Class 12 Chemistry Chapter 4 The D and F Blocks Elements

Key concept: Element Identification

a) Cr
[Solution Description] The element with the electronic configuration $3d^5 4s^1$ is chromium (Cr). This is an exception to the typical filling order due to the stability provided by a half-filled d-subshell.

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Key concept: Configuration Exceptions

d) Palladium
[Solution Description] The typical electronic configuration for d-block elements is $(n-1)d^{1-10}ns^{1-2}$. However, there are exceptions due to stability associated with half-filled and completely filled orbitals. Palladium (Pd) is one such element where instead of having a $4d^95s^1$ configuration, it achieves a stable state through full filling of its 4d orbital resulting in $4d^{10}5s^0$.

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Key concept: Physical Properties

d) Assertion is false, but Reason is true.

[Solution Description] Assertion: The dark purple color of potassium permanganate is not due to unpaired electrons but rather due to charge transfer between manganese and oxygen atoms within the permanganate ion.

Reason: The intense color is indeed related to light absorption by the crystal structure, which aligns with the assertion about its physical property but does not explain the role of unpaired electrons as mentioned in the assertion.

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Key concept: Textile Bleaching

d) Bleaching
[Solution Description] In the textile industry, potassium permanganate is primarily used for bleaching purposes due to its strong oxidizing properties, which help in removing unwanted colors and impurities.

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Key concept: Catalytic Reactions

b) Adsorption of reactant
[Solution Description] In nickel-catalyzed hydrogenation, the critical step involves the adsorption of hydrogen molecules onto the nickel surface. Here, the 3d and 4s electrons of nickel play a crucial role in forming bonds with the hydrogen atoms. This electron sharing weakens the H-H bond, facilitating the addition of hydrogen to unsaturated bonds in the reactants, such as alkenes or alkynes, thereby completing the hydrogenation process.

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Key concept: Reaction Prediction

c) Zinc $(Zn^{2+}/Zn)$

[Solution Description] Only metals with a standard reduction potential lower than that of hydrogen ($H^+/H_2$) can liberate hydrogen gas from dilute acids. Metals like zinc, which have more negative $E^\circ$ values compared to $0 \, V$ for hydrogen, are capable of this reaction.

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Key concept: Basic Position

c) Middle section
[Solution Description] The d-block elements are situated between the s- and p-blocks in the periodic table, occupying the middle section.

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Key concept: Trends in Density

c) Hafnium (Hf) and Zirconium (Zr)

[Solution Description] Lanthanoid contraction refers to the gradual decrease in ionic radii and atomic radii of the lanthanoids from La to Lu, leading to lesser variation in size among elements of subsequent periods. As a result, elements like hafnium (Hf) and zirconium (Zr), which are in the same group but different periods, exhibit similar sizes and therefore similar densities.

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Key concept: Basic Uses

a) As an oxidant in organic reactions
[Solution Description] Potassium permanganate is used as an oxidizing agent in organic chemistry and for bleaching textiles, due to its strong oxidizing properties.

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Key concept: Properties of Interstitial Compounds

b) High melting point
[Solution Description] Interstitial compounds typically have high melting points due to the strong metallic bonding present in their structure. Small atoms like hydrogen, carbon, or nitrogen fit into the lattice of the metal, which enhances the strength and stability of these compounds.

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Key concept: Nuclear Energy, Preparation and Properties of Strong Oxidising Agents

b) Europium due to its high neutron absorption
[Solution Description] Boron is frequently used in nuclear reactor control rods because of its excellent ability to absorb neutrons without undergoing fission itself. Though boron is not an inner transition metal, in terms of nuclear chemistry connections, europium is an inner transition metal known for similar properties regarding neutron capture. In the context of strong oxidizing agents like potassium dichromate, which involves chromium—a transition metal—such materials are essential for driving reactions where electron transfer and redox stability are critical, somewhat analogous to the stability required in managing nuclear reactions.

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Key concept: Consequences

c) Similar radii of elements like Zr and Hf
[Solution Description] Lanthanoid contraction refers to the regular decrease in atomic and ionic sizes across the lanthanoid series. This contraction occurs due to the imperfect shielding of nuclear charge by 4f electrons, which leads to a more significant effective nuclear charge experienced by the outer electrons. As a result, there is a similarity in radii between elements of the second and third transition series, particularly Zr and Hf.

