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

Elements form ions by losing the electrons found in their outermost energy level, and for Europium, which has an atomic number of 63, this process involves the removal of two electrons from the $6s$ orbital. Consequently, the neutral atom's ground state configuration of $[\mathrm{Xe}]\,4f^{7}\,6s^{2}$ yields the $\mathrm{Eu}^{2+}$ ion by losing these outer electrons, resulting in the electronic configuration $[\mathrm{Xe}]\,4f^{7}$.

Magnetic behavior in metallic oxides is primarily determined by the presence of unpaired electrons within the $d$-orbitals of the constituent metal ions. Unlike ${SiO}_{2}$, ${MgO}$, and ${Na}_{2} {O}$, which are diamagnetic due to the lack of such unpaired electrons, manganese is a $d$-block element that frequently retains them in its stable oxidation states. With an electronic configuration of ${ }_{25} {Mn}=[ {Ar}] 3 d^{5} 4 s^{2}$, manganese forms ${Mn}^{2+}$ ions with five unpaired electrons and ${Mn}^{3+}$ ions with four, which allows ${Mn}_{3} {O}_{4}$ to display the observed magnetic property.

Determining the oxidation states of these compounds involves assigning a value of -2 to oxygen and solving for the charge of each metal atom. This calculation results in +6 for (a) $\mathrm{CrO}_3$, +5 for (d) $\mathrm{V}_2\mathrm{O}_5$, +4 for (c) $\mathrm{MnO}_2$, +3 for (b) $\mathrm{Fe}_2\mathrm{O}_3$, and +1 for (e) $\mathrm{Cu}_2\mathrm{O}$. Sorting these values from highest to lowest produces the order (a) > (d) > (c) > (b) > (e).

The standard reduction potential of a chemical species indicates its tendency to gain electrons, where a large positive value signifies that the higher oxidation state acts as a strong oxidizing agent. The reaction $\text{Ce}^{4+} + e^{-} \rightarrow \text{Ce}^{3+}$ is characterized by an $E^{\circ}$ of $1.74\ \text{V}$, which makes the first statement factually accurate. Although $\text{Ce}^{4+}$ features a $[Xe] 4f^0$ electronic configuration, its high reduction potential indicates that it is not stable in this state; rather, it functions as a strong oxidant that readily reverts to the $\text{Ce}^{3+}$ form. Since the metal is more stable in the $\text{Ce}^{3+}$ oxidation state, the claim that $\text{Ce}^{4+}$ is more stable is erroneous.

The spin-only magnetic moment of a transition metal ion is determined by the number of unpaired electrons $n$ in its $d$-orbitals, calculated using the relationship $\mu = \sqrt{n(n+2)}$ BM.

For $\mathrm{Ti}^{3+}$, the electronic configuration is [Ar] $3d^1$, which means $n=1$. Substituting this into the formula gives $\mu = \sqrt{1(1+2)}$, which results in 1.73 BM. For $\mathrm{V}^{2+}$, the configuration is [Ar] $3d^3$, resulting in $n=3$, so the magnetic moment is $\mu = \sqrt{3(3+2)}$, or 3.87 BM. Finally, $\mathrm{Sc}^{3+}$ has the configuration [Ar] $3d^0 4s^0$, meaning it has no unpaired electrons, so the magnetic moment is 0 BM.

The formation of interstitial hydrides follows a distinct trend across the d-block, primarily governed by the exclusion of elements in groups 7, 8, and 9 from this process. Because Manganese, Iron, and Cobalt fall within this so-called hydride gap, they are unable to form these structures under standard conditions. Chromium, occupying group 6, resides just before this inactive region and is the only metal in the given set capable of facilitating the insertion of hydrogen atoms into its crystal lattice.

The fundamental principle governing the nature of transition metal oxides is that metallic character decreases as the metal's oxidation state increases. Consequently, transition metals in lower oxidation states tend to form basic oxides that readily react with acids to produce salt solutions. In $\mathrm{V}_2\mathrm{O}_3$, vanadium is present in the +3 oxidation state, which aligns with basic properties. Likewise, the chromium in CrO exists in the +2 oxidation state, which reinforces its classification as a basic oxide.

