CUET UG Chemistry Booster Test – 3 Electronic Configuration

High enthalpy of atomization arises because a greater number of d-electrons participate in metallic bonding; therefore 1 matches c. Paramagnetism results from the presence of unpaired d-electrons, so 2 matches a. Lanthanoid contraction is caused by the poor shielding effect of one 4f electron by another, leading to increased effective nuclear charge across the series; hence 3 matches d. Catalytic activity of transition elements is closely related to their ability to exhibit multiple oxidation states through participation of both (n−1)d and ns electrons in bonding and redox processes, so 4 matches b. Therefore, the correct matching is 1-c, 2-a, 3-d and 4-b.

The shielding effect of a d-electron on an ns electron is relatively weak because d-orbitals do not shield nuclear charge very effectively. Therefore, Statement 1 is correct. As electrons are added to the 3d subshell across the first transition series, they partially shield the 4s electrons from the increasing nuclear charge. As a result, atomic radii decrease less rapidly than expected, making Statement 2 correct. Statement 3 is incorrect because ns electrons do not completely shield d-electrons. Statement 4 is also incorrect because ns electrons are generally removed before (n−1)d electrons during ion formation. Therefore, only Statements 1 and 2 are correct.

Zinc possesses the electronic configuration [Ar]3d¹⁰4s². Upon ionization, the two 4s electrons are removed, producing Zn²⁺ with a highly stable d¹⁰ configuration. This stability favours ion formation and contributes to the negative electrode potential. However, because the d-subshell is completely filled, there are no unpaired d-electrons available to contribute effectively to strong metallic bonding. Consequently, Zinc exhibits a relatively low enthalpy of atomization compared with many transition metals. The combination of a stable d¹⁰ ionic state and weaker metallic bonding explains the observed behaviour.

The process Cu(s) → Cu²⁺(aq) + 2e⁻ involves the loss of two electrons by a copper atom. According to the modern definition of oxidation, any process involving loss of electrons is classified as oxidation. The oxidation state of copper increases from 0 in metallic copper to +2 in Cu²⁺ ions. This electron-loss process is fundamental to electrochemistry and explains the behaviour of copper in redox reactions. Therefore, the reaction is correctly identified as oxidation.

Chromium has the expected configuration [Ar]3d⁴4s² according to the Aufbau principle. However, one electron from the 4s orbital is promoted to the 3d orbital, resulting in the actual configuration [Ar]3d⁵4s¹. This arrangement provides additional stability because a half-filled d-subshell possesses greater exchange energy and symmetrical electron distribution. As a result, Chromium exhibits one of the most important exceptions to the expected electronic configurations of transition elements. Therefore, the element showing the Cr anomaly is Chromium.

The standard electrode potential of copper is influenced by several energetic factors, including atomization enthalpy, ionization enthalpy and hydration enthalpy. To convert metallic copper into aqueous Cu²⁺ ions, significant energy is required to atomize the metal and remove electrons. Although hydration of Cu²⁺ releases energy, this hydration enthalpy is not sufficiently large to compensate completely for the high energy required in the earlier steps. Consequently, the overall process is less favourable than for many other transition metals, giving copper its characteristic positive standard electrode potential. Therefore, hydration enthalpy is the factor that fails to offset the energy requirement adequately.

Mn²⁺ possesses the configuration 3d⁵, a half-filled d-subshell that provides exceptional stability due to maximum exchange energy. Cr³⁺ has a 3d³ configuration, corresponding to a half-filled t₂g set in an octahedral environment, which is also highly stable. Cr²⁺ and Mn³⁺ both possess d⁴ configurations and are less stable. Among them, Cr²⁺ is relatively more stable than Mn³⁺ because Mn³⁺ readily undergoes reduction to Mn²⁺, which possesses the highly stable d⁵ arrangement. Therefore, the decreasing order of stability is Mn²⁺ > Cr³⁺ > Cr²⁺ > Mn³⁺.

Zinc has the electronic configuration [Ar]3d¹⁰4s². After losing two 4s electrons, Zn²⁺ acquires the highly stable configuration [Ar]3d¹⁰. The third ionization process requires removal of an electron from the completely filled 3d subshell. Since fully filled subshells possess extra stability due to symmetrical electron distribution and exchange-energy effects, a large amount of energy is required. Consequently, the third ionization enthalpy of Zinc is exceptionally high compared with neighbouring elements.

Palladium is unique among transition elements because its ground-state electronic configuration is [Kr]4d¹⁰5s⁰, making Statement 3 correct. Zinc, Cadmium and Mercury are the final members of the 3d, 4d and 5d series respectively, making Statement 4 correct. Statement 2 is incorrect because Mercury generally retains a filled d¹⁰ configuration in its common oxidation states. Statement 1 is also incorrect because Group 12 elements do not exhibit transition-metal behaviour due to empty p-orbitals; their chemistry is dominated by filled d-subshells. Therefore, only Statements 3 and 4 are correct.

