition metal) is bonded to a number of surrounding ions or neutral molecules, which are called ligands. These complexes have distinct properties from the constituent ions or molecules.

Examples include [Fe(CN)$_6$]$^{3-}$, [Fe(CN)$_6$]$^{4-}$, [Cu(NH$_3$)$_4$]$^{2+}$, and [PtCl$_4$]$^{2-}$. The formation and properties of complex compounds are studied in detail in Unit 9.

The ability of transition metals to form a large variety of complexes is attributed to three main factors:

  1. Relatively smaller size of the metal ions.
  2. High ionic charges (or high charge density) on the metal ions.
  3. Availability of vacant d orbitals of appropriate energy to accept electron pairs from the ligands (for coordinate covalent bond formation).

Catalytic Properties

Many transition metals and their compounds exhibit significant catalytic activity, meaning they can increase the rate of chemical reactions without being consumed themselves. This property is closely related to two features of transition elements:

  1. Their ability to exist in multiple oxidation states.
  2. Their capacity to form complex compounds.

Examples of transition metals or their compounds used as catalysts include Vanadium(V) oxide (V$_2$O$_5$) in the Contact Process for sulfuric acid production, finely divided Iron (Fe) in the Haber's Process for ammonia synthesis, and Nickel (Ni) in catalytic hydrogenation reactions.

In heterogeneous catalysis (where the catalyst is in a different phase, typically solid, than the reactants), surface catalysis by transition metals involves the formation of temporary bonds between the reactant molecules and the atoms on the catalyst surface. First-row transition metals utilise their 3d and 4s electrons for this bonding. This interaction helps in two ways:

Furthermore, the ability of transition metal ions to change their oxidation states allows them to participate in the reaction pathway by forming intermediate compounds, making them more effective catalysts. For instance, Iron(III) ions (Fe$^{3+}$) catalyse the reaction between iodide ions (I$^-$) and persulphate ions (S$_2$O$_8^{2-}$):

$2\text{I}^-\text{(aq)} + \text{S}_2\text{O}_8^{2-}\text{(aq)} \rightarrow \text{I}_2\text{(aq)} + 2\text{SO}_4^{2-}\text{(aq)}$

This reaction is believed to occur via the following steps involving the interconversion of iron oxidation states:

$2\text{Fe}^{3+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + \text{I}_2\text{(aq)}$

$2\text{Fe}^{2+}\text{(aq)} + \text{S}_2\text{O}_8^{2-}\text{(aq)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2\text{SO}_4^{2-}\text{(aq)}$

The Fe$^{3+}$ catalyst is regenerated at the end of the reaction cycle.

Formation Of Interstitial Compounds

Interstitial compounds are formed when small non-metal atoms like Hydrogen (H), Carbon (C), or Nitrogen (N) get trapped and occupy the voids or interstitial spaces within the crystal lattice of a metal. These compounds are well-known for transition metals because their crystal lattices often have appropriate sized interstitial holes.

These compounds are typically non-stoichiometric, meaning the ratio of the non-metal atom to the metal atom is not fixed and can vary within a range (e.g., TiC$_{0.98}$ or TiH$_{1.7}$). They do not conform to typical ionic or covalent bonding rules; their bonding is complex and includes contributions from metallic bonding.

Examples include TiC, Mn$_4$N, Fe$_3$H, VH$_{0.56}$, and TiH$_{1.7}$. The formulas reflect the composition range rather than fixed valency.

Interstitial compounds formed by transition metals possess several characteristic physical and chemical properties that differ significantly from the parent metals:

  1. They have high melting points, often considerably higher than those of the pure transition metals.
  2. They are remarkably hard; some metal borides, for instance, are nearly as hard as diamond.
  3. They generally retain metallic conductivity, although it may be lower than the parent metal.
  4. They are typically very chemically inert, being resistant to attack by acids, bases, and other reagents.

Alloy Formation

An alloy is a material composed of two or more metallic elements, or a metal and a non-metal, that are mixed together. Transition metals readily form alloys because they have similar atomic sizes and other metallic characteristics. Alloys are formed by metals whose atomic radii differ by no more than approximately 15 percent, allowing atoms of one metal to substitute for atoms of the other in the crystal lattice, forming homogeneous solid solutions.

Alloys formed by transition metals are often harder and have higher melting points than the constituent pure metals.

Some of the most important alloys are ferrous alloys, which are based on iron. Adding transition metals like Chromium (Cr), Vanadium (V), Tungsten (W), Molybdenum (Mo), and Manganese (Mn) to iron produces various types of steel, including stainless steel, which are crucial for construction and manufacturing.

Alloys involving transition metals and non-transition metals are also industrially significant. Examples include Brass (an alloy of Copper and Zinc) and Bronze (an alloy of Copper and Tin).