The spin-only magnetic moment of a metal complex is determined by the number of unpaired electrons, $n$, according to the formula $\mu = \sqrt{n(n+2)}$. For the $\mathrm{Ce}^{4+}$ ion, the $4f^0$ configuration results in zero unpaired electrons and a magnetic moment of 0 BM, while the $\mathrm{Eu}^{3+}$ ion possesses a $4f^6$ configuration with six unpaired electrons, corresponding to a magnetic moment of $\sqrt{6(6+2)} = 6.93$ BM. The gadolinium species exhibits a $4f^7 5d^1$ configuration with eight unpaired electrons, yielding a magnetic moment of $\sqrt{8(8+2)} = 8.94$ BM. Comparing these values confirms that the increasing order of magnetic moments is A < C < B.

Paramagnetic behavior and the presence of colour in transition metal ions result directly from having unpaired electrons that allow for electronic transitions and magnetic interaction. Specifically, $\mathrm{Cu}^{2+}$ holds a $[\mathrm{Ar}]\,3d^{9}\,4s^{0}$ configuration with one unpaired electron, $\mathrm{Cr}^{3+}$ holds a $[\mathrm{Ar}]\,3d^{3}\,4s^{0}$ configuration with three unpaired electrons, and $\mathrm{Sc}^{+}$ holds a $[\mathrm{Ar}]\,3d^{1}\,4s^{1}$ configuration with two unpaired electrons. Because every ion in this group contains these unpaired electrons, they are both paramagnetic and coloured.

Density is governed by the relationship $\rho = \frac{m}{V}$, where atomic mass typically increases and metallic radius decreases across the transition series, generally causing density to rise. When comparing the specific values for these elements, zinc is the lightest at 7.13 g/cm³, followed by chromium at 7.19 g/cm³, iron at 7.8 g/cm³, cobalt at 8.7 g/cm³, and copper at 8.9 g/cm³. This trend establishes an increasing order of $\mathrm{Zn} < \mathrm{Cr} < \mathrm{Fe} < \mathrm{Co} < \mathrm{Cu}$.

To determine the electronic configuration of a cation, electrons must first be removed from the orbital with the highest principal quantum number, as these are the most loosely held. For neutral gadolinium, which has an atomic number of 64, the ground state configuration is $[\mathrm{Xe}] 4f^7 5d^1 6s^2$. Removing two electrons to form $\mathrm{Gd}^{2+}$ requires emptying the $6s$ subshell, resulting in the configuration $[\mathrm{Xe}] 4f^7 5d^1$. Consequently, the 4f subshell remains unchanged, containing exactly 7 electrons.

Lanthanoids predominantly exhibit the $+3$ oxidation state, and the formation of $MO_2$ oxides requires the metal to reach a $+4$ oxidation state. Among the lanthanoid series, only Pr, Nd, Tb, and Dy are capable of forming $MO_2$ type oxides due to their ability to achieve this specific $+4$ state. Yb does not form such an oxide because it does not attain the $+4$ oxidation state, differentiating it from the other elements listed.

Coordination chemistry is central to both biological systems and synthetic applications, where specific metal ions define the unique properties of these compounds. Chlorophyll functions as a coordination complex of magnesium, while Vitamin B12 is characterized by a central cobalt atom. In practical applications, the anticancer drug cis-platin, $[ {Pt}( {NH}_{3})_{2} {Cl}_{2}]$, relies on platinum, whereas the Grubbs catalyst is defined by the presence of ruthenium.

To determine the outermost electronic configuration of transition metal ions, one must track how electrons are removed from neutral atoms, where $4s$ electrons are generally lost before $3d$ electrons. Chromium, with an atomic number of 24, has a neutral ground state of $[Ar] 3d^5 4s^1$, meaning the $\mathrm{Cr}^{+}$ ion formed by removing one electron retains an $[Ar] 3d^5$ configuration. Manganese, with atomic number 25, has a neutral ground state of $[Ar] 3d^5 4s^2$, and since the $\mathrm{Mn}^{2+}$ ion loses its two $4s$ electrons, it also results in an identical $[Ar] 3d^5$ configuration.

