Q.Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Electron Configuration
Electron Configuration: Where Do Electrons Actually Live?
Imagine a school building. Students don't just wander randomly — they sit in specific classrooms, on specific floors, in specific rows. Electrons in an atom behave similarly. They don't buzz around the nucleus chaotically. They occupy specific energy levels (floors), sublevels (classrooms), and orbitals (seats).
The electron configuration is simply the address system that tells you exactly which "seats" every electron in an atom is sitting in.
The Intuition: Why Can't Electrons Sit Anywhere?
Two big rules force electrons into this orderly arrangement:
- Energy matters. Electrons want to be as close to the nucleus as possible (lowest energy). The first "floor" (n=1) is the most comfortable. Higher floors cost more energy.
- No crowding. A famous rule called the Pauli Exclusion Principle says: no two electrons in the same atom can have the exact same set of four quantum numbers. In plain language: each orbital (seat) can hold at most two electrons, and they must spin in opposite directions.
So electrons fill up from the bottom floor upward, like students filling a theatre from the front row back.
The Precise Statement
Electron configuration is the distribution of electrons of an atom or molecule in atomic orbitals. It is written as a sequence of:
- Principal quantum number n (the energy level: 1, 2, 3...)
- Sublevel letter (s, p, d, f) — tells you the shape of the orbital
- Superscript — the number of electrons in that sublevel
For example, the configuration of carbon (6 electrons) is:
1s22s22p2
This reads: "Two electrons in the 1s orbital, two in the 2s orbital, and two in the 2p orbitals."
The Filling Order: The Aufbau Principle
Electrons don't fill levels in simple numerical order. Here's the actual sequence (memorise this — it's exam gold):
1s→2s→2p→3s→3p→4s→3d→4p→5s→4d→5p→6s→4f→5d→6p→7s→5f→6d→7p
Notice: 4s fills before 3d. This is because the 4s orbital is actually lower in energy than 3d. This trips up many students.
Use the diagonal rule (Madelung's rule) to remember the order: draw arrows diagonally across the (n+ℓ) chart. Or just remember the mnemonic: "Silly People Don't Fail" for the sublevel order within each shell.
How to Write Any Configuration (Step-by-Step)
Let's do iron (Fe, atomic number 26).
Step 1: Know the total electrons = 26.
Step 2: Follow the filling order, counting electrons as you go:
- 1s2 (2 used, 24 left)
- 2s2 (4 used, 22 left)
- 2p6 (10 used, 16 left)
- 3s2 (12 used, 14 left)
- 3p6 (18 used, 8 left)
- 4s2 (20 used, 6 left)
- 3d6 (26 used, 0 left)
Step 3: Write it in order of increasing n (standard notation):
1s22s22p63s23p63d64s2
Many textbooks write configurations in order of filling (4s before 3d), but IUPAC standard lists them by principal quantum number n (3d before 4s). Check which convention your exam uses. For CBSE/ICSE, write in order of increasing n: 1s,2s,2p,3s,3p,3d,4s,4p...
The Three Golden Rules (Memorise These)
| Rule | What it says | Why it matters |
|---|---|---|
| Aufbau Principle | Electrons fill lowest energy orbitals first | Determines the order of filling |
Why this formula?
Electron Configuration: Why the Rules Work
Let's build this from the ground up — not just what the rules are, but why they exist.
The Core Question
Why do electrons arrange themselves in specific shells, subshells, and orbitals — and not just pile up anywhere?
The answer lies in three fundamental principles, each rooted in physics and quantum mechanics.
1. The Aufbau Principle: Why "Build Up" in Order?
What it says: Electrons fill orbitals from lowest to highest energy.
Why it holds: Nature seeks the lowest possible energy state (ground state). An atom is most stable when its electrons occupy the lowest available energy levels.
Think of it like water flowing downhill — electrons "fall" into the lowest energy orbitals first.
The energy ordering (for multi-electron atoms) is:
1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s...
Why this order? It comes from the (n+ℓ) rule:
- n = principal quantum number (shell)
- ℓ = azimuthal quantum number (subshell: s=0, p=1, d=2, f=3)
Orbitals fill in order of increasing (n+ℓ). If two have the same (n+ℓ), the one with lower n fills first.