Some Important Compounds Of Transition Elements

Oxides And Oxoanions Of Metals

Transition metal oxides are typically formed by reacting the metals directly with oxygen at high temperatures. Most transition metals, with the exception of Scandium, form monoxides with the formula MO, which are ionic in nature.

As discussed earlier, the highest oxidation state observed in oxides often corresponds to the group number, up to Manganese (Group 7). For example, Sc$_2$O$_3$ (+3), TiO$_2$ (+4), V$_2$O$_5$ (+5), CrO$_3$ (+6), and Mn$_2$O$_7$ (+7).

Beyond Group 7, achieving the highest possible oxidation state in simple oxides becomes less common. For instance, the highest stable oxide of iron is Fe$_2$O$_3$ (+3), not FeO$_3$ (+6).

Besides simple oxides, transition metals can form oxocations that stabilise higher oxidation states, such as VO$_2^+$ (V in +5 state), VO$^{2+}$ (V in +4 state), and TiO$^{2+}$ (Ti in +4 state).

A notable trend in the oxides of transition metals is the relationship between oxidation state and acidic character. As the oxidation number of the metal increases, the ionic character of the oxide decreases, and the acidic character becomes predominant. For example, Mn$_2$O$_7$ (+7) is a covalent green oil and is acidic, forming permanganic acid (HMnO$_4$). CrO$_3$ (+6) is also acidic, forming chromic acid (H$_2$CrO$_4$) and dichromic acid (H$_2$Cr$_2$O$_7$).

Vanadium oxides show a gradual change in acidity with increasing oxidation state: V$_2$O$_3$ (+3) is basic, V$_2$O$_4$ (+4) is less basic, and V$_2$O$_5$ (+5) is amphoteric (reacts with both acids and bases), although mainly acidic. V$_2$O$_5$ dissolves in bases to form vanadates (like VO$_4^{3-}$) and in strong acids to form oxocations (like VO$_2^+$ or VO$^{2+}$ salts).

Similarly, CrO (+2) is basic, while Cr$_2$O$_3$ (+3) is amphoteric.

Potassium Dichromate K2Cr2O7

Potassium dichromate (K$_2$Cr$_2$O$_7$) is an important chemical widely used in the leather industry and as an oxidising agent in organic synthesis and volumetric analysis.

Dichromates are typically produced from chromates. Chromates are obtained by fusing chromite ore (FeCr$_2$O$_4$) with sodium or potassium carbonate in the presence of air:

$4\text{FeCr}_2\text{O}_4\text{(chromite)} + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \xrightarrow{\text{fusion}} 8\text{Na}_2\text{CrO}_4\text{(yellow)} + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2$

The resulting yellow solution of sodium chromate (Na$_2$CrO$_4$) is filtered and then acidified, usually with sulfuric acid. Acidification converts the chromate ions to orange dichromate ions:

$2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7\text{(orange)} + 2\text{Na}^+ + \text{H}_2\text{O}$

Sodium dichromate (Na$_2$Cr$_2$O$_7 \cdot 2\text{H}_2\text{O}$) can be crystallised from this solution. Since sodium dichromate is more soluble in water than potassium dichromate, potassium dichromate (K$_2$Cr$_2$O$_7$) is usually prepared by treating the sodium dichromate solution with potassium chloride. Due to its lower solubility, potassium dichromate precipitates out:

$\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7\text{(orange crystals)} + 2\text{NaCl}$

Chromates and dichromates exist in equilibrium in aqueous solution, and their interconversion is dependent on the pH of the solution. In acidic conditions, chromate ions (CrO$_4^{2-}$) are converted to dichromate ions (Cr$_2$O$_7^{2-}$). In alkaline conditions, dichromate ions are converted back to chromate ions. The oxidation state of Chromium in both chromate and dichromate ions is +6.

Interconversion reactions:

$2\text{CrO}_4^{2-}\text{(yellow)} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-}\text{(orange)} + \text{H}_2\text{O}$

The chromate ion (CrO$_4^{2-}$) has a tetrahedral structure. The dichromate ion (Cr$_2$O$_7^{2-}$) consists of two tetrahedra sharing a common oxygen atom, forming a bent Cr-O-Cr bridge with a bond angle of 126°.

Structures of chromate and dichromate ions

(The image would show the structures: CrO$_4^{2-}$ as a tetrahedron with Cr at the center and O at vertices; Cr$_2$O$_7^{2-}$ as two such tetrahedra connected by a shared oxygen atom, with the Cr-O-Cr angle shown as 126°).

Sodium and potassium dichromates are powerful oxidising agents. Sodium dichromate is more soluble and often used in organic chemistry. Potassium dichromate is frequently used as a primary standard in volumetric analysis.

