D And F Block Elements – JEE Main ChemistryPractice Questions & PYQs

The oxidation state of chromium in chromyl chloride (\text{CrO}_2\text{Cl}_2$) is calculated using the equation$x + 2(-2) + 2(-1) = 0$, which yields$x = +6$. With an atomic number of 24, the neutral chromium configuration is$[\text{Ar}] 3d^5 4s^1$, so the$\text{Cr}^{6+}$ion possesses an electronic configuration of$[\text{Ar}] 3d^0$, resulting \in zero d-electrons. Evaluating the given options,$\text{Fe}^{3+}$retains five d-electrons ([\text{Ar}] 3d^5$), while $\text{V}^{4+}$ and $\text{Ti}^{3+}$ each retain one d-electron ([\text{Ar}] 3d^1$). The manganese ion$\text{Mn}^{7+}$has an atomic number of 25 with a ground state configuration of$[\text{Ar}] 3d^5 4s^2$, and the loss of seven electrons leaves the configuration as$[\text{Ar}] 3d^0$, which matches the zero d-electrons found in chromyl chloride.

The reaction between iodide ions and hydrogen peroxide generates molecular iodine, denoted as compound 'A', as the experiment proceeds. To observe the rate of this process, the indicator 'X', which is starch, is added to the solution because it has a specific affinity for the produced iodine. When starch combines with the iodine molecules, it forms a distinct blue complex that serves as a clear visual marker for the presence of iodine. This interaction allows for the direct monitoring of the $I_2$ concentration in the reaction mixture, where $I^-$ reacts with $H_2O_2$ to drive the process.

An oxidizing agent functions by gaining electrons, meaning its chemical strength is defined by its inherent tendency to undergo reduction. Lanthanides typically favor a positive oxidation state of three as their most stable configuration, which causes species in higher oxidation states to strive to return to this favorable electronic arrangement. As a result, $\mathrm{Ce}^{4+}$ and $\mathrm{Tb}^{4+}$ act as strong oxidizing agents by readily accepting electrons to achieve that more stable state.

Nessler's reagent is technically classified as potassium tetraiodomercurate(II), a coordination complex used primarily to detect the presence of ammonia. This chemical species is accurately represented by the formula $\mathrm{K}_2\mathrm{HgI}_4$, which indicates the coordination of four iodide ions around the central mercury atom, balanced by two potassium cations.

The spin-only magnetic moment of a metal ion is determined by the formula $\mu_s = \sqrt{n(n+2)}$ BM, where $n$ represents the number of unpaired electrons in the ion. Because the value of the magnetic moment depends strictly on $n$, ions containing an equal number of unpaired electrons will possess an identical spin-only magnetic moment.

The electronic configurations of these ions in the gaseous state reveal the number of unpaired electrons for each. The $V^{3+}$ ion has a configuration of $[Ar] 3d^2$, which contains 2 unpaired electrons. The $Cr^{3+}$ ion follows with a configuration of $[Ar] 3d^3$, resulting in 3 unpaired electrons. The $Fe^{2+}$ ion has a configuration of $[Ar] 3d^6$, leading to 4 unpaired electrons. Finally, the $Ni^{3+}$ ion has a configuration of $[Ar] 3d^7$, which also results in 3 unpaired electrons. Given these values, $Cr^{3+}$ and $Ni^{3+}$ both contain 3 unpaired electrons, meaning there are 2 metal ions in the list with a similar spin-only magnetic moment.

The ability of a metal to displace hydrogen gas from mineral acids depends on its standard oxidation potential, where only metals with positive values relative to hydrogen can spontaneously undergo the reaction. Upon reviewing the provided data, $E_{\mathrm{Mn}/\mathrm{Mn}^{+2}} = 1.18$, $E_{\mathrm{Zn}/\mathrm{Zn}^{+2}} = 0.76$, and $E_{\mathrm{Ni}/\mathrm{Ni}^{+2}} = 0.25$, all of which are positive and indicate the ability to liberate hydrogen. In contrast, the oxidation potential for copper is $E_{\mathrm{Cu}/\mathrm{Cu}^{+2}} = -0.34$, and because this value is negative, it lacks the necessary thermodynamic drive to displace hydrogen ions, making it the metal with the least tendency to liberate $\mathrm{H}_2$.

Liberating hydrogen from dilute mineral acids requires a species to act as a reducing agent, characterized by a negative standard reduction potential. While the other ions in the set can participate in such reactions, $\mathrm{Co}^{3+}$ behaves as a robust oxidizing agent with a positive reduction potential of $E^\circ_{\mathrm{Co}^{3+}/\mathrm{Co}^{2+}} = +1.97\, \mathrm{V}$, preventing it from liberating hydrogen gas.