Example: 4s has (4+0)=4, 3d has (3+2)=5. So 4s fills before 3d — even though 4s is a higher shell number.
2. Pauli Exclusion Principle: Why Only Two Per Orbital?
What it says: No two electrons in an atom can have the same set of all four quantum numbers.
Why it holds: This is a fundamental law of quantum mechanics — electrons are fermions (spin-1/2 particles). Fermions obey the Pauli exclusion principle, which arises from the antisymmetry of the wavefunction.
The four quantum numbers:
- n (shell)
- ℓ (subshell shape)
- mℓ (orbital orientation)
- ms (spin: +21 or −21)
Since only ms can differ for electrons in the same orbital, maximum 2 electrons per orbital — one spin-up (↑) and one spin-down (↓).
Key result: The s subshell (ℓ=0, one orbital) holds 2 electrons. The p subshell (ℓ=1, three orbitals) holds 6 electrons. The d subshell (ℓ=2, five orbitals) holds 10 electrons.
3. Hund's Rule: Why Spread Out First?
What it says: Within a subshell, electrons occupy empty orbitals singly before pairing up — and all unpaired electrons have parallel spins.
Why it holds: Electrons repel each other (Coulomb repulsion). By occupying different orbitals, they stay farther apart, reducing repulsion energy.
The spin alignment (all parallel) comes from exchange energy — a quantum mechanical effect where parallel spins have a slightly lower energy state due to wavefunction symmetry.
Example for carbon (1s22s22p2):
- Correct: ↑ | ↑ | (two unpaired, parallel)
- Wrong: ↑↓ | | (paired in one orbital — higher repulsion)
The Big Picture: Why These Three Rules Together? …
Concept: Electron Configuration – filling order follows the Aufbau principle (n+l rule) and the (n-1)d, (n-2)f blocks fill after the ns orbital.
Reasoning steps:
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Atomic number 61 (Promethium, Pm):
After Xe (54 electrons), the next 7 electrons go into 6s² and then 4f.
Configuration: [Xe]6s24f5 (since 4f fills before 5d).
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Atomic number 91 (Protactinium, Pa):
After Rn (86 electrons), the next 5 electrons fill 7s² and then 5f², with one electron going into 6d¹ (anomaly due to stability).
Configuration: [Rn]7s25f26d1.
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Atomic number 101 (Mendelevium, Md):
After Rn (86), the next 15 electrons fill 7s², 5f¹³, and then 6d⁰ (no 6d electron).
Configuration: [Rn]7s25f13.
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Atomic number 109 (Meitnerium, Mt): …
The key is to follow the Aufbau principle (n+l rule) and account for the special stability of half-filled and fully-filled orbitals. The configurations are: 61: [Xe]4f56s2; 91: [Rn]5f26d17s2; 101: [Rn]5f137s2; 109: [Rn]5f146d77s2.
Why Electron Configuration Works This Way
Electrons fill orbitals in order of increasing energy, not just increasing principal quantum number n. The rule is: an orbital with lower (n+l) fills first; if two have the same (n+l), the one with lower n fills first. This is the Aufbau principle, and it explains why the 4f subshell fills after 6s, and 5f after 7s.
For elements beyond lanthanum (atomic number 57), the 4f orbitals begin to fill. Similarly, beyond actinium (89), the 5f orbitals fill. But there are exceptions — half-filled (f⁷) and fully-filled (f¹⁴) subshells are extra stable, so sometimes an electron from the s-orbital moves into the f-orbital to achieve that stability.
Let’s work through each element.
1. Atomic number 61 — Promethium (Pm)
The nearest noble gas is xenon (Xe, Z=54). That gives us a core of [Xe].
Remaining electrons: 61−54=7 electrons.
The filling order after Xe is: 6s (2 electrons), then 4f (up to 14 electrons), then 5d, then 6p.
So we put 2 electrons into 6s: 6s2.
That leaves 7−2=5 electrons. These go into the 4f subshell: 4f5.