In acidic solution, the dichromate ion is reduced to the Chromium(III) ion (Cr$^{3+}$). The half-reaction for its oxidising action is:

$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 14\text{H}^+\text{(aq)} + 6e^- \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)} \quad (E^{\circ} = +1.33 \text{ V})$

Acidified potassium dichromate can oxidise various substances. Examples of half-reactions for typical reducing agents include:

To obtain the balanced ionic equation for the reaction between dichromate and a reducing agent, the dichromate half-reaction is combined with the half-reaction of the reducing agent, balancing the number of electrons exchanged. For example, with Fe$^{2+}$:

$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 14\text{H}^+\text{(aq)} + 6\text{Fe}^{2+}\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 6\text{Fe}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}$

Potassium Permanganate KMnO4

Potassium permanganate (KMnO$_4$) is another powerful oxidising agent. It is prepared by the fusion of Manganese dioxide (MnO$_2$, pyrolusite ore) with an alkali metal hydroxide (like KOH) and an oxidising agent such as KNO$_3$ or air. This process yields dark green potassium manganate (K$_2$MnO$_4$).

$2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \rightarrow 2\text{K}_2\text{MnO}_4\text{(green)} + 2\text{H}_2\text{O}$

The green manganate solution is then treated in a neutral or acidic solution where it undergoes disproportionation to produce the purple permanganate:

$3\text{MnO}_4^{2-}\text{(green)} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^-\text{(purple)} + \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O}$

Alternatively, potassium permanganate can be prepared commercially by alkaline oxidative fusion of MnO$_2$ followed by electrolytic oxidation of the manganate(VI) ions in alkaline solution:

$\text{MnO}_2 \xrightarrow{\text{Fused with KOH, Oxidised with air or KNO}_3} \text{MnO}_4^{2-}\text{(manganate ion)}$

$\text{MnO}_4^{2-}\text{(manganate)} \xrightarrow{\text{Electrolytic oxidation in alkaline solution}} \text{MnO}_4^-\text{(permanganate ion)}$

In the laboratory, potassium permanganate can be prepared by oxidising Manganese(II) ions (Mn$^{2+}$) using peroxodisulphate ions (S$_2$O$_8^{2-}$):

$2\text{Mn}^{2+}\text{(aq)} + 5\text{S}_2\text{O}_8^{2-}\text{(aq)} + 8\text{H}_2\text{O(l)} \rightarrow 2\text{MnO}_4^-\text{(aq)} + 10\text{SO}_4^{2-}\text{(aq)} + 16\text{H}^+\text{(aq)}$

Potassium permanganate forms dark purple (almost black) crystals. It is isostructural with KClO$_4$. It has moderate solubility in water (6.4 g per 100 g water at 293 K) and decomposes when heated above 513 K:

$2\text{KMnO}_4\text{(s)} \xrightarrow{\Delta} \text{K}_2\text{MnO}_4\text{(s)} + \text{MnO}_2\text{(s)} + \text{O}_2\text{(g)}$

Potassium permanganate is known for its intense colour and its magnetic properties (diamagnetic with temperature-dependent weak paramagnetism), which are explained by molecular orbital theory.

Both the manganate (MnO$_4^{2-}$) and permanganate (MnO$_4^-$) ions have a tetrahedral structure. Pi ($\pi$) bonding is involved, formed by the overlap of oxygen p orbitals with manganese d orbitals. The green manganate ion (MnO$_4^{2-}$) is paramagnetic because it has one unpaired electron. The purple permanganate ion (MnO$_4^-$) is diamagnetic as it has no unpaired electrons.

Potassium permanganate is a strong oxidising agent, but its reduction product depends on the pH of the solution:

The high positive reduction potential in acidic medium (+1.52 V) suggests that acidic permanganate should be a very strong oxidising agent, capable of oxidising water ($2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$, $E^{\circ} = +1.23 \text{ V}$). While thermodynamically favourable, the reaction with water is kinetically very slow unless catalysed by Mn$^{2+}$ ions or at elevated temperatures.

Important oxidising reactions of KMnO$_4$:

1. In Acid Solutions:

2. In Neutral or Faintly Alkaline Solutions:

Note: Permanganate titrations cannot be performed accurately in the presence of hydrochloric acid (HCl) because HCl is oxidised to chlorine gas by permanganate.

Uses: Beyond its extensive use in analytical chemistry for titrations, potassium permanganate is a versatile oxidising agent in preparative organic chemistry. Its strong oxidising power is also utilised for bleaching materials like wool, cotton, silk, and other textile fibres, and for decolorising oils.



The Lanthanoids

The f-block of the periodic table is divided into two series: the Lanthanoids and the Actinoids. The lanthanoids comprise the fourteen elements that follow Lanthanum (La) in the periodic table (Cerium, Ce, to Lutetium, Lu).

