NCERT Solutions for Class 12 Chemistry Chapter 4 – d- and f- Block Elements (English Medium)
🧪 Get NCERT Solutions for Class 12 Chemistry Chapter 4 – d- and f- Block Elements with accurate answers and step-by-step explanations in an exam-friendly format. This chapter focuses on transition elements (d-block) and inner transition elements (f-block), including electronic configuration, oxidation states, colour & magnetic properties, complex formation, catalytic behaviour, lanthanoid contraction, and key trends of Sc–Zn and lanthanoids/actinoids. Helpful for CBSE boards and NEET/JEE revision. ✅
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Q.1: Write down the electronic configuration of:
(i) Cr3+
(ii) Pm3+
(iii) Cu+
(iv) Ce4+
(v) Co2+
(vi) Lu2+
(vii) Mn2+
(viii) Th4+
(i) Cr3+
(ii) Pm3+
(iii) Cu+
(iv) Ce4+
(v) Co2+
(vi) Lu2+
(vii) Mn2+
(viii) Th4+
Answer
📌 Rule (quick)
: When forming cations, electrons are removed first from the highest n shell (for transition metals: remove ns before (n−1)d).
How to write (example): let say ‘Cr’
24Cr = 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1 or [Ar] 3d5 4s1
24Cr3+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d3 or [Ar] 3d3
How to write (example): let say ‘Cr’
24Cr = 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1 or [Ar] 3d5 4s1
24Cr3+ = 1s2, 2s2, 2p6, 3s2, 3p6, 3d3 or [Ar] 3d3
(i) Cr3+
👉 Cr: [Ar] 3d5 4s1
✅ Cr3+ = [Ar] 3d3
👉 Cr: [Ar] 3d5 4s1
✅ Cr3+ = [Ar] 3d3
(ii) Pm3+
👉 Pm: [Xe] 4f5 6s2
✅ Pm3+ = [Xe] 4f4
👉 Pm: [Xe] 4f5 6s2
✅ Pm3+ = [Xe] 4f4
(iii) Cu+
👉 Cu: [Ar] 3d10 4s1
✅ Cu+ = [Ar] 3d10
👉 Cu: [Ar] 3d10 4s1
✅ Cu+ = [Ar] 3d10
(iv) Ce4+
👉 Ce: [Xe] 4f1 5d1 6s2
✅ Ce4+ = [Xe]
👉 Ce: [Xe] 4f1 5d1 6s2
✅ Ce4+ = [Xe]
(v) Co2+
👉 Co: [Ar] 3d7 4s2
✅ Co2+ = [Ar] 3d7
👉 Co: [Ar] 3d7 4s2
✅ Co2+ = [Ar] 3d7
(vi) Lu2+
👉 Lu: [Xe] 4f14 5d1 6s2
✅ Lu2+ = [Xe] 4f14 5d1
👉 Lu: [Xe] 4f14 5d1 6s2
✅ Lu2+ = [Xe] 4f14 5d1
(vii) Mn2+
👉 Mn: [Ar] 3d5 4s2
✅ Mn2+ = [Ar] 3d5
👉 Mn: [Ar] 3d5 4s2
✅ Mn2+ = [Ar] 3d5
(viii) Th4+
👉 Th: [Rn] 6d2 7s2
✅ Th4+ = [Rn]
👉 Th: [Rn] 6d2 7s2
✅ Th4+ = [Rn]
Explanation (simple)
🤔 Transition metals lose electrons from
4s first
,
then
3d
.📌 For lanthanides/actinides, electrons are removed from the outermost shell (like 6s/7s) first, then from d/f if needed.
Did You Know?
✅ Cu+ has a very stable
3d10
configuration, which is why Cu+ compounds are often stable in certain conditions.✅ Ce4+ = [Xe] is a noble-gas configuration, so Ce4+ is a strong oxidising state and is commonly found in compounds like ceric salts.
Q.2: Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
Answer
👉 Mn2+ compounds are more stable because Mn2+ has a
half-filled 3d5
configuration, which is extra stable.🤔 So, oxidising Mn2+ to Mn3+ (3d4) breaks this stability and requires a lot of energy.
Explanation (Step by Step)
📌 Electronic configurations
• Mn (Z = 25):
[Ar] 3d5 4s2
✅ Mn2+ = [Ar] 3d5 (half-filled, very stable)
✅ Mn2+ = [Ar] 3d5 (half-filled, very stable)
• Fe (Z = 26):
[Ar] 3d6 4s2
✅ Fe2+ = [Ar] 3d6 (not half-filled)
✅ Fe2+ = [Ar] 3d6 (not half-filled)
👉 Mn2+ → Mn3+ + e−
This changes 3d5 → 3d4 , so stability decreases.
📌 इसलिए Mn2+ को oxidise करना मुश्किल होता है (3rd ionisation energy is high) .
This changes 3d5 → 3d4 , so stability decreases.