No special stability is reached here (f⁷ would be half-filled, but we only have 5), so no exception occurs.
Configuration: [Xe]4f56s2
For lanthanides (Z=58 to 71), the 4f subshell fills after 6s. The 5d orbital is usually empty or has at most 1 electron in this series — only exceptions are La, Ce, Gd, and Lu.
2. Atomic number 91 — Protactinium (Pa)
Nearest noble gas: radon (Rn, Z=86). Core: [Rn].
Remaining electrons: 91−86=5 electrons.
After Rn, the filling order is: 7s (2), then 5f (14), then 6d (10), then 7p.
First, 2 electrons go into 7s: 7s2.
That leaves 5−2=3 electrons. According to the Aufbau order, the next orbital is 5f. So we would expect 5f3.
But here’s the catch: for protactinium, the 5f and 6d orbitals are very close in energy. Experimentally, the configuration is [Rn]5f26d17s2, not [Rn]5f37s2. Why? Because having one electron in the 6d orbital (which is slightly lower in energy for Pa) is more stable than putting all three into 5f.
A common mistake is to blindly follow the Aufbau order for actinides. The 5f and 6d orbitals are very close in energy, and for elements like Pa, U, Np, and Cm, you get 6d electrons. Always check the actual configuration — don’t assume the simple filling order holds.
Configuration: [Rn]5f26d17s2
3. Atomic number 101 — Mendelevium (Md)
Core: [Rn] (Z=86).
Remaining electrons: 101−86=15 electrons. …
Method: Aufbau Principle with (n + ℓ) Rule
This is the standard method for writing ground-state electron configurations. It uses the order of increasing orbital energy determined by the sum (n+ℓ) — and for equal sums, by lower n first.
Steps
- Identify the atomic number (Z) — this equals the total number of electrons in a neutral atom.
- Follow the Aufbau order (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p...).
- Fill each subshell to its maximum capacity:
- s: 2 electrons
- p: 6 electrons
- d: 10 electrons
- f: 14 electrons
- Stop when the total electrons equal Z.
- Write in order of increasing principal quantum number n (standard notation), not in filling order.
Solutions
Atomic number 61 — Promethium (Pm)
- Z = 61
- Fill: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f⁵
- Final configuration (by n): 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f5 5s2 5p6 6s2
Atomic number 91 — Protactinium (Pa)
- Z = 91
- Fill: ... up to 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f² 6d¹
- Final configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 5f2 6s2 6p6 6d1 7s2
Note: Pa is an anomaly — the 5f and 6d are very close in energy. The above is the accepted ground state.
Atomic number 101 — Mendelevium (Md)
- Z = 101
- Fill: ... up to 7s² 5f¹³
- Final configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 5f13 6s2 6p6 7s2
Atomic number 109 — Meitnerium (Mt)
- Z = 109
- Fill: ... up to 7s² 5f¹⁴ 6d⁷
- Final configuration: …
🧠 The Core Idea First
Electronic configuration follows the Aufbau principle (fill lowest energy orbitals first), Hund’s rule (maximize unpaired spins), and the Pauli exclusion principle. But for elements beyond atomic number 57 (La), the energy ordering of orbitals changes due to nuclear charge and shielding effects.
The correct filling order is:
1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
✗ Common Mistake #1: Forgetting the f-block (lanthanides & actinides)
The error:
Students write configurations for atomic numbers 61, 91, 101, 109 as if they are normal d-block elements, skipping the f-subshell entirely.
Example of wrong answer for Z=61:
[Xe]6s24f1 ✗ (a 57-electron count — and not even real La, whose actual configuration is [Xe]5d16s2)
Why it happens:
They memorise the order but forget that after La (Z=57), the 4f subshell starts filling — not 5d.
How to avoid:
- Remember: Lanthanides (Z=58 to 71) fill the 4f subshell.
- Actinides (Z=90 to 103) fill the 5f subshell.
- Use the n + ℓ rule to confirm: 4f has n+ℓ = 4+3 = 7, 5d has 5+2 = 7, but 4f is lower in energy because of lower n.
