Elements with partially filled d-orbitals in atom or commonly occurring oxidation state. d-block: Groups 3-12. Show variable oxidation states, paramagnetism, color, catalytic activity, complex formation.
-- NCERT Class 12 Chemistry, Ch. 4, p. 89Electronic Config Transition
8 MCQs9-step worked example
Source: NCERT The d and f Block ElementsOfficial key: NTA-verifiedLast updated: 7 Oct 2026
Try this first
Which general outer-orbital configuration does NCERT give for the d-block elements (apart from the stated exception)?
- A.(n-1)d¹⁻¹⁰ ns² only
- B.(n-1)d¹⁻¹⁰ np¹⁻²
- C.(n-1)d¹⁻¹⁰ ns¹⁻²
- D.nd¹⁻¹⁰ (n+1)s¹⁻²
Tap to see the answer
Answer: C. NCERT Class 12 Chemistry, Chapter 4, page 90 gives the general outer configuration as (n-1)d¹⁻¹⁰ ns¹⁻², with one or two electrons in the outermost ns orbital.
A is wrong: A is wrong because ns² only would leave out Cr, Cu and most 4d elements, which have a single ns electron (page 90 and Table 4.1).
B is wrong: B is wrong because the outermost orbital filled in a transition element is ns, not np; the d orbitals of the penultimate level receive the electrons (page 90).
D is wrong: D is wrong because it names the wrong shells; the d orbitals belong to the penultimate level (n-1), and the s orbital to the outermost level n (page 90).
Electronic Config Transition, explained for NEET
The trap: Students fill 4s before 3d and write Cr as 3d⁴ 4s² and Cu as 3d⁹ 4s². NCERT prints the opposite: Cr is 3d⁵ 4s¹ and Cu is 3d¹⁰ 4s¹. A question that hands you Table 4.1 values expects you to read the exceptions, not to rebuild them from the Aufbau order.
The general rule. The d-block sits between the s- and p-blocks. In these elements the d orbitals of the penultimate energy level receive electrons, giving four rows: 3d, 4d, 5d and 6d. NCERT gives the general outer configuration as (n-1)d¹⁻¹⁰ ns¹⁻², "except for Pd where its electronic configuration is 4d¹⁰5s⁰" (NCERT Class 12 Chemistry, Chapter 4, page 90).
Why exceptions occur. There is "very little energy difference between (n-1)d and ns orbitals", and half-filled and completely filled sets of orbitals are relatively more stable. For Cr the 3d–4s gap is small enough that an electron moves into 3d (3d⁵ 4s¹, not 3d⁴ 4s²). For Cu it gives 3d¹⁰ 4s¹, not 3d⁹ 4s² (page 90).
Reading Table 4.1. The 3d row runs Sc 3d¹4s², Ti 3d²4s², V 3d³4s², Cr 3d⁵4s¹, Mn 3d⁵4s², Fe 3d⁶4s², Co 3d⁷4s², Ni 3d⁸4s², Cu 3d¹⁰4s¹, Zn 3d¹⁰4s² (page 90). In the 4d row Pd is 4d¹⁰5s⁰, and several others (Nb, Mo, Tc, Ru, Rh, Ag) have a single 5s electron (page 91).
Who counts as a transition element. Zn, Cd, Hg and Cn have (n-1)d¹⁰ns², and their d orbitals are completely filled "in the ground state as well as in their common oxidation states". So they are "not regarded as transition elements". Scandium counts, because its ground state has an incomplete 3d (3d¹) (page 91).
Watch-out: the exceptions are read from the table. Do not try to predict every one from half-filled and filled-shell stability.
How do you solve a Electronic Config Transition question? A worked example
- 1
Given
An element has atomic number Z = 24. The argon core holds 18 electrons. NCERT's general outer configuration for transition elements is (n-1)d¹⁻¹⁰ ns¹⁻², with exceptions for the Cr and Cu cases.
- 2
Required
The ground-state outer configuration of this element, and the reason it differs from the "expected" one.
- 3
Concept
After the argon core, the outer electrons go into 3d and 4s. The 3d and 4s energies are close, and half-filled sets of orbitals are relatively more stable (NCERT Class 12 Chemistry, Chapter 4, page 90).
- 4
Formula
Z = 18 (argon core) + (number of 3d electrons) + (number of 4s electrons).
- 5
Substitution
Outer electrons = 24 − 18 = 6. Expected split: 3d⁴ 4s². Check against the Z identity: 18 + 4 + 2 = 24.
- 6
Calculation
Moving one 4s electron into 3d gives 3d⁵ 4s¹, which has a half-filled 3d set. Check: 18 + 5 + 1 = 24. The counts 18, 5 and 1 are counting integers, so they do not enter any significant-figure count.
- 7
Final answer
3d⁵ 4s¹, which is chromium. NCERT prints exactly this on page 90, "instead of 3d⁴4s²".
- 8
Common trap
Writing 3d⁴ 4s² because "4s fills first". The electron count balances either way, so the error does not show up in a Z check; only the exception rule catches it.
- 9
Similar NEET-style question
Write the outer configuration of copper (Z = 29). (Answer: outer electrons = 29 − 18 = 11. The expected split 3d⁹ 4s² has 9 + 2 = 11, but the completely filled set is more stable, so NCERT gives 3d¹⁰ 4s¹: 18 + 10 + 1 = 29, page 90.)
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Can you answer these Electronic Config Transition MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which general outer-orbital configuration does NCERT give for the d-block elements (apart from the stated exception)?
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Chemistry, Chapter 4, page 90 gives the general outer configuration as (n-1)d¹⁻¹⁰ ns¹⁻², with one or two electrons in the outermost ns orbital.
Why A is wrong: A is wrong because ns² only would leave out Cr, Cu and most 4d elements, which have a single ns electron (page 90 and Table 4.1).
Why B is wrong: B is wrong because the outermost orbital filled in a transition element is ns, not np; the d orbitals of the penultimate level receive the electrons (page 90).
Why D is wrong: D is wrong because it names the wrong shells; the d orbitals belong to the penultimate level (n-1), and the s orbital to the outermost level n (page 90).
MCQ 2Easy RecallPractice
Which element is named by NCERT as the exception to the (n-1)d¹⁻¹⁰ ns¹⁻² formula, with the configuration 4d¹⁰5s⁰?
Show answer and why every option is right or wrong
Answer: A. NCERT Class 12 Chemistry, Chapter 4, page 90 states the formula "except for Pd where its electronic configuration is 4d¹⁰5s⁰".
Why B is wrong: B is wrong because Pt is a 5d element with 5d⁹6s¹ in Table 4.1 (page 91), which still has an ns electron.
Why C is wrong: C is wrong because Ag is 4d¹⁰5s¹ in Table 4.1 (page 91); it has one 5s electron, so it fits the general formula.
Why D is wrong: D is wrong because Cd is 4d¹⁰5s² in Table 4.1 (page 91); it fits the formula too.
MCQ 3Easy RecallPractice
What is the ground-state outer configuration of chromium (Z = 24) according to NCERT?
Show answer and why every option is right or wrong
Answer: D. NCERT Class 12 Chemistry, Chapter 4, page 90: Cr has 3d⁵4s¹ "instead of 3d⁴4s²", because the 3d–4s energy gap is small.
Why A is wrong: A is wrong because 3d⁴4s² is exactly the configuration NCERT says Cr does not have (page 90).
Why B is wrong: B is wrong because 4s is not emptied; Table 4.1 gives Cr one 4s electron (page 90).
Why C is wrong: C is wrong because 3d⁵4s² is the Table 4.1 entry for Mn (Z = 25), not Cr (page 90).
MCQ 4Direct ApplicationPractice
From Table 4.1, which pair of 3d-series elements both have a single 4s electron in the ground state?
Show answer and why every option is right or wrong
Answer: B. Table 4.1 (NCERT Class 12 Chemistry, Chapter 4, page 90) lists 4s¹ only for Cr (3d⁵4s¹) and Cu (3d¹⁰4s¹) in the first series.
Why A is wrong: A is wrong because Sc is 3d¹4s² and Ti is 3d²4s² in Table 4.1; both have two 4s electrons (page 90).
Why C is wrong: C is wrong because Mn is 3d⁵4s² and Zn is 3d¹⁰4s²; both have two 4s electrons (page 90).
Why D is wrong: D is wrong because Fe is 3d⁶4s² and Co is 3d⁷4s²; both have two 4s electrons (page 90).
MCQ 5Direct ApplicationPractice
Which element of the 3d series has the ground-state outer configuration 3d⁷ 4s² in Table 4.1?
Show answer and why every option is right or wrong
Answer: D. Table 4.1 (NCERT Class 12 Chemistry, Chapter 4, page 90) gives Co (Z = 27) as 3d⁷4s². Counting from the argon core, 18 + 7 + 2 = 27.