Cu⁺ undergoes disproportionation in aqueous solution to form Cu²⁺ and Cu, so 1 matches b. Zinc commonly exhibits only the +2 oxidation state and its d-electrons do not significantly participate in bonding, so 2 matches d. Cadmium is the final member of the 4d series and is not considered a true transition metal because of its d¹⁰ configuration, so 3 matches a. Mercury possesses a filled 5d¹⁰ configuration and is unique among metals because it exists as a liquid at room temperature; therefore 4 matches c. Hence the correct matching is 1-b, 2-d, 3-a and 4-c.

In the actinoid series, the energies of the 5f, 6d and 7s orbitals are very close to one another. Because of this energy similarity, electrons from all three types of orbitals can participate in chemical bonding and ion formation. As a result, actinoids exhibit a much wider range of oxidation states compared with lanthanoids. For example, uranium commonly shows oxidation states ranging from +3 to +6. This variability is a characteristic feature of actinoid chemistry and arises directly from the comparable energies of the valence orbitals. Therefore, the correct answer is a greater range of oxidation states.

Atomic radius represents the size of an atom and is measured as a distance. Since atomic dimensions are extremely small, the picometre (pm), equal to 10⁻¹² metre, is used as the standard unit. Trends in atomic radii across the 3d, 4d and 5d series help explain the effects of shielding, effective nuclear charge and lanthanoid contraction. Therefore, the correct unit for atomic radius is picometre (pm).

Scandium commonly forms Sc³⁺ because removal of three electrons produces the stable noble-gas configuration [Ar], making Sc²⁺ relatively unstable. Therefore, Statement 1 is correct. Manganese exhibits oxidation states ranging from +2 to +7, making Statement 2 correct. Titanium most commonly exists in the +4 oxidation state and Ti(IV) compounds are generally more stable than Ti(III) and Ti(II) compounds; therefore Statement 4 is correct. Statement 3 is incorrect because Zinc usually exhibits only the +2 oxidation state and retains its stable d¹⁰ configuration. Hence Statements 1, 2 and 4 are correct.

In the complex [Pt(NH₃)₂Cl₂], ammonia is a neutral ligand and is named ammine. Chloride ligands are named chlorido. Since the complex is neutral, the oxidation state of platinum is calculated as +2 because the two chloride ligands contribute a total charge of –2. According to IUPAC nomenclature, ligands are named alphabetically before the metal. Therefore, the correct name is Diamminedichloridoplatinum (II). This complex is commonly known as cisplatin or transplatin depending on the spatial arrangement of ligands.

Between the 4d and 5d transition series lies the lanthanoid series, where electrons are progressively added to the 4f orbitals. The shielding effect of 4f electrons is poor, causing a steady increase in effective nuclear charge across the lanthanoid series. This leads to a significant contraction in atomic size known as lanthanoid contraction. As a consequence, the 5d transition elements become much smaller than expected and their atomic radii become very similar to those of the corresponding 4d elements. Therefore, lanthanoid contraction is responsible for the close similarity in atomic radii between the two series.

The atomic numbers of the given elements are: Cerium (58), Lutetium (71), Thorium (90) and Lawrencium (103). When arranged in decreasing order of atomic number, the element with the highest atomic number must appear first. Therefore, Lawrencium (103) comes before Thorium (90), followed by Lutetium (71) and Cerium (58). Both Lutetium and Cerium belong to the lanthanoid series, while Thorium and Lawrencium belong to the actinoid series. Thus, the correct decreasing order is Lr > Th > Lu > Ce.

Transition elements possess ns and (n−1)d electrons of comparable energies. During ion formation, ns electrons are generally removed first, followed by d-electrons if higher oxidation states are formed. Since electrons are removed one at a time, successive oxidation states often differ by one unit. For example, iron commonly exhibits +2 and +3 oxidation states, while manganese exhibits oxidation states from +2 to +7. Therefore, the oxidation states of transition elements typically differ by unity.

In acidic medium, permanganate ion (MnO₄⁻) acts as a powerful oxidizing agent. It oxidizes Fe²⁺ ions to Fe³⁺ according to the reaction: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O Since Fe²⁺ loses an electron and its oxidation state increases from +2 to +3, the process occurring to iron is oxidation. Meanwhile, permanganate itself is reduced from Mn(VII) to Mn(II). Therefore, the correct classification with respect to Fe²⁺ is oxidation.

The passage clearly states that, unlike the p-block where lower oxidation states become more stable down a group because of the inert pair effect, the d-block shows the opposite trend. In Group 6, Chromium(VI) is less stable than Molybdenum(VI) and Tungsten(VI). Consequently, Mo(VI) and W(VI) compounds are relatively stable, whereas Cr(VI) compounds readily undergo reduction. This demonstrates that higher oxidation states become increasingly stable as one moves down the d-block group.

Chromium(VI) compounds such as dichromate are powerful oxidizing agents because the +6 oxidation state is not as stable for chromium as it is for molybdenum and tungsten. As a result, Cr(VI) readily gains electrons and is reduced to lower oxidation states such as Cr(III), which are thermodynamically more stable. This strong tendency to undergo reduction enables Cr(VI) species to oxidize other substances. In contrast, Mo(VI) and W(VI) are already relatively stable and therefore do not exhibit comparable oxidizing power. Hence, Cr(VI) acts as a strong oxidizing agent because it prefers reduction to a more stable lower oxidation state.