NCERT Solutions · Class 11 Chemistry Structure of Atom

The de Broglie wavelength formula needs the electron's momentum, not its kinetic energy directly — so the first job is to turn K.E. into momentum before touching de Broglie's equation at all.Step $\displaystyle 1$ — get the speed from the kinetic energy.Kinetic energy: \(\displaystyle KE = \tfrac{1}{2}mv^2 \), where \(\displaystyle m \) is the mass of the electron and \(\displaystyle v \) is its speed. Rearranging for \(\displaystyle v \):\[v = \sqrt{\dfrac{2\,KE}{m}} \]Substituting \(\displaystyle KE = 3.0 \times 10^{-25}\ \text{J} \) and \(\displaystyle m = 9.1 \times 10^{-31}\ \text{kg} \):\[v = \sqrt{\dfrac{2 \times 3.0 \times 10^{-25}\ \text{J}}{9.1 \times 10^{-31}\ \text{kg}}} = \sqrt{6.593 \times 10^{5}\ \text{m}^2\text{s}^{-2}} \](A joule is \(\displaystyle \text{kg}\,\text{m}^2\text{s}^{-2} \), so dividing by kg leaves \(\displaystyle \text{m}^2\text{s}^{-2}\) — exactly what needs a square root to become a speed.)\[v \approx 812\ \text{m s}^{-1} \]Step $\displaystyle 2$ — convert speed into momentum.Momentum: \(\displaystyle p = mv \), so\[p = (9.1 \times 10^{-31}\ \text{kg})(812\ \text{m s}^{-1}) \approx 7.39 \times 10^{-28}\ \text{kg m s}^{-1} \]This is the step people skip — plugging \(\displaystyle KE \) straight into the de Broglie formula. De Broglie's equation runs on momentum, so \(\displaystyle K.E. \) has to be converted to \(\displaystyle v \), and then to \(\displaystyle p = mv \), first.Step $\displaystyle 3$ — apply de Broglie's equation.De Broglie's relation: \(\displaystyle \lambda = \dfrac{h}{p} \), where \(\displaystyle h \) is Planck's constant, \(\displaystyle 6.626 \times 10^{-34}\ \text{J s} \), and \(\displaystyle p \) is momentum.\[\lambda = \dfrac{6.626 \times 10^{-34}\ \text{J s}}{7.39 \times 10^{-28}\ \text{kg m s}^{-1}} \]Since \(\displaystyle 1\ \text{J s} = 1\ \text{kg m}^2\text{s}^{-1} \), the kg and one factor of \(\displaystyle \text{s}^{-1} \) cancel, leaving units of metres — confirming this is a length, as a wavelength must be.\[\lambda \approx 8.97 \times 10^{-7}\ \text{m} \]The mass and kinetic energy were each given to two significant figures, so this is rounded once, at the end, to three significant figures (the extra figure is kept because \(\displaystyle h \) itself is known far more precisely) — not rounded at every intermediate step, which would drift the answer.\[\lambda \approx 8.97 \times 10^{-7}\ \text{m} = 897\ \text{nm} \]Answer: \(\displaystyle \lambda \approx 8.97 \times 10^{-7}\ \text{m} \) ($\displaystyle 897$ nm)

Isoelectronic means the same NUMBER of electrons — so count them, and the counting rule is: electrons = atomic number \(\displaystyle Z\) minus the charge. A positive charge means electrons were removed, a negative charge means electrons were added. Nothing else about the species matters here: not its mass, not whether it is a metal, not the size of its charge.