Although Lanthanum (Z=57) is formally a d-block element, its properties are very similar to those of the lanthanoids. Therefore, it is conventionally discussed along with the lanthanoids, and the general symbol Ln is often used to represent any element in this series, including Lanthanum.

Lanthanoids show greater similarity among themselves compared to the variation seen among elements in a typical transition series. This is mainly because their characteristic properties are governed by the filling of the deeply buried 4f orbitals, while the outer electron configuration (5d$^x$ 6s$^2$) remains relatively constant.

They predominantly exhibit a single stable oxidation state (+3), which allows for the study of how small, progressive changes in size and nuclear charge affect the properties of otherwise similar elements.

Electronic Configurations

Atoms of lanthanoid elements generally have a 6s$^2$ electron configuration in their outermost shell, while the occupancy of the inner 4f and occasionally 5d levels varies.

The most stable oxidation state for all lanthanoids is +3. The electronic configurations of the tripositive ions (Ln$^{3+}$) are primarily of the form 4f$^n$, where n increases from 1 (for Ce$^{3+}$) to 14 (for Lu$^{3+}$) with increasing atomic number.

Atomic NumberNameSymbolElectronic configuration (M)Electronic configuration (Ln$^{2+}$)Electronic configuration (Ln$^{3+}$)Electronic configuration (Ln$^{4+}$)
57LanthanumLa5d$^1$6s$^2$5d$^1$4f$^0$
58CeriumCe4f$^1$5d$^1$6s$^2$4f$^2$4f$^1$4f$^0$
59PraseodymiumPr4f$^3$6s$^2$4f$^3$4f$^2$4f$^1$
60NeodymiumNd4f$^4$6s$^2$4f$^4$4f$^3$4f$^2$
61PromethiumPm4f$^5$6s$^2$4f$^5$4f$^4$
62SamariumSm4f$^6$6s$^2$4f$^6$4f$^5$
63EuropiumEu4f$^7$6s$^2$4f$^7$4f$^6$
64GadoliniumGd4f$^7$5d$^1$6s$^2$4f$^7$5d$^1$4f$^7$
65TerbiumTb4f$^9$6s$^2$4f$^9$4f$^8$4f$^7$
66DysprosiumDy4f$^{10}$6s$^2$4f$^{10}$4f$^9$4f$^8$
67HolmiumHo4f$^{11}$6s$^2$4f$^{11}$4f$^{10}$
68ErbiumEr4f$^{12}$6s$^2$4f$^{12}$4f$^{11}$
69ThuliumTm4f$^{13}$6s$^2$4f$^{13}$4f$^{12}$
70YtterbiumYb4f$^{14}$6s$^2$4f$^{14}$4f$^{13}$
71LutetiumLu4f$^{14}$5d$^1$6s$^2$4f$^{14}$5d$^1$4f$^{14}$

Atomic And Ionic Sizes

A defining characteristic of lanthanoids is the overall decrease in atomic and ionic radii from Lanthanum (Z=57) to Lutetium (Z=71). This phenomenon is known as the Lanthanoid contraction.

The decrease in atomic radii (derived from metallic structures) is not entirely smooth, but the decrease in the radii of the tripositive ions (Ln$^{3+}$) is more regular.

The lanthanoid contraction is caused by the imperfect shielding of the nuclear charge by the 4f electrons. As the atomic number increases across the series, the nuclear charge increases by one unit at each step, and an electron is added to a 4f orbital. Electrons within the same subshell do not shield each other very effectively. Specifically, the shielding provided by one 4f electron for another electron is less effective than that of a d electron. This results in a gradual increase in the effective nuclear charge experienced by the outer electrons, pulling the electron cloud closer to the nucleus and causing a contraction in size.

Atomic NumberNameSymbolRadii/pm (Ln)Radii/pm (Ln$^{3+}$)
57LanthanumLa187106
58CeriumCe183103
59PraseodymiumPr182101
60NeodymiumNd18199
61PromethiumPm18198
62SamariumSm18096
63EuropiumEu19995
64GadoliniumGd18094
65TerbiumTb17892
66DysprosiumDy17791
67HolmiumHo17689
68ErbiumEr17588
69ThuliumTm17487
70YtterbiumYb17386
71LutetiumLu
Trend in ionic radii of lanthanoids

(The image would show a graph depicting the decrease in ionic radii of Ln$^{3+}$ ions from La to Lu).

The lanthanoid contraction has significant implications for the chemistry of the elements that follow the lanthanoids in the periodic table, specifically the third (5d) transition series. The cumulative contraction causes the atomic radii of the 5d transition metals (from Hafnium, Hf, onwards) to be very similar to those of their corresponding elements in the 4d series. For example, Zr (4d, 160 pm) and Hf (5d, 159 pm) have nearly identical radii due to the lanthanoid contraction. This similarity in size accounts for their natural co-occurrence and the difficulty encountered in separating them.

