NCERT Solutions · Class 11 Chemistry Classification of Elements and Periodicity in Properties

Electron gain enthalpy turns positive the moment you're adding an electron to something that is already negatively charged.Electron gain enthalpy \(\displaystyle \Delta_{eg}H \) is the energy change when a gaseous atom (or ion) accepts an extra electron. A negative value means energy is released — the process is favourable, because the electron is attracted into the atom. A positive value means energy must be supplied to force the electron in.Write out the two steps separately, because they are two different processes with two different amounts of charge:Step $\displaystyle 1$ — first electron gain: \[\text{O(g)} + e^- \longrightarrow \text{O}^-\text{(g)} \qquad \Delta_{eg}H_1 = -141\ \text{kJ mol}^{-1} \] Here a neutral oxygen atom, with its full nuclear charge unshielded by any extra electron cloud, pulls the incoming electron in. This releases energy, so \(\displaystyle \Delta_{eg}H_1 \) is negative (exothermic).Step $\displaystyle 2$ — second electron gain: \[\text{O}^-\text{(g)} + e^- \longrightarrow \text{O}^{2-}\text{(g)} \qquad \Delta_{eg}H_2 = +780\ \text{kJ mol}^{-1} \] This time the electron is not approaching a neutral atom — it is approaching \(\displaystyle \text{O}^- \), a species that is already negatively charged. The incoming electron and the existing negative charge on \(\displaystyle \text{O}^- \) repel each other electrostatically.The aside people miss: don't picture the second electron reacting with "oxygen" again — it is reacting with an anion. Once the species being added to already carries negative charge, the interaction with a further electron is repulsive, not attractive, no matter which element it is. To push the electron on against that repulsion, energy has to be put in from outside rather than being given out. That is exactly what a positive \(\displaystyle \Delta_{eg}H \) means.So the second electron gain enthalpy of oxygen is not just "less negative" than the first — it flips sign entirely and becomes positive, while the first is negative. The overall two-step process (forming \(\displaystyle \text{O}^{2-} \) from \(\displaystyle \text{O} \)) is therefore net endothermic even though it is exactly this doubly-negative oxide ion, \(\displaystyle \text{O}^{2-} \), that is stable in ionic solids like MgO — because the lattice energy released on forming the crystal more than compensates for this endothermic electron-addition step.Answer: The second electron gain enthalpy of O is positive (unlike the first, which is negative). Adding the second electron means adding it to the already-negative \(\displaystyle \text{O}^- \) ion, and the incoming electron is repelled by that existing negative charge, so energy must be supplied rather than released.

Electron gain enthalpy belongs to a lone, isolated atom; electronegativity belongs to an atom that is already bonded to another atom. These two ideas get mixed up because both describe "how much an atom wants electrons," but they answer that question in completely different situations, and one of them is a real number you can measure while the other is only a relative comparison.1. What electron gain enthalpy isElectron gain enthalpy, \(\displaystyle \Delta_{eg}H \), is defined for a single, isolated, gaseous atom taking up one extra electron: \[X(g) + e^- \rightarrow X^-(g) \] Here \(\displaystyle X(g) \) is one mole of the element as a free gas-phase atom (not bonded to anything), and \(\displaystyle \Delta_{eg}H \) is the enthalpy change of this one reaction — energy released is written negative, energy absorbed is written positive.This is a genuine, measurable thermodynamic quantity. It has fixed units, kJ/mol, and a fixed sign convention. Chlorine, for example, has a large negative electron gain enthalpy because \(\displaystyle Cl(g) + e^- \rightarrow Cl^-(g) \) releases a lot of energy — the atom has a strong pull for that extra electron when it is sitting alone in the gas phase.2. What electronegativity isElectronegativity is not a reaction and it is not something you can isolate an atom to measure. It is the tendency of an atom that is already part of a molecule, in a covalent bond, to pull the shared pair of bonding electrons toward itself.Because it describes a pull within a bond rather than a single physical process, it cannot be measured directly the way an enthalpy can. It is only ever assigned as a relative number on an arbitrary scale — the common one is Pauling's scale, where fluorine is fixed at $\displaystyle 4.0$ and every other element is ranked against it. Electronegativity therefore has no units and no defined zero; it only tells you which of two bonded atoms pulls harder, not "how many kJ" of pulling is happening.3. The basic difference, stated directlyThe aside people miss: it is easy to treat these as two words for the same "electron-hungriness," but they differ on every count that matters — Answer: Electron gain enthalpy is the measurable enthalpy change (in kJ/mol) when an isolated gaseous atom gains an extra electron, \(\displaystyle X(g)+e^-\rightarrow X^-(g) \); electronegativity is the relative, unit-less tendency of an atom that is already bonded within a molecule to attract the shared pair of bonding electrons toward itself, and it can only be compared on an arbitrary scale (e.g., Pauling), not measured directly like an enthalpy.

Electronegativity is not a fixed label stamped on an atom the way atomic number is — it depends on the hybridisation state of the atom and on what it is bonded to, so a single number cannot be "the" electronegativity of nitrogen in every compound.Electronegativity is defined as the tendency of an atom, when present in a molecule, to attract the shared pair of bonding electrons towards itself. The word "in a molecule" matters here: unlike ionisation enthalpy or electron gain enthalpy, which are properties of an isolated atom or ion, electronegativity is a property of an atom as bonded — it is not measured in isolation, and its value changes with the atom's bonding environment.Why it changes: percentage of s-character. The pull an atom exerts on a shared pair depends on how close, on average, its bonding electrons sit to the nucleus. An \(\displaystyle s\) orbital is closer to (and lower in energy than) a \(\displaystyle p\) orbital of the same shell, so a hybrid orbital with more \(\displaystyle s\)-character holds its electrons more tightly and pulls the bonding pair harder. The s-character of the common hybridisations is:\[sp \; (50\% \; s) \;>\; sp^2 \;(33\%\; s) \;>\; sp^3\;(25\%\;s) \]So, for the same atom, electronegativity rises as the hybridisation goes \(\displaystyle sp^3 \to sp^2 \to sp\).Applying this to nitrogen. Nitrogen does not always use the same hybrid orbital to bond: Because electronegativity tracks s-character, the nitrogen of a nitrile (\(\displaystyle sp\)) is more electronegative than the nitrogen of pyridine (\(\displaystyle sp^2\)), which in turn is more electronegative than the nitrogen of ammonia (\(\displaystyle sp^3\)) — even though it is "the same element" in each case. The value $\displaystyle 3.0$ on the Pauling scale that periodic tables quote for nitrogen is essentially an average, representative figure (close to nitrogen's more common, roughly \(\displaystyle sp^3\), bonding state); it is not a universal constant that holds unchanged across every compound nitrogen forms.The aside people miss: electronegativity is often mistaken for an atomic constant like ionisation enthalpy because both are tabulated as single numbers per element. But ionisation enthalpy refers to a free gaseous atom, while electronegativity explicitly refers to an atom bonded within a molecule — so it must, and does, shift with the hybridisation and the identity of the atoms it is bonded to.**Answer: The statement is not correct. Electronegativity is not an invariant atomic property; it depends on the state of hybridisation of the atom and on the nature of the other atoms bonded to it. The quoted value of $\displaystyle 3.0$ (Pauling scale) is an average value for nitrogen. In practice, nitrogen's electronegativity increases with the s-character of its hybrid orbital — it is lowest when nitrogen is \(\displaystyle sp^3\)-hybridised (as in \(\displaystyle NH_3\)), higher when \(\displaystyle sp^2\)-hybridised, and highest when \(\displaystyle sp\)-hybridised (as in a nitrile, \(\displaystyle -C\equiv N\)) — so a single value of $\displaystyle 3.0$ cannot hold for nitrogen in all its compounds.