📌 इसलिए Mn2+ को oxidise करना मुश्किल होता है (3rd ionisation energy is high) .
👉 Fe2+ → Fe3+ + e−
This changes 3d6 → 3d5 .
✅ Now Fe3+ becomes half-filled (3d5) and more stable.
📌 इसलिए Fe2+ आसानी से oxidise होकर Fe3+ बन जाता है.
This changes 3d6 → 3d5 .
✅ Now Fe3+ becomes half-filled (3d5) and more stable.
📌 इसलिए Fe2+ आसानी से oxidise होकर Fe3+ बन जाता है.
Did You Know?
✅ Half-filled
(3d5)
and fully-filled
(3d10)
d-subshells are unusually stable due to symmetry and exchange energy, which strongly affects oxidation states in transition metals.
Q.3: Explain briefly how +2 oxidation state becomes more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer
👉 In the first half of the 3d series (from Ti to Mn), the
+2 oxidation state
becomes increasingly stable because these metals form M2+ ions by losing two
4s electrons
,
and the resulting 3d configuration moves steadily toward the extra-stable
half-filled 3d5
arrangement (maximum stability at
Mn2+
).
Explanation (Step by Step)
📌 Key point
: For first-row transition metals, the
4s electrons
are lost first.
👉 M (…4s23dn) → M2+ (…3dn) + 2e−
So, the +2 state generally corresponds to the 3d electrons left behind.
So, the +2 state generally corresponds to the 3d electrons left behind.
🤔 As atomic number increases from Sc (21) to Mn (25), the number of 3d electrons increases, and in the
+2 state
the d-electron count gets closer to
3d5 (half-filled)
,
which is specially stable.
d-electrons in +2 state (first half) 📌
• Sc2+ : 3d1 (rare/unstable)
• Ti2+ : 3d2
• V2+ : 3d3
• Cr2+ : 3d4 (Cr is special in neutral state, but +2 tends toward stable patterns)
• Mn2+ : 3d5 ✅ half-filled (most stable)
• Sc2+ : 3d1 (rare/unstable)
• Ti2+ : 3d2
• V2+ : 3d3
• Cr2+ : 3d4 (Cr is special in neutral state, but +2 tends toward stable patterns)
• Mn2+ : 3d5 ✅ half-filled (most stable)
👉 Therefore, from Ti → V → Cr → Mn, the
+2 state
becomes more stable because the 3d subshell is getting closer to
half-filled (3d5)
,
giving extra stability (exchange energy + symmetry).
📌 Why Sc doesn’t show +2 easily?
👉 Sc usually prefers +3 because Sc3+ becomes [Ar] , a noble-gas configuration, which is very stable.
👉 Sc usually prefers +3 because Sc3+ becomes [Ar] , a noble-gas configuration, which is very stable.
Did You Know?
✅ Mn2+
(3d5)
is exceptionally stable, so many manganese(II) salts are quite common.👉 This half-filled stability is also why Fe3+ (3d5) is often more stable than Fe2+ (3d6) in oxidising conditions.
Q.4: To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
Answer
👉 Electronic configuration plays a
major role
in deciding which oxidation states are more stable, especially when an ion attains a
half-filled (3d5)
or
fully-filled (3d10)
d-subshell.📌 But it is not the only factor . Stability also depends on lattice enthalpy, hydration enthalpy, ionisation enthalpy, ligand field stabilisation (complex formation), and the reaction medium .
👉 So, configuration gives a strong trend , but real stability is decided by both electronic + energetic factors .
Explanation (with examples)
1) Half-filled and fully-filled d-subshell = extra stability 📌
🤔 3d5 and 3d10 arrangements are unusually stable due to symmetry + exchange energy .
🤔 3d5 and 3d10 arrangements are unusually stable due to symmetry + exchange energy .
Example A: Mn
• Mn: [Ar] 3d5 4s2
• Mn2+ = [Ar] 3d5 ✅ (half-filled, very stable)
👉 That’s why Mn2+ compounds resist oxidation to Mn3+ (which would be 3d4 , less stable).
• Mn: [Ar] 3d5 4s2
• Mn2+ = [Ar] 3d5 ✅ (half-filled, very stable)
👉 That’s why Mn2+ compounds resist oxidation to Mn3+ (which would be 3d4 , less stable).
Example B: Fe
• Fe2+ = [Ar] 3d6
• Fe3+ = [Ar] 3d5 ✅ (half-filled)
👉 इसलिए Fe2+ अक्सर oxidise होकर Fe3+ बन जाता है, क्योंकि Fe3+ gets the stable 3d5 configuration.
• Fe2+ = [Ar] 3d6
• Fe3+ = [Ar] 3d5 ✅ (half-filled)
👉 इसलिए Fe2+ अक्सर oxidise होकर Fe3+ बन जाता है, क्योंकि Fe3+ gets the stable 3d5 configuration.
