🧪 Chemistry · Class 12 · NEET & JEE
d- and f-Block Elements - Practice Questions with Answers 75 free MCQs on d- and f-Block Elements, each with its own worked answer and explanation. Transition metals (d-block) and lanthanides/actinides (f-block). Known for coloured compounds, variable oxidation states, complex formation, and catalytic properties. Frequently tested in JEE and NEET.
Take the timed d- and f-Block Elements chapterwise test → 75 practice questions on d- and f-Block Elements , sorted Easy → Hard. Try each one first, then open its answer page for the worked explanation. Want the full theory first? Read the d- and f-Block Elements notes .
Colours of Common Transition Metal Ions (Aqueous) Cu²⁺ blue Fe²⁺ pale green Fe³⁺ brown-yellow Mn²⁺ pale pink Ni²⁺/Cr³⁺ green Colour arises from d-d electron transitions - d⁰ ions like Sc³⁺, Ti⁴⁺ are colourless Aqueous solutions of transition metal ions show characteristic colours caused by electrons absorbing visible light to jump between split d-orbitals; the exact colour depends on the metal, its oxidation state, and surrounding ligands.
Easy - 25 questions Q1.
The general outer electronic configuration of d-block (transition) elements is:
A (n-1)d<sup>1-10</sup>ns<sup>0-2</sup>B ns<sup>1-2</sup>C (n-2)f<sup>1-14</sup>ns<sup>2</sup>D ns<sup>2</sup>np<sup>1-6</sup>Show answer & explanation →
Q2.
Which of the following ions is colourless in aqueous solution?
A Cu<sup>2+</sup>B Sc<sup>3+</sup>C Ni<sup>2+</sup>D Co<sup>2+</sup>Show answer & explanation →
Q3.
Poor shielding by which electrons is chiefly responsible for the lanthanoid contraction?
A 3d electronsB 4p electronsC 4f electronsD 5d electronsShow answer & explanation →
Q7.
Which property is characteristic of transition metals?
A Variable oxidation statesB Fixed single oxidation stateC No colour in compoundsD Non-magneticShow answer & explanation →
Q8.
What colour is the KMnO<sub>4</sub> (potassium permanganate) solution?
A Purple/violetB YellowC GreenD ColourlessShow answer & explanation →
Q9.
Which transition metal is used as a catalyst in the Haber process?
A Iron (Fe)B Platinum (Pt)C Vanadium (V)D Nickel (Ni)Show answer & explanation →
Q10.
Which transition metal is used in the catalytic converter in cars?
A Platinum (Pt) and Palladium (Pd)B Iron, valued for its low cost in oxidation catalysisC Zinc, commonly used as a sacrificial anode in galvanisingD Copper, prized for its high electrical conductivity in wiringShow answer & explanation →
Q11.
The colour of transition metal compounds is due to:
A d-d electron transitions absorbing visible lightB Their characteristically high melting points compared to main-group saltsC Their predominantly ionic character in the solid latticeD Their general paramagnetic or ferromagnetic behaviourShow answer & explanation →
Q13.
The electronic configuration of Cu is [Ar]3d<sup>10</sup> 4s<sup>1</sup> (anomalous). This is because:
A Fully filled d<sup>10</sup> is extra stableB It has odd atomic number under usual circumstancesC 3d orbitals are empty according to most researchersD It is a noble gas in the majority of cases studiedShow answer & explanation →
Q14.
Which transition metal is the best conductor of electricity?
A Silver (Ag)B Copper (Cu)C Gold (Au)D Aluminium (Al)Show answer & explanation →
Q15.
The lanthanides are also called:
A Rare earth metalsB Alkaline earth metalsC Transition metalsD Noble metalsShow answer & explanation →
Q16.
Which element is called the 'king of metals' due to its nobility?
A Gold (Au)B Silver (Ag)C Platinum (Pt)D Iron (Fe)Show answer & explanation →
Q17.
Rust is formed when iron reacts with:
A Oxygen and water (moist air)B Carbon dioxide gas dissolved in dry airC Nitrogen gas present in the surrounding atmosphereD Dilute hydrochloric acid splashed onto the surfaceShow answer & explanation →
Q19.
K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> (potassium dichromate) has which colour?
A OrangeB PurpleC BlueD GreenShow answer & explanation →
Q21.
Transition metals generally have high melting points because:
A Strong metallic bonding involving both s and d electronsB A significant degree of ionic character within the metallic latticeC Their unusually large atomic masses compared to main-group metalsD Their underlying noble gas electron configurationShow answer & explanation →
Q22.
Which element is used as the filament in incandescent light bulbs?
A Tungsten (W)B IronC MolybdenumD NickelShow answer & explanation →
Q24.
Uranium is used in:
A Nuclear fuel and nuclear weaponsB Industrial catalysts for hydrogenation reactionsC Stainless steel alloys requiring corrosion resistanceD Permanent magnets used in electric motorsShow answer & explanation →
Medium - 25 questions Q26.
Among Ti<sup>3+</sup>, V<sup>3+</sup>, Cr<sup>3+</sup> and Mn<sup>2+</sup>, the ion with the highest spin-only magnetic moment is:
A Ti<sup>3+</sup>B Mn<sup>2+</sup>C Cr<sup>3+</sup>D V<sup>3+</sup>Show answer & explanation →
Q27.
Manganese exhibits the maximum number of oxidation states (+2 to +7) in the 3d series because it:
A has the smallest atomic radius in its periodB has a fully filled 3d subshellC can use both its 3d and 4s electrons for bondingD is the most electronegative 3d metalShow answer & explanation →
Q29.
When acidified KMnO<sub>4</sub> is reduced to Mn<sup>2+</sup>, the number of electrons gained per MnO<sub>4</sub><sup>-</sup> ion is:
Show answer & explanation →
Q30.
Copper is the only 3d metal that does not liberate hydrogen from dilute acids mainly because it has:
A a very high hydration enthalpy and atomisation energyB a positive standard reduction potential (E° of Cu<sup>2+</sup>/Cu)C a completely filled 4s orbitalD no unpaired electronsShow answer & explanation →
Q31.
The electronic configuration of Cr is [Ar]3d<sup>5</sup> 4s<sup>1</sup> rather than [Ar]3d<sup>4</sup> 4s<sup>2</sup> because:
A Half-filled d<sup>5</sup> is extra stable due to exchange energy and symmetryB The 3d subshell is usually filled largely before the 4s subshellC Chromium is a diamagnetic element with little unpaired electronsD The 4s subshell must usually retain exactly two electrons under most conditions encounteredShow answer & explanation →
Q32.
Crystal field theory (CFT) explains colour and magnetic properties. In an octahedral field, d orbitals split into:
A t<sub>2g</sub> (lower energy, 3 orbitals) and eg (higher energy, 2 orbitals)B Two perfectly equal-energy sets of orbitals with no splitting at allC Three separate sets of orbitals all at different energy levelsD An unpredictable, random pattern of orbital energiesShow answer & explanation →
Q33.
A complex with 3 unpaired electrons is:
A Paramagnetic (attracted to magnetic field)B Diamagnetic and therefore weakly repelled by a magnetic fieldC Completely non-magnetic with zero interaction with a fieldD Antiferromagnetic with perfectly cancelling adjacent spinsShow answer & explanation →
Q34.
Which species is diamagnetic?
A Zn<sup>2+</sup> ([Ar]3d<sup>10</sup>, all paired)B Fe<sup>3+</sup> ([Ar]3d<sup>5</sup>) in typical laboratory settingsC Cu<sup>2+</sup> ([Ar]3d<sup>9</sup>) under usual circumstancesD Mn<sup>2+</sup> ([Ar]3d<sup>5</sup>) according to most researchersShow answer & explanation →
Q35.
What is the spin-only magnetic moment formula?
A mu = sqrt(n(n+2)) Bohr magnetons, where n = number of unpaired electronsB mu = n Bohr magnetons, scaling linearly with unpaired electron countC mu = n<sup>2</sup> Bohr magnetons, scaling with the square of electron countD mu = 2n Bohr magnetons, simply doubling the unpaired electron countShow answer & explanation →
Q36.
The Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> ion in acidic solution is a strong oxidising agent because:
A Cr is reduced from +6 to +3, releasing 3e- per Cr atom (6e- total per formula unit)B The dichromate ion is itself generally an acidic species in solution in the majority of cases studiedC Chromium is oxidised from its elemental state of 0 up to +6 as widely reportedD No actual electron transfer occurs during this redox reaction in standard practiceShow answer & explanation →
Q37.
Lanthanide contraction refers to:
A Progressive decrease in atomic/ionic radii from La to Lu due to poor shielding by 4f electronsB A steady increase in both atomic and ionic radii moving from La all the way to LuC The spontaneous radioactive decay process observed uniformly across the whole lanthanide seriesD The gradual stepwise loss of all fourteen 4f electrons moving from La through to LuShow answer & explanation →
Q38.
The consequence of lanthanide contraction is that:
A 5d elements (Period 6) are nearly the same size as 4d elements (Period 5) in the same groupB All fifteen lanthanide elements end up with completely identical chemical propertiesC The entire d-block of the periodic table physically contracts in sizeD The lanthanide elements become progressively larger across the seriesShow answer & explanation →
Q39.
Which transition metal shows the highest oxidation state (+8)?
A Osmium (OsO<sub>4</sub>) and Ruthenium (RuO<sub>4</sub>)B Iron, which is most stable in the +2 and +3 oxidation statesC Manganese, whose highest common oxidation state is +7D Chromium, whose highest common oxidation state is +6Show answer & explanation →
Q40.
In the variable oxidation states of Mn, the most stable in acidic aqueous solution is:
A Mn<sup>2+</sup>B Mn<sup>3+</sup>C MnO<sub>4</sub><sup>-</sup>D MnO<sub>4</sub><sup>2-</sup>Show answer & explanation →
Q41.
Which complex shows the most intense colour in crystal field theory?
A Complex with large crystal field splitting (strong field ligands)B A complex showing only a small crystal field splitting energyC Any diamagnetic complex regardless of its splitting energyD A complex with an empty d<sup>0</sup> configuration and no d electrons to exciteShow answer & explanation →
Q42.
The spectrochemical series arranges ligands by:
A Increasing field strength: I- < Br- < Cl- < F- < OH- < H<sub>2</sub>O < NH<sub>3</sub> < en < CN-B Decreasing atomic mass of the donor atom across the ligand series under most conditions encounteredC Charge on the ligand alone, independent of donor atom identity as frequently observed in practiceD Physical size of the ligand alone, independent of field strength in many documented casesShow answer & explanation →
Q43.
Interstitial compounds of transition metals with C, N, or H are:
A Hard, high melting point, conduct electricity but chemically inert (e.g., TiC, WC)B Simple ionic compounds held together by electrostatic lattice forces according to conventional understandingC Discrete molecular compounds with well-defined low melting points in routine practiceD Readily soluble compounds that dissolve largely in water overall in most casesShow answer & explanation →
Q44.
Which test distinguishes Fe<sup>2+</sup> from Fe<sup>3+</sup>?
A KSCN gives blood-red with Fe<sup>3+</sup> (FeSCN2+); no colour with Fe<sup>2+</sup>; K<sub>4</sub>[Fe(CN)<sub>6</sub>] gives Turnbull's blue with Fe<sup>3+</sup>B Both Fe<sup>2+</sup> and Fe<sup>3+</sup> are said to give an identical colour reaction with every common test reagent under typical conditionsC A simple pH measurement taken of the dissolved iron salt solution alone according to standard textbooksD A standard flame test performed by observing the colour of the emitted light in general practice as frequently describedShow answer & explanation →
Q45.
Catalytic activity of transition metals is due to:
A Ability to change oxidation states and provide active surface for adsorptionB Their generally high molecular or atomic mass compared to other metalsC Their predominantly ionic character in the bulk metallic stateD Their underlying noble gas electron configurationShow answer & explanation →
Q46.
The actinide series fills which orbitals?
A 5f orbitals (from Ac, Z=89 to Lr, Z=103)B 4f orbitals, the same set filled by the lanthanide seriesC 6d orbitals, filled instead in the subsequent transactinide seriesD 7s orbitals, which are already filled before the actinides beginShow answer & explanation →
Q47.