Atomic size is a tug-of-war between the nuclear charge pulling electrons in and electron–electron repulsion pushing them apart — changing the electron count without changing the nuclear charge tips that balance.The nucleus of an atom has a fixed number of protons, so its nuclear charge \(\displaystyle Z \) does not change when the atom gains or loses an electron. What changes is the number of electrons around that same nucleus, and with it, two things: how strongly each electron is, on average, pulled inward (the effective nuclear charge, \(\displaystyle Z_{eff} \)), and how much the electrons repel each other.(a) When an atom gains an electron (forms an anion)The number of protons stays the same, but one more electron is added to the outermost shell. Both effects work in the same direction: the electron cloud expands. This is why an anion is always larger than the neutral atom it came from — for example, a fluorine atom (\(\displaystyle \sim 72\ \text{pm} \)) becomes the much bigger fluoride ion \(\displaystyle \text{F}^- \) (\(\displaystyle \sim 136\ \text{pm} \)), because the same $\displaystyle 9$ protons must now hold onto $\displaystyle 10$ electrons instead of 9.A common mistake here is to think the ion is bigger only because "it has one more shell" — for \(\displaystyle \text{F} \to \text{F}^- \) there is no new shell at all; the growth comes purely from weaker per-electron pull and greater repulsion.(b) When an atom loses an electron (forms a cation)The number of protons again stays the same, but one electron is removed, usually from the outermost occupied subshell. Both effects — stronger pull per electron, and sometimes an entire shell disappearing — make the electron cloud contract. This is why a cation is always smaller than the neutral atom it came from: sodium (\(\displaystyle \sim 186\ \text{pm} \)) shrinks to \(\displaystyle \text{Na}^+ \) (\(\displaystyle \sim 95\ \text{pm} \)) because losing the single \(\displaystyle 3s \) electron leaves behind only the tightly-held \(\displaystyle n = 1, 2 \) shells, now pulled in by $\displaystyle 11$ protons instead of being shielded by a \(\displaystyle 3s \) electron.Answer: In both cases the nuclear charge is unchanged, so the size change comes entirely from the electron–electron balance: gaining an electron increases repulsion and lowers the effective nuclear charge per electron, so the electron cloud expands and the anion is larger than the atom; losing an electron reduces repulsion and raises the effective nuclear charge per electron (and can remove an entire outer shell), so the electron cloud contracts and the cation is smaller than the atom.

Ionization enthalpy depends on the nuclear charge and the electron arrangement around the nucleus — not on the number of neutrons.Isotopes of an element are atoms with the same atomic number (same number of protons, hence the same nuclear charge \(\displaystyle Z\)) and the same number of electrons, but different numbers of neutrons (different mass number \(\displaystyle A\)).The first ionization enthalpy, \(\displaystyle \Delta_iH_1\), is the energy needed for the process\[\text{X(g)} \rightarrow \text{X}^+\text{(g)} + e^- \]and its size is governed by two things: Neither of these depends on the number of neutrons. Extra neutrons add mass to the nucleus, but they carry no charge, so they do not change \(\displaystyle Z_{\text{eff}}\), and since the number of electrons is unchanged, the electronic configuration (and hence the size and shape of the electron cloud) is also unchanged.This is the point students often trip on: it is tempting to think a "heavier" isotope should hold its electrons more tightly, but chemical/electronic properties like ionization enthalpy are controlled by charge and electron arrangement, not by mass. (A tiny mass effect on ionization energy does exist in very precise physics — because the nucleus is not infinitely heavy compared to the electron — but it is far too small to matter at this level and is not what the question is testing.)Since all isotopes of an element have identical proton number, identical electron count, and identical electronic configuration, they present the same effective nuclear charge to the electron being removed.**Answer: The first ionization enthalpies of the isotopes of an element are expected to be (essentially) the same, because ionization enthalpy depends on the nuclear charge and the electronic configuration of the atom, and isotopes of the same element have identical atomic number (protons) and identical number of electrons — they differ only in the number of neutrons, which does not affect these factors.

The core difference is electron loss versus electron gain — everything else about metals and non-metals follows from that one fact.Across a period, atomic size shrinks and the nuclear pull on the outer electrons grows stronger, which pushes ionization enthalpy up and pulls electron gain enthalpy down (more negative). Metals sit on the low-ionization-enthalpy side of this trend; non-metals sit on the high-electron-affinity side. That single positional fact drives every difference below.1. Ionization enthalpy \(\displaystyle \Delta_i H \) \(\displaystyle \Delta_i H \) is the energy needed to remove the outermost electron from an isolated gaseous atom, \(\displaystyle M(g) \rightarrow M^+(g) + e^- \).Metals have low ionization enthalpy — their valence electrons are loosely held (few electrons in the outer shell, larger atomic size, more shielding), so they lose electrons easily to form cations. Non-metals have high ionization enthalpy — the outer electrons are close to a nearly-full shell and strongly attracted by the nucleus, so removing an electron costs a lot of energy.A step people skip: ionization enthalpy is about a gaseous, isolated atom — it says nothing about the solid's melting point or hardness, which is a separate (metallic-bonding) property.2. Electron gain enthalpy \(\displaystyle \Delta_{eg} H \) \(\displaystyle \Delta_{eg} H \) is the energy released when an electron is added to a gaseous atom, \(\displaystyle M(g) + e^- \rightarrow M^-(g) \).Non-metals have large negative (highly exothermic) electron gain enthalpy — adding an electron completes or nearly completes a stable octet, so the atom "wants" the electron and releases energy doing so. Metals have small negative or even positive electron gain enthalpy — adding an electron to an atom that would rather lose electrons is energetically unfavourable.3. Electronegativity Electronegativity is the tendency of an atom, when bonded, to pull shared electron density toward itself. Non-metals are highly electronegative (F is the highest at $\displaystyle 4.0$ on the Pauling scale); metals are electropositive (they push electron density away rather than pulling it in). This is why, in a bond between a metal and a non-metal, the electron pair ends up almost entirely on the non-metal — the bond becomes ionic.4. Type of ion / compound formed Because metals lose electrons and non-metals gain them, metals form cations (\(\displaystyle Na \rightarrow Na^+ + e^- \)) and non-metals form anions (\(\displaystyle Cl + e^- \rightarrow Cl^- \)). A metal and a non-metal together therefore typically form an ionic compound (e.g., \(\displaystyle NaCl \)), held together by electrostatic attraction between the oppositely charged ions; two non-metals together form a covalent compound (e.g., \(\displaystyle Cl_2, CO_2 \)) by sharing electrons, since neither side can afford to fully give up an electron.5. Nature of oxides Metal oxides are predominantly basic (some, like \(\displaystyle Al_2O_3 \), are amphoteric) — they react with acids to form salt and water, e.g., \(\displaystyle MgO + 2HCl \rightarrow MgCl_2 + H_2O \). Non-metal oxides are predominantly acidic — they react with bases, e.g., \(\displaystyle CO_2 + 2NaOH \rightarrow Na_2CO_3 + H_2O \). This basic-versus-acidic split is a direct chemical consequence of electropositive versus electronegative character.6. Physical properties Metals are typically lustrous, malleable (can be hammered into sheets), ductile (can be drawn into wires), and good conductors of heat and electricity, because their loosely held valence electrons form a delocalised "electron sea" that moves freely through the lattice. Non-metals (in the solid state) are typically brittle, non-lustrous, and poor conductors (with exceptions such as graphite), because their electrons are localised in covalent bonds rather than free to move.7. Reducing versus oxidizing behaviour Because metals lose electrons readily, they act as reducing agents in reactions (they get oxidized while reducing something else). Because non-metals gain electrons readily, they act as oxidizing agents (they get reduced while oxidizing something else).Answer: Metals have low ionization enthalpy and are electropositive, so they lose electrons to form cations; they act as reducing agents, form basic oxides, and are typically lustrous, malleable, ductile solids that conduct heat and electricity well (via delocalised electrons). Non-metals have high ionization enthalpy and large negative electron gain enthalpy and are electronegative, so they gain electrons to form anions; they act as oxidizing agents, form acidic oxides, and are typically brittle, non-lustrous, poor conductors. Metal + non-metal combinations tend to form ionic compounds; non-metal + non-metal combinations form covalent compounds.

The number of electrons in an atom's outermost (valence) shell is what fixes its group, and the group is what fixes whether it gains, loses, or simply sits at a certain physical state. Work each part off the valence-shell configuration, not off memorised element names.(a) An element with five electrons in the outer shellGroup $\displaystyle 15$ (the nitrogen family — N, P, As, Sb, Bi) has the general valence configuration \(\displaystyle ns^2\,np^3 \), where \(\displaystyle n \) is the outermost shell number, \(\displaystyle s \) and \(\displaystyle p \) name the subshells, and the superscripts count electrons in each. Adding the superscripts, \(\displaystyle 2+3=5 \), so every member of this group carries five electrons in its outer shell.Nitrogen is the clearest example: \(\displaystyle Z=7 \), configuration \(\displaystyle 1s^2\,2s^2\,2p^3 \). Its outermost shell (\(\displaystyle n=2 \)) holds \(\displaystyle 2s^2\,2p^3 = 5 \) electrons.(b) An element that would tend to lose two electronsThis is a Group $\displaystyle 2$ element (the alkaline earth metals — Be, Mg, Ca, Sr, Ba), general configuration \(\displaystyle ns^2 \). An atom loses electrons when doing so gets it to the nearest noble-gas configuration cheaply — and for these elements that means shedding just the two \(\displaystyle ns^2 \) electrons.Take calcium: \(\displaystyle Z=20 \), configuration \(\displaystyle 1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,4s^2 \). Losing the two \(\displaystyle 4s \) electrons leaves \(\displaystyle \text{Ca}^{2+} \) with the argon configuration \(\displaystyle 1s^2\,2s^2\,2p^6\,3s^2\,3p^6 \) — a filled shell, which is why the loss happens readily (Group $\displaystyle 2$ metals have low second ionisation enthalpies relative to what a further electron loss would cost). Metals lose electrons; the mistake to avoid is reaching for a non-metal here because the question says "lose."(c) An element that would tend to gain two electronsThis is a Group $\displaystyle 16$ element (the oxygen family — O, S, Se, Te), general configuration \(\displaystyle ns^2\,np^4 \). With six electrons already in the valence shell, only two more are needed to complete the octet, so these atoms gain electrons rather than lose them — the opposite pull from part (b).Oxygen illustrates it: \(\displaystyle Z=8 \), configuration \(\displaystyle 1s^2\,2s^2\,2p^4 \). Gaining two electrons gives \(\displaystyle \text{O}^{2-} \) with configuration \(\displaystyle 1s^2\,2s^2\,2p^6 \), the neon configuration. The rule of thumb: elements with $\displaystyle 1$–$\displaystyle 3$ valence electrons (metals, left side) lose electrons; elements with $\displaystyle 5$–$\displaystyle 7$ valence electrons (non-metals, right side) gain electrons — both moving toward the nearest noble gas.(d) The group with a metal, a non-metal, a liquid, and a gas togetherThis is Group $\displaystyle 17$, the halogens: F, Cl, Br, I, At. Down this one group, physical state changes steadily with increasing atomic size and strengthening interatomic forces: So Group $\displaystyle 17$ alone spans gas, liquid, non-metallic solid, and metal — no other group covers all four states across its members. The aside worth flagging: it is tempting to look for this variety across a period instead of a group, because state usually feels like a period-driven property; here it is the group that supplies the full spread.Answer: (a) Nitrogen family, Group $\displaystyle 15$ (e.g., N: \(\displaystyle 2s^2 2p^3\), $\displaystyle 5$ outer electrons); (b) Group $\displaystyle 2$ alkaline earth metals (e.g., Ca, config \(\displaystyle ns^2\), loses $\displaystyle 2$ e⁻ to reach noble-gas configuration); (c) Group $\displaystyle 16$ oxygen family (e.g., O, config \(\displaystyle ns^2np^4\), gains $\displaystyle 2$ e⁻ to complete octet); (d) Group $\displaystyle 17$, the halogens — F and Cl are gases, Br is a liquid, I is a non-metallic solid, and At shows metallic character.