Oxidation States

The most common and stable oxidation state for lanthanoids is +3. However, some lanthanoids can occasionally exhibit +2 and +4 oxidation states, particularly in solid compounds or specific solutions.

These variations from the typical +3 state are often linked to the extra stability associated with achieving empty (4f$^0$), half-filled (4f$^7$), or completely filled (4f$^{14}$) f subshells. This contributes to irregularities in trends like ionisation enthalpies and oxidation states.

In essence, the availability of stable f$^0$, f$^7$, or f$^{14}$ configurations drives some lanthanoids to adopt +2 or +4 states besides their common +3 state, and these states often exhibit strong reducing or oxidising properties as they tend to return to the +3 state.

General Characteristics

Lanthanoids share many similar physical and chemical properties:

Many trivalent lanthanoid ions (Ln$^{3+}$) are coloured in both solid compounds and aqueous solutions, with the exception of La$^{3+}$ (4f$^0$) and Lu$^{3+}$ (4f$^{14}$) which are colourless. The colour is due to f-f electronic transitions, similar to d-d transitions in transition metals. However, the absorption bands for f-f transitions are characteristically narrow because the 4f orbitals are deeply shielded from the external environment.

Lanthanoid ions (except those with f$^0$ or f$^{14}$ configurations, like La$^{3+}$, Ce$^{4+}$, Yb$^{2+}$, Lu$^{3+}$) are paramagnetic due to the presence of unpaired electrons in the 4f orbitals.

The first ionisation enthalpies (IE1) of lanthanoids are around 600 kJ mol$^{-1}$, and the second (IE2) are about 1200 kJ mol$^{-1}$, comparable to Calcium. Variations in the third ionisation enthalpies (IE3) are influenced by the stability of empty, half-filled, or completely filled 4f orbitals, similar to the exchange energy effects in d-block elements. Lanthanum, Gadolinium (4f$^7$), and Lutetium (4f$^{14}$) show abnormally low third ionisation enthalpies, reflecting the stability of their resulting f$^0$, f$^7$, and f$^{14}$ configurations in the +3 state.

In terms of chemical behaviour, the earlier members of the lanthanoid series are quite reactive, similar to Calcium. As the atomic number increases, their reactivity decreases, and they behave more like Aluminium. The standard electrode potentials (E°) for the half-reaction Ln$^{3+}$(aq) + 3e$^-$ $\rightarrow$ Ln(s) are generally in the range of –2.2 V to –2.4 V, with Europium (Eu) being an exception at –2.0 V. This indicates they are quite electropositive metals.

General reactions of lanthanoids (Ln):

The hydroxides, Ln(OH)$_3$, are distinct compounds (not just hydrated oxides) and are basic in nature, similar to alkaline earth metal hydroxides.

Diagram summarising chemical reactions of lanthanoids

(The image would depict a central 'Ln' symbol with arrows pointing outwards to indicate reactions with H$_2$, C, N$_2$, H$_2$O, acids, O$_2$, S, halogens, showing the resulting compounds).

The primary application of lanthanoids is in the production of alloy steels, particularly for plates and pipes. A notable alloy is mischmetall, which is composed of about 95% lanthanoid metals and 5% iron, with small amounts of other elements. Mischmetall is used in Magnesium-based alloys for items like bullets and shell casings, as well as for lighter flints.

Mixed oxides of lanthanoids serve as catalysts in processes such as petroleum cracking. Some individual lanthanoid oxides are used as phosphors, materials that emit light when energised, finding applications in television screens and fluorescent surfaces.



The Actinoids

The Actinoids form the second series of inner transition elements, comprising the fourteen elements that follow Actinium (Ac) in the periodic table (Thorium, Th, to Lawrencium, Lr).

Actinium (Z=89) is formally a d-block element, similar to Lanthanum for the lanthanoids. However, due to its close resemblance to the actinoids, it is typically included in discussions of this series.

A significant characteristic of actinoids is that all of them are radioactive. The earlier members of the series have relatively long half-lives, making them easier to study and handle in larger quantities. However, the later actinoids are highly unstable with very short half-lives, sometimes only a few minutes. These later elements can only be prepared and studied in extremely small quantities, which makes their chemistry more challenging to investigate thoroughly.

The chemistry of actinoids is considerably more complex than that of lanthanoids, largely because they exhibit a wider range of oxidation states.

Electronic Configurations

Actinoid elements are generally believed to have a 7s$^2$ electron configuration in their outermost shell, with variable occupancy of the inner 5f and 6d subshells.

