A transition element is defined as an element whose atom or common ions have an incomplete d-subshell. Sc shows incomplete 3d occupancy in its states while Zn2+ has a completely filled 3d10 configuration. Hence (b).
Cr has electronic configuration [Ar] 3d5 4s1: the penultimate 3d subshell is half-filled (3d5) and the valence 4s is half-filled (4s1). Hence Cr (a).
The most negative M2+/M standard potential (most reducing metal) among the choices is titanium (Ti → Ti2+ + 2e−), Ti has a very negative potential (~ −1.6 V), more negative than Zn or Mn; Cu is positive. Hence (a).
V3+ (V, Z=23) has configuration 3d2 → 2 unpaired electrons. Ni2+ (Ni, Z=28) has 3d8 → in the free/typical complex form it has 2 unpaired electrons. Ti3+ is 3d1 (1 unpaired), Fe3+ is 3d5 (5 unpaired), Cr3+ is 3d3 (3 unpaired). Hence Ni2+ (c).
Mn2+ is d5 (high spin) with n = 5 unpaired electrons. Spin-only magnetic moment μ = √[n(n+2)] = √(5×7) = √35 ≈ 5.92 BM. Hence (a).
Catalytic activity is largely due to the ability of transition metals to undergo changes in oxidation state during the reaction cycle, facilitating electron transfer. (Unfilled d orbitals also help adsorption, but variable oxidation states are the main reason.) Hence (c).
Oxidising power increases from lower to higher oxidation state of central atom and from less powerful to more powerful oxidants: VO2+ (V IV) is weakest, dichromate (CrVI) is stronger, permanganate (MnVII) is strongest. So VO2+ < Cr2O7^2- < MnO4^- (a).
Oxalic acid (H2C2O4) is oxidized by KMnO4 in acid solution to CO2: C2O4^2- → 2 CO2. Hence (b).
K2Cr2O7 is the preferred primary standard for volumetric analysis (stable, non-hygroscopic). Na2Cr2O7 is not preferred over K2Cr2O7. The other statements about colour changes and precipitation on H2S (sulfur formation giving milky appearance) are acceptable. Hence (b) is not true.
In acidic medium MnO4^- is reduced to Mn^2+ (MnO4^- + 8H^+ + 5e^- → Mn^2+ + 4H2O). So (b).
Reaction: Cr2O7^{2-} + 14H^+ + 6I^- → 2Cr^{3+} + 3I2 + 7H2O. One mole dichromate liberates 3 mol I2. Hence (c).
FeC2O4 contains Fe^{2+} (gives 1 e^-) and C2O4^{2-} (gives 2 e^-) → total 3 e^- lost per mole. KMnO4 (MnO4^- → Mn^{2+}) accepts 5 e^- per mole. So moles KMnO4 = 5/3 per 5 FeC2O4 etc. For 1 mole FeC2O4: required = 3/5 = 0.6 mol. Hence (c).
Statements (a), (b) and (d) are correct. (c) is false: not all lanthanoids are "much more reactive" than aluminium; reactivity varies and the blanket statement is incorrect. Hence (c).
Yb^{2+} has configuration 4f^{14} (all 4f electrons paired) and is diamagnetic. Eu^{2+} is 4f^7 (paramagnetic), Ce^{2+} and Sm^{2+} have unpaired f-electrons. Hence (b).
The most common oxidation state of lanthanoids is +3 (Ln^{3+}). Hence (d).
Ce^{4+} acts as an oxidizing agent because it is readily reduced to the more stable Ce^{3+} state. Both statements are true and the reason explains the assertion. Hence (a).
The most common oxidation state for actinides is +3 (An^{3+}), analogous to the lanthanoids. Hence (b).
Neptunium (Np) and some transuranium elements such as Pu and Am are known to exhibit high oxidation states up to +7 under certain conditions. Hence (a).
Basicity decreases across the lanthanoid series (La → Lu) because ionic radius decreases; thus La(OH)3 is more basic than Lu(OH)3. Statement (a) is incorrect. The other statements are correct (c refers to chemical similarity to transition metals, and (d) is due to lanthanoid contraction). Hence (a).