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Key concept: Stability Analysis

c) Due to half-filled electronic configuration

[Solution Description] $Fe^{3+}$ has an electronic configuration of $[\text{Ar}] 3d^5$, which is half-filled and hence more stable due to exchange energy stabilization. In contrast, $Fe^{2+}$ with $[\text{Ar}] 3d^6$ does not have this stability feature, making $Fe^{3+}$ more stable towards further oxidation.

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Key concept: Density Trend

c) Copper (Cu)
[Solution Description] Density generally increases across a period due to an increase in atomic mass and decrease in atomic volume. Among transition metals from titanium to copper, copper has a much higher density compared to others.

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Key concept: Stability of Oxidation States

a) Due to the half-filled stability of +2 state
[Solution Description] The +3 oxidation state of manganese is not very stable due to the instability of the electronic configuration that results from the removal of a single electron from a half-filled d subshell. Half-filled and completely filled subshells are particularly stable, so deviations from these configurations make the oxidation states less stable.

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Key Concepts:Observed vs Calculated

a) Ni$^{2+}$
[Solution Description] Nickel ($ Ni^{2+} $) has an electron configuration of $[Ar]\,3d^8$. It exhibits significant deviations from calculated magnetic moments due to considerable spin-orbit coupling effects, resulting in higher observed values compared to calculated ones using the spin-only approach. Other options may not have such significant spin-orbit coupling effects.

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Key concept: Colour Change with Oxidation State

a) Co$^{2+}$ to Co$^{3+}$
[Solution Description] Co$^{2+}$ is typically pink in solution while Co$^{3+}$ appears pale yellow. Upon oxidation, Co$^{2+}$ is converted to Co$^{3+}$, resulting in the observed colour change.

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Key concept: Energy Level Analysis

b) Due to similar energy levels of (n-1)d and ns orbitals
[Solution Description] Transition elements have electrons in both the (n-1)d and ns orbitals. The energy levels of these orbitals are very close, allowing for flexible electron exchange which leads to multiple oxidation states.

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Key concept: Reactivity

c) They have low ionisation enthalpies
[Solution Description] Actinoids are highly reactive due to their partially filled f orbitals which can easily participate in chemical reactions. When finely divided, their increased surface area enhances interaction with reactants such as boiling water and non-metals.

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Key concept: Element Classification

c) Assertion is true, but Reason is false.

[Solution Description] The assertion is true because f-block elements are indeed referred to as inner transition metals due to their position in the periodic table and their unique electronic configurations. However, the reason provided is false because f-block elements do not have partially filled d-orbitals; instead, they have partially filled f-orbitals. Hence, while the assertion is true, the reason does not correctly describe it.

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Key concept: Application of Concepts

c) A complex compound that forms colored ions
[Solution Description] The compound $[Fe(CN)_6]^{4-}$ is a complex compound, which exhibits the characteristic property of forming colored ions due to d-d transitions in its electronic configuration. In transition metal complexes, the presence of d orbitals allows electrons to absorb visible light, causing them to appear colored.

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Key concept: Element Comparison

a) Zr and Hf
[Solution Description] Due to the lanthanoid contraction, the radii of Zr (zirconium) and Hf (hafnium) become nearly identical. This happens because the additional f-electrons do not effectively shield the increased nuclear charge, resulting in a radius for Hf that is very close to that of Zr despite its higher atomic number.

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Key concept: Irregularities

c) Zinc (Zn)
[Solution Description] The standard electrode potentials $(E^\circ)$ for transition metals show certain irregularities, particularly for Mn, Ni, and Zn. Typically, these metals have more negative $E^\circ$ values than what might be anticipated from general trends because of specific electronic configurations or other energetic factors. Among them, zinc (Zn), due to its fully filled $d^{10}$ configuration, exhibits this behavior.

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Key Concept:Complex Formation

d) Availability of vacant d orbitals
[Solution Description] Transition elements readily form complex compounds due to their small ionic size and the availability of vacant d orbitals which can accommodate electron pairs donated by ligands, facilitating complex formation.