Reactivity here means two different things for the two groups: group $\displaystyle 1$ atoms react by LOSING an electron, group $\displaystyle 17$ atoms react by GAINING one — so the same trend down a group (increasing atomic size) pushes them in opposite directions.Group $\displaystyle 1$: reactivity is the ease of losing the outer electronAlkali metals have the configuration \(\displaystyle ns^1 \). To react, an atom must give up this lone outer electron to form \(\displaystyle M^+ \): \[M(g) \rightarrow M^+(g) + e^- \] The metal is acting as a reducing agent, and how easily it does this is measured by the ionization enthalpy, \(\displaystyle \Delta_i H \) — the energy needed to pull the outermost electron off a gaseous atom. A low ionization enthalpy means the electron leaves easily, so the metal is more reactive.Going down the group, Li → Na → K → Rb → Cs, a new shell is added each time, so: A far-away, well-shielded electron is held less tightly, so \(\displaystyle \Delta_i H \) falls steadily down the group. Losing the electron becomes progressively easier, so reactivity increases: Li < Na < K < Rb < Cs.Group $\displaystyle 17$: reactivity is the ease of gaining an electronHalogens have the configuration \(\displaystyle ns^2\,np^5 \) — one electron short of a stable octet. To react, an atom must pick up an electron to form \(\displaystyle X^- \): \[X(g) + e^- \rightarrow X^-(g) \] Here the atom acts as an oxidizing agent, and the relevant quantity is the electron gain enthalpy, \(\displaystyle \Delta_{eg}H \) — the energy released when a gaseous atom accepts an electron. A more negative \(\displaystyle \Delta_{eg}H \) means the electron is pulled in more readily, so the atom is more reactive.This is the step people mix up with group $\displaystyle 1$: for group $\displaystyle 1$ you want ionization enthalpy to go DOWN for reactivity to go up; for group $\displaystyle 17$ you want electron gain enthalpy to be MORE negative for reactivity to go up — opposite requirements, from opposite processes.Going down group $\displaystyle 17$, F → Cl → Br → I, atomic size again increases and the extra inner shells shield the nucleus, so an incoming electron is attracted less strongly even though the nuclear charge is bigger. Electron gain becomes less exothermic (less favourable), so accepting an electron gets progressively harder, and reactivity decreases: F > Cl > Br > I.A short aside on fluorine. By electron gain enthalpy alone, chlorine actually attracts an extra electron slightly more strongly than fluorine does — fluorine's 2p subshell is so small that the incoming electron suffers noticeable repulsion from the electrons already there, making \(\displaystyle \Delta_{eg}H \) for F slightly less negative than for Cl. Yet fluorine is still the most reactive halogen overall, because reactivity is decided by the whole reaction, not electron gain enthalpy alone: the F–F bond is unusually weak (easy to break) and the \(\displaystyle F^- \) ion is very strongly hydrated (releases a lot of energy in solution). Both effects favour fluorine so strongly that it wins out despite the electron-gain-enthalpy hiccup.Answer: Group $\displaystyle 1$ metals react by losing their one valence electron; down the group (Li → Cs) atomic size and shielding increase, so ionization enthalpy falls and the electron is lost more easily — reactivity rises: Li < Na < K < Rb < Cs. Group $\displaystyle 17$ non-metals react by gaining one electron; down the group (F → I) atomic size increases in the same way, so the nucleus attracts an incoming electron less strongly, electron gain enthalpy becomes less favourable, and the electron is gained less easily — reactivity falls: F > Cl > Br > I. The two orders run opposite ways because "reactive" means losing an electron in one group and gaining one in the other, and increasing atomic size makes the first easier but the second harder.

