NEET (UG)

The d- and f-Block Elements

Transition metals, their characteristic properties, lanthanoids and actinoids for NEET

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Module 1

Transition (d-Block) Elements

Electronic Configuration and General TrendsTopic 1

The d-block elements, or transition elements, occupy the middle of the periodic table (groups 3–12) and are defined as elements whose atoms or stable ions have partly filled d orbitals. Their general electronic configuration is $(n-1)d^{1-10}\,ns^{1-2}$, where the penultimate $d$ shell fills up across each series. By this strict definition zinc, cadmium and mercury ($d^{10}$) are not true transition elements, although they are studied with the block — a classic NEET point.

Two configurations break the expected order because half-filled and fully-filled $d$ sub-shells are extra stable: chromium is $3d^5\,4s^1$ (not $3d^4\,4s^2$) and copper is $3d^{10}\,4s^1$ (not $3d^9\,4s^2$). These anomalies are frequently tested, so commit them to memory.

The transition metals share characteristic physical trends. They have high melting and boiling points and high densities, owing to strong metallic bonding involving both $d$ and $s$ electrons. Their atomic radii decrease at first across a series, then stay almost constant, and rise slightly at the end — a much smaller variation than in the main groups, because added $d$ electrons shield the increasing nuclear charge fairly well.

Most transition metals also form alloys readily (their similar atomic sizes let one metal replace another in the lattice), display a range of oxidation states, and behave as good catalysts — properties explored in the next topic. The first transition series (Sc to Zn, filling $3d$) is the most important for NEET, and being fluent with writing the configuration of any first-series atom or ion (remembering to remove $4s$ electrons before $3d$ when forming cations) underpins the whole chapter.

Figure — Electronic Configuration and General Trends
FeatureTransition elements
General config$(n-1)d^{1-10}\,ns^{1-2}$
Cr / Cu anomaly$3d^5 4s^1$ / $3d^{10} 4s^1$
Melting point / densityhigh (strong metallic bonding)
Ion formationremove $ns$ electrons before $(n-1)d$
Worked Examples
1

Write the electronic configuration of $\text{Fe}$ (Z = 26) and the ion $\text{Fe}^{3+}$.

Show solution

$\text{Fe}: [\text{Ar}]\,3d^6\,4s^2$. To form $\text{Fe}^{3+}$, remove the two $4s$ electrons first, then one $3d$: $\text{Fe}^{3+}: [\text{Ar}]\,3d^5$ (a stable half-filled set).

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Why is the configuration of chromium $3d^5\,4s^1$ rather than $3d^4\,4s^2$?

Show solution

A half-filled $3d^5$ set together with a half-filled $4s^1$ gives extra exchange-energy stability, so one $4s$ electron shifts into $3d$. Copper is similar: $3d^{10}\,4s^1$.

✎ Self-Check — 5 questions0 / 5
Q1.

The general configuration of transition elements is:

Explanation: Partly filled $(n-1)d$ with $ns$ electrons.
Q2.

The electronic configuration of chromium is:

Explanation: Half-filled stability gives $3d^5 4s^1$.
Q3.

When a transition metal forms a cation, electrons are removed first from:

Explanation: $ns$ electrons are lost before $(n-1)d$.
Q4.

Which is NOT regarded as a true transition element?

Explanation: Zn is $d^{10}$ in all states — not partly filled.
Q5.

Transition metals generally have:

Explanation: Strong metallic bonding gives high m.p. and density.

NEET tip: memorise Cr ($3d^5 4s^1$) and Cu ($3d^{10} 4s^1$). For ions, remove $4s$ before $3d$. Zn/Cd/Hg ($d^{10}$) are not true transition elements.

Oxidation States, Colour, Magnetism, Catalysis and Key CompoundsTopic 2

The most distinctive chemistry of the transition metals follows from their $d$ electrons. Because the $(n-1)d$ and $ns$ orbitals are close in energy, electrons from both can be used in bonding, giving variable oxidation states. Manganese is the classic example, showing every state from $+2$ to $+7$; the highest states appear in oxides and oxoanions (e.g. $\text{MnO}_4^-$). The maximum oxidation state usually rises to the middle of a series and then falls. This variability is a guaranteed NEET topic.

Transition metal ions are typically coloured, because a partly filled $d$ sub-shell allows d–d electronic transitions: the ion absorbs one wavelength of visible light to promote a $d$ electron and transmits the complementary colour. Ions with empty ($\text{Sc}^{3+}$, $d^0$) or completely filled ($\text{Zn}^{2+}$, $d^{10}$) $d$ orbitals have no such transition and are colourless — a popular NEET discriminator.

Their unpaired $d$ electrons make most transition metal ions paramagnetic. The (spin-only) magnetic moment is $\mu = \sqrt{n(n+2)}\ \text{BM}$, where $n$ is the number of unpaired electrons; more unpaired electrons give a larger moment. NEET often asks you to count unpaired electrons in an ion and compute $\mu$.

Transition metals and their compounds are excellent catalysts (using variable oxidation states to form intermediates, e.g. Fe in the Haber process, $\text{V}_2\text{O}_5$ in the Contact process), form interstitial compounds (small atoms like H, C, N occupy lattice holes, making the metal harder), and readily form complex (coordination) compounds. Two compounds are studied in detail: potassium permanganate ($\text{KMnO}_4$) and potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$), both powerful oxidising agents whose preparation, structure and reactions are regular NEET fare.

Figure — Oxidation States, Colour, Magnetism, Catalysis and Key Compounds
PropertyReason / point
Variable oxidation statesclose $(n-1)d$ & $ns$ energies (Mn $+2$ to $+7$)
Colour$d$–$d$ transitions ($d^0$, $d^{10}$ colourless)
Magnetic moment$\mu = \sqrt{n(n+2)}\ \text{BM}$
Key compounds$\text{KMnO}_4$, $\text{K}_2\text{Cr}_2\text{O}_7$ (oxidisers)
Worked Examples
1

Calculate the spin-only magnetic moment of $\text{Fe}^{2+}$ ($3d^6$).

Show solution

$3d^6$ has 4 unpaired electrons. $\mu = \sqrt{n(n+2)} = \sqrt{4(6)} = \sqrt{24} \approx 4.9\ \text{BM}$.

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Why is $\text{Sc}^{3+}$ colourless while $\text{Cu}^{2+}$ is blue?

Show solution

$\text{Sc}^{3+}$ is $3d^0$ — it has no $d$ electrons, so no $d$–$d$ transition is possible and it is colourless. $\text{Cu}^{2+}$ is $3d^9$ with a partly filled $d$ sub-shell, allowing a $d$–$d$ transition that gives its blue colour.

✎ Self-Check — 5 questions0 / 5
Q1.

Variable oxidation states in transition metals arise because:

Explanation: Close $(n-1)d$ and $ns$ energies allow both to bond.
Q2.

The colour of transition metal ions is due to:

Explanation: Partly filled $d$ allows $d$–$d$ transitions.
Q3.

Which ion is colourless?

Explanation: $\text{Zn}^{2+}$ is $d^{10}$ — no $d$–$d$ transition.
Q4.

The spin-only magnetic moment is given by:

Explanation: $\mu = \sqrt{n(n+2)}\ \text{BM}$.
Q5.

Iron acts as a catalyst in the:

Explanation: Fe catalyses ammonia synthesis (Haber process).

NEET tip: colour & paramagnetism need partly filled $d$ ($d^0$/$d^{10}$ = colourless, diamagnetic). $\mu=\sqrt{n(n+2)}$ BM. Highest oxidation states show in oxoanions ($\text{MnO}_4^-$, $\text{Cr}_2\text{O}_7^{2-}$).

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Module 2

The f-Block: Lanthanoids and Actinoids

Lanthanoids and the Lanthanoid ContractionTopic 3

The f-block elements, or inner transition elements, are placed separately at the bottom of the periodic table and are characterised by the filling of $f$ orbitals. The first series, the lanthanoids (cerium to lutetium, often taken with lanthanum), involves the progressive filling of the $4f$ subshell; their general configuration is $[\text{Xe}]\,4f^{1-14}\,5d^{0-1}\,6s^2$.

The lanthanoids are remarkably similar to one another in chemistry. Their most common and stable oxidation state is $+3$, although a few show $+2$ or $+4$ when these give a stable $f^0$, $f^7$ or $f^{14}$ configuration (e.g. $\text{Ce}^{4+}$ from $f^0$, $\text{Eu}^{2+}$ from $f^7$). Most lanthanoid ions are coloured and paramagnetic owing to unpaired $4f$ electrons, and the metals are reactive, silvery, and used in alloys such as misch metal (in lighter flints) and as magnets.

The defining feature of the series is the lanthanoid contraction: a steady, small decrease in atomic and ionic radius from one lanthanoid to the next across the series. It occurs because the $4f$ electrons being added shield the increasing nuclear charge very poorly (the $4f$ orbitals are diffuse and inner), so the effective nuclear charge felt by the outer electrons rises and pulls them in. NEET regularly tests both the cause and the consequences.

The consequences of the lanthanoid contraction are far-reaching: the elements of the second and third transition series become almost identical in size (for example zirconium and hafnium, or niobium and tantalum), which makes these pairs very hard to separate; it also accounts for the similar properties of the lanthanoids themselves and influences the basicity of their hydroxides (which decreases as size decreases). Understanding this single contraction, its cause and its size-related consequences is the highest-yield idea from the f-block for NEET.

Figure — Lanthanoids and the Lanthanoid Contraction
FeatureLanthanoids
Configuration$[\text{Xe}]\,4f^{1-14}\,5d^{0-1}\,6s^2$
Common oxidation state$+3$ (some $+2$/$+4$)
Lanthanoid contractionpoor $4f$ shielding to steady size decrease
ConsequenceZr ≈ Hf in size (hard to separate)
Worked Examples
1

What causes the lanthanoid contraction?

Show solution

The $4f$ electrons added across the series shield the nuclear charge poorly (they are diffuse, inner orbitals). So effective nuclear charge increases steadily, pulling the electron cloud inward and causing a gradual decrease in atomic and ionic size.

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Why are zirconium and hafnium difficult to separate?

Show solution

Because of the lanthanoid contraction, the size of Hf (a 5d element following the lanthanoids) is almost the same as that of Zr (the 4d element above it). Their nearly identical sizes give them very similar chemical properties, making separation difficult.

✎ Self-Check — 5 questions0 / 5
Q1.

The lanthanoids are characterised by filling of the:

Explanation: Lanthanoids fill the $4f$ subshell.
Q2.

The most common oxidation state of lanthanoids is:

Explanation: $+3$ is the most stable state.
Q3.

The lanthanoid contraction is due to poor shielding by:

Explanation: Diffuse $4f$ electrons shield poorly.
Q4.

A consequence of the lanthanoid contraction is:

Explanation: 4d/5d pairs become nearly the same size.
Q5.

Across the lanthanoid series, atomic size:

Explanation: Steady decrease — the contraction.

NEET tip: lanthanoid contraction = poor $4f$ shielding to steady size decrease to 4d/5d pairs (Zr/Hf, Nb/Ta) nearly identical and hard to separate. Common state $+3$.

Actinoids and Comparison with LanthanoidsTopic 4

The second f-block series, the actinoids (thorium to lawrencium, taken with actinium), involves the filling of the $5f$ subshell, with the general configuration $[\text{Rn}]\,5f^{1-14}\,6d^{0-1}\,7s^2$. Unlike the lanthanoids, all actinoids are radioactive, and the elements beyond uranium (the transuranium elements) are synthetic, made artificially in nuclear reactions.

A major difference from the lanthanoids is that the actinoids show a much greater range of oxidation states. While $+3$ is common, states up to $+6$ and $+7$ are seen in the early actinoids (for example uranium commonly shows $+6$ in $\text{UO}_2^{2+}$). This wider variability arises because the $5f$, $6d$ and $7s$ orbitals are close in energy and the $5f$ electrons are less tightly held and more available for bonding than the $4f$ electrons of the lanthanoids.

There is an analogous actinoid contraction — a steady decrease in size across the series — which, like the lanthanoid contraction, is due to the poor shielding of the nuclear charge, here by $5f$ electrons. The actinoid contraction is slightly greater from element to element than the lanthanoid contraction because $5f$ electrons shield even more poorly.

For NEET, the key comparison points to remember are: both series show a $+3$ state and a contraction caused by poor f-orbital shielding; but actinoids are all radioactive, show more and higher oxidation states, and have $5f$ electrons that are more chemically available than the lanthanoids' $4f$ electrons. Actinoids also have important applications — uranium and plutonium as nuclear fuels, americium in smoke detectors — which NEET sometimes references. Holding these similarities and differences clearly, together with the lanthanoid contraction from the previous topic, covers the f-block questions you are likely to face.

Figure — Actinoids and Comparison with Lanthanoids
FeatureActinoids vs Lanthanoids
Orbital filled$5f$ (vs $4f$)
Radioactivityall radioactive (lanthanoids mostly not)
Oxidation stateswider range, up to $+6$/$+7$
Contractionactinoid contraction (slightly greater)
Worked Examples
1

Give two ways in which actinoids differ from lanthanoids.

Show solution

(1) All actinoids are radioactive, whereas most lanthanoids are not. (2) Actinoids show a wider range of (and higher) oxidation states (up to $+6$/$+7$) because their $5f$ electrons are more available for bonding than the lanthanoids' $4f$ electrons.

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What is the actinoid contraction?

Show solution

It is the steady decrease in atomic and ionic size across the actinoid series, caused by the poor shielding of the nuclear charge by the $5f$ electrons — analogous to, but slightly greater than, the lanthanoid contraction.

✎ Self-Check — 5 questions0 / 5
Q1.

The actinoids involve the filling of the:

Explanation: Actinoids fill the $5f$ subshell.
Q2.

Compared with lanthanoids, actinoids show:

Explanation: $5f$ electrons are more available, giving more states.
Q3.

All actinoids are:

Explanation: Every actinoid is radioactive.
Q4.

Elements beyond uranium are called:

Explanation: Synthetic transuranium elements.
Q5.

The common oxidation state shared by both series is:

Explanation: $+3$ is common to both lanthanoids and actinoids.

NEET tip: actinoids = $5f$, all radioactive, more/higher oxidation states (up to $+6$/$+7$); lanthanoids = $4f$, mostly $+3$. Both share a $+3$ state and an f-shielding contraction.

Quick Revision — The d- and f-Block Elements

  • Transition (d-block) elements: partly filled $d$ orbitals; general configuration $(n-1)d^{1-10}\,ns^{1-2}$.
  • Anomalies: Cr is $3d^5 4s^1$ and Cu is $3d^{10} 4s^1$ (extra stability of half/fully-filled $d$).
  • Variable oxidation states arise from similar energies of $(n-1)d$ and $ns$ electrons (Mn shows $+2$ to $+7$).
  • Coloured ions: due to $d$–$d$ electronic transitions (need partly filled $d$; $\text{Sc}^{3+}$, $\text{Zn}^{2+}$ are colourless).
  • Magnetic moment: $\mu = \sqrt{n(n+2)}\ \text{BM}$ ($n$ = unpaired electrons) — paramagnetism rises with unpaired electrons.
  • Other traits: good catalysts, form alloys, interstitial compounds, and complex ions; high melting points & densities.
  • Important compounds: $\text{KMnO}_4$ and $\text{K}_2\text{Cr}_2\text{O}_7$ (strong oxidisers).
  • f-block: lanthanoids show the lanthanoid contraction (steady size decrease) explaining similarity of 4d/5d elements; actinoids are radioactive with more variable oxidation states.

Frequently Asked Questions

Why do transition elements show variable oxidation states?
Because the energies of the (n-1)d and ns orbitals are very close, electrons from both can take part in bonding. So transition metals can lose different numbers of electrons and exhibit several oxidation states, such as manganese from +2 to +7.
Why are most transition metal ions coloured?
Their partly filled d orbitals allow d-d electronic transitions: an electron absorbs a particular wavelength of visible light and the complementary colour is seen. Ions with empty or completely filled d orbitals, like Sc3+ and Zn2+, are colourless.
What is the lanthanoid contraction and why is it important?
The lanthanoid contraction is the steady decrease in atomic and ionic size across the lanthanoid series, caused by the poor shielding of the nuclear charge by the diffuse 4f electrons. It makes the second and third transition series elements (like Zr and Hf) almost identical in size, so they are hard to separate.
Why do transition metals act as good catalysts?
Their variable oxidation states let them form intermediate compounds with reactants and provide alternative low-energy pathways, while their ability to adsorb reactants on their surface (in heterogeneous catalysis) also speeds reactions. Iron in the Haber process and vanadium pentoxide in the Contact process are examples.
Why is the magnetic moment of a transition metal ion calculated from unpaired electrons only?
For most transition metal ions the magnetism is mainly due to the spin of unpaired electrons (the spin-only contribution), so the magnetic moment is calculated as the square root of n(n+2) Bohr magnetons, where n is the number of unpaired electrons. More unpaired electrons means a larger moment.

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