2) Completely filled d-subshell makes Cu+ stable 📌
Copper (Cu):
• Cu: [Ar] 3d10 4s1
• Cu+ = [Ar] 3d10 ✅ (fully-filled, very stable)
👉 That’s why Cu+ is often stabilised in solids and complexes (though in aqueous solution it can disproportionate depending on conditions).
Copper (Cu):
• Cu: [Ar] 3d10 4s1
• Cu+ = [Ar] 3d10 ✅ (fully-filled, very stable)
👉 That’s why Cu+ is often stabilised in solids and complexes (though in aqueous solution it can disproportionate depending on conditions).
3) Configuration explains trends across early 3d series (Sc → Mn)
👉 +2 state is formed by losing two 4s electrons, giving 3dn.
As n increases toward 3d5 , the +2 state becomes more stable (Ti2+ < V2+ < … < Mn2+) .
👉 +2 state is formed by losing two 4s electrons, giving 3dn.
As n increases toward 3d5 , the +2 state becomes more stable (Ti2+ < V2+ < … < Mn2+) .
4) But configuration alone cannot explain everything 🤔
📌 Sometimes a “less obvious” oxidation state becomes stable because of energetic support:
• High oxidation states like V5+, Cr6+, Mn7+ are stabilised in oxides/oxoanions (VO43−, CrO42−, MnO4−) because strong M–O bonds and high lattice/hydration energies compensate the ionisation cost.
👉 This is not just configuration; it’s overall thermodynamics.
📌 Sometimes a “less obvious” oxidation state becomes stable because of energetic support:
• High oxidation states like V5+, Cr6+, Mn7+ are stabilised in oxides/oxoanions (VO43−, CrO42−, MnO4−) because strong M–O bonds and high lattice/hydration energies compensate the ionisation cost.
👉 This is not just configuration; it’s overall thermodynamics.
Did You Know?
✅ The most “favoured” oxidation states in 3d metals often correlate with reaching
3d0, 3d5, or 3d10
(empty, half-filled, full).👉 But in real chemistry, ligands and the medium can “lock in” an oxidation state by stabilising it through bonding and crystal/complex formation.
Q.5: What may be the stable oxidation state of the transition element with the following d-electron configurations in the ground state of their atoms:
3d3, 3d5, 3d8 and 3d4?
Answer (Probable stable oxidation states)
(i) 3d3
(example: V = 3d34s2)
👉 Stable oxidation states: +2, +3, +4, +5
📌 (Vanadium commonly shows all these states in compounds.)
👉 Stable oxidation states: +2, +3, +4, +5
📌 (Vanadium commonly shows all these states in compounds.)
(ii) 3d5
(example: Cr = 3d54s1 / Mn = 3d54s2)
👉 If it is Cr (3d54s1): +3, +4, +5, +6
👉 If it is Mn (3d54s2): +2, +4, +6, +7
📌 So 3d5 can support several stable oxidation states, but +2 (Mn2+) and +3 (Cr3+) are especially common due to extra stability patterns.
👉 If it is Cr (3d54s1): +3, +4, +5, +6
👉 If it is Mn (3d54s2): +2, +4, +6, +7
📌 So 3d5 can support several stable oxidation states, but +2 (Mn2+) and +3 (Cr3+) are especially common due to extra stability patterns.
(iii) 3d8
(example: Ni = 3d84s2)
👉 Stable oxidation states: +2 (most common) , sometimes +3 (in complexes)
📌 Ni2+ corresponds to a very common and stable state in salts and complexes.
👉 Stable oxidation states: +2 (most common) , sometimes +3 (in complexes)
📌 Ni2+ corresponds to a very common and stable state in salts and complexes.
(iv) 3d4
🤔 Note: In the first-row transition elements, 3d4 is not a common ground-state configuration because of extra stability of half-filled subshells.
📌 Example: Chromium is 3d54s1 , not 3d44s2 .
👉 But if an element/ion effectively has d4, it may show oxidation states like +2, +3, +4 (depending on the element and ligands).
🤔 Note: In the first-row transition elements, 3d4 is not a common ground-state configuration because of extra stability of half-filled subshells.
📌 Example: Chromium is 3d54s1 , not 3d44s2 .
👉 But if an element/ion effectively has d4, it may show oxidation states like +2, +3, +4 (depending on the element and ligands).
Explanation (Simple logic)
📌 Transition metals lose
4s electrons first
,
then
3d
.👉 So the number of d-electrons left after losing electrons helps decide which oxidation states are relatively stable.
✅ Extra stability is often seen when the ion reaches d0 , d5 (half-filled) , or d10 (fully-filled) .
Did You Know?
✅ Many oxidation states become “more stable” in complexes because ligands can stabilise a particular d-configuration (this is why +3 states are common in coordination chemistry).✅ The “missing” 3d4 ground state in chromium is a classic example of half-filled (3d5) stability.