Which transition metal is used in the Contact process as a catalyst?
A Vanadium(V) oxide (V<sub>2</sub>O<sub>5</sub>)B Iron, the catalyst used instead in the Haber process for ammoniaC Platinum, the catalyst used instead in the Ostwald process for nitric acidD Nickel, the catalyst used instead in the hydrogenation of vegetable oilsShow answer & explanation →
Q48.
Transition metals form alloys easily because:
A Similar atomic radii allow atoms to substitute for each other in the metallic latticeB Strong ionic bonding forms directly between the two different metal atoms in most textbook accountsC Every transition metal happens to share an identical valence electron count during normal conditionsD All transition metals possess exactly two valence electrons each as generally observedShow answer & explanation →
Q49.
The colour change of K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> from orange to yellow on adding NaOH is due to:
A Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> (orange, acidic) converting to CrO<sub>4</sub><sup>2-</sup> (yellow, basic)B A reduction of chromium from the +6 to the +3 oxidation stateC An oxidation of chromium to an even higher oxidation state than +6D A simple temperature change caused by mixing the two solutionsShow answer & explanation →
Q50.
The Ziegler-Natta catalyst contains which transition metal?
A Titanium (TiCl<sub>4</sub>) with organoaluminium compoundB Iron, paired with an organoaluminium co-catalyst insteadC Cobalt, paired with an organoaluminium co-catalyst insteadD Platinum, paired with an organoaluminium co-catalyst insteadShow answer & explanation →
Hard - 25 questions Q51.
Which of the following transition-metal ions has the greatest number of unpaired electrons?
A Ti<sup>3+</sup>B Mn<sup>2+</sup>C Fe<sup>2+</sup>D Ni<sup>2+</sup>Show answer & explanation →
Q52.
The actinoid contraction is greater than the lanthanoid contraction because the:
A 5f orbitals are completely filledB actinoids experience a much smaller effective nuclear chargeC 5f orbitals shield the nuclear charge even more poorly than 4fD actinoids are all diamagneticShow answer & explanation →
Q53.
Cerium commonly shows a stable +4 oxidation state chiefly because Ce<sup>4+</sup>:
A is strongly hydratedB is a strong reducing agentC has a stable exactly half-filled 4f subshellD attains the stable 4f<sup>0</sup> (noble gas) configurationShow answer & explanation →
Q54.
Which pair of ions is isoelectronic and therefore has the same spin-only magnetic moment?
A Ti<sup>3+</sup> and V<sup>3+</sup>B Mn<sup>2+</sup> and Fe<sup>3+</sup>C Cu<sup>2+</sup> and Zn<sup>2+</sup>D Cr<sup>3+</sup> and Fe<sup>3+</sup>Show answer & explanation →
Q55.
The standard reduction potential E°(Mn<sup>3+</sup>/Mn<sup>2+</sup>) is unusually high (very positive) because:
A Mn<sup>3+</sup> already has a stable half-filled configurationB Mn<sup>3+</sup> is strongly hydratedC Mn<sup>2+</sup> (d<sup>5</sup>) has an extra-stable half-filled 3d subshellD Mn<sup>2+</sup> is a strong oxidising agentShow answer & explanation →
Q56.
Using crystal field theory, predict whether [Co(NH<sub>3</sub>)<sub>6</sub>]3+ is high spin or low spin:
A Low spin: NH<sub>3</sub> is a strong field ligand, large Delta causes electron pairing (t<sub>2g</sub><sup>6</sup>, 0 unpaired)B High spin, since NH<sub>3</sub> behaves as a weak field ligand toward cobalt(III) under most conditions encounteredC Paramagnetic with four unpaired electrons distributed across t<sub>2g</sub> and eg as frequently observed in practiceD An outcome that cannot be predicted from crystal field theory in many documented cases according to conventional understandingShow answer & explanation →
Q57.
The crystal field stabilisation energy (CFSE) for d<sup>3</sup> in an octahedral field is:
A -1.2 Delta_o (three electrons each in t<sub>2g</sub>, each contributing -0.4 Delta_o)B Zero, since a d<sup>3</sup> configuration shows no net stabilisation in routine practiceC +1.2 Delta_o, a positive destabilisation for this configuration overallD -0.8 Delta_o, corresponding instead to a d2 configuration in most casesShow answer & explanation →
Q58.
The Jahn-Teller effect in Cu<sup>2+</sup> complexes (d<sup>9</sup>) causes:
A Distortion of the octahedral geometry (elongation along z-axis due to unequal occupation of eg orbitals)B No distortion, leaving the geometry as a perfectly regular octahedron in every case under typical conditionsC A complete rearrangement of the ligands into an largely trigonal prismatic geometry instead according to standard textbooksD An overall increase in complex stability while the octahedral geometry stays perfectly regular in general practiceShow answer & explanation →
Q59.
Which of the following complexes does NOT show Jahn-Teller distortion?
A [Co(NH<sub>3</sub>)<sub>6</sub>]3+ (d<sup>6</sup>, low spin, t<sub>2g</sub><sup>6</sup> = symmetric)B [Cu(H<sub>2</sub>O)6]2+ (d<sup>9</sup>)C [Mn(H<sub>2</sub>O)6]3+ (d<sup>4</sup> high spin)D [CrF6]3- (d<sup>3</sup>)Show answer & explanation →
Q60.
The 18-electron rule in organometallics states that:
A Stable transition metal complexes have 18 electrons in the valence shell (sum of metal d electrons + ligand electrons)B Every transition metal atom is said to contain exactly 18 neutrons within its nucleus as frequently described in most textbook accountsC A maximum of 18 separate individual ligands can ever coordinate to one single metal centre during normal conditionsD Transition metals are said to be capable of exhibiting up to 18 distinct oxidation states as generally observed in typical laboratory settingsShow answer & explanation →
Q61.
Fe(CO)<sub>5</sub> is a stable organometallic compound. The oxidation state of Fe in Fe(CO)<sub>5</sub> is:
A 0 (CO is a neutral ligand)B 2+, since each CO ligand is treated as a -2/5 charge donorC 3+, matching iron's most common oxidation state in its saltsD -2, as iron is formally reduced by the five CO ligandsShow answer & explanation →
Q62.
The trans-influence in square planar complexes refers to:
A Weakening of the bond trans to a strong trans-influencing ligand (strong sigma donor weakens trans bond)B A simple shift in the colour the square planar complex visibly displays under usual circumstances according to most researchersC A measurable change observed in the complex's overall magnetic moment instead in the majority of cases studiedD The complete absence of any measurable effect whatsoever on the trans bond strength as widely reportedShow answer & explanation →
Q63.
Which of the following metal ions will form colourless complexes?
A Sc3+ (d<sup>0</sup>) and Ti4+ (d<sup>0</sup>) and Zn<sup>2+</sup> (d<sup>10</sup>)B Fe<sup>3+</sup> (d<sup>5</sup>), which gives pale yellow complexes from a spin-forbidden d-d transitionC Cu<sup>2+</sup> (d<sup>9</sup>), which gives characteristically blue complexesD Ni<sup>2+</sup> (d<sup>8</sup>), which gives characteristically green complexesShow answer & explanation →
Q64.
The effective atomic number (EAN) rule is satisfied by [Ni(CO)<sub>4</sub>]. Ni has which electron count in this complex?
A 18 (Ni: 10 d+4s electrons; 4 CO donate 2e each = 8; total 10+8=18)B 16, the electron count typical of square planar d<sup>8</sup> complexes insteadC 20, an electron count that would exceed the stable noble gas totalD 12, an electron count well below the stable 18-electron configurationShow answer & explanation →
Q65.
Lanthanide ions (Ln3+) are less coloured than transition metals because:
A 4f orbitals are well-shielded and inner; f-f transitions are Laporte-forbidden and very weakB Lanthanide ions largely lack any d orbitals in their electron configuration in standard practiceC Lanthanide ions are uniformly diamagnetic with little unpaired electrons under most conditions encounteredD Lanthanide ions characteristically adopt unusually high oxidation states as frequently observed in practiceShow answer & explanation →
Q66.
Nuclear fission of 235U is initiated by:
A Slow (thermal) neutrons captured by the nucleusB High-energy fast neutrons exclusively, since slow neutrons cannot be capturedC High-energy gamma ray photons absorbed by the nucleusD Alpha particles fired directly at the uranium nucleusShow answer & explanation →
Q67.
The trans effect in square planar Pt(II) complexes is used to synthesise:
A Cisplatin [cis-Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>] by exploiting that Cl- has stronger trans-labilising effect than NH<sub>3</sub>B A largely random, loosely controlled mixture of cis and trans geometric isomers in many documented casesC Mainly the trans isomer of the platinum complex with little cis product formed according to conventional understandingD Octahedral platinum complexes formed instead of the intended square planar product in routine practiceShow answer & explanation →
Q68.
Ferromagnetism (as in iron) occurs because:
A Unpaired electrons in adjacent atoms align parallel in magnetic domains (quantum exchange coupling)B A significant degree of ionic character present throughout the bulk metal lattice in many documented casesC The simple presence of a very large total number of electrons within each metal atom according to conventional understandingD The metal's characteristically high melting point compared to other transition metals in routine practiceShow answer & explanation →
Q69.
What is the difference between lanthanides and actinides in terms of chemical behaviour?
A Actinides show greater variety of oxidation states (due to 5f, 6d and 7s close in energy) vs lanthanides (predominantly +3)B There is said to be essentially no meaningful chemical difference between the two series overall in most cases under typical conditionsC Actinides are said to be largely non-radioactive elements, unlike the well-known lanthanides according to standard textbooksD Lanthanides are said to characteristically show a far wider range of oxidation states than actinides in general practiceShow answer & explanation →
Q70.
The Bohr magneton (BM) is the unit of magnetic moment. The spin magnetic moment is given by mu = sqrt(n(n+2)). For Fe<sup>3+</sup> (d<sup>5</sup>, high spin), mu =
A 5.92 BM (n=5 unpaired electrons)B 3.87 BM, the value corresponding instead to three unpaired electronsC 1.73 BM, the value corresponding instead to a single unpaired electronD 2.83 BM, the value corresponding instead to two unpaired electronsShow answer & explanation →
Q71.
Which transition metal complex is used as an anti-cancer drug?
A Cisplatin [cis-Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>]B K2[PtCl<sub>4</sub>] as frequently describedC [Co(NH<sub>3</sub>)<sub>6</sub>]Cl3 in most textbook accountsD [Fe(CN)<sub>6</sub>]4- during normal conditionsShow answer & explanation →
Q72.
Why do transition metals and their compounds act as catalysts for many industrial reactions?
A They have accessible multiple oxidation states that allow electron transfer cycles, and they can adsorb and activate substratesB Catalytic activity is said to generally correlate with how expensive each particular metal happens to be as generally observedC Their generally large atomic weights are what is said to account for their catalytic ability in typical laboratory settingsD Every transition metal catalyst is said to necessarily be an inherently paramagnetic species under usual circumstances according to most researchersShow answer & explanation →
Q73.
The crystal field splitting in octahedral (Delta_o) vs tetrahedral (Delta_t) field: which is larger?
A Octahedral (Delta_o = 9/4 Delta_t); octahedral complexes have much larger splittingB Tetrahedral splitting, which exceeds octahedral splitting for the same ligandC Both geometries give numerically identical splitting energiesD The relative splitting depends solely on the ligand and not on geometryShow answer & explanation →
Q74.
In the disproportionation of MnO<sub>4</sub><sup>2-</sup> in acidic solution:
A 3MnO<sub>4</sub><sup>2-</sup> + 4H+ → 2MnO<sub>4</sub><sup>-</sup> + MnO<sub>2</sub> + 2H<sub>2</sub>O (Mn goes from +6 to +7 and +4)B Manganese remains largely in the +6 oxidation state throughoutC Manganese is largely reduced down to the +2 oxidation state during normal conditionsD No reaction occurs when MnO<sub>4</sub><sup>2-</sup> is acidified as generally observedShow answer & explanation →
Q75.
The magnetic moment of a complex ion can be used to determine:
A Number of unpaired electrons (and thus high-spin vs low-spin configuration)B The exact charge carried by each coordinated ligandC The metal's oxidation state read off directly without any calculationD The precise colour the complex will display in solutionShow answer & explanation →