Determining the electron count in an ion requires recognizing that electrons are removed from the outermost shell before inner subshells during ionization. A neutral holmium atom with atomic number 67 possesses the electronic configuration $[Xe] 4f^{11} 6s^{2}$. To create the $Ho^{3+}$ ion, three electrons are lost, beginning with the two electrons in the $6s$ orbital and continuing with one electron from the $4f$ subshell to arrive at the final configuration of $[Xe] 4f^{10}$. This process leaves 10 electrons occupying the $4f$ subshell.

The chemical behavior of lanthanides is defined by the high stability of the +3 oxidation state. Since the $Ce^{4+}$ ion strongly tends to gain electrons to achieve this stable configuration, $CeO_2$ functions as an effective oxidizing agent capable of oxidizing substances such as alcohols, aldehydes, and ketones. In contrast, the $Eu^{2+}$ ion present in $EuSO_4$ readily loses an electron to reach the more stable +3 state, which characterizes the compound as a strong reducing agent in aqueous solution.

The borax bead test relies on the distinct oxidizing and reducing nature of different flame zones to determine the oxidation state of transition metals. In statement I, the claim is incorrect because cupric metaborate is characteristically blue, not colourless; it is the cuprous metaborate formed after reduction in a luminous flame that is colourless. Similarly, statement II is flawed because the reaction between copper sulphate and boric anhydride in a non-luminous flame yields cupric metaborate ($\mathrm{Cu}(\mathrm{BO}_2)_2$), not cuprous metaborate.

The spin-only magnetic moment is calculated using the formula $\mu = \sqrt{n(n+2)}$ BM, where $n$ represents the number of unpaired electrons. For an element with an atomic number of 25, known as manganese, the neutral electronic configuration is $[\text{Ar}]\,4s^2\,3d^5$. As a divalent ion in an aqueous solution, manganese loses its two $4s$ electrons to become $[\text{Ar}]\,3d^5$, which results in 5 unpaired electrons. Substituting 5 for $n$ in the formula $\mu = \sqrt{n(n+2)}$ gives $\mu = \sqrt{5(5+2)} = \sqrt{35}$, yielding 5.92 BM.

Magnetic behavior in elements is governed by the presence of unpaired electrons, meaning a species must have completely paired electrons to be classified as diamagnetic. Ytterbium, which has an atomic number of 70, maintains a ground-state configuration of $[\mathrm{Xe}] 4f^{14} 6s^{2}$. Upon forming a +2 ion, it loses the two $6s$ electrons to leave the stable $[\mathrm{Xe}] 4f^{14}$ configuration, where the fully filled $4f$ subshell ensures there are no unpaired electrons, confirming its diamagnetic nature.

A disproportionation reaction occurs when an element in an intermediate oxidation state undergoes simultaneous oxidation and reduction, allowing it to transition to both higher and lower oxidation states. In the set containing $\mathrm{ClO}_2^{-}$, $\mathrm{Cl}_2$, and $\mathrm{Mn}^{3+}$, every species meets this requirement because they are not restricted by maximum or minimum oxidation limits. Specifically, the chlorine in $\mathrm{ClO}_2^{-}$ at a +3 state can shift to -1 and +5 oxidation states, while neutral $\mathrm{Cl}_2$ similarly divides into $\mathrm{Cl}^{-}$ and $\mathrm{ClO}_3^{-}$. Likewise, $\mathrm{Mn}^{3+}$ is unstable in aqueous media, spontaneously forming $\mathrm{Mn}^{2+}$ and $\mathrm{MnO}_2$ where manganese resides in the +4 state. Other sets fail this criterion because they include species like $\mathrm{MnO}_4^{-}$, where the metal is already at its highest possible oxidation state and cannot be oxidized further.