The fourteen electrons are formally added to the 5f orbitals. While Thorium (Z=90) has a configuration [Rn] 6d$^2$ 7s$^2$ (or 5f$^0$ 6d$^2$ 7s$^2$), the filling of 5f orbitals effectively begins from Protactinium (Z=91) onwards and is completed at element 103, Lawrencium (Z=103).

Similar to lanthanoids, there are irregularities in the electronic configurations of some actinoids. These deviations are again related to the enhanced stability associated with the empty (5f$^0$), half-filled (5f$^7$), and completely filled (5f$^{14}$) 5f orbitals.

Examples of configurations showing these stabilities include Americium (Am), [Rn] 5f$^7$ 7s$^2$, and Curium (Cm), [Rn] 5f$^7$ 6d$^1$ 7s$^2$.

Although the 5f orbitals resemble the 4f orbitals in their angular wave functions, the 5f orbitals are less effectively shielded from the nucleus and protrude further outwards compared to the 4f orbitals. Consequently, the 5f electrons are less 'buried' and can participate in chemical bonding to a much greater extent than the 4f electrons.

Atomic NumberNameSymbolElectronic configuration (M)Electronic configuration (M$^{3+}$)Electronic configuration (M$^{4+}$)
89ActiniumAc6d$^1$7s$^2$5f$^0$
90ThoriumTh6d$^2$7s$^2$5f$^1$5f$^0$
91ProtactiniumPa5f$^2$6d$^1$7s$^2$5f$^2$5f$^1$
92UraniumU5f$^3$6d$^1$7s$^2$5f$^3$5f$^2$
93NeptuniumNp5f$^4$6d$^1$7s$^2$5f$^4$5f$^3$
94PlutoniumPu5f$^6$7s$^2$5f$^5$5f$^4$
95AmericiumAm5f$^7$7s$^2$5f$^6$5f$^5$
96CuriumCm5f$^7$6d$^1$7s$^2$5f$^7$5f$^6$
97BerkeliumBk5f$^9$7s$^2$5f$^8$5f$^7$
98CaliforniumCf5f$^{10}$7s$^2$5f$^9$5f$^8$
99EinsteniumEs5f$^{11}$7s$^2$5f$^{10}$5f$^9$
100FermiumFm5f$^{12}$7s$^2$5f$^{11}$5f$^{10}$
101MendeleviumMd5f$^{13}$7s$^2$5f$^{12}$5f$^{11}$
102NobeliumNo5f$^{14}$7s$^2$5f$^{13}$5f$^{12}$
103LawrenciumLr5f$^{14}$6d$^1$7s$^2$5f$^{14}$5f$^{13}$

Ionic Sizes

Actinoids also exhibit a gradual decrease in the size of their atoms and M$^{3+}$ ions across the series, similar to the lanthanoids. This effect is known as the Actinoid contraction.

However, the contraction from one element to the next in the actinoid series is generally greater than in the lanthanoid series. This larger contraction is attributed to the even poorer shielding provided by the 5f electrons compared to the 4f electrons. The 5f electrons are less effective at shielding the increasing nuclear charge, leading to a stronger pull on the outer electrons and a more pronounced decrease in size across the series.

Atomic NumberNameSymbolRadii/pm (M$^{3+}$)Radii/pm (M$^{4+}$)
89ActiniumAc111
90ThoriumTh99
91ProtactiniumPa96
92UraniumU10393
93NeptuniumNp10192
94PlutoniumPu10090
95AmericiumAm9989
96CuriumCm9988
97BerkeliumBk9887
98CaliforniumCf9886
99EinsteniumEs
100FermiumFm
101MendeleviumMd
102NobeliumNo
103LawrenciumLr

Oxidation States

Actinoids display a greater range and variability of oxidation states compared to lanthanoids. This is attributed, in part, to the relatively comparable energies of the 5f, 6d, and 7s electron shells, allowing more electrons to be involved in bonding.

AcThPaUNpPuAmCmBkCfEsFmMdNoLr
Oxidation States 33, 43, 4, 53, 4, 5, 63, 4, 5, 6, 73, 4, 5, 6, 73, 4, 5, 63, 43, 4333333

The most common oxidation state for actinoids is +3, similar to lanthanoids. However, elements in the first half of the series frequently exhibit higher oxidation states.

For example, the maximum oxidation state increases from +4 in Thorium (Th) to +5 in Protactinium (Pa), +6 in Uranium (U), and +7 in Neptunium (Np) and Plutonium (Pu). This trend reverses in the later elements, with the maximum oxidation state decreasing.

While actinoids have more compounds in the +3 state than in the +4 state, both +3 and +4 ions of actinoids tend to undergo hydrolysis in aqueous solution.

Due to the uneven distribution and variation in oxidation states, particularly between the early and later actinoids, it is often less straightforward to systematically review their chemistry based solely on oxidation states compared to the more consistent +3 state of lanthanoids.

