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d- and f-Block Elements - Practice Questions with Answers

60 free MCQs on d- and f-Block Elements with worked answers and explanations. 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.

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Below are 60 practice questions on d- and f-Block Elements, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the d- and f-Block Elements notes.

Colours of Common Transition Metal Ions (Aqueous)Cu²⁺ blueFe²⁺ pale greenFe³⁺ brown-yellowMn²⁺ pale pinkNi²⁺/Cr³⁺ greenColour 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 - 20 questions

Q1.

Transition metals are found in which block of the periodic table?

  • A d-block
  • B s-block
  • C p-block
  • D f-block
Show answer & explanation

Answer: A. d-block

Why: Transition metals are d-block elements (Groups 3-12) where the d orbitals are progressively filled.

Q2.

Which property is characteristic of transition metals?

  • A Variable oxidation states
  • B Fixed single oxidation state
  • C No colour in compounds
  • D Non-magnetic
Show answer & explanation

Answer: A. Variable oxidation states

Why: Transition metals show multiple oxidation states because d electrons can be lost at different energies.

Q3.

What colour is the KMnO<sub>4</sub> (potassium permanganate) solution?

  • A Purple/violet
  • B Yellow
  • C Green
  • D Colourless
Show answer & explanation

Answer: A. Purple/violet

Why: KMnO<sub>4</sub> is deep purple due to charge-transfer in the MnO<sub>4</sub><sup>-</sup> ion.

Q4.

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

Answer: A. Iron (Fe)

Why: Iron (with K2O and Al<sub>2</sub>O<sub>3</sub> promoters) is the catalyst in the Haber process for ammonia synthesis.

Q5.

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 catalysis
  • C Zinc, commonly used as a sacrificial anode in galvanising
  • D Copper, prized for its high electrical conductivity in wiring
Show answer & explanation

Answer: A. Platinum (Pt) and Palladium (Pd)

Why: Catalytic converters use Pt, Pd, and Rh to oxidise CO and reduce NOx to non-toxic gases.

Q6.

The colour of transition metal compounds is due to:

  • A d-d electron transitions absorbing visible light
  • B Their characteristically high melting points compared to main-group salts
  • C Their predominantly ionic character in the solid lattice
  • D Their general paramagnetic or ferromagnetic behaviour
Show answer & explanation

Answer: A. d-d electron transitions absorbing visible light

Why: Partially filled d orbitals allow d-d electronic transitions; the absorbed wavelength corresponds to visible light, producing colour.

Q7.

Which transition metal is liquid at room temperature?

  • A Mercury (Hg)
  • B Zinc
  • C Iron
  • D Nickel
Show answer & explanation

Answer: A. Mercury (Hg)

Why: Mercury is the only metal that is liquid at room temperature; it freezes at -39°C.

Q8.

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 stable
  • B It has odd atomic number under usual circumstances
  • C 3d orbitals are empty according to most researchers
  • D It is a noble gas in the majority of cases studied
Show answer & explanation

Answer: A. Fully filled d<sup>10</sup> is extra stable

Why: A fully filled d<sup>10</sup> sub-shell has extra stability; Cu 'borrows' one electron from 4s to achieve [Ar]3d<sup>10</sup> 4s<sup>1</sup>.

Q9.

Which transition metal is the best conductor of electricity?

  • A Silver (Ag)
  • B Copper (Cu)
  • C Gold (Au)
  • D Aluminium (Al)
Show answer & explanation

Answer: A. Silver (Ag)

Why: Silver has the highest electrical conductivity of all metals; copper is second and most commonly used (cheaper).

Q10.

The lanthanides are also called:

  • A Rare earth metals
  • B Alkaline earth metals
  • C Transition metals
  • D Noble metals
Show answer & explanation

Answer: A. Rare earth metals

Why: Lanthanides (La-Lu) are also known as rare earth metals, though most are not actually rare.

Q11.

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

Answer: A. Gold (Au)

Why: Gold is called the king of metals for its nobility (resists corrosion), lustre, and rarity.

Q12.

Rust is formed when iron reacts with:

  • A Oxygen and water (moist air)
  • B Carbon dioxide gas dissolved in dry air
  • C Nitrogen gas present in the surrounding atmosphere
  • D Dilute hydrochloric acid splashed onto the surface
Show answer & explanation

Answer: A. Oxygen and water (moist air)

Why: Rusting: 4Fe + 3O<sub>2</sub> + 6H<sub>2</sub>O → 4Fe(OH)3; this dehydrates to Fe2O<sub>3</sub>·xH2O (rust) over time.

Q13.

Which transition metal is essential for haemoglobin?

  • A Iron (Fe<sup>2+</sup>)
  • B Zinc
  • C Copper
  • D Manganese
Show answer & explanation

Answer: A. Iron (Fe<sup>2+</sup>)

Why: Haemoglobin contains Fe<sup>2+</sup> in its haem group, which binds O<sub>2</sub> for oxygen transport in blood.

Q14.

K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> (potassium dichromate) has which colour?

  • A Orange
  • B Purple
  • C Blue
  • D Green
Show answer & explanation

Answer: A. Orange

Why: K<sub>2</sub>Cr<sub>2</sub>O<sub>7</sub> is orange due to the Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> ion; in alkaline conditions it converts to yellow CrO<sub>4</sub><sup>2-</sup>.

Q15.

Which transition metal is used to galvanise iron?

  • A Zinc (Zn)
  • B Tin (Sn)
  • C Nickel (Ni)
  • D Chromium (Cr)
Show answer & explanation

Answer: A. Zinc (Zn)

Why: Galvanising coats iron with zinc; zinc acts as a sacrificial anode protecting iron from corrosion.

Q16.

Transition metals generally have high melting points because:

  • A Strong metallic bonding involving both s and d electrons
  • B A significant degree of ionic character within the metallic lattice
  • C Their unusually large atomic masses compared to main-group metals
  • D Their underlying noble gas electron configuration
Show answer & explanation

Answer: A. Strong metallic bonding involving both s and d electrons

Why: Participation of both 4s and 3d electrons in metallic bonding gives transition metals stronger metallic bonds and higher melting points than s-block metals.

Q17.

Which element is used as the filament in incandescent light bulbs?

  • A Tungsten (W)
  • B Iron
  • C Molybdenum
  • D Nickel
Show answer & explanation

Answer: A. Tungsten (W)

Why: Tungsten has the highest melting point of all metals (3422°C) and can withstand filament temperatures without melting.

Q18.

Which f-block elements are radioactive?

  • A All actinides
  • B All lanthanides
  • C Mainly uranium
  • D Mainly thorium
Show answer & explanation

Answer: A. All actinides

Why: All actinides (Z=89-103) are radioactive. The lanthanides include only one radioactive element (promethium, Pm).

Q19.

Uranium is used in:

  • A Nuclear fuel and nuclear weapons
  • B Industrial catalysts for hydrogenation reactions
  • C Stainless steel alloys requiring corrosion resistance
  • D Permanent magnets used in electric motors
Show answer & explanation

Answer: A. Nuclear fuel and nuclear weapons

Why: Uranium (235U) undergoes fission and is used as nuclear reactor fuel; enriched uranium is used in nuclear weapons.

Q20.

Which transition metal is used in stainless steel?

  • A Chromium (Cr)
  • B Copper
  • C Zinc
  • D Tin
Show answer & explanation

Answer: A. Chromium (Cr)

Why: Stainless steel is an alloy of iron with at least 10.5% chromium; the Cr forms a passive oxide layer preventing rust.

Medium - 20 questions

Q21.

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 symmetry
  • B The 3d subshell is usually filled largely before the 4s subshell
  • C Chromium is a diamagnetic element with little unpaired electrons
  • D The 4s subshell must usually retain exactly two electrons under most conditions encountered
Show answer & explanation

Answer: A. Half-filled d<sup>5</sup> is extra stable due to exchange energy and symmetry

Why: A half-filled 3d<sup>5</sup> has all electrons unpaired with maximum exchange energy, giving it extra stability; one 4s electron 'migrates' to 3d.