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Key concept: Oxidation State and Ligands

b) +2
[Solution Description] Let’s denote the oxidation state of nickel by x. Cyanide $(CN^-)$ has a charge of -1. Since there are four cyanide ions, their total charge is -4. The overall charge of the complex is -2, which can be represented by:
$x + 4(-1) = -2$
Solving for $x$,
$x – 4 = -2$
$x = 2$
Therefore, the oxidation state of nickel in this complex is +2.

b) +2
[Solution Description] Let’s denote the oxidation state of nickel by x. Cyanide $(CN^-)$ has a charge of -1. Since there are four cyanide ions, their total charge is -4. The overall charge of the complex is -2, which can be represented by:
$x + 4(-1) = -2$
Solving for $x$,
$x – 4 = -2$
$x = 2$
Therefore, the oxidation state of nickel in this complex is +2.

b) +2
[Solution Description] Let’s denote the oxidation state of nickel by x. Cyanide $(CN^-)$ has a charge of -1. Since there are four cyanide ions, their total charge is -4. The overall charge of the complex is -2, which can be represented by:
$x + 4(-1) = -2$
Solving for $x$,
$x – 4 = -2$
$x = 2$
Therefore, the oxidation state of nickel in this complex is +2.

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Key concept: Electronic Configuration

a) $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$
[Solution Description] The general electronic configuration for lanthanoids is represented by $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$. This reflects the filling of 4f orbitals.

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Key concept: Composition Elements

b) Hydrogen, Carbon, Nitrogen
[Solution Description] Small atoms like Hydrogen (H), Carbon (C), and Nitrogen (N) are commonly trapped in the crystal lattices of metals to form interstitial compounds.

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Key concept: Oxidation States

c) +2

[Solution Description] Manganese exhibits several oxidation states ranging from +2 to +7. In its compounds, the +2 and +4 states are common, but Mn in the +2 state is the most stable and frequently encountered due to its electron configuration ([Ar] $3d^5 4s^2$, losing two electrons gives a stable half-filled d-subshell).

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Key concept: Irregularity Explanation

d) Due to stable electronic configurations

[Solution Description] Mn, Ni, and Zn exhibit more negative $E^\circ$ values than expected due to specific stability factors associated with their electronic configurations. For Mn, it is the half-filled d-sublevel 4(d^5)$, for Ni, it is a stable $\text{Ni}^{2+}$ state, and for Zn, it has a filled d-sublevel $(d^{10})$. These stable electronic configurations lead to more energy being released upon ionisation, thus making the $E^\circ$ values more negative.

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Key concept: Oxidation State Distribution

c) Assertion is true, but Reason is false.

[Solution Description] – Actinoids do exhibit a variety of oxidation states due to their electron configurations involving 5f, 6d, and 7s orbitals. This allows +3, +4, and sometimes higher oxidation states.

– However, the reason provided is incorrect. The 5f electrons in actinoids actually provide poor shielding compared to the 4f electrons in lanthanoids. As a result, 5f electrons are more exposed to nuclear charge, making this Reason false.

– Therefore, while both aspects of the assertion that actinoids have varied oxidation states are true, the reason given is not correct.

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Key concept: Electronic Configuration Details

c) $[Rn] 5f^{10} 7s^2$

[Solution Description] The general electronic configuration of actinoids is $[Rn] 5f^{n}6d^m7s^2$. For Californium, the atomic number is 98. This means it has 98 electrons to fill. It falls in the series where $5f$ subshell can be occupied by up to 10 electrons. Thus, the electronic configuration for Californium is $[Rn] 5f^{10}7s^2$.

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Key concept: Colour and Ligand Relationship

b) It determines the frequency of light absorbed
[Solution Description] The nature of the ligand affects the splitting of d orbitals in the transition metal ion. Different ligands cause different extents of d orbital splitting, which changes the energy gap between these orbitals. The specific frequency of light that is absorbed corresponds to this energy difference, thus affecting the perceived colour of the complex.

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Key concept: Paramagnetic vs Diamagnetic

b) $Ni^{2+}$
[Solution Description] A substance is paramagnetic if it has unpaired electrons. Looking at the given ions:
$- Zn^{2+}$: Electron configuration is $[\text{Ar}] \, 3d^{10}$; it has no unpaired electrons, so it is diamagnetic.
$- Ni^{2+}$: Electron configuration is $[\text{Ar}] \, 3d^8$; it has two unpaired electrons, so it is paramagnetic.
$- Cu^{+}$: Electron configuration is $[\text{Ar}] \, 3d^{10}$; it has no unpaired electrons, so it is diamagnetic.
$- Sc^{3+}$: Electron configuration is $[\text{Ar}]$; it has no unpaired electrons, so it is diamagnetic.