Definition: Transition elements (transition metals) are elements in which the differentiating electron enters a d-sub-shell, and they form atoms or ions with an incomplete d-subshell (e.g. Ti, V, Cr, Mn, Fe, Co, Ni, Cu). Examples: Fe (iron), Co (cobalt), Ni (nickel), Cr (chromium).
Transition metals are d-block elements whose atoms or common ions have incompletely filled d-subshells. Examples: Fe, Co, Ni, Cr.
The 4d series (Y to Cd) has valence electrons in 5s and 4d orbitals. Early 4d metals (Y, Zr, Nb, Mo) commonly exhibit high positive oxidation states (+3, +4, +5, +6) because removal of 5s and 4d electrons is energetically feasible and the 4d orbitals can participate in bonding. Middle and late 4d metals show lower and more stable oxidation states (e.g. Ru, Rh, Pd commonly +2, +3; Ag +1; Cd +2) because increasing nuclear charge lowers the energy of 4d orbitals and pairing energies and relativistic effects favour lower states. Transition-metal multiple bonding and availability of d orbitals allow a variety of oxidation states; stability of a particular state depends on ionization energies, lattice/hydration energies and ligand stabilization.
4d-series elements show a range of oxidation states from +1 to +6 (rarely up to +7/+8 for some species), with lower oxidation states more common for the heavier members.
Inner transition elements occupy the f-block of the periodic table. Lanthanoids are 4f-series elements (La–Lu) in which 4f orbitals are being filled; actinoids are 5f-series elements (Ac–Lr) in which 5f orbitals are being filled. They are usually shown separately at the bottom of the periodic table due to similar chemical properties within each series and to keep the table compact.
Inner transition elements are elements in which the differentiating (last) electron enters an f-orbital; they comprise the lanthanoids (4f) and actinoids (5f).
Electron configuration justification: lanthanoids have general configuration [Xe]4f1–14 5d0–1 6s2 (4f filling), actinoids have [Rn]5f1–14 6d0–1 7s2 (5f filling). Because the f-orbitals are interior (inner) and produce a separate pattern of properties, these series are shown as two separate rows (f-block) below the main body. Their chemistry (similar ionic radii, common oxidation states like +3, gradual contraction across series) also supports separate placement.
Lanthanoids and actinoids are placed as separate f-block rows because their valence electrons fill 4f and 5f subshells respectively; they are chemically similar within each series and their inclusion inline would disrupt the periodic table's structure.
Actinides are elements in which the 5f subshell is being filled. They lie below the lanthanoids in the f-block and exhibit multiple oxidation states (commonly +3 to +6, and for some elements +7, +4 etc.), strong radioactivity, and significant involvement of 5f orbitals in bonding. Examples include Th (Z = 90), U (Z = 92), Pu (Z = 94).
Actinides are the 5f-series inner transition elements (elements 89–103). Examples: thorium (Th), uranium (U), plutonium (Pu).
Steps (typical industrial route): 1. Fusion of ore with alkali: 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2 (Chromite fused with Na2CO3 gives sodium chromate.) 2. Conversion to dichromate by acidification: 2Na2CrO4 + 2HCl → Na2Cr2O7 + 2NaCl + H2O (equivalently acidify to get H2CrO4 which condenses to Cr2O7^2−) 3. Exchange to potassium salt: Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl Overall product: K2Cr2O7 (potassium dichromate), which can be isolated by crystallization.
Potassium dichromate is prepared from chromite (FeCr2O4) by alkaline fusion to give chromate, followed by acidification and conversion to dichromate and metathesis with K+.
Reason: As protons are added across the lanthanoid series, the 4f electrons do not shield the increasing nuclear charge effectively, so outer electrons are drawn closer and radii decrease. Important consequences: - Chemical similarity of subsequent 5d elements with 4d congeners (e.g. Zr/Hf similarity). - Gradual decrease in ionic radii affects coordination numbers and bond strengths. - Basicity of M(OH)3 decreases from La(OH)3 to Lu(OH)3 because smaller, more polarizing cations make hydroxide more acidic. - Affects separation (difficult) and spectral/chemical properties of lanthanoids.