Q22.

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 all
  • C Three separate sets of orbitals all at different energy levels
  • D An unpredictable, random pattern of orbital energies
Show answer & explanation

Answer: A. t<sub>2g</sub> (lower energy, 3 orbitals) and eg (higher energy, 2 orbitals)

Why: An octahedral ligand field splits d orbitals into lower-energy t<sub>2g</sub> (dxy, dxz, dyz) and higher-energy eg (dx2-y2, dz<sup>2</sup>) sets.

Q23.

A complex with 3 unpaired electrons is:

  • A Paramagnetic (attracted to magnetic field)
  • B Diamagnetic and therefore weakly repelled by a magnetic field
  • C Completely non-magnetic with zero interaction with a field
  • D Antiferromagnetic with perfectly cancelling adjacent spins
Show answer & explanation

Answer: A. Paramagnetic (attracted to magnetic field)

Why: Unpaired electrons generate magnetic dipoles; paramagnetic substances are attracted to external magnetic fields.

Q24.

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 settings
  • C Cu<sup>2+</sup> ([Ar]3d<sup>9</sup>) under usual circumstances
  • D Mn<sup>2+</sup> ([Ar]3d<sup>5</sup>) according to most researchers
Show answer & explanation

Answer: A. Zn<sup>2+</sup> ([Ar]3d<sup>10</sup>, all paired)

Why: Zn<sup>2+</sup> has completely filled 3d<sup>10</sup> configuration with all electrons paired, making it diamagnetic.

Q25.

What is the spin-only magnetic moment formula?

  • A mu = sqrt(n(n+2)) Bohr magnetons, where n = number of unpaired electrons
  • B mu = n Bohr magnetons, scaling linearly with unpaired electron count
  • C mu = n<sup>2</sup> Bohr magnetons, scaling with the square of electron count
  • D mu = 2n Bohr magnetons, simply doubling the unpaired electron count
Show answer & explanation

Answer: A. mu = sqrt(n(n+2)) Bohr magnetons, where n = number of unpaired electrons

Why: The spin-only magnetic moment: mu = sqrt(n(n+2)) BM. For n=1: mu=1.73 BM; n=2: 2.83 BM; n=3: 3.87 BM; n=4: 4.90 BM; n=5: 5.92 BM.

Q26.

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 studied
  • C Chromium is oxidised from its elemental state of 0 up to +6 as widely reported
  • D No actual electron transfer occurs during this redox reaction in standard practice
Show answer & explanation

Answer: A. Cr is reduced from +6 to +3, releasing 3e- per Cr atom (6e- total per formula unit)

Why: Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> + 14H+ + 6e- → 2Cr<sup>3+</sup> + 7H<sub>2</sub>O; it accepts 6 electrons per Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> unit, making it a powerful 6-electron oxidant in acid.

Q27.

Lanthanide contraction refers to:

  • A Progressive decrease in atomic/ionic radii from La to Lu due to poor shielding by 4f electrons
  • B A steady increase in both atomic and ionic radii moving from La all the way to Lu
  • C The spontaneous radioactive decay process observed uniformly across the whole lanthanide series
  • D The gradual stepwise loss of all fourteen 4f electrons moving from La through to Lu
Show answer & explanation

Answer: A. Progressive decrease in atomic/ionic radii from La to Lu due to poor shielding by 4f electrons

Why: As 4f electrons are added across the lanthanide series, poor shielding by 4f orbitals allows increasing nuclear charge to contract the atom/ion progressively.

Q28.

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 group
  • B All fifteen lanthanide elements end up with completely identical chemical properties
  • C The entire d-block of the periodic table physically contracts in size
  • D The lanthanide elements become progressively larger across the series
Show answer & explanation

Answer: A. 5d elements (Period 6) are nearly the same size as 4d elements (Period 5) in the same group

Why: Lanthanide contraction causes Period 6 transition metals (5d) to be nearly the same size as their Period 5 (4d) counterparts, affecting their chemistry significantly.

Q29.