Correct for Z=61 (Promethium, Pm):
[Xe]6s24f5 ✓
✗ Common Mistake #2: Misplacing the 5f and 6d orbitals for Z=91 and 101
The error:
For Z=91 (Protactinium), students write [Rn]7s25f3 ✗ — but the actual configuration has a 5f² 6d¹ arrangement.
Why it happens:
They assume the 5f subshell fills strictly after 7s, ignoring that 6d can be slightly lower in energy for early actinides.
How to avoid:
- For early actinides (Th, Pa, U, Np), the 6d orbital may get one electron before 5f fills completely.
- Memorise the exceptions for Pa (Z=91): [Rn]7s25f26d1 ✓
- For later actinides (Am onwards), 5f fills normally.
Correct for Z=91 (Protactinium):
[Rn]7s25f26d1 ✓
Correct for Z=101 (Mendelevium, Md):
[Rn]7s25f13 ✓ (no 6d electron here)
✗ Common Mistake #3: Forgetting the d-block exception for Z=109
The error:
For Z=109 (Meitnerium, Mt), students write [Rn]7s25f146d7 — the same electrons, but listed in filling order. This is a presentation slip rather than wrong chemistry: the convention is to list subshells in order of increasing n, so present it as [Rn]5f146d77s2.
Why it happens:
They write orbitals in filling order (7s before 6d) but forget that in the periodic table, we write by increasing n (principal quantum number), not filling order.
How to avoid: …
- TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQQ.In which of the following, elements are correctly arranged in the decreasing order of atomic radii? (A) Eu > Ce > Gd > Ho (B) Ce > Eu > Gd > Ho (C) Gd > Eu > Ce > Ho (D) Ho > Ce > Eu > Gd
›Reveal solutionSolution
The key idea is that atomic radii in the lanthanide series generally decrease from left to right (lanthanide contraction), but europium (Eu) has an anomalously large radius due to its half-filled 4f⁷ shell. The correct decreasing order is Eu > Ce > Gd > Ho, which corresponds to option (A).
Why This Approach Works
Atomic radii in the lanthanide series (Ce to Lu) follow a well-known trend called the lanthanide contraction: as nuclear charge increases across the series, the 4f electrons are poorly shielded, so the effective nuclear pull on the outer electrons increases, causing a steady decrease in atomic radius. However, there are two important exceptions: europium (Eu) and ytterbium (Yb). Eu has a half-filled 4f⁷ subshell, which is especially stable and causes the atom to be slightly larger than its neighbors. Similarly, Yb has a filled 4f¹⁴ subshell. So, when ordering radii, we must account for these "bumps."
The problem asks for decreasing order among four specific lanthanides: Ce (cerium, atomic number 58), Eu (europium, 63), Gd (gadolinium, 64), and Ho (holmium, 67). We need to place them from largest to smallest radius.
Step-by-Step Reasoning
-
Recall the general trend across the lanthanide series.
From left to right (increasing atomic number), atomic radii decrease. So, in a "normal" trend, Ce (Z=58) would be the largest, then Eu (Z=63), then Gd (Z=64), then Ho (Z=67) would be the smallest. But this ignores the special stability of Eu.
-
Identify the anomaly for europium.
Europium has the electron configuration [Xe] 4f⁷ 6s². The half-filled 4f subshell is exceptionally stable, which reduces the effective nuclear charge felt by the outer 6s electrons. This makes the atomic radius of Eu larger than that of its immediate neighbors, including Ce. In fact, Eu’s radius is even larger than that of Ce, despite Ce being to the left.
-
Compare Eu and Ce directly.
Experimental data (e.g., metallic radii in picometers) show:
- Ce: ~181.8 pm
- Eu: ~204.2 pm
- Gd: ~180.2 pm
- Ho: ~176.6 pm Clearly, Eu is the largest of the four, followed by Ce, then Gd, then Ho.
-
Check the options against this order.
The correct decreasing order is: Eu > Ce > Gd > Ho.
- Option (A) says exactly that.