Therefore, $Ni^{2+}$ is paramagnetic.

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Key concept: Configuration Comparison

c) Ni and Pd

[Solution Description] Both Ni and Pd have electronic configurations where all d-orbitals are filled, with electrons in the s-orbital. Specifically, Ni has the configuration $3d^{10}4s^0$, while Pd has $4d^{10}5s^0$. This means they both have a noble gas-like d-subshell.

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Key concept: Transition Metal Alloys

c) Chromium
[Solution Description] Chromium is a transition metal that is commonly used in the production of high-strength steel alloys. Its inclusion enhances the alloy’s strength, hardness, and resistance to corrosion.

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Key concept: Spin-Only Magnetic Moment

b) 5.92 BM
[Solution Description] The spin-only magnetic moment is calculated using the formula

\[
\mu_s = \sqrt{n(n+2)}
\]

where n is the number of unpaired electrons. For

\[
\text{Fe}^{3+}_4
\]

, the electronic configuration is $d\^5$, with 5 unpaired electrons. Therefore,

n = 5
\[
\mu_s = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92
\]

Thus, the spin-only magnetic moment is approximately 5.92 BM (Bohr Magneton).

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Key concept: Structure Identification

c) 126°
[Solution Description] The dichromate ion ($Cr_2O_7^{2-}$) consists of two tetrahedral chromate ions joined by an oxygen bridge with a characteristic bond angle between the chromium atoms through the bridging oxygen. The Cr–O–Cr bond angle in the dichromate ion is approximately 126°.

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Key concept: Density Trends

c) It increases

[Solution Description] Generally, as you move from titanium to copper across the transition series, the atomic mass increases while the atomic size remains relatively constant or slightly decreases. This leads to an increase in density. Therefore, the density of these elements tends to increase.

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Key concept: Oxidation States

d) Manganese (Mn)

[Solution Description] Manganese (Mn) exhibits the highest number of oxidation states among transition metals, ranging from -3 to +7. This diversity in oxidation states is due to its ability to use electrons from both 3d and 4s orbitals in bonding.

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Key concept: Bond Formation

c) d and s orbitals
[Solution Description] In transition metal catalysis at solid surfaces, 3d and 4s electrons play significant roles in forming bonds with reactant molecules, facilitating the catalytic process.

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Key concept: Variable Oxidation States

c) As a result of incomplete d-orbital filling.
[Solution Description] Iron exhibits variable oxidation states primarily because of the involvement of its 3d and 4s electrons in bonding. The electronic configuration of iron is $$[\text{Ar}]\,3d^6\,4s^2$$. In different chemical environments, Fe can lose either two or three electrons to form +2 or +3 oxidation states respectively. This variability arises due to the relatively close energies of the 3d and 4s orbitals, allowing for both configurations to be energetically feasible.

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Key concept: Shielding Effect

b) Poor shielding by 5f electrons
[Solution Description] Actinoid contraction is caused by poor shielding effect of the 5f electrons. These electrons are not effective in reducing nuclear charge experienced by outer electrons, resulting in a decrease in atomic size across the series.

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Key concept: Role of d-block elements

d) Nickel

[Solution Description] In hydrogenation reactions, alkenes are converted to alkanes using a catalyst. A common catalyst used in these reactions is nickel ($Ni$). Nickel provides active sites that facilitate the adsorption of hydrogen gas and the alkene, allowing the hydrogen atoms to add across the double bond effectively.

As such, nickel is widely used for this purpose due to its effectiveness and cost-efficiency compared to other metals.

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Key concept: Reaction with Acids

d) Lanthanoid chloride

[Solution Description] Lanthanoids typically react with acids to form their corresponding salts and hydrogen gas. When reacting with dilute HCl, a lanthanoid forms lanthanoid chloride and hydrogen gas. The reaction can be represented as:

$$2Ln + 6HCl \rightarrow 2LnCl_3 + 3H_2$$

where $Ln$ represents any lanthanoid metal.