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 states
  • C Manganese, whose highest common oxidation state is +7
  • D Chromium, whose highest common oxidation state is +6
Show answer & explanation

Answer: A. Osmium (OsO<sub>4</sub>) and Ruthenium (RuO<sub>4</sub>)

Why: Os and Ru can reach +8 in their tetroxides (OsO<sub>4</sub>, RuO<sub>4</sub>), the highest oxidation states among transition metals.

Q30.

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

Answer: A. Mn<sup>2+</sup>

Why: Mn<sup>2+</sup> ([Ar]3d<sup>5</sup>) is the most stable oxidation state in acidic conditions due to the extra stability of the half-filled d<sup>5</sup> configuration.

Q31.

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 energy
  • C Any diamagnetic complex regardless of its splitting energy
  • D A complex with an empty d<sup>0</sup> configuration and no d electrons to excite
Show answer & explanation

Answer: A. Complex with large crystal field splitting (strong field ligands)

Why: Stronger field ligands cause larger splitting (Delta), allowing d-d transitions of higher energy, giving more intense (and often deeper) colour.

Q32.

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 encountered
  • C Charge on the ligand alone, independent of donor atom identity as frequently observed in practice
  • D Physical size of the ligand alone, independent of field strength in many documented cases
Show answer & explanation

Answer: A. Increasing field strength: I- < Br- < Cl- < F- < OH- < H<sub>2</sub>O < NH<sub>3</sub> < en < CN-

Why: The spectrochemical series ranks ligands from weak field (causing small Delta, spin-free) to strong field (large Delta, spin-paired) based on their crystal field splitting ability.

Q33.

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 understanding
  • C Discrete molecular compounds with well-defined low melting points in routine practice
  • D Readily soluble compounds that dissolve largely in water overall in most cases
Show answer & explanation

Answer: A. Hard, high melting point, conduct electricity but chemically inert (e.g., TiC, WC)

Why: Interstitial compounds trap small atoms (C, N, H) in the holes of the metallic lattice; they are harder, higher melting, and less reactive than the parent metal.

Q34.

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 conditions
  • C A simple pH measurement taken of the dissolved iron salt solution alone according to standard textbooks
  • D A standard flame test performed by observing the colour of the emitted light in general practice as frequently described
Show answer & explanation

Answer: 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>

Why: KSCN (potassium thiocyanate) forms an intensely blood-red complex [Fe(SCN)]2+ with Fe<sup>3+</sup>; Fe<sup>2+</sup> gives no colour with KSCN.

Q35.

Catalytic activity of transition metals is due to:

  • A Ability to change oxidation states and provide active surface for adsorption
  • B Their generally high molecular or atomic mass compared to other metals
  • C Their predominantly ionic character in the bulk metallic state
  • D Their underlying noble gas electron configuration
Show answer & explanation

Answer: A. Ability to change oxidation states and provide active surface for adsorption

Why: Transition metals catalyse reactions by: (1) adsorbing reactants on their surface (heterogeneous) or (2) changing oxidation state to shuttle electrons (homogeneous).

Q36.

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 series
  • C 6d orbitals, filled instead in the subsequent transactinide series
  • D 7s orbitals, which are already filled before the actinides begin
Show answer & explanation

Answer: A. 5f orbitals (from Ac, Z=89 to Lr, Z=103)

Why: Actinides (Z=89-103) progressively fill 5f orbitals, analogous to lanthanides filling 4f.

Q37.

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 ammonia
  • C Platinum, the catalyst used instead in the Ostwald process for nitric acid
  • D Nickel, the catalyst used instead in the hydrogenation of vegetable oils
Show answer & explanation

Answer: A. Vanadium(V) oxide (V<sub>2</sub>O<sub>5</sub>)

Why: V<sub>2</sub>O<sub>5</sub> catalyses the oxidation of SO<sub>2</sub> to SO<sub>3</sub> in the Contact process for H<sub>2</sub>SO<sub>4</sub> manufacture.

Q38.