The block tells you which subshell is being filled last — s, p, d, or f — and "outer" configuration means you write only that subshell plus the ones already complete below it, using \(\displaystyle n\) for the outermost shell.Recall how the periodic table's four blocks are defined: each block is named for the subshell that receives the last electron as you build up the atom (the Aufbau order). So to write a "general" configuration for a block, ask: which subshell is filling, and how many electrons can it hold?s-blockThe last electron enters the \(\displaystyle ns\) subshell, and an s subshell holds at most $\displaystyle 2$ electrons. These are the alkali metals (Group $\displaystyle 1$, one outer electron) and alkaline earth metals (Group $\displaystyle 2$, two outer electrons), so the outer shell can have $\displaystyle 1$ or $\displaystyle 2$ electrons in it: \[ns^{1-2} \] Here \(\displaystyle n\) is the outermost (valence) shell number, and the superscript range \(\displaystyle 1\text{–}2\) means "$\displaystyle 1$ for Group $\displaystyle 1$, $\displaystyle 2$ for Group $\displaystyle 2$" — it is not one atom having both at once.p-blockThe last electron enters the \(\displaystyle np\) subshell, but the \(\displaystyle ns\) subshell below it is already full ($\displaystyle 2$ electrons) because s always fills before p in the same shell. A p subshell holds at most $\displaystyle 6$ electrons, giving Groups $\displaystyle 13$–$\displaystyle 18$: \[ns^{2}\,np^{1-6} \] The aside to catch here: helium (\(\displaystyle 1s^2\)) looks s-block by its configuration, but it is placed with the p-block (Group $\displaystyle 18$) because it behaves as a noble gas — configuration alone doesn't always fix the block placement for helium.d-block (transition elements)Here the electron being added last goes into the \(\displaystyle (n-1)d\) subshell — one shell inside the outermost one — while the outermost \(\displaystyle ns\) subshell was already filled at the previous element. A d subshell holds up to $\displaystyle 10$ electrons, and \(\displaystyle ns\) can be $\displaystyle 2$ (the normal case) or drop to $\displaystyle 0$ or $\displaystyle 1$ for a few exceptions (like Cr, Cu) where a half-filled or fully-filled d subshell is extra stable and pulls an electron out of \(\displaystyle ns\): \[(n-1)d^{1-10}\,ns^{0-2} \] The step people get wrong: the d electrons are one shell number behind the outermost shell — it is \(\displaystyle (n-1)d\), not \(\displaystyle nd\) — even though this is called the outermost/valence region of a transition metal.f-block (inner transition elements)The last electron goes into the \(\displaystyle (n-2)f\) subshell — two shells inside the outermost one — while \(\displaystyle (n-1)d\) and \(\displaystyle ns\) are already occupied from the elements before it. An f subshell holds up to $\displaystyle 14$ electrons: \[(n-2)f^{1-14}\,(n-1)d^{0-1}\,ns^{2} \] Again the subshell being filled sits two shells behind the label \(\displaystyle n\); this is what makes the lanthanoids and actinoids so similar to each other across a row — the differentiating electron is buried two shells deep and barely affects chemical behaviour, which depends mainly on the outer \(\displaystyle ns^2\).Answer: s-block: \(\displaystyle ns^{1-2}\); p-block: \(\displaystyle ns^{2}\,np^{1-6}\); d-block: \(\displaystyle (n-1)d^{1-10}\,ns^{0-2}\); f-block: \(\displaystyle (n-2)f^{1-14}\,(n-1)d^{0-1}\,ns^{2}\), where \(\displaystyle n\) is the number of the outermost shell.