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Key concept: Oxidation State and Stability

a) $\text{Mn}_2\text{O}_3$
[Solution Description] Manganese(III) oxide $(\text{Mn}_2\text{O}_3)$, cobalt(II) chloride $(\text{CoCl}_2)$, iron(II) sulfate $(\text{FeSO}_4)$, and copper(I) iodide $(\text{CuI})$ are given. Among these, $\text{Mn}_2\text{O}_3$ is less stable as $\text{Mn}^{3+}$ tends to disproportionate into $\text{Mn}^{2+}$ and $\text{Mn}^{4+}$.

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Key concept: Oxidation State Range

b) +2 to +7
[Solution Description] Manganese exhibits a wide range of oxidation states from +2 to +7. This variety arises because manganese can lose different numbers of electrons based on its chemical environment and bonding conditions, allowing it to participate in numerous redox reactions.

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Key Concept: Reaction Mechanism, Kinetics of Redox Reactions

c) $Fe^{3+}$
[Solution Description] In acidic media, $KMnO_4$ reduces from $MnO_4^-$ to $Mn^{2+}$. During the oxidation of ferrous ion ($Fe^{2+}$) by $KMnO_4$, $Fe^{3+}$ is an intermediate product before final reduction of $MnO_4^-$. Excess permanganate ensures all $Fe^{2+}$ is oxidized to ferric ion $(Fe^{3+})$ without forming any further intermediates.

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Key concept: Oxidation State Variability

c) Manganese
[Solution Description] Manganese exhibits a wide range of oxidation states from +2 to +7, thanks to its ability to lose electrons from both the 3d and 4s orbitals. This makes it capable of forming compounds in various oxidation states.

Key concept: Transition Element Identification

b) Zn

[Solution Description] Zn is not considered a transition element as it does not have partly filled d orbitals in any of its oxidation states.

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Key concept: Configuration Comparison

d) Chromium (Cr)
[Solution Description] The electronic configurations of Chromium (Cr) and Copper (Cu) deviate from the typical pattern because of their unusual arrangements that help achieve extra stability. Cr has a configuration of $[Ar] \, 3d^5 \, 4s^1$, which deviates to stabilize the half-filled d subshell. Similarly, Cu with $[Ar] \, 3d^{10} \, 4s^1$ stabilizes its full d subshell. These deviations are known exceptions among transition metals.

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Key concept: Oxidation State Range

d) Neptunium
[Solution Description] The tendency of an element to exhibit a certain oxidation state depends on the relative stability of its electronic configuration across different energy levels. In actinoids, elements like uranium and neptunium can exhibit high oxidation states because their 5f, 6d, and 7s orbitals have similar energies, allowing electrons to be removed relatively easily. Among the given options, Neptunium is known for exhibiting higher oxidation states such as +7.

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Key concept: Surface Catalysis

b) 3d and 4s
[Solution Description] When performing surface catalysis, first-row transition metals typically use their 3d and 4s electrons for forming bonds with reactant molecules.

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Key concept: Actinoid vs Lanthanoid, Catalytic Mechanism, Industrial Applications

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Actinoids indeed exhibit complex catalytic behaviors due to their ability to change between a variety of oxidation states, facilitated by the involvement of 5f orbitals. This property makes them versatile in forming multiple coordination complexes which can be useful in industrial catalysis. On the other hand, lanthanoids primarily utilize 4f orbitals, which are less involved in bonding, limiting their catalytic versatility compared to actinoids. Hence, both statements are true and the reason correctly explains the assertion that actinoids have more complex catalytic behavior than lanthanoids.

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Key concept: Spin-Only Formula

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] To determine the correctness of both the assertion and reason, we begin by calculating the magnetic moment using the provided spin-only formula, $\mu = \sqrt{n(n+2)}$, where n is the number of unpaired electrons.

Step 1: Identify the value of n. According to the assertion, there are three unpaired electrons ($n = 3$).

Step 2: Substitute $n = 3$ into the formula:

$$\mu = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \text{ BM}$$

The calculated magnetic moment matches the given assertion value of approximately 3.87 BM, verifying that both the Assertion and Reason are true, and the reason provides the correct explanation.

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