The electronic configuration of $\mathrm{Co}^{3+}$ is $[\mathrm{Ar}] 3d^6$, meaning it possesses 4 unpaired electrons. The spin-only magnetic moment is defined by $\mu = \sqrt{n(n+2)}$ B.M., where $n$ denotes the number of unpaired electrons. Substituting $n = 4$ into this equation gives $\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.92\, \mathrm{B.M.}$, which leads to 5 when rounded to the nearest integer.

Isoelectronic species are defined as chemical entities that possess an identical number of electrons, regardless of the identity of the parent nuclei. To determine if a given pair meets this criterion, one calculates the total electron count by subtracting the net ionic charge from the atomic number of the neutral atom. For the pair $\mathrm{Tb}^{2+}$ and $\mathrm{Tm}^{4+}$, this calculation reveals that $\mathrm{Tb}^{2+}$ has $65 - 2 = 63$ electrons, while $\mathrm{Tm}^{4+}$ has $69 - 4 = 65$ electrons. Because these electron counts are not equivalent, the two species are not isoelectronic.

Permanganate ions act as versatile oxidizing agents, with their specific reduction products dictated by the pH of the reaction environment. In neutral or faintly alkaline conditions, the manganese atom within the permanganate ion, $\mathrm{MnO}_4^{-}$, initiates the reaction with an oxidation state of +7. As it oxidizes thiosulphate, the manganese itself is reduced to form the insoluble solid manganese dioxide, $\mathrm{MnO}_2$, where the manganese atom now holds an oxidation state of +4. Calculating the difference between these initial and final states reveals that the overall change in the oxidation number of manganese is 3.

The reaction between silver nitrate and a coordination complex results in the precipitation of $\mathrm{AgCl}$ only for the chloride ions located outside the coordination sphere. Because two equivalents of $\mathrm{AgCl}$ are produced, exactly two chloride ions must be present as counter ions outside the coordination bracket. Expressing the complex as $[\mathrm{Co}(\mathrm{NH}_3)_x\mathrm{Cl}_{(3-2)}]\mathrm{Cl}_2$, the cobalt atom requires a total of six ligands to fulfill its coordination geometry. Assigning the one remaining chloride ion as a ligand leaves five sites to be filled by ammonia, establishing that $x$ equals 5 to form the complex $[\mathrm{Co}(\mathrm{NH}_3)_5\mathrm{Cl}]\mathrm{Cl}_2$.

Disproportionation is a redox process in which a single chemical species is simultaneously oxidized and reduced. When Manganese (VI) ions undergo this transformation in an acidic medium, they generate the permanganate ion ($MnO_4^-$) and manganese dioxide ($MnO_2$). Within these products, manganese exists in the +7 oxidation state in $MnO_4^-$ and the +4 oxidation state in $MnO_2$. Taking the difference between these two oxidation states results in $7 - 4 = 3$.

Enthalpy of hydration for transition metal ions depends on both the ionic radius and the Crystal Field Stabilization Energy (CFSE). While hydration enthalpy generally becomes more exothermic with smaller ionic radii, the electronic configuration is decisive; specifically, $\mathrm{Mn}^{2+}$ has a $d^5$ configuration with zero CFSE, whereas ions such as $\mathrm{Cr}^{2+}$, $\mathrm{Fe}^{2+}$, and $\mathrm{Co}^{2+}$ gain additional stabilization. Consequently, the observed hydration enthalpies are -1925 kJ/mole for $\mathrm{Cr}^{2+}$, -1862 kJ/mole for $\mathrm{Mn}^{2+}$, -1998 kJ/mole for $\mathrm{Fe}^{2+}$, and -2079 kJ/mole for $\mathrm{Co}^{2+}$, making the value for $\mathrm{Mn}^{2+}$ the least negative among the set.

The validity of these statements hinges on specific catalytic and bonding properties in inorganic chemistry. Statement I is false because $\mathrm{Fe}^{3+}$ is not used as a catalyst for the oxidation of $\mathrm{I}^{-}$ to $\mathrm{I}_2$. Statement II is also false because, while the manganate ion ($\mathrm{MnO}_4^{2-}$) is indeed paramagnetic, its bonding is defined by $d\pi - p\pi$ overlap between the metal center and the ligands, which contradicts the claim of $p\pi - p\pi$ bonding.

Zinc is an amphoteric metal, which means it reacts with strong bases to yield hydrogen gas. When treated with an excess of aqueous alkali, the metal undergoes an oxidation-reduction reaction resulting in the formation of the soluble complex ion known as tetrahydroxozincate(II), as shown by the equation $\mathrm{Zn} + 2\,\mathrm{OH}^- + 2\,\mathrm{H}_2\mathrm{O} \longrightarrow [\mathrm{Zn}(\mathrm{OH})_4]^{2-} + \mathrm{H}_2\uparrow$. Although the anhydrous zincate ion $\mathrm{ZnO}_2^{2-}$ is sometimes cited in chemical conventions, the presence of the aqueous medium ensures that the stable species formed is the hydrated complex ion, $[\mathrm{Zn}(\mathrm{OH})_4]^{2-}$.

The colour of transition metal ions generally arises from d-d transitions, a process that requires the presence of unpaired electrons to absorb light from the visible region. Because $\mathrm{Sc}^{3+}$ and $\mathrm{Zn}^{2+}$ contain no unpaired electrons in their d-orbitals, they cannot facilitate these electronic transitions and therefore appear colourless. In contrast, ions such as $\mathrm{Cu}^{2+}$, $\mathrm{Ti}^{3+}$, $\mathrm{V}^{2+}$, and $\mathrm{Mn}^{2+}$ possess unpaired electrons, leading to the absorption of specific wavelengths and the manifestation of colour.

The arrangement of electrons in an atom typically follows the Aufbau principle, yet many heavy transition metals like Platinum, with an atomic number of 78, exhibit exceptional configurations to achieve greater stability. After accounting for the 54 electrons within the Xenon noble gas core, the remaining 24 electrons are expected to fill the orbitals in a way that minimizes energy, but Platinum deviates from the standard pattern to reach a more favorable electronic state. This results in an configuration where one electron occupies the $6s$ orbital while the $5d$ subshell holds nine, yielding the stable arrangement of $[\mathrm{Xe}] 4f^{14} 5d^9 6s^1$.

Acidified manganate, $MnO_4^{2-}$, undergoes a disproportionation reaction to form permanganate, $MnO_4^-$, and manganese dioxide, $MnO_2$, as described by the equation $3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2O$ Manganese transitions from the 6 oxidation state to the 7 oxidation state \in the permanganate ion and the 4 state \in manganese dioxide, identifying the permanganate ion as the product with the higher oxidation state. Given that the $Mn^{7+}$ ion has an electronic configuration of $3d^0$, there are no unpaired electrons, leading to a spin-only magnetic moment of 0 B.M.

Thermal decomposition and specific redox interactions dictate the gaseous products evolved in these chemical transformations. Ammonium dichromate undergoes self-oxidation-reduction when heated, liberating nitrogen gas through $(\mathrm{NH}_4)_2\mathrm{Cr}_2\mathrm{O}_7 \xrightarrow{\Delta} \mathrm{N}_2 + 4\,\mathrm{H}_2\mathrm{O} + \mathrm{Cr}_2\mathrm{O}_3$, while the strong oxidizing action of potassium permanganate on hydrochloric acid releases chlorine gas via $\mathrm{KMnO}_4 + \mathrm{HCl} \rightarrow \mathrm{KCl} + \mathrm{MnCl}_2 + \mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O}$. Furthermore, aluminium reacts with sodium hydroxide to displace hydrogen gas as shown by $\mathrm{Al} + \mathrm{NaOH} + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Na}(\mathrm{Al}(\mathrm{OH})_4) + \mathrm{H}_2$, and sodium nitrate thermally dissociates into oxygen gas following the equation $2\,\mathrm{NaNO}_3 \xrightarrow{\Delta} 2\,\mathrm{NaNO}_2 + \mathrm{O}_2$.

The spin-only magnetic moment of a transition metal ion is governed by the number of unpaired electrons according to the formula $\mu = \sqrt{n(n+2)}$ Bohr magnetons, meaning the value decreases as the number of unpaired electrons decreases. Comparing the provided ions, $V^{2+}$ contains 3d3 with 3 unpaired electrons, while both $Cr^{2+}$ and $Fe^{2+}$ have 3d4 and 3d6 configurations containing 4 unpaired electrons, respectively. The $Ni^{2+}$ ion uniquely possesses a 3d8 configuration with only 2 unpaired electrons. Consequently, because it has the lowest number of unpaired electrons among these species, it exhibits the lowest spin-only magnetic moment.

Lanthanoid elements predominantly exist in the $+3$ oxidation state, yet electronic stability can drive some to deviate from this norm. Cerium, having the electronic configuration $[Xe] 4f^1 5d^1 6s^2$, can easily lose four electrons to form $Ce^{+4}$, resulting in the exceptionally stable noble gas configuration of $[Xe]$. This attainment of a stable closed-shell electronic arrangement makes the $+4$ oxidation state a favorable configuration for Cerium, distinguishing it from other lanthanoids that strictly adhere to the $+3$ state.