- Option (B) says Ce > Eu > Gd > Ho — wrong because Eu is larger than Ce. …
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- TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQQ.In dipositive zinc ion, the number of electrons with l=1, m=0 is x, and number of electrons with l=2, m=−1 is y. The sum of x,y is equal to (A) 6 (B) 4 (C) 8 (D) 5
›Reveal solutionSolution
We first determine the electron configuration of Zn2+ by removing electrons from the outermost shell of neutral zinc. Then, we count electrons based on the given quantum numbers l and ml for p-subshells (l=1) and d-subshells (l=2). The sum of x and y is 6.
The problem asks us to find the total number of electrons in a dipositive zinc ion (Zn2+) that satisfy specific conditions on their azimuthal (l) and magnetic (ml) quantum numbers. To do this, we need to understand the electron configuration of Zn2+ and how quantum numbers describe the state of an electron.
The four quantum numbers (n,l,ml,ms) uniquely describe an electron in an atom:
- n (principal quantum number) defines the electron shell and energy level.
- l (azimuthal or angular momentum quantum number) defines the subshell and the shape of the orbital.
- l=0 corresponds to an s-subshell.
- l=1 corresponds to a p-subshell.
- l=2 corresponds to a d-subshell.
- l=3 corresponds to an f-subshell.
- ml (magnetic quantum number) defines the orientation of the orbital in space. For a given l, ml can take any integer value from −l to +l, including 0.
- For l=0 (s-subshell), ml=0 (1 orbital).
- For l=1 (p-subshell), ml=−1,0,+1 (3 orbitals).
- For l=2 (d-subshell), ml=−2,−1,0,+1,+2 (5 orbitals).
- ms (spin quantum number) describes the intrinsic angular momentum (spin) of the electron, which can be +1/2 or −1/2. Each orbital can hold a maximum of two electrons, one with ms=+1/2 and one with ms=−1/2.
We will first determine the electron configuration of Zn2+ and then count the electrons based on the given l and ml values.
-
Determine the electron configuration of neutral Zinc (Zn).
Zinc (Zn) has an atomic number Z=30, meaning a neutral zinc atom has 30 electrons. Following the Aufbau principle, its electron configuration is:
1s22s22p63s23p64s23d10
-
Determine the electron configuration of dipositive Zinc ion (Zn2+).
To form a Zn2+ ion, two electrons must be removed from the neutral zinc atom. Electrons are always removed from the outermost shell (the shell with the highest principal quantum number, n) first. In the configuration of neutral Zn, the 4s orbital has n=4, which is higher than n=3 for the 3d orbital. Therefore, the two electrons are removed from the 4s orbital.
The electron configuration of Zn2+ is:
1s22s22p63s23p63d10
-
Calculate x, the number of electrons with l=1,ml=0.
The condition l=1 refers to p-subshells. In the Zn2+ configuration, we have electrons in the 2p6 and 3p6 subshells. …
- TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQQ.Ionization enthalpies of five successive elements in a period are 899, 801, 1086, 1402 and 1314 kJ mol−1 respectively. The corresponding elements are (A) Li, Be, B, C, N (B) Be, B, C, N, O (C) B, C, N, O, F (D) C, N, O, F, Ne
›Reveal solutionSolution
The pattern of ionization enthalpies — a drop from the first to the second element, then a steady rise — matches the trend across period 2 from Be to O. The correct sequence is Be, B, C, N, O, which is option (B).
The key to this problem is understanding how ionization enthalpy varies across a period. As we move from left to right, the nuclear charge increases, pulling the valence electrons tighter, so ionization enthalpy generally rises. But there are two famous exceptions: the drop from group 2 to group 13 (Be to B) and from group 15 to group 16 (N to O). These dips happen because of changes in electron configuration — from a filled s-orbital to a p-orbital with one electron (easier to remove), and from a half-filled p-subshell to one with a paired electron (also easier to remove).
Here the given values are: 899, 801, 1086, 1402, 1314 kJ mol−1. Notice the second value (801) is lower than the first (899) — that’s the first dip. Then the values climb: 1086, 1402, and then a slight drop to 1314. That second drop matches the N-to-O exception. So the sequence must start with a group 2 element, then group 13, then group 14, then group 15, then group 16. In period 2, that’s Be, B, C, N, O.
Let’s walk through it step by step.
