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

The molecule $\mathrm{Mn}_2 \mathrm{O}_7$ adopts a structure consisting of two $\mathrm{MnO}_3$ tetrahedra joined by a single shared oxygen atom. Within this geometry, every Manganese atom is connected to three distinct oxygen atoms via double bonds, while the remaining bond for each Manganese connects to the shared central oxygen. Since each of the two Manganese centers features three such terminal bonds, the molecule contains a total of 6 $\mathrm{Mn} = \mathrm{O}$ bonds.

The ability of a chemical species to act as an oxidizing agent is directly determined by its standard reduction potential, where a more positive value indicates a stronger tendency to accept electrons. Comparing the potential values for the given ions, we have $E^{\circ}_{\mathrm{Mn}^{3+}/\mathrm{Mn}^{2+}} = +1.57\;\mathrm{V}$, $E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = +0.77\;\mathrm{V}$, $E^{\circ}_{\mathrm{Ti}^{3+}/\mathrm{Ti}^{2+}} = -0.37\;\mathrm{V}$, and $E^{\circ}_{\mathrm{Cr}^{3+}/\mathrm{Cr}^{2+}} = -0.41\;\mathrm{V}$. Because $\mathrm{Mn}^{3+}$ possesses the most positive reduction potential, it exhibits the greatest ease of reduction, confirming it as the strongest oxidizing agent in the series.

The electronic stability of lanthanide ions is often determined by the extra exchange energy gained from half-filled or completely-filled subshells. Terbium, which has an atomic number of 65, exhibits a neutral ground state configuration of $4f^9 6s^2$; upon forming the $\text{Tb}^{4+}$ ion, it reduces to a stable $4f^7$ configuration. Similarly, Ytterbium, with an atomic number of 70, starts with a ground state of $4f^{14} 6s^2$, where the removal of two electrons to form $\text{Yb}^{2+}$ leaves the $4f^{14}$ subshell intact, satisfying the condition for a completely-filled state.

The chemical process begins with the oxidation of $\mathrm{FeCr}_2\mathrm{O}_4$ in the presence of $\mathrm{Na}_2\mathrm{CO}_3$ to produce sodium chromate, $\mathrm{Na}_2\mathrm{CrO}_4$, which serves as product A. In the presence of acidic conditions, two chromate ions undergo a condensation reaction to form the dichromate ion, $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, identified here as product B. The structure of $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ consists of two $\mathrm{CrO}_4$ tetrahedra joined by a single oxygen atom, which acts as a bridge between the two chromium centers. Because the total oxygen count in the dichromate ion is seven and one oxygen atom is shared as this bridge, the number of terminal oxygen atoms-those attached to only one chromium atom-is calculated as $7 - 1$, resulting in 6 terminal oxygen atoms.

Disproportionation reactions involve the simultaneous oxidation and reduction of the same reactant species. In an acidic environment, the unstable manganate ion, $\text{MnO}_4^{2-}$, produces the permanganate ion, $\text{MnO}_4^-$, where manganese is in the +7 oxidation state, and manganese dioxide, $\text{MnO}_2$, where manganese is in the +4 oxidation state. Since the question specifies that compound $B$ has a lower oxidation state than compound $A$, $B$ must correspond to $\text{MnO}_2$. The electronic configuration of the $\text{Mn}^{4+}$ ion is $3d^3$, which contains $n = 3$ unpaired electrons. Substituting this value into the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$ results in $\sqrt{3(5)}$, or $\sqrt{15}$, which evaluates to approximately 3.87 BM. Rounding this result to the nearest integer yields 4.

Transition metal oxides exhibit a clear trend where basicity decreases as the metal's oxidation state increases. Among the provided compounds, $V_2O_3$ contains vanadium in the +3 oxidation state, while $V_2O_4$ and $V_2O_5$ contain vanadium in the +4 and +5 oxidation states, respectively. Because acidity correlates positively with higher oxidation states, $V_2O_3$ stands out as the most basic compound in this set.