\[\begin{aligned} \mathrm{Na^+} &: Z = 11,\ \text{charge } +1 &&\Rightarrow 11 - 1 = 10\ e^- \\ \mathrm{Mg^{2+}} &: Z = 12,\ \text{charge } +2 &&\Rightarrow 12 - 2 = 10\ e^- \\ \mathrm{K^+} &: Z = 19,\ \text{charge } +1 &&\Rightarrow 19 - 1 = 18\ e^- \\ \mathrm{Ca^{2+}} &: Z = 20,\ \text{charge } +2 &&\Rightarrow 20 - 2 = 18\ e^- \\ \mathrm{S^{2-}} &: Z = 16,\ \text{charge } -2 &&\Rightarrow 16 + 2 = 18\ e^- \\ \mathrm{Ar} &: Z = 18,\ \text{neutral} &&\Rightarrow 18\ e^- \end{aligned} \]The step people get wrong is the sign. \(\displaystyle \mathrm{S^{2-}}\) has GAINED two electrons, so you add: \(\displaystyle 16 + 2 = 18\), not \(\displaystyle 16 - 2\). And \(\displaystyle \mathrm{Ca^{2+}}\) is easy to file with \(\displaystyle \mathrm{Mg^{2+}}\) because both carry \(\displaystyle +2\) — but the charge is not the electron count. Calcium starts eight electrons further along than magnesium, so \(\displaystyle \mathrm{Ca^{2+}}\) ends up with $\displaystyle 18$, not 10.Grouping by the counts: Answer: two isoelectronic sets — \(\displaystyle \mathrm{Na^+}\) and \(\displaystyle \mathrm{Mg^{2+}}\) with $\displaystyle 10$ electrons each, and \(\displaystyle \mathrm{K^+}\), \(\displaystyle \mathrm{Ca^{2+}}\), \(\displaystyle \mathrm{S^{2-}}\) and \(\displaystyle \mathrm{Ar}\) with $\displaystyle 18$ electrons each.

Ionic charge changes the electron count, not the proton count — the number of electrons an atom has equals its atomic number \(\displaystyle Z\), and gaining or losing electrons shifts that count by the charge, while the number of protons (which fixes which element it is) stays fixed.(i) Electronic configurations of the ionsStart from the parent atom's electron count (its atomic number \(\displaystyle Z\), since a neutral atom has electrons = protons), then add electrons for a negative charge or remove electrons for a positive charge.(a) H⁻ — Hydrogen (\(\displaystyle Z=1\)) has $\displaystyle 1$ electron: \(\displaystyle 1s^1\). The ion H⁻ carries an extra electron (charge −$\displaystyle 1$ means one electron gained), giving $\displaystyle 2$ electrons total. \[\text{H}^- : 1s^2 \](b) Na⁺ — Sodium (\(\displaystyle Z=11\)) has $\displaystyle 11$ electrons: \(\displaystyle 1s^2\,2s^2\,2p^6\,3s^1\). The ion Na⁺ has lost $\displaystyle 1$ electron (charge +$\displaystyle 1$), leaving $\displaystyle 10$ electrons — the aside worth flagging here: losing the charge's worth of electrons means dropping the outermost electron first (the lone \(\displaystyle 3s^1\)), not any arbitrary one. \[\text{Na}^+ : 1s^2\,2s^2\,2p^6 \](c) O²⁻ — Oxygen (\(\displaystyle Z=8\)) has $\displaystyle 8$ electrons: \(\displaystyle 1s^2\,2s^2\,2p^4\). The ion O²⁻ has gained $\displaystyle 2$ electrons (charge −$\displaystyle 2$), giving \(\displaystyle 8+2=10\) electrons. \[\text{O}^{2-} : 1s^2\,2s^2\,2p^6 \](d) F⁻ — Fluorine (\(\displaystyle Z=9\)) has $\displaystyle 9$ electrons: \(\displaystyle 1s^2\,2s^2\,2p^5\). The ion F⁻ has gained $\displaystyle 1$ electron (charge −$\displaystyle 1$), giving \(\displaystyle 9+1=10\) electrons. \[\text{F}^- : 1s^2\,2s^2\,2p^6 \]Notice Na⁺, O²⁻ and F⁻ all end up with $\displaystyle 10$ electrons and the same configuration as neon — this is exactly why they're called isoelectronic species, even though they came from different parent atoms with different numbers of protons.