General Characteristics And Comparison With Lanthanoids

Actinoid metals generally have a silvery appearance. They exhibit a wider variety of crystal structures compared to lanthanoids. This structural variability is linked to the more significant irregularities in their metallic radii across the series.

Actinoids are highly reactive metals, especially when in a finely divided state. They react with boiling water to form a mixture of oxide and hydride, and readily combine with most non-metals at moderate temperatures.

Dilute hydrochloric acid attacks all actinoid metals. However, nitric acid has less effect on most actinoids, likely due to the formation of a protective oxide layer on the surface (passivation). Alkaline solutions generally do not react with actinoid metals.

The magnetic properties of actinoids are more complex than those of lanthanoids. While the general trend in magnetic susceptibility with the number of unpaired 5f electrons is somewhat similar to the lanthanoids, the actinoids typically show lower magnetic moment values.

Evidence suggests that the ionisation enthalpies of the early actinoids are lower than those of the early lanthanoids. This is expected because the 5f orbitals, when they begin to be filled, are less penetrating into the inner electron core than the 4f orbitals. This means the 5f electrons are more effectively shielded from the nucleus compared to the 4f electrons in corresponding lanthanoids.

Because the outer electrons in actinoids are less tightly bound, they are more readily available for participation in chemical bonding, contributing to the wider range of oxidation states observed.

Comparing actinoids and lanthanoids:

CharacteristicLanthanoidsActinoids
Electronic Configuration[Xe] 4f$^{0-14}$ 5d$^{0-1}$ 6s$^2$[Rn] 5f$^{0-14}$ 6d$^{0-1}$ 7s$^2$
Primary Oxidation State+3 (most stable)+3 (most common, but wider range of higher states exist)
Other Oxidation StatesOccasionally +2 and +4 (linked to f$^0$, f$^7$, f$^{14}$ stability)More common +4, +5, +6, +7, especially in early members (due to comparable energy levels of 5f, 6d, 7s)
ContractionLanthanoid contraction (gradual decrease in size)Actinoid contraction (greater decrease in size per element due to poorer 5f shielding)
Shielding4f electrons provide better shielding than 5f5f electrons provide poorer shielding than 4f
Bonding Involvement4f electrons are deeply buried, less involved in bonding beyond +3 state5f electrons are less buried, more involved in bonding, contributing to higher oxidation states
RadioactivityMostly non-radioactive (except Pm)All are radioactive
ReactivityGenerally reactive, similar to alkaline earth metals (early members) or Al (later members)Highly reactive, particularly when finely divided
Chemical SimilarityMore similar to each other, especially in +3 stateShow greater variability, similarity mainly among elements in the second half of the series

The behaviour similar to the close-knit properties of lanthanoids is more apparent in the second half of the actinoid series. However, even the early actinoids show similarities with each other and gradual property variations without necessarily changing oxidation state.

Both the lanthanoid and actinoid contractions have lasting effects on the sizes and properties of the elements that succeed them in their respective periods (5d and 6d transition series). The lanthanoid contraction is currently considered more significant in its impact because the chemistry of elements after the actinoids (transactinide elements) is much less understood due to their extreme radioactivity and short lifetimes.

Example 8.10. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.

Answer:

Cerium (Z = 58) is the member of the lanthanoid series that is well known to exhibit the +4 oxidation state (Ce$^{4+}$), besides the common +3 state. This +4 state corresponds to a stable 4f$^0$ electron configuration.

Intext Question 8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer:

Actinoid contraction refers to the decrease in atomic and ionic size across the actinoid series. It is more pronounced than the lanthanoid contraction because the 5f electrons provide less effective shielding of the increasing nuclear charge compared to the 4f electrons in the lanthanoids. The 5f orbitals are less effectively "buried" within the electron core. As the atomic number increases and electrons are added to the 5f orbitals, the increased nuclear charge is less effectively counterbalanced by the repulsion from other 5f electrons. This results in a stronger net attraction on the outer electrons, causing a larger contraction in size from one element to the next in the actinoid series compared to the lanthanoid series.



Some Applications Of D- And F-Block Elements

D-block and f-block elements and their compounds have numerous important applications:



Intext Questions



Question 8.1. Silver atom has completely filled d orbitals ($4d^{10}$) in its ground state. How can you say that it is a transition element?

Answer:

Question 8.2. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol$^{-1}$. Why?

Answer:

Question 8.3. Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why?

Answer:

Question 8.4. The $E^\ominus(M^{2+}/M)$ value for copper is positive (+0.34V). What is possible reason for this? (Hint: consider its high $\Delta_aH^\ominus$ and low $\Delta_{hyd}H^\ominus$)

Answer:

Question 8.5. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

Answer:

Question 8.6. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Answer:

Question 8.7. Which is a stronger reducing agent $Cr^{2+}$ or $Fe^{2+}$ and why ?