Transition metals form alloys easily because:

  • A Similar atomic radii allow atoms to substitute for each other in the metallic lattice
  • B Strong ionic bonding forms directly between the two different metal atoms in most textbook accounts
  • C Every transition metal happens to share an identical valence electron count during normal conditions
  • D All transition metals possess exactly two valence electrons each as generally observed
Show answer & explanation

Answer: A. Similar atomic radii allow atoms to substitute for each other in the metallic lattice

Why: Transition metal atoms are of similar size; they can replace each other in metal lattices without distorting the crystal structure, forming alloys with tunable properties.

Q39.

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 state
  • C An oxidation of chromium to an even higher oxidation state than +6
  • D A simple temperature change caused by mixing the two solutions
Show answer & explanation

Answer: 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)

Why: Cr<sub>2</sub>O<sub>7</sub><sup>2-</sup> + 2OH- → 2CrO<sub>4</sub><sup>2-</sup> + H<sub>2</sub>O. In acidic conditions dichromate (orange) is dominant; in basic conditions chromate (yellow) predominates.

Q40.

The Ziegler-Natta catalyst contains which transition metal?

  • A Titanium (TiCl<sub>4</sub>) with organoaluminium compound
  • B Iron, paired with an organoaluminium co-catalyst instead
  • C Cobalt, paired with an organoaluminium co-catalyst instead
  • D Platinum, paired with an organoaluminium co-catalyst instead
Show answer & explanation

Answer: A. Titanium (TiCl<sub>4</sub>) with organoaluminium compound

Why: Ziegler-Natta catalyst: TiCl<sub>4</sub> + Al(C<sub>2</sub>H<sub>5</sub>)<sub>3</sub> (triethylaluminium), used for stereospecific polymerisation of alkenes.

Hard - 20 questions

Q41.

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 encountered
  • C Paramagnetic with four unpaired electrons distributed across t<sub>2g</sub> and eg as frequently observed in practice
  • D An outcome that cannot be predicted from crystal field theory in many documented cases according to conventional understanding
Show answer & explanation

Answer: 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)

Why: Co3+ has d<sup>6</sup> configuration; NH<sub>3</sub> is a strong field ligand causing large Delta, making electron pairing energetically favourable: t<sub>2g</sub><sup>6</sup> e<sub>g</sub><sup>0</sup> (0 unpaired), diamagnetic.

Q42.

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 practice
  • C +1.2 Delta_o, a positive destabilisation for this configuration overall
  • D -0.8 Delta_o, corresponding instead to a d2 configuration in most cases
Show answer & explanation

Answer: A. -1.2 Delta_o (three electrons each in t<sub>2g</sub>, each contributing -0.4 Delta_o)

Why: In an octahedral field, t<sub>2g</sub> electrons each lower energy by 0.4 Δo; for d<sup>3</sup> (all in t<sub>2g</sub>): CFSE = 3 × (-0.4)Δo = -1.2Δo.

Q43.

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 conditions
  • C A complete rearrangement of the ligands into an largely trigonal prismatic geometry instead according to standard textbooks
  • D An overall increase in complex stability while the octahedral geometry stays perfectly regular in general practice
Show answer & explanation

Answer: A. Distortion of the octahedral geometry (elongation along z-axis due to unequal occupation of eg orbitals)

Why: Cu<sup>2+</sup> (d<sup>9</sup>) has an unequal eg occupancy (eg: one in dz<sup>2</sup>, one in dx2-y2... or vice versa); this unequal occupation causes a tetragonal distortion (Jahn-Teller elongation along z).

Q44.

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

Answer: D. [CrF6]3- (d<sup>3</sup>)

Why: [CrF6]3- has d<sup>3</sup> configuration: t<sub>2g</sub><sup>3</sup> (one electron in each t<sub>2g</sub> orbital); this is symmetric and shows no Jahn-Teller distortion. [Co(NH<sub>3</sub>)<sub>6</sub>]3+ d<sup>6</sup> low spin is also symmetric but for different reasons. d<sup>3</sup> and high-spin d<sup>5</sup> are always symmetric.

Q45.