Why A is wrong: A is wrong because Fe (Z = 26) is 3d⁶4s²; reading one column to the left of Co gives the wrong d count (page 90).
Why B is wrong: B is wrong because Ni (Z = 28) is 3d⁸4s²; one column to the right gives the wrong d count (page 90).
Why C is wrong: C is wrong because Mn (Z = 25) is 3d⁵4s² (page 90).
MCQ 6Direct ApplicationPractice
On what ground is zinc (Z = 30) not regarded as a transition element?
Show answer and why every option is right or wrong
Answer: A. NCERT Class 12 Chemistry, Chapter 4, page 91 (Example 4.1): zinc has completely filled d orbitals (3d¹⁰) in its ground state as well as in its oxidised state, so it is not regarded as a transition element.
Why B is wrong: B is wrong because the 3d orbitals of Zn are full, not empty (3d¹⁰ in Table 4.1, page 90).
Why C is wrong: C is wrong because Zn has two 4s electrons (4s²), so 4s is full, not half-filled (page 90).
Why D is wrong: D is wrong because 3d¹⁰4s¹ is copper's configuration; Zn is 3d¹⁰4s² (page 90).
MCQ 7CalculationPractice
Using the general formula NCERT gives for the elements that are not regarded as transition elements, which 4d-series element in Table 4.1 is excluded?
Show answer and why every option is right or wrong
Answer: B. NCERT Class 12 Chemistry, Chapter 4, page 91 says Zn, Cd, Hg and Cn have the formula (n-1)d¹⁰ns² and are not regarded as transition elements. Table 4.1 shows Cd as 4d¹⁰5s², which is the only 4d entry with a full d set and two s electrons.
Why A is wrong: A is wrong because Pd is 4d¹⁰5s⁰ (page 90). It does not fit the (n-1)d¹⁰ns² formula, and NCERT does not list it among the excluded elements.
Why C is wrong: C is wrong because Ag is 4d¹⁰5s¹ (page 91), not 4d¹⁰5s², so it does not fit the excluded formula.
Why D is wrong: D is wrong because Y is 4d¹5s² (page 91) and has an incomplete d set, like Sc in the 3d row.
MCQ 8Concept TrapPractice
Which reason does NCERT give for Cr being 3d⁵4s¹ rather than 3d⁴4s²?
Show answer and why every option is right or wrong
Answer: C. NCERT Class 12 Chemistry, Chapter 4, page 90: there is very little energy difference between (n-1)d and ns orbitals, and half and completely filled sets of orbitals are relatively more stable.
Why A is wrong: A is wrong because it reverses the idea; NCERT says the 3d–4s gap is small, not large (page 90).
Why B is wrong: B is wrong because the stability NCERT names is for half-filled and completely filled sets of orbitals (here 3d⁵), not for a filled s subshell (page 90).
Why D is wrong: D is wrong because most 3d elements, for example Mn 3d⁵4s² and Fe 3d⁶4s², have two 4s electrons (page 90).
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What to remember before solving Electronic Config Transition questions
4 NCERT lines
In general the electronic configuration of outer orbitals of these elements is (n-1)d1– 10ns1–2except for Pd where its electronic configuration is 4d105s0. However, this generalisation has several exceptions because of very little energy difference between (n-1)d and ns orbitals. Furthermore, half and completely filled sets of orbitals are relatively more stable.
-- NCERT Class 12 Chemistry, Ch. 4, p. 90For example, consider the case of Cr, which has 3d5 4s1 configuration instead of 3d44s2; the energy gap between the two sets (3d and 4s) of orbitals is small enough to prevent electron entering the 3d orbitals. Similarly in case of Cu, the configuration is 3d104s1 and not 3d94s2.
-- NCERT Class 12 Chemistry, Ch. 4, p. 90The electronic configurations of outer orbitals of Zn, Cd, Hg and Cn are represented by the general formula (n-1)d10ns2. The orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d1), it is regarded as a transition element.
-- NCERT Class 12 Chemistry, Ch. 4, p. 91Electronic Config Transition: NEET previous year questions (PYQs) with answers
15 questions in The d and f Block Elements
No question in our NEET 2020–2025 set targets this topic directly.
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More in The d and f Block Elements: 3 exam traps and mistakes · 3 formulas · 3 question patterns from its other lessons.
Sources
NCERT refs: Class 12 Chemistry Chapter 4, p.90
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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