(ii) Atomic number from the outermost subshellThe subshell given is the last one filled. Using the building-up (Aufbau) order \(\displaystyle 1s,2s,2p,3s,3p,4s,\dots\), every subshell before it in that order must already be full. The atomic number is the sum of all electrons in every filled subshell plus the electrons named in the outermost one — total electrons = \(\displaystyle Z\) because the atom is neutral.(a) 3s¹ — Subshells filled before 3s, in order, are \(\displaystyle 1s^2\,2s^2\,2p^6\). Adding the outermost \(\displaystyle 3s^1\): \[Z = 2+2+6+1 = 11 \] Atomic number $\displaystyle 11$ (sodium).(b) 2p³ — Subshells before 2p are \(\displaystyle 1s^2\,2s^2\). Adding \(\displaystyle 2p^3\): \[Z = 2+2+3 = 7 \] Atomic number $\displaystyle 7$ (nitrogen).(c) 3p⁵ — Subshells before 3p are \(\displaystyle 1s^2\,2s^2\,2p^6\,3s^2\). Adding \(\displaystyle 3p^5\): \[Z = 2+2+6+2+5 = 17 \] Atomic number $\displaystyle 17$ (chlorine).The trap to watch here: 3p⁵ does not mean "$\displaystyle 3$ and p and $\displaystyle 5$" combine some other way — it means the 3p subshell holds $\displaystyle 5$ of its $\displaystyle 6$ possible electrons, and every subshell that fills before 3p (by energy order, not just by lower shell number) has to be counted too, including 3s before 3p.(iii) Identify the atom from a noble-gas-core configurationHere \(\displaystyle [\text{He}]\), \(\displaystyle [\text{Ne}]\), \(\displaystyle [\text{Ar}]\) are shorthand for that noble gas's full electron count — a filled inner shell structure that itself has a fixed number of electrons. Add the noble gas's electron count to the additional electrons listed after it; the total again equals \(\displaystyle Z\).(a) [He] 2s¹ — Helium has $\displaystyle 2$ electrons. Adding \(\displaystyle 2s^1\): \[Z = 2 + 1 = 3 \] \(\displaystyle Z=3\) is lithium (Li).(b) [Ne] 3s² 3p³ — Neon has $\displaystyle 10$ electrons. Adding \(\displaystyle 3s^2\,3p^3\) (\(\displaystyle 2+3=5\) electrons): \[Z = 10 + 5 = 15 \] \(\displaystyle Z=15\) is phosphorus (P).(c) [Ar] 4s² 3d¹ — Argon has $\displaystyle 18$ electrons. Adding \(\displaystyle 4s^2\,3d^1\) (\(\displaystyle 2+1=3\) electrons): \[Z = 18 + 3 = 21 \] \(\displaystyle Z=21\) is scandium (Sc). The aside worth naming: even though 3d is written after 4s here, 3d still belongs to the third shell — the order in which subshells are written follows the shell-then-subshell convention, while the order in which they fill follows the Aufbau energy order (4s before 3d), and both facts are needed to read this notation correctly.Answer: (i) H⁻: \(\displaystyle 1s^2\); Na⁺: \(\displaystyle 1s^2 2s^2 2p^6\); O²⁻: \(\displaystyle 1s^2 2s^2 2p^6\); F⁻: \(\displaystyle 1s^2 2s^2 2p^6\). (ii) 3s¹ → Z = $\displaystyle 11$; 2p³ → Z = $\displaystyle 7$; 3p⁵ → Z = 17. (iii) [He]2s¹ → Z = $\displaystyle 3$ (Li); [Ne]3s²3p³ → Z = $\displaystyle 15$ (P); [Ar]4s²3d¹ → Z = $\displaystyle 21$ (Sc).