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 accounts
  • C A maximum of 18 separate individual ligands can ever coordinate to one single metal centre during normal conditions
  • D Transition metals are said to be capable of exhibiting up to 18 distinct oxidation states as generally observed in typical laboratory settings
Show answer & explanation

Answer: A. Stable transition metal complexes have 18 electrons in the valence shell (sum of metal d electrons + ligand electrons)

Why: The 18-electron rule (EAN rule): stable organometallic complexes have 18 electrons in the metal valence shell (like a noble gas configuration for the metal).

Q46.

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 donor
  • C 3+, matching iron's most common oxidation state in its salts
  • D -2, as iron is formally reduced by the five CO ligands
Show answer & explanation

Answer: A. 0 (CO is a neutral ligand)

Why: CO is a neutral ligand with no charge; Fe(CO)<sub>5</sub> has no overall charge, so Fe must be in the 0 oxidation state.

Q47.

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 researchers
  • C A measurable change observed in the complex's overall magnetic moment instead in the majority of cases studied
  • D The complete absence of any measurable effect whatsoever on the trans bond strength as widely reported
Show answer & explanation

Answer: A. Weakening of the bond trans to a strong trans-influencing ligand (strong sigma donor weakens trans bond)

Why: Trans-influence: strong sigma-donor ligands (CO, CN-, H-) weaken and lengthen the metal-ligand bond trans to them by competing for the same metal orbital.

Q48.

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 transition
  • C Cu<sup>2+</sup> (d<sup>9</sup>), which gives characteristically blue complexes
  • D Ni<sup>2+</sup> (d<sup>8</sup>), which gives characteristically green complexes
Show answer & explanation

Answer: A. Sc3+ (d<sup>0</sup>) and Ti4+ (d<sup>0</sup>) and Zn<sup>2+</sup> (d<sup>10</sup>)

Why: Colourless complexes form with d<sup>0</sup> (no d electrons for d-d transitions) and d<sup>10</sup> (completely filled, no d-d transitions). Sc3+ and Zn<sup>2+</sup> are classic examples.

Q49.

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 instead
  • C 20, an electron count that would exceed the stable noble gas total
  • D 12, an electron count well below the stable 18-electron configuration
Show answer & explanation

Answer: A. 18 (Ni: 10 d+4s electrons; 4 CO donate 2e each = 8; total 10+8=18)

Why: Ni(0) has 10 valence electrons (3d<sup>10</sup> 4s0 after redistribution); 4 CO × 2e = 8e; total = 18. Ni(CO)<sub>4</sub> obeys the 18e rule.

Q50.

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 weak
  • B Lanthanide ions largely lack any d orbitals in their electron configuration in standard practice
  • C Lanthanide ions are uniformly diamagnetic with little unpaired electrons under most conditions encountered
  • D Lanthanide ions characteristically adopt unusually high oxidation states as frequently observed in practice
Show answer & explanation

Answer: A. 4f orbitals are well-shielded and inner; f-f transitions are Laporte-forbidden and very weak

Why: f-f electronic transitions are both Laporte-forbidden and spin-forbidden; they give pale absorptions. Also, 4f orbitals are inner and shielded from ligand field effects.

Q51.

Nuclear fission of 235U is initiated by:

  • A Slow (thermal) neutrons captured by the nucleus
  • B High-energy fast neutrons exclusively, since slow neutrons cannot be captured
  • C High-energy gamma ray photons absorbed by the nucleus
  • D Alpha particles fired directly at the uranium nucleus
Show answer & explanation

Answer: A. Slow (thermal) neutrons captured by the nucleus

Why: 235U is fissile and undergoes fission when it absorbs a thermal (slow, ~0.025 eV) neutron, splitting into fission fragments + 2-3 fast neutrons + energy.

Q52.

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 cases
  • C Mainly the trans isomer of the platinum complex with little cis product formed according to conventional understanding
  • D Octahedral platinum complexes formed instead of the intended square planar product in routine practice
Show answer & explanation

Answer: 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>

Why: Trans effect: by starting with [PtCl<sub>4</sub>]2- and adding NH<sub>3</sub>, Cl- labilises the trans position; the second NH<sub>3</sub> goes cis because of the kinetic trans effect, giving cisplatin.