The period number is just the value of \(\displaystyle n \) in the given configuration (the outermost shell being filled); the group number is worked out by a different counting rule for each block, and the whole trap in this question is applying the wrong block's rule.Before assigning positions, fix the four rules: In every case, the period number equals the highest \(\displaystyle n \) appearing in the configuration.(i) \(\displaystyle ns^{2}np^{4} \) for \(\displaystyle n=3 \): this is \(\displaystyle 3s^{2}3p^{4} \). Writing out the full configuration, \(\displaystyle 1s^{2}2s^{2}2p^{6}3s^{2}3p^{4} \), gives $\displaystyle 16$ electrons, i.e. sulfur.Period \(\displaystyle = n = 3 \).This is a \(\displaystyle p \)-block configuration, so group \(\displaystyle = 10 + (2+4) = 16 \).Position: Period $\displaystyle 3$, Group $\displaystyle 16$ (the oxygen family) — the element is sulfur, S.(ii) \(\displaystyle (n-1)d^{2}ns^{2} \) for \(\displaystyle n=4 \): this is \(\displaystyle 3d^{2}4s^{2} \). Full configuration \(\displaystyle [\mathrm{Ar}]3d^{2}4s^{2} \), Z = $\displaystyle 22$, titanium.Period \(\displaystyle = n = 4 \).This is a \(\displaystyle d \)-block configuration, so group \(\displaystyle = (\text{electrons in } 3d) + (\text{electrons in } 4s) = 2 + 2 = 4 \).This is the step people get wrong here — it is tempting to reuse the p-block "+$\displaystyle 10$" rule on a d-block count and get group $\displaystyle 16$ by mistake. The +$\displaystyle 10$ rule only applies once the p-subshell is the one being filled; for a d-block element you just add up the \(\displaystyle (n-1)d\) and \(\displaystyle ns\) electrons directly, with no offset.Position: Period $\displaystyle 4$, Group $\displaystyle 4$ — the element is titanium, Ti, in the first transition series.(iii) \(\displaystyle (n-2)f^{7}(n-1)d^{1}ns^{2} \) for \(\displaystyle n=6 \): this is \(\displaystyle 4f^{7}5d^{1}6s^{2} \). Full configuration \(\displaystyle [\mathrm{Xe}]4f^{7}5d^{1}6s^{2} \), Z = $\displaystyle 64$, gadolinium.Period \(\displaystyle = n = 6 \).This is an \(\displaystyle f \)-block configuration (the \(\displaystyle 4f \) subshell is occupied), so it belongs to the lanthanoid series and, by the rule above, sits in the group $\displaystyle 3$ slot — not group $\displaystyle 17$ or any number obtained by counting \(\displaystyle 7+1+2 \) electrons and adding 10. The f-block elements of a given period all share that one group-$\displaystyle 3$ position in the main table; what distinguishes them from each other is only the period ($\displaystyle 6$ for lanthanoids, $\displaystyle 7$ for actinoids) and their place along the separately-drawn f-block row, not a unique group number.Position: Period $\displaystyle 6$, Group $\displaystyle 3$, f-block (lanthanoid series) — the element is gadolinium, Gd.Answer: (i) \(\displaystyle 3s^23p^4\) is period $\displaystyle 3$, group $\displaystyle 16$ (sulfur, S). (ii) \(\displaystyle 3d^24s^2\) is period $\displaystyle 4$, group $\displaystyle 4$ (titanium, Ti). (iii) \(\displaystyle 4f^75d^16s^2\) is period $\displaystyle 6$, group $\displaystyle 3$, in the f-block lanthanoid series (gadolinium, Gd).