The azimuthal quantum number \(\displaystyle l \) is fixed by the orbital's letter, and \(\displaystyle l \) can never exceed \(\displaystyle n-1 \).Each orbital letter corresponds to a specific value of the azimuthal (subsidiary) quantum number \(\displaystyle l \):\[l = 0 \to s, \quad l = 1 \to p, \quad l = 2 \to d, \quad l = 3 \to f, \quad l = 4 \to g \]So a g orbital requires \(\displaystyle l = 4 \).The rule people slip on: for a given principal quantum number \(\displaystyle n \), the allowed values of \(\displaystyle l \) run only from \(\displaystyle 0 \) to \(\displaystyle n-1 \) — not up to \(\displaystyle n \). This comes from the quantum mechanical solution of the Schrödinger equation for the hydrogen atom, and it is a hard constraint, not a convention.For a g orbital to exist for some shell \(\displaystyle n \), that shell must permit \(\displaystyle l = 4 \), which means\[n - 1 \geq 4 \]\[n \geq 5 \]Checking it directly: for \(\displaystyle n = 4 \), the allowed \(\displaystyle l \) values are \(\displaystyle 0, 1, 2, 3 \) (s, p, d, f only) — no g orbital yet. For \(\displaystyle n = 5 \), the allowed \(\displaystyle l \) values become \(\displaystyle 0, 1, 2, 3, 4 \) (s, p, d, f, g) — the g orbital appears for the first time.So the smallest principal quantum number that permits \(\displaystyle l = 4 \), and therefore a g orbital, is \(\displaystyle n = 5 \).Answer: The lowest value of \(\displaystyle n \) that allows g orbitals to exist is \(\displaystyle n = 5 \).

The orbital label "3d" already encodes two of the three quantum numbers — n and l are read directly off the name, and only \(\displaystyle m_l \) needs the range rule.Step $\displaystyle 1$: the principal quantum number \(\displaystyle n \).\(\displaystyle n \) tells you the shell (energy level) and is simply the number written in front of the orbital letter. The label is "3d", so\[n = 3 \]Step $\displaystyle 2$: the azimuthal (angular momentum) quantum number \(\displaystyle l \).\(\displaystyle \mathrm{ l }\) tells you the subshell — its allowed values for a given \(\displaystyle n \) run \(\displaystyle 0, 1, 2, \dots, (n-1)\) and each value has a letter code:\[l = 0 \to s, \quad l = 1 \to p, \quad l = 2 \to d, \quad l = 3 \to f \]The orbital is a d orbital, so\[l = 2 \]A common slip is writing \(\displaystyle l = 3 \) for "d" by miscounting the sequence \(\displaystyle s, p, d, f \) as starting from $\displaystyle 1$ instead of $\displaystyle 0$ — \(\displaystyle l \) always starts at $\displaystyle 0$ for the s subshell, so d is the third letter but the second value of \(\displaystyle l \).Step $\displaystyle 3$: the magnetic quantum number \(\displaystyle m_l \).\(\displaystyle m_l \) fixes which orbital within the subshell (its orientation in space), and for a given \(\displaystyle l \) it takes every integer value from \(\displaystyle -l \) to \(\displaystyle +l \), including zero:\[m_l = -l, -(l-1), \dots, 0, \dots, (l-1), l \]With \(\displaystyle l = 2 \):\[m_l = -2, -1, 0, +1, +2 \]This is the step people rush: \(\displaystyle m_l \) is not a single number here — a d subshell has five orbitals (\(\displaystyle 2l+1 = 5 \)), so all five values of \(\displaystyle m_l \) are valid answers, not just one of them. The question asks for the possible values precisely because \(\displaystyle n \) and \(\displaystyle l \) alone don't tell you which of the five d-orbitals the electron sits in.Answer: \(\displaystyle n = 3 \), \(\displaystyle l = 2 \), and \(\displaystyle m_l = -2, -1, 0, +1, \) or \(\displaystyle +2 \).

In a neutral atom, the number of protons always equals the number of electrons.(i) Number of protonsAn atom carries no net charge, so the positive charge from the protons must exactly cancel the negative charge from the electrons:\[Z = \text{number of protons} = \text{number of electrons} \]Here the atom is given as neutral (just "an atom," not an ion) with $\displaystyle 29$ electrons, so\[Z = 29 \]Number of protons = 29.This also fixes the mass number, since mass number \(\displaystyle A\) is protons plus neutrons, \(\displaystyle A = Z + n\):\[A = 29 + 35 = 64 \]So this is the isotope with $\displaystyle 29$ protons and mass number $\displaystyle 64$ — that is, \(\displaystyle ^{64}_{29}\mathrm{Cu}\), copper-64. (The number of neutrons, $\displaystyle 35$, tells you which isotope of the \(\displaystyle Z=29\) element this is; it plays no role in the electronic configuration, which depends only on the number of electrons.)(ii) Electronic configurationWith \(\displaystyle Z = 29\), the atom has $\displaystyle 29$ electrons to place into subshells in order of increasing energy (the Aufbau principle), respecting the Pauli exclusion principle ($\displaystyle 2$ electrons per orbital) and Hund's rule.Filling subshells in the usual order \(\displaystyle 1s, 2s, 2p, 3s, 3p, 4s, 3d, \dots\) and counting electrons as you go:\[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2\,3d^7 \]Count check: \(\displaystyle 2+2+6+2+6+2+7 = 27\) — that's only $\displaystyle 27$, not $\displaystyle 29$, so this filling is wrong. Redo it correctly, keeping a running total:\[1s^2(2) \;\; 2s^2(4) \;\; 2p^6(10) \;\; 3s^2(12) \;\; 3p^6(18) \;\; 4s^2(20) \;\; 3d^9(29) \]This naive Aufbau filling gives\[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^9\,4s^2 \]This is the step people get wrong. A completely filled \(\displaystyle d\) subshell (\(\displaystyle d^{10}\)) is noticeably more stable than a \(\displaystyle d^9\) configuration, because a fully filled set of orbitals has extra symmetry and more favorable electron-electron exchange energy. So one electron shifts from the \(\displaystyle 4s\) orbital into the \(\displaystyle 3d\) subshell to complete it, even though \(\displaystyle 4s\) would otherwise be filled first:\[3d^9\,4s^2 \;\longrightarrow\; 3d^{10}\,4s^1 \]The true ground-state configuration is therefore\[1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1 \]Check the electron count: \(\displaystyle 2+2+6+2+6+10+1 = 29\), matching the given $\displaystyle 29$ electrons.**Answer: Number of protons = $\displaystyle 29$; electronic configuration = \(\displaystyle 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^1\) (the element is copper, with the \(\displaystyle 3d^9 4s^2\) Aufbau prediction corrected to \(\displaystyle 3d^{10}4s^1\) for the extra stability of a fully filled \(\displaystyle 3d\) subshell).

A positive charge on a molecule means it has lost electrons — the number of protons never changes, only the electron count drops by the size of the charge. The species here are \(\displaystyle \mathrm{H_2^+} \), \(\displaystyle \mathrm{H_2} \), and \(\displaystyle \mathrm{O_2^+} \), and each is found the same way: count the electrons in the neutral molecule, then add or remove electrons for any charge.Rule $\displaystyle 1$ — electrons in a neutral atom. For a neutral atom, the number of electrons equals the atomic number \(\displaystyle Z \) (this is what makes the atom electrically neutral: protons = electrons = \(\displaystyle Z \)). Rule $\displaystyle 2$ — electrons in a neutral molecule. A neutral molecule made of two identical atoms has twice the per-atom electron count, since nothing has been added or removed — the atoms have simply bonded.\[\text{electrons in } \mathrm{X_2} = 2 \times Z(\mathrm{X}) \]Rule $\displaystyle 3$ — electrons in a cation. A species written with a \(\displaystyle + \) charge has lost that many electrons relative to the neutral species (this is the step people get backwards — a positive charge means fewer electrons, not more, because it's electrons, not protons, that are being removed).\[\text{electrons in cation} = \text{electrons in neutral species} - (\text{charge}) \]Working through each species\(\displaystyle \mathrm{H_2} \): Two hydrogen atoms, each contributing $\displaystyle 1$ electron: \[\text{electrons} = 2 \times Z(\mathrm{H}) = 2 \times 1 = 2 \text{ electrons} \]\(\displaystyle \mathrm{H_2^+} \): Start from neutral \(\displaystyle \mathrm{H_2} \) ($\displaystyle 2$ electrons) and remove $\displaystyle 1$ electron for the \(\displaystyle +1 \) charge: \[\text{electrons} = 2 - 1 = 1 \text{ electron} \]\(\displaystyle \mathrm{O_2^+} \): First find neutral \(\displaystyle \mathrm{O_2} \), two oxygen atoms each contributing $\displaystyle 8$ electrons: \[\text{electrons in } \mathrm{O_2} = 2 \times Z(\mathrm{O}) = 2 \times 8 = 16 \text{ electrons} \] Then remove $\displaystyle 1$ electron for the \(\displaystyle +1 \) charge on \(\displaystyle \mathrm{O_2^+} \): \[\text{electrons} = 16 - 1 = 15 \text{ electrons} \]Answer: \(\displaystyle \mathrm{H_2^+} \) has $\displaystyle 1$ electron, \(\displaystyle \mathrm{H_2} \) has $\displaystyle 2$ electrons, and \(\displaystyle \mathrm{O_2^+} \) has $\displaystyle 15$ electrons.

The azimuthal quantum number \(\displaystyle l \) is capped by the shell it lives in, and the magnetic quantum number \(\displaystyle ml \) is capped by \(\displaystyle l \) itself — get these two ranges right and every part of this question falls out.The rules connecting the three quantum numbers are: (i) n = $\displaystyle 3$: possible values of \(\displaystyle l \) and \(\displaystyle m_l \)Since \(\displaystyle l \) runs from \(\displaystyle 0 \) to \(\displaystyle (n-1) \), with \(\displaystyle n = 3 \): \[l = 0, 1, 2 \]Now take each \(\displaystyle l \) in turn and list its \(\displaystyle m_l \) values, remembering \(\displaystyle m_l \) runs from \(\displaystyle -l \) to \(\displaystyle +l \): The step people skip is treating \(\displaystyle l \) and \(\displaystyle m_l \) as independent — they are not; \(\displaystyle m_l \) is only ever read off after \(\displaystyle l \) is fixed, one subshell at a time.(ii) Quantum numbers ( \(\displaystyle l \) and \(\displaystyle m_l \) ) for the 3d orbitalThe label "3d" already fixes two numbers: \(\displaystyle n = 3 \), and "d" means \(\displaystyle l = 2 \).For \(\displaystyle l = 2 \), the allowed \(\displaystyle m_l \) values run from \(\displaystyle -l \) to \(\displaystyle +l \): \[m_l = -2, -1, 0, +1, +2 \]So the 3d subshell is described by \(\displaystyle l = 2 \) with \(\displaystyle m_l = -2, -1, 0, +1, +2 \) — five orbitals in total, matching \(\displaystyle (2l+1) = 2(2)+1 = 5 \).(iii) Which of 1p, 2s, 2p, 3f are possible orbitals?Check each one against the rule \(\displaystyle l \le (n-1) \), i.e. \(\displaystyle l \) can never reach \(\displaystyle n \) or exceed it. The trap in this part is reading the subshell letter as just a name; each letter is a specific \(\displaystyle l \) value, and that value has to fit inside \(\displaystyle 0 \) to \(\displaystyle (n-1) \) for the given \(\displaystyle n \) or the orbital simply is not a valid state.**Answer: (i) For n = $\displaystyle 3$: l = $\displaystyle 0$, $\displaystyle 1$, $\displaystyle 2$, with \(\displaystyle m_l = 0 \) for l = $\displaystyle 0$; \(\displaystyle m_l = -1, 0, +1 \) for l = $\displaystyle 1$; \(\displaystyle m_l = -2, -1, 0, +1, +2 \) for l = 2. (ii) For the 3d orbital, l = $\displaystyle 2$ and \(\displaystyle m_l = -2, -1, 0, +1, +2 \). (iii) Of the given orbitals, 2s and 2p are possible; 1p and 3f are not possible.

The letter in an orbital's name is read off the azimuthal quantum number \(\displaystyle l\), not off \(\displaystyle n\). Every orbital is labelled by two things: the principal quantum number \(\displaystyle n\), written as the number in front, and the subshell letter, which comes from a fixed correspondence with \(\displaystyle l\):\[l = 0 \rightarrow s, \qquad l = 1 \rightarrow p, \qquad l = 2 \rightarrow d, \qquad l = 3 \rightarrow f \]Here \(\displaystyle n\) fixes the shell (the orbital's size and energy level) and \(\displaystyle l\) fixes the shape of the orbital (spherical, dumbbell, etc.), so the full symbol is written as \(\displaystyle nl\) — the value of \(\displaystyle n\) followed by the letter for \(\displaystyle l\).A step people rush past: \(\displaystyle l\) can only run from \(\displaystyle 0\) to \(\displaystyle n-1\) for a given \(\displaystyle n\), but that constraint doesn't change how you name the orbital once you're given a valid \(\displaystyle (n, l)\) pair — it only tells you which pairs are allowed to exist. You still just read \(\displaystyle l\) off the table above.Now applying this to each pair:(a) \(\displaystyle n = 1, \; l = 0\) \(\displaystyle l = 0\) is the letter s, and \(\displaystyle n = 1\), so the orbital is \(\displaystyle 1s\).(b) \(\displaystyle n = 3, \; l = 1\) \(\displaystyle l = 1\) is the letter p, and \(\displaystyle n = 3\), so the orbital is \(\displaystyle 3p\).(c) \(\displaystyle n = 4, \; l = 2\) \(\displaystyle l = 2\) is the letter d, and \(\displaystyle n = 4\), so the orbital is \(\displaystyle 4d\).(d) \(\displaystyle n = 4, \; l = 3\) \(\displaystyle l = 3\) is the letter f, and \(\displaystyle n = 4\), so the orbital is \(\displaystyle 4f\).Answer: (a) 1s, (b) 3p, (c) 4d, (d) 4f.