Lanthanoid contraction is the steady decrease in ionic/atomic radii of the lanthanoid elements with increasing atomic number due to poor shielding by 4f electrons. Effects: similar sizes of 4d and 5d elements, decrease in basicity of lanthanoid hydroxides across the series, increased densities and melting points, and influence on complex formation and separation chemistry.
Because the question as supplied contains damaged OCR (several fragments unclear), only unambiguous standard equations are given here: (a) MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O. (f) Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl (metathesis to give potassium dichromate). If you provide the original clear reaction fragments for parts (b)–(e) I will complete them precisely. (Confidence for the two given equations: high; for the rest: cannot reconstruct.)
The provided text is ambiguous/OCR-corrupted; I can confidently supply common completions for typical reactions: (a) MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O; (f) Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl. For the remaining items please supply the clear original reactions so I can complete them exactly.
Characteristics: small atoms (e.g. H in PdH, C in Fe3C (cementite), N in TiN) occupy octahedral or tetrahedral interstices of metal lattices. Properties include high hardness, high melting points, often metallic conductivity (but sometimes reduced), and altered magnetic/mechanical properties. They are typically formed by transition metals that have large interstitial sites and strong bonding to small atoms.
Interstitial compounds are non-stoichiometric compounds formed when small atoms (H, C, N, B, etc.) occupy interstitial sites in a metal lattice without disrupting the metal framework.
Electronic configurations: - Ti neutral [Ar] 3d2 4s2. Ti3+ → [Ar] 3d1 ⇒ 1 unpaired electron. Spin-only magnetic moment μ = √(n(n+2)) μB where n = number of unpaired e−. So μ = √(1×3) = √3 ≈ 1.73 BM. - Mn neutral [Ar] 3d5 4s2. Mn2+ → [Ar] 3d5 ⇒ 5 unpaired electrons. μ = √(5×7) = √35 ≈ 5.92 BM.
Ti3+: 1 unpaired electron, μspin-only = √[1(1+2)] = √3 = 1.73 BM. Mn2+: 5 unpaired electrons, μspin-only = √[5(5+2)] = √35 = 5.92 BM.
Ce (Z = 58) neutral configuration commonly written [Xe] 4f1 5d1 6s2. Removing four electrons (to Ce4+) removes 6s, 5d and 4f electrons to give Ce4+ = [Xe]. Co (Z = 27) neutral: [Ar] 3d7 4s2. Co2+ removes the 4s electrons to give [Ar] 3d7.
Ce4+: [Xe]. Co2+: [Ar] 3d7.
Reasoning: The M2+ state corresponds to removal of the two 4s electrons. As we move from Sc to Mn, additional electrons enter 3d orbitals which are poor shields; the effective nuclear charge felt by outer electrons increases, so the energy required to remove a third electron (a 3d electron) rises. Therefore the +2 state (after removal of 4s electrons) becomes relatively more stable compared with +3. Also, exchange stabilization and the growing pairing energy effects for certain d-electron counts favor retention of the +2 state in the first half of the series.
+2 states become increasingly stable across the first half of the 3d series because the third ionization energy (to form M3+) increases as nuclear charge and 3d electron binding increase; thus removing only two electrons (giving M2+) is relatively easier and more stable.
Electronic argument: Fe3+ is d5 (half-filled), which has extra exchange stabilization. Energetically, formation of Fe3+ costs ionization energy but is compensated by large hydration/lattice energies and the stability of d5. As a result, Fe3+ often predominates in oxidizing conditions; however, in reducing environments Fe2+ may be present. The standard potential Fe3+/Fe2+ (+0.77 V) indicates Fe3+ is a reasonably strong oxidizing agent and thermodynamically favored under oxidizing conditions.
Fe3+ is generally more stable than Fe2+ because Fe3+ has a half-filled d5 configuration (relatively stable) and ionization/hydration factors favor the trivalent state in many contexts.
Qualitative factors: - Ionization energy: as atomic number increases, IE3 generally increases, making oxidation to M3+ less favorable (more positive E°) or more favorable depending on balance. - Hydration energy: smaller M3+ ions have larger hydration enthalpies, stabilizing M3+ and making E° more positive. - Electronic configuration: particularly stable configurations (e.g., d5 for Mn2+ or half-filled/filled cases) raise or lower E° accordingly. For example, Mn2+ (d5) is relatively stable, so Mn3+/Mn2+ has a large positive potential (Mn3+ is a strong oxidant in aqueous solution) because Mn2+ is especially stabilized. Thus E° is the result of competition between ionization and solvation/ligand stabilization energies and electronic stability.
E°(M3+/M2+) varies across the 3d series because of changes in third ionization energy, ionic radius (hence hydration energy), electronic configuration (stability of dN vs dN−1), and ligand/hydration stabilization; these combined factors cause a non-linear trend with peaks where d-electron configurations are especially stable.
Key points: - Electronic: lanthanoids fill 4f (tightly bound, poor shielding); actinoids fill 5f (more spatially extended, more shielding variability). - Oxidation states: lanthanoids largely show +3 (some +2, +4 in a few); actinoids display multiple stable oxidation states (from +3 up to +6 or higher for some elements, e.g. U, Np, Pu). - Radioactivity: actinoids are radioactive; lanthanoids mostly stable (except promethium). - Chemical bonding: 5f orbitals participate more in bonding, so actinoids show greater covalency and complex chemistry. - Contraction: both show contraction, but actinoid contraction (element-to-element) is generally larger.
Both are f-block inner transition series: lanthanoids (4f) and actinoids (5f). Similarities: gradual filling of f-orbitals, contraction across series, common +3 oxidation state, comparable chemical behaviour within each series. Differences: actinoids show greater range of oxidation states, are all radioactive (many highly so), 5f electrons are more involved in bonding (more covalent character), and actinoid contraction per element is larger.
Electronic stability explanation: both Cr2+ and Mn3+ have the same d configuration (d4), but their tendencies depend on the relative stabilities of adjacent oxidation states. For Cr, the +3 state (d3) is significantly more stable than +2, so Cr2+ tends to lose an electron → strong reducing behavior. For Mn, the +2 state (d5) is especially stable (half-filled), so Mn3+ tends to gain an electron → strong oxidizing behaviour. Standard potentials reflect this: Cr3+/Cr2+ is negative (Cr2+ reducing) while Mn3+/Mn2+ is positive (Mn3+ oxidizing under many conditions).
Cr2+ (d4) is a strong reducing agent because it is readily oxidized to the stable Cr3+ (d3). Mn3+ (d4) is strongly oxidizing because it is readily reduced to the particularly stable Mn2+ (d5, half-filled).
Trend details: - Overall rise across the period because Z increases while shielding does not increase proportionally. - Irregularities: Cr (3d5 4s1) and Cu (3d10 4s1) have unusually low first IE relative to the trend because removal of the single 4s electron leads to relatively stable d configurations (half-filled or filled d subshells), making the 1st IE slightly lower than expected. - The increase is less steep than in s- and p-block because added electrons occupy 3d which shield poorly and increase electron-electron repulsion and pairing effects.
Ionization enthalpies (first IE) across the 3d series generally increase from Sc to Zn due to increasing nuclear charge, but there are small irregularities (notably at Cr and Cu) due to subshell electron configurations (half-filled or filled stability).
Explanation: As protons are added across the actinoid series the poorly shielding 5f electrons do not compensate the increased nuclear charge, causing a stronger effective pull on outer electrons and greater reduction in ionic radii. Additionally, 5f orbitals are more spatially extended and subject to relativistic effects which influence bonding and contraction; overall the element-to-element decrease in size (actinoid contraction) tends to be larger and less uniform than the lanthanoid contraction.
Actinoid contraction per element is greater because 5f electrons shield the increasing nuclear charge even less effectively than 4f electrons, and relativistic effects and stronger penetration of 5f orbitals enhance the contraction.
As lanthanoid ionic radius decreases across the series (lanthanoid contraction), cations become more polarizing and increase the acidity of coordinated hydroxide. La3+ (largest) polarizes OH− less, so La(OH)3 behaves as a stronger base compared with Lu(OH)3 whose smaller, more polarizing Lu3+ makes the hydroxide more acidic (less basic).
La(OH)3 is more basic than Lu(OH)3 because La3+ has a larger ionic radius and is less polarizing, so its hydroxide is less covalent and more basic.
Electronic configurations: - Eu2+ → [Xe]4f7 (half-filled f-shell gives extra exchange/stability). - Ce2+ → would be [Xe]4f2 (not specially stable); Ce prefers +4 (empty 4f) or +3. Thus Eu2+ is unusually stable among lanthanoids due to the half-filled 4f7 arrangement.
Eu2+ is more stable because Eu2+ has a half-filled 4f7 configuration, which is particularly stable; Ce2+ lacks such stabilization and Ce commonly stabilizes +4.
Because the lanthanoid contraction reduces the expected increase in atomic/ionic size for Hf (5d) relative to Zr (4d), both elements have very similar ionic radii and hence very similar coordination chemistry, oxidation states (commonly +4), complex stability, and chemical reactivity. This is a classical example of the effect of lanthanoid contraction on periodic trends.
Zr and Hf exhibit similar properties because of the lanthanoid contraction: addition of 14 lanthanoid elements between them causes Hf's 5d orbitals to be similar in radial extent to Zr's 4d orbitals, producing nearly identical ionic radii and chemical behaviour.
Reducing strength correlates with tendency to be oxidized. For the couples: Cr3+ + e– ⇌ Cr2+ , E° = –0.41 V and Fe3+ + e– ⇌ Fe2+ , E° = +0.77 V. The oxidation potential (reverse sign) for Cr2+ → Cr3+ + e– is +0.41 V and for Fe2+ → Fe3+ + e– is –0.77 V, so Cr2+ is more easily oxidized and hence a stronger reducing agent. Also Cr2+ (d4) is less stable than Cr3+ (d3), favouring oxidation.
Cr2+
Cu2+ + 2e– ⇌ Cu (E° ≈ +0.34 V). The positive E° arises because neutral copper (electron configuration [Ar]3d10 4s1 → after reduction effectively 3d10) corresponds to a very stable filled 3d10 shell; gaining electrons to reach this stable configuration and favourable hydration/ lattice energies make the reduction thermodynamically favourable, giving a positive standard potential.
Because metallic Cu attains a stable filled d10 configuration and Cu2+ reduction is thermodynamically favourable
The 3d and 4s orbitals have comparable energies so different numbers of d and s electrons can be lost to form ions, producing variable oxidation states. Early 3d elements show low positive states (e.g. Ti +2,+3,+4), middle ones show a wide range (V up to +5, Cr up to +6, Mn up to +7), and later ones are commonly +2 and +3 (Fe, Co, Ni) with Cu showing +1,+2 and Zn predominantly +2. The relative stability depends on electron configuration (half‑filled or filled d subshells), ionisation energies and bonding/covalency.
3d elements show multiple oxidation states due to similar energies of 3d and 4s electrons; highest states up to group number; trend peaks at Mn.
Cu commonly forms Cu+ because Cu+ has the electron configuration 3d10 (a stable filled d‑subshell). Loss of the single 4s electron yields this stable d10 state, making the +1 oxidation state relatively stable for copper.
Copper (Cu)
Cr: [Ar]3d5 4s1 — removal of the single 4s electron is relatively easy and produces the extra-stable half‑filled 3d5 configuration. Zn: [Ar]3d10 4s2 — the valence electrons are more strongly bound due to the higher effective nuclear charge and completely filled 3d10 subshell, so Zn has a larger first ionisation enthalpy than Cr.
Because Cr has a single 4s electron that is weakly held and its removal gives the stable half‑filled 3d5 configuration; Zn has a filled 3d10 and higher effective nuclear charge.
Transition metals have partially filled d orbitals whose electrons can delocalise and overlap with s electrons, producing strong metallic bonds. The presence of several unpaired d electrons increases bond strength and cohesive energy, resulting in high melting points and boiling points.
Because of strong metallic bonding from delocalised d and s electrons and many unpaired d electrons leading to high cohesive energy.