To determine the spin-only magnetic moment, we must identify the electronic state of the vanadium ion in $V_2O_3$. The $V^{+3}$ ion results in a $3d^2$ configuration, which leaves two unpaired electrons. Applying the spin-only formula $\mu = \sqrt{n(n+2)}$, where $n$ represents the number of unpaired electrons, we calculate $\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ B.M. Rounding this value to the nearest integer provides a result of 3.

Metals are generally characterized by strong metallic bonding that results in high melting points, yet this specific group contains notable exceptions where those forces are relatively weak. While mercury exists as a liquid at -38.83 °C and both cesium and gallium melt at the extremely low temperatures of 28.44 °C and 29.76 °C, respectively, silver maintains a far more robust atomic arrangement. This stronger metallic structure necessitates significantly higher thermal energy to overcome, resulting in a melting point of 961.8 °C.

Permanganate acts as a powerful oxidizing agent that reduces to manganese dioxide when present in a neutral or alkaline environment. In the presence of thiosulphate, this reaction follows the balanced stoichiometry of $8 \mathrm{MnO}_4^{-} + 3 \mathrm{S}_2\mathrm{O}_3^{2-} + \mathrm{H}_2\mathrm{O} \rightarrow 8 \mathrm{MnO}_2 + 6 \mathrm{SO}_4^{2-} + 2 \mathrm{OH}^{-}$. The conversion of the thiosulphate species throughout this process demonstrates that the sulfur atoms are oxidized to the sulphate ion, $\mathrm{SO}_4^{2-}$.

Among the manganese fluorides listed, $\mathrm{MnF}_3$ serves as the strongest oxidizing agent because the Mn(III) center is highly inclined to accept an electron to reduce to the more stable Mn(II) state. Manganese has an atomic number of 25, meaning the Mn(III) ion possesses an electronic configuration of $[Ar] 3d^4$, which results in 4 unpaired electrons. Calculating the spin-only magnetic moment using the formula $\mu = \sqrt{n(n+2)}$ requires substituting 4 for $n$, which results in the equation $\mu = \sqrt{4(4+2)} = \sqrt{24}$. This calculates to approximately 4.89 B.M., and rounding this value to the nearest integer yields 5.

Although the $\text{Ce}^{4+}$ ion achieves a stable noble gas electronic configuration, its high positive charge makes it energetically eager to reduce back to the $\text{Ce}^{3+}$ state. This tendency is evidenced by the standard reduction potential of the reaction $\text{Ce}^{4+} + e^- \rightarrow \text{Ce}^{3+}$, which stands at +1.61 V. Because the ion acts as a strong electron acceptor, it is classified as a potent oxidizing agent that facilitates redox reactions by gaining electrons.

To determine the oxidation state of the central metal in these ions, one balances the total charge of the polyatomic species against the known oxidation state of oxygen, which is -2. For the chromate ion $\mathrm{CrO}_4^{2-}$ and the dichromate ion, this calculation confirms that chromium exists in the +6 oxidation state in both cases. Since the oxidation states of chromium are identical in both species, the difference between them is exactly zero.

Stability among lanthanoids is largely determined by the pursuit of favorable electronic arrangements, specifically those featuring half-filled subshells. Europium (Eu) with atomic number 63 possesses the electronic configuration $[\mathrm{Kr}] 4f^7 6s^2$, and it attains a highly stable state by losing two electrons from the $6s$ orbital to form the $Eu^{2+}$ ion. This results in a half-filled $4f$ subshell, a configuration that offers substantial energetic stability and makes the divalent form of Europium the most stable among those given.

The central metal ion in the complex $[\mathrm{MnBr}_6]^{4-}$ is manganese in the +2 oxidation state because each bromide ligand carries a charge of -1. The $\mathrm{Mn}^{2+}$ ion has an electronic configuration of $3d^5$. Since bromide is a weak-field ligand, the crystal field splitting is small, resulting in a high-spin state with 5 unpaired electrons in the $t_{2g}^3 e_g^2$ arrangement. Using the spin-only magnetic moment formula $\mu = \sqrt{n(n+2)}$, where $n$ represents the 5 unpaired electrons, the calculation becomes $\mu = \sqrt{5(5+2)}$, which simplifies to $\mu = \sqrt{35}$, or approximately 5.91 B.M. Rounding this result to the nearest integer gives 6.

The electronic configuration of copper dictates its chemical behavior, with $\text{Cu}^{2+}$ possessing a $d^9$ configuration and one unpaired electron, which necessitates paramagnetism, while $\text{Cu}^+$ contains a filled $d^{10}$ subshell that inhibits electronic transitions and renders its complexes generally colorless. In aqueous environments, $\text{Cu}^+$ is unstable and prone to spontaneous disproportionation into $\text{Cu}^{2+}$ and metallic copper, confirming that it is easily oxidized. Because Fehling's solution utilizes $\text{Cu}^{2+}$ ions as the active oxidizing agent instead of $\text{Cu}^+$, exactly three of the provided statements are accurate.

Potassium permanganate functions as a strong oxidizing agent when placed in an acidic solution, facilitating a reduction reaction where the manganese atom transitions from the permanganate ion, $\mathrm{MnO}_4^-$, to the manganous ion, $\mathrm{Mn}^{2+}$. The oxidation state of manganese in the permanganate ion is $+7$, while it is $+2$ in the resulting manganous form. This shift represents a net decrease of 5 in the oxidation number during the reaction.

The stability of iron halides is governed by the relative oxidizing strength of the ferric ion, $\mathrm{Fe}^{3+}$, compared to the reducing power of the halide anion. While $\mathrm{Fe}^{2+}$ salts are stable across all halogens, $\mathrm{Fe}^{3+}$ acts as a strong enough oxidant to promote the reaction $2\,\mathrm{Fe}^{3+}+2\,\mathrm{I}^{-}\text{longrightarrow} 2\,\mathrm{Fe}^{2+}+\mathrm{I}_{2}$ in the presence of iodide. Since the reducing capacity of halide ions follows the order $\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}>\mathrm{F}^{-}$, the iodide ion is capable of reducing iron(III) to iron(II), which precludes the formation of $\mathrm{FeI}_{3}$. As a result, iron(II) halides can form with fluorine, chlorine, bromine, and iodine, whereas iron(III) halides are restricted to fluorine, chlorine, and bromine.

To determine the number of electrons in the p-orbitals of a Vanadium atom, we distribute its 23 electrons into atomic orbitals according to the Aufbau principle, resulting in the configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2$. Looking specifically at the subshells designated with the letter 'p', we identify the $2p$ and $3p$ orbitals. Because both of these subshells are fully occupied with 6 electrons each, we simply combine these values to find a total of 12 electrons.

When potassium permanganate is heated at 513 K, it undergoes thermal decomposition to produce potassium manganate, manganese dioxide, and oxygen as shown in the reaction: $2\text{KMnO}_4 \xrightarrow{513\text{ K}} \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2$ The potassium manganate product contains manganese \in the $+6$ oxidation state, which results in an electronic configuration that leaves a single unpaired electron in its d-orbital. Because of this unpaired electron, the compound is paramagnetic and characteristically green in color.

The ability of a metal salt to produce a coloured solution in water depends on the electronic configuration of its metal ion, specifically whether it contains unpaired electrons to facilitate electronic transitions. Among the given options, $\mathrm{CuCl}_2$ dissolves readily to form $\mathrm{Cu}^{2+}$ ions, which possess unpaired electrons and impart a distinct blue colour to the solution. Conversely, $\mathrm{ZnCl}_2$ dissolves to form colourless $\mathrm{Zn}^{2+}$ ions because they lack unpaired electrons, while both $\mathrm{AgCl}$ and $\mathrm{Cu}_2\mathrm{Cl}_2$ remain insoluble in water.

The dichromate ion, $\mathrm{Cr}_2\mathrm{O}_7^{2-}$, is constructed from two $\mathrm{CrO}_4$ tetrahedra that share a common corner oxygen atom. Due to the presence of lone pairs on this bridging oxygen atom, repulsive forces push the chromium centers apart, forcing the $\mathrm{Cr}-\mathrm{O}-\mathrm{Cr}$ linkage into a bent, non-linear geometry. Because the two tetrahedral units attached to the central oxygen are identical, the overall arrangement of the linkage is perfectly symmetrical.