-
Identify the first dip. The first two values are 899 and 801. Since 801 < 899, the second element has a lower ionization enthalpy than the first. This is the classic Be-to-B drop: Be has a filled 2s2 configuration (stable), while B has 2s22p1 — that single p-electron is easier to remove. So the first element must be Be (group 2) and the second must be B (group 13).
-
Check the rise after the dip. After 801, the values increase to 1086 and then to 1402. That’s consistent with moving from B (group 13) to C (group 14) to N (group 15). Each step adds a proton, increasing nuclear charge, and the electron is removed from the same shell, so ionization enthalpy rises. …
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- TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQQ.The number of electrons with (n+l) values equal to 3, 4 and 5 in an element with atomic number (z) 24 are respectively (n=principal quantum number and l=azimuthal quantum number) (A) 7, 8, 5 (B) 6, 8, 6 (C) 8, 7, 5 (D) 8, 8, 5
›Reveal solutionSolution
For chromium (Z=24), the electrons are distributed according to the Aufbau principle with the exception of a half-filled 3d⁵ subshell. Counting electrons by (n+l) values: (n+l)=3 gives 8 electrons, (n+l)=4 gives 7 electrons, (n+l)=5 gives 5 electrons. The correct option is (C).
The key idea is that the sum n+l determines the order of filling orbitals (the Aufbau principle). For a given atom, we need to list all occupied orbitals, group them by their n+l value, and count the electrons in each group. Chromium (Z=24) has a well-known electronic configuration exception: instead of 3d44s2, it is 3d54s1 to achieve a half-filled d-subshell.
Let’s work through it step by step.
- Write the full electronic configuration of chromium (Z=24). The order of filling is: 1s, 2s, 2p, 3s, 3p, 4s, then 3d. Normally, we would have 1s22s22p63s23p64s23d4. But chromium is an exception: one electron from 4s moves to 3d, giving a half-filled 3d⁵. So the actual configuration is:
1s22s22p63s23p64s13d5
Total electrons: 2+2+6+2+6+1+5=24.
-
List each subshell with its (n, l) and compute n+l.
- 1s: n=1,l=0 → n+l=1
- 2s: n=2,l=0 → n+l=2
- 2p: n=2,l=1 → n+l=3
- 3s: n=3,l=0 → n+l=3
- 3p: n=3,l=1 → n+l=4
- 4s: n=4,l=0 → n+l=4
- 3d: n=3,l=2 → n+l=5
-
Group the subshells by their n+l value and count electrons.
- n+l=3: includes 2p (6 electrons) and 3s (2 electrons). Total = 6+2=8.
- n+l=4: includes 3p (6 electrons) and 4s (1 electron). Total = 6+1=7. …
- TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQQ.The maximum number of orbitals present in n=4 energy level of an atom and the maximum number of electrons with spin value +21 in the same orbitals are _______, respectively. (A) 16, 5 (B) 16, 7 (C) 16, 9 (D) 16, 16
›Reveal solutionSolution
For the n=4 energy level, the total number of orbitals is 16, and the maximum number of electrons with spin +21 in those orbitals is 16 as well — each orbital can hold one electron of each spin. The correct option is (D).
The key idea here is that the number of orbitals in a given energy level is determined by the sum of all possible ml values across all subshells, while the maximum number of electrons with a specific spin is simply half the total electron capacity — because each orbital holds exactly two electrons of opposite spin.
Let’s break it down.
-
Number of orbitals in the n=4 level
For a principal quantum number n, the possible azimuthal quantum numbers are l=0,1,2,…,n−1. So for n=4, we have:
- l=0 (4s subshell): ml=0 → 1 orbital
- l=1 (4p subshell): ml=−1,0,+1 → 3 orbitals
- l=2 (4d subshell): ml=−2,−1,0,+1,+2 → 5 orbitals
- l=3 (4f subshell): ml=−3,−2,−1,0,+1,+2,+3 → 7 orbitals
Total orbitals = 1+3+5+7=16.
A quick formula: the number of orbitals in the nth level is n2, so 42=16 — same result.
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Maximum number of electrons with spin +21 …
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