Answer:

Question 8.9. Explain why $Cu^+$ ion is not stable in aqueous solutions?

Answer:

Question 8.10. Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer:



Exercises



Question 8.1. Write down the electronic configuration of:

(i) $Cr^{3+}$

(ii) $Pm^{3+}$

(iii) $Cu^{+}$

(iv) $Ce^{4+}$

(v) $Co^{2+}$

(vi) $Lu^{2+}$

(vii) $Mn^{2+}$

(viii) $Th^{4+}$

Answer:

Question 8.2. Why are $Mn^{2+}$ compounds more stable than $Fe^{2+}$ towards oxidation to their +3 state?

Answer:

Question 8.3. Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Answer:

Question 8.4. To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Answer:

Question 8.5. What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : $3d^3$, $3d^5$, $3d^8$ and $3d^4$?

Answer:

Question 8.6. 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.

Answer:

Question 8.7. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Answer:

Question 8.8. What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Answer:

Question 8.9. In what way is the electronic configuration of the transition elements different from that of the non transition elements?

Answer:

Question 8.10. What are the different oxidation states exhibited by the lanthanoids?

Answer:

Question 8.11. Explain giving reasons:

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

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

(iii) The transition metals generally form coloured compounds.

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

Answer:

Question 8.12. What are interstitial compounds? Why are such compounds well known for transition metals?

Answer:

Question 8.13. How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.

Answer:

Question 8.14. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Answer:

Question 8.15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i) iodide

(ii) iron(II) solution and

(iii) $H_2S$

Answer:

Question 8.16. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) $SO_2$ and (iii) oxalic acid? Write the ionic equations for the reactions.

Answer:

Question 8.17. For $M^{2+}/M$ and $M^{3+}/M^{2+}$ systems the $E^\ominus$ values for some metals are as follows:

$Cr^{2+}/Cr \quad -0.9V \quad \quad Cr^{3}/Cr^{2+} \quad -0.4 V$

$Mn^{2+}/Mn \quad -1.2V \quad \quad Mn^{3+}/Mn^{2+} \quad +1.5 V$

$Fe^{2+}/Fe \quad -0.4V \quad \quad Fe^{3+}/Fe^{2+} \quad +0.8 V$

Use this data to comment upon:

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

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

Answer:

Question 8.18. Predict which of the following will be coloured in aqueous solution? $Ti^{3+}$, $V^{3+}$, $Cu^{+}$, $Sc^{3+}$, $Mn^{2+}$, $Fe^{3+}$ and $Co^{2+}$. Give reasons for each.

Answer:

Question 8.19. Compare the stability of +2 oxidation state for the elements of the first transition series.

Answer:

Question 8.20. 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.

Answer:

Question 8.21. How would you account for the following:

(i) Of the $d^4$ species, $Cr^{2+}$ 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.

Answer:

Question 8.22. What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.

Answer:

Question 8.23. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Answer:

Question 8.24. Calculate the number of unpaired electrons in the following gaseous ions: $Mn^{3+}$, $Cr^{3+}$, $V^{3+}$ and $Ti^{3+}$. Which one of these is the most stable in aqueous solution?

Answer:

Question 8.25. 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.

Answer:

Question 8.26. Indicate the steps in the preparation of:

(i) $K_2Cr_2O_7$ from chromite ore.

(ii) $KMnO_4$ from pyrolusite ore.

Answer:

Question 8.27. What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

Answer:

Question 8.28. 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.

Answer:

Question 8.29. 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.

Answer:

Question 8.30. 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.

Answer:

Question 8.31. Use Hund’s rule to derive the electronic configuration of $Ce^{3+}$ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.

Answer:

Question 8.32. 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.

Answer:

Question 8.33. Compare the chemistry of the actinoids with that of lanthanoids with reference to:

(i) electronic configuration

(ii) oxidation states and

(iii) chemical reactivity.

Answer:

Question 8.34. Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

Answer:

Question 8.35. 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.

Answer:

Question 8.36. Write down the number of 3d electrons in each of the following ions: $Ti^{2+}$, $V^{2+}$, $Cr^{3+}$, $Mn^{2+}$, $Fe^{2+}$, $Fe^{3+}$, $Co^{2+}$, $Ni^{2+}$ and $Cu^{2+}$. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

Answer:

Question 8.37. Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Answer:

Question 8.38. What can be inferred from the magnetic moment values of the following complex species ?

Example Magnetic Moment (BM)
$K_4[Mn(CN)_6]$ 2.2
$[Fe(H_2O)_6]^{2+}$ 5.3
$K_2[MnCl_4]$ 5.9

Answer: