Class 12 Chemistry · Original EZT practice
The d- and f-Block Elements (Class 12 Chapter 4): summary and practice
The d- and f-block chapter runs the middle of the periodic table: 3d configurations with their Cr and Cu exceptions, variable oxidation states, coloured ions and spin-only magnetic moments, the workhorse compounds K2Cr2O7 and KMnO4, and the lanthanoids with their famous contraction. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Position and electronic configurations
The d-block fills the large middle section of the table where d orbitals are being occupied. The 3d series runs Sc to Zn with two famous exceptions: chromium takes 3d5 4s1 and copper takes 3d10 4s1, because half-filled and filled d subshells carry extra stability.
2. General properties and trends
Across each series the atomic and ionic sizes shrink slightly then level off, ionisation enthalpies rise irregularly, and oxidation states multiply - manganese alone runs from +2 to +7. The standard electrode potentials and the stability of the higher states trend along with them through sections 4.3.1 to 4.3.8.
3. Magnetism, colour and complexes
Unpaired d electrons make ions paramagnetic with spin-only moment mu = sqrt(n(n+2)) BM, and d-d transitions make them coloured. The same partly filled d orbitals form complex compounds, drive catalysis, and host small atoms in interstitial compounds and alloys (4.3.9 to 4.3.14).
4. Important compounds
Section 4.4 spotlights two oxidising agents: potassium dichromate K2Cr2O7 with chromium in +6, and potassium permanganate KMnO4 with manganese in +7. Both owe their power to the metal's highest oxidation states.
5. Lanthanoids, actinoids and applications
The lanthanoids fill 4f and show the lanthanoid contraction - a steady shrinking of radii whose knock-on effect makes the 4d and 5d pairs (like Zr and Hf) almost identical in size. The actinoids fill 5f with a wider spread of oxidation states, and the chapter closes with applications from steels to catalysts.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Spin-only moment of a d4 ion
Choose the method: mu = sqrt(n(n+2)) BM with n unpaired electrons.
EZT practice question
Find the spin-only magnetic moment of an ion with 4 unpaired d electrons.
Show the worked solution
- n = 4.
- mu = sqrt(4 x (4 + 2)) = sqrt(24).
Answer: mu = 4.90 BM (approximately).
Check: More unpaired electrons, larger moment - d5 at 5.92 BM is the 3d series maximum.
Watch for this: Count UNPAIRED electrons only - a d4 ion is not automatically 4 unpaired if it is low-spin, but in this chapter's high-spin examples it is.
Now try it yourself
Spin-only moment of a d5 ion?
Check your answer
5.92 BM.
sqrt(5 x 7) = sqrt(35) = 5.92 BM.
Spin-only moment of a d3 ion
Choose the method: mu = sqrt(n(n+2)) BM.
EZT practice question
Find the spin-only magnetic moment of a d3 ion such as Cr3+.
Show the worked solution
- n = 3 unpaired electrons.
- mu = sqrt(3 x 5) = sqrt(15).
Answer: mu = 3.87 BM (approximately).
Check: The moment tracks unpaired count, not the total d electrons - a paired d4 would score lower than d3.
Watch for this: Do not plug the d electron count straight in - pair them up first, then count the unpaired.
Now try it yourself
Spin-only moment of a d1 ion?
Check your answer
1.73 BM.
sqrt(1 x 3) = sqrt(3) = 1.73 BM.
Assign an oxidation state
Choose the method: Sum of oxidation states equals the overall charge.
EZT practice question
Find the oxidation state of Mn in KMnO4.
Show the worked solution
- K is +1, each O is -2.
- +1 + Mn + 4 x (-2) = 0.
- Mn = 8 - 1.
Answer: Mn is +7 - its maximum, which is why permanganate oxidises so aggressively.
Check: The same arithmetic gives Cr = +6 in K2Cr2O7: 2 x 1 + 2Cr + 7 x (-2) = 0, so 2Cr = 12.
Watch for this: Oxygen is -2 here, not -1 - peroxides are the exception, not these compounds.
Now try it yourself
Oxidation state of Cr in K2Cr2O7?
Check your answer
+6.
2 + 2Cr - 14 = 0 gives Cr = +6.
Write a 3d configuration
Choose the method: Fill 4s before 3d, except half-filled and filled d subshells win.
EZT practice question
Write the ground-state electronic configuration of chromium (Z = 24) and explain it.
Show the worked solution
- Expected: [Ar] 3d4 4s2.
- A half-filled d subshell is more stable.
- One 4s electron moves into 3d.
Answer: Cr is [Ar] 3d5 4s1, not 3d4 4s2.
Check: Copper (Z = 29) does the same trick to reach 3d10 4s1 - the two classic exceptions sit side by side in the series.
Watch for this: Do not write 3d4 4s2 for Cr - the half-filled d5 arrangement is the observed ground state.
Now try it yourself
Configuration of copper (Z = 29)?
Check your answer
[Ar] 3d10 4s1.
A filled d10 subshell beats the expected 3d9 4s2.
Configure a transition-metal ion
Choose the method: Remove 4s electrons BEFORE 3d when forming cations.
EZT practice question
Write the electronic configuration of Fe2+ (Fe, Z = 26).
Show the worked solution
- Fe atom: [Ar] 3d6 4s2.
- Cations lose the outermost (4s) electrons first.
- Remove both 4s electrons.
Answer: Fe2+ is [Ar] 3d6.
Check: Fe3+ then is [Ar] 3d5 - the half-filled configuration that makes +3 iron so stable.
Watch for this: The fill order and the strip order differ - 4s fills first but also EMPTIES first.
Now try it yourself
Configuration of Fe3+?
Check your answer
[Ar] 3d5.
Remove 4s2 then one 3d electron: 3d5, half-filled and stable.
Explain lanthanoid contraction
Choose the method: Poor 4f shielding shrinks radii steadily across the series.
EZT practice question
What is lanthanoid contraction and why does it make zirconium and hafnium so similar?
Show the worked solution
- Across the lanthanoids, 4f electrons shield the growing nuclear charge poorly.
- Radii shrink steadily from La to Lu.
- Hf comes right after the lanthanoids, so its radius is pulled down to Zr's size.
Answer: The steady decrease in lanthanoid atomic and ionic radii; it leaves Hf (4f14 5d2) almost the same size as Zr (4d2), so the pair separate with difficulty and share chemistry.
Check: The same contraction explains why the 5d metals have nearly the radii of their 4d partners rather than larger ones.
Watch for this: The contraction is gradual, element by element - not a sudden drop at one point in the series.
Now try it yourself
Which block do the actinoids fill?
Check your answer
5f.
Actinoids follow actinium filling the 5f orbitals, mirroring the 4f lanthanoids.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Spin-only moment?
Show answer and reason
mu = sqrt(n(n+2)) BM.
n counts unpaired d electrons.
2. Cr configuration?
Show answer and reason
[Ar] 3d5 4s1.
A half-filled d subshell is extra stable.
3. Forming cations?
Show answer and reason
Remove 4s before 3d.
4s fills first and empties first.
4. Lanthanoid contraction?
Show answer and reason
Steady radius decrease across the 4f series.
4f electrons shield the nuclear charge poorly.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Find the spin-only moment of a d5 ion.
- Find the oxidation state of Cr in K2Cr2O7.
- Write the electronic configuration of Fe3+ (Z = 26).
Know the rule but lose track of the steps?
Bring your book edition, your school topic list and a question you attempted. Ask a tutor to explain the choice of method and test it with a different problem. Agree the lesson topics, schedule and fee before starting.
For cost context, see home tuition fees in Delhi NCR. Agree the topics, schedule and fee with your chosen tutor.
Questions about this practice page
- Why do transition metals show variable oxidation states?
- The 3d and 4s electrons lie close in energy, so different numbers of them can take part in bonding. Manganese runs from +2 to +7; the middle of the series shows the widest spread of states.
- Why are transition-metal compounds coloured?
- Partly filled d orbitals split in the field of surrounding ligands, and electrons jump the gap by absorbing visible light - the d-d transition. The colour you see is the light left over. Ions with d0 or d10 configurations, like Sc3+ or Zn2+, stay colourless.
- What is lanthanoid contraction?
- The steady shrinking of atomic and ionic radii from lanthanum to lutetium, caused by 4f electrons shielding the rising nuclear charge poorly. Its famous consequence: the 5d elements end up nearly the same size as their 4d partners, so Zr and Hf are almost chemically identical.
- Are zinc, cadmium and mercury typical transition metals?
- No. They keep full d10 configurations in their common states, so they skip the signature behaviour - no variable oxidation states, no coloured ions, no d-d magnetism. The chapter treats them as the edge of the block, not its heart.
- Are these the NCERT exercise questions for the d- and f-Block (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.