Q53.

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 cases
  • C The simple presence of a very large total number of electrons within each metal atom according to conventional understanding
  • D The metal's characteristically high melting point compared to other transition metals in routine practice
Show answer & explanation

Answer: A. Unpaired electrons in adjacent atoms align parallel in magnetic domains (quantum exchange coupling)

Why: Ferromagnetism arises from quantum exchange interactions that cause parallel alignment of unpaired electron spins in magnetic domains, producing a permanent magnetic moment.

Q54.

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 conditions
  • C Actinides are said to be largely non-radioactive elements, unlike the well-known lanthanides according to standard textbooks
  • D Lanthanides are said to characteristically show a far wider range of oxidation states than actinides in general practice
Show answer & explanation

Answer: A. Actinides show greater variety of oxidation states (due to 5f, 6d and 7s close in energy) vs lanthanides (predominantly +3)

Why: In lanthanides, 4f is much lower in energy than 5d and 6s; predominant state is +3. In actinides, 5f/6d/7s are close in energy, enabling multiple stable oxidation states (+2 to +7).

Q55.

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 electrons
  • C 1.73 BM, the value corresponding instead to a single unpaired electron
  • D 2.83 BM, the value corresponding instead to two unpaired electrons
Show answer & explanation

Answer: A. 5.92 BM (n=5 unpaired electrons)

Why: Fe<sup>3+</sup> has d<sup>5</sup> high-spin configuration with 5 unpaired electrons: mu = sqrt(5×7) = sqrt(35) = 5.92 BM.

Q56.

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 described
  • C [Co(NH<sub>3</sub>)<sub>6</sub>]Cl3 in most textbook accounts
  • D [Fe(CN)<sub>6</sub>]4- during normal conditions
Show answer & explanation

Answer: A. Cisplatin [cis-Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>]

Why: Cisplatin (cis-Pt(NH<sub>3</sub>)<sub>2</sub>Cl<sub>2</sub>) is the most widely used platinum-based anticancer drug; it cross-links DNA strands, preventing cell division.

Q57.

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 substrates
  • B Catalytic activity is said to generally correlate with how expensive each particular metal happens to be as generally observed
  • C Their generally large atomic weights are what is said to account for their catalytic ability in typical laboratory settings
  • D Every transition metal catalyst is said to necessarily be an inherently paramagnetic species under usual circumstances according to most researchers
Show answer & explanation

Answer: A. They have accessible multiple oxidation states that allow electron transfer cycles, and they can adsorb and activate substrates

Why: Transition metals can accept and donate electrons (cycling oxidation states) and adsorb molecules on their surfaces or form unstable intermediates, lowering activation energy.

Q58.

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 splitting
  • B Tetrahedral splitting, which exceeds octahedral splitting for the same ligand
  • C Both geometries give numerically identical splitting energies
  • D The relative splitting depends solely on the ligand and not on geometry
Show answer & explanation

Answer: A. Octahedral (Delta_o = 9/4 Delta_t); octahedral complexes have much larger splitting

Why: Delta_o/Delta_t = 9/4; octahedral fields cause about 2.25 times more splitting than tetrahedral fields due to 6 vs 4 ligands and geometry.

Q59.

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 throughout
  • C Manganese is largely reduced down to the +2 oxidation state during normal conditions
  • D No reaction occurs when MnO<sub>4</sub><sup>2-</sup> is acidified as generally observed
Show answer & explanation

Answer: 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)

Why: Manganate (Mn +6) is unstable in acid and disproportionates to permanganate (Mn +7) and MnO<sub>2</sub> (Mn +4).

Q60.

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 ligand
  • C The metal's oxidation state read off directly without any calculation
  • D The precise colour the complex will display in solution
Show answer & explanation

Answer: A. Number of unpaired electrons (and thus high-spin vs low-spin configuration)

Why: By measuring the magnetic moment and using mu = sqrt(n(n+2)), we can determine n (unpaired electrons), which reveals the spin state and d electron configuration.