Every quantum number has its own allowed range, and that range is fixed by the quantum numbers that come before it — check \(\displaystyle n\) first, then \(\displaystyle l\), then \(\displaystyle m_l\), then \(\displaystyle m_s\).The four rules, named before we use them: \[l = 0, 1, 2, \ldots, (n-1) \] The trap here is thinking \(\displaystyle l\) can go up to \(\displaystyle n\) — it stops one short of \(\displaystyle n\), at \(\displaystyle n-1\). \[m_l = -l, \ldots, 0, \ldots, +l \] Now check each set against these rules, in order.(a) \(\displaystyle n = 0, l = 0, m_l = 0, m_s = +\tfrac{1}{2}\) — not possible. The very first check fails: \(\displaystyle n\) must be a positive integer, and \(\displaystyle n = 0\) is not allowed. It doesn't matter that \(\displaystyle l\), \(\displaystyle m_l\), and \(\displaystyle m_s\) look fine individually — an invalid \(\displaystyle n\) invalidates the whole set.(b) \(\displaystyle n = 1, l = 0, m_l = 0, m_s = -\tfrac{1}{2}\) — possible. For \(\displaystyle n = 1\), the allowed values of \(\displaystyle l\) are \(\displaystyle 0\) to \(\displaystyle n - 1 = 0\), so \(\displaystyle l = 0\) is the only option and it fits. For \(\displaystyle l = 0\), the only allowed \(\displaystyle m_l\) is \(\displaystyle 0\), which fits. \(\displaystyle m_s = -\tfrac{1}{2}\) is one of the two allowed spin values. This set describes a \(\displaystyle 1s\) electron with spin down.(c) \(\displaystyle n = 1, l = 1, m_l = 0, m_s = +\tfrac{1}{2}\) — not possible. For \(\displaystyle n = 1\), \(\displaystyle l\) must lie between \(\displaystyle 0\) and \(\displaystyle n - 1 = 0\) — so the only legal value is \(\displaystyle l = 0\). Here \(\displaystyle l = 1\), one more than the maximum allowed. This is exactly the trap named above: \(\displaystyle l\) cannot equal \(\displaystyle n\), only go up to \(\displaystyle n - 1\).(d) \(\displaystyle n = 2, l = 1, m_l = 0, m_s = -\tfrac{1}{2}\) — possible. For \(\displaystyle n = 2\), \(\displaystyle l\) can be \(\displaystyle 0\) or \(\displaystyle 1\) (up to \(\displaystyle n - 1 = 1\)), so \(\displaystyle l = 1\) is allowed. For \(\displaystyle l = 1\), \(\displaystyle m_l\) can be \(\displaystyle -1, 0, +1\), so \(\displaystyle m_l = 0\) fits. \(\displaystyle m_s = -\tfrac{1}{2}\) is a valid spin value. This set describes a \(\displaystyle 2p\) electron.(e) \(\displaystyle n = 3, l = 3, m_l = -3, m_s = +\tfrac{1}{2}\) — not possible. For \(\displaystyle n = 3\), \(\displaystyle l\) can only go up to \(\displaystyle n - 1 = 2\), i.e. \(\displaystyle l = 0, 1, 2\). Here \(\displaystyle l = 3\) again sets \(\displaystyle l = n\), which is one past the allowed maximum. (Once \(\displaystyle l\) itself is invalid, the stated \(\displaystyle m_l = -3\) is moot — it would only have been legal for an \(\displaystyle l\) of \(\displaystyle 3\) or more, which was never a valid \(\displaystyle l\) here in the first place.)(f) \(\displaystyle n = 3, l = 1, m_l = 0, m_s = +\tfrac{1}{2}\) — possible. For \(\displaystyle n = 3\), the allowed \(\displaystyle l\) values are \(\displaystyle 0, 1, 2\), so \(\displaystyle l = 1\) fits. For \(\displaystyle l = 1\), \(\displaystyle m_l\) can be \(\displaystyle -1, 0, +1\), so \(\displaystyle m_l = 0\) fits. \(\displaystyle m_s = +\tfrac{1}{2}\) is valid. This set describes a \(\displaystyle 3p\) electron.Collecting the results: (a) fails on \(\displaystyle n\), while (c) and (e) both fail the same way — each sets \(\displaystyle l\) equal to \(\displaystyle n\) instead of stopping at \(\displaystyle n - 1\). Sets (b), (d), and (f) satisfy every rule and correspond to real electrons (\(\displaystyle 1s\), \(\displaystyle 2p\), \(\displaystyle 3p\)).Answer: (a), (c), and (e) are not possible — (a) because \(\displaystyle n = 0\) is not allowed (n must be a positive integer); (c) and (e) because each has \(\displaystyle l = n\), while \(\displaystyle l\) can only range from \(\displaystyle 0\) to \(\displaystyle n-1\). Sets (b), (d), and (f) are all valid.