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

A big first ionization enthalpy tells you an atom hates losing an electron; a very negative electron gain enthalpy tells you it loves gaining one — read the table as a story about each atom's electron count, not as six unrelated numbers.Here \(\displaystyle \Delta_iH_1 \) is the energy needed to pull the first electron off a neutral gaseous atom, \(\displaystyle \Delta_iH_2 \) is the energy needed to pull a second electron off the resulting +$\displaystyle 1$ ion, and \(\displaystyle \Delta_{eg}H \) is the energy change when the neutral atom gains one electron (negative = energy is released = the atom "wants" the electron; positive = energy must be put in = the atom resists).Step $\displaystyle 1$ — sort metals from non-metals using \(\displaystyle \Delta_iH_1 \). Metals give up an electron easily, so they have low \(\displaystyle \Delta_iH_1 \); non-metals hold on tightly, so they have high \(\displaystyle \Delta_iH_1 \). Step $\displaystyle 2$ — the size of the jump \(\displaystyle \Delta_iH_2/\Delta_iH_1 \) tells you how many electrons come off easily.A big jump means: once the first electron is gone, the ion has a full (noble-gas-like) shell, so removing a second electron is far harder — that is alkali-metal (Group $\displaystyle 1$) behaviour, forming \(\displaystyle M^+ \). A small, comparable jump means both electrons come off with similar difficulty — that is alkaline-earth (Group $\displaystyle 2$) behaviour, forming \(\displaystyle M^{2+} \).\[\frac{\Delta_iH_2}{\Delta_iH_1}:\quad \text{I} = \frac{7300}{520}\approx 14,\qquad \text{II} = \frac{3051}{419}\approx 7.3,\qquad \text{VI} = \frac{1451}{738}\approx 2.0 \]I and II show a huge jump (≈$\displaystyle 14$× and ≈$\displaystyle 7$×) — both are Group $\displaystyle 1$ metals, each stopping at \(\displaystyle M^+ \). VI shows only a ~$\displaystyle 2$× jump — both electrons are lost with similar (moderate) ease, the classic Group $\displaystyle 2$ signature, stopping at \(\displaystyle M^{2+} \).Step $\displaystyle 3$ — among the non-metals, the sign and size of \(\displaystyle \Delta_{eg}H \) separates "wants an electron" from "already has enough." (A slip people make here: don't count V as "a non-metal that just doesn't react" — a positive \(\displaystyle \Delta_{eg}H \) is a completely different chemical story, filled shell, from a small negative one, which is still an atom short of a filled shell.)Step $\displaystyle 4$ — among the metals, lower \(\displaystyle \Delta_iH_1 \) means a bigger atom holding its outer electron more loosely, i.e. a more reactive metal.I ($\displaystyle 520$) and II ($\displaystyle 419$) are both Group-$\displaystyle 1$-type metals from Step $\displaystyle 2$, but II needs less energy to lose its electron than I does. Down a group, atomic radius increases and the outer electron sits farther from the nucleus, so ionization enthalpy falls — II is therefore the larger, more metallic, more reactive atom: II is the most reactive metal — part (b).Step $\displaystyle 5$ — which metal forms the ionic \(\displaystyle MX_2 \), and which forms the covalent \(\displaystyle MX \)?VI is the Group-$\displaystyle 2$-type metal from Step $\displaystyle 2$ (loses two electrons with comparable, moderate energy cost each time), so it forms a stable \(\displaystyle M^{2+} \) ion and an ordinary ionic halide of formula \(\displaystyle MX_2 \): VI is the metal forming the stable binary halide \(\displaystyle MX_2 \) — part (e).That leaves I and II, the two Group-$\displaystyle 1$-type metals, to explain part (f). Both stop at \(\displaystyle M^+ \) and would normally give an ionic \(\displaystyle MX \) salt — but I has the higher \(\displaystyle \Delta_iH_1 \) of the two ($\displaystyle 520$ vs. $\displaystyle 419$), and within the same group ionization enthalpy is higher precisely when the atom (and hence the resulting cation) is smaller. A small, highly charged cation has a high charge density and strongly distorts (polarizes) the electron cloud of the halide ion sitting next to it — Fajans' rule — and enough polarization turns what would be an ionic bond into one with substantial covalent character. Because I is the smaller of the two \(\displaystyle M^+ \)-forming metals, its halide \(\displaystyle MX \) is the one with dominant covalent character (this is exactly why, for instance, lithium halides behave more like covalent compounds — soluble in organic solvents — while the halides of bigger alkali metals stay firmly ionic): I is the metal forming the predominantly covalent halide \(\displaystyle MX \) — part (f).(As a check, these six rows are in fact the standard data-book values for Li, K, F, I, He and Mg respectively — I = Li, II = K, III = F, IV = I, V = He, VI = Mg — which is exactly the classification the reasoning above arrives at independently.)Answer: (a) V — highest ionization enthalpy and a positive electron gain enthalpy mark a noble gas, the least reactive element. (b) II — the lowest ionization enthalpy among the Group-$\displaystyle 1$-type metals, so the most reactive metal. (c) III — the most negative electron gain enthalpy, so the strongest pull on an extra electron, the most reactive non-metal. (d) IV — still a genuine non-metal (negative electron gain enthalpy) but with a smaller pull on an electron than III, so the least reactive non-metal. (e) VI — the Group-$\displaystyle 2$-type metal (comparable, moderate first and second ionization enthalpies) forms the stable ionic halide \(\displaystyle MX_2 \). (f) I — the smaller of the two Group-$\displaystyle 1$-type metals (higher \(\displaystyle \Delta_iH_1 \) than II), whose small, more polarizing \(\displaystyle M^+ \) ion gives its halide \(\displaystyle MX \) predominantly covalent character.

The formula of a binary compound is fixed by one rule: the total positive charge on the metal atoms must exactly cancel the total negative charge on the non‑metal atoms.To use that rule you need, for each element, the charge it settles into when it empties or fills its outer shell. For a representative (main‑group) element that charge is read straight off the periodic table, using the old style group labels IA–VIIA where the group number itself equals the number of valence electrons: A short aside on the mistake people make here: if you instead use the modern $\displaystyle 1$–$\displaystyle 18$ numbering, group $\displaystyle 13$ is not "charge +13." For p‑block groups you must subtract $\displaystyle 10$ (or count valence electrons directly) to get the real charge — group $\displaystyle 13$ gives \(\displaystyle +3\), not \(\displaystyle +13\).Once both charges are known, formulas are built by the criss‑cross rule: the number of one atom in the formula equals the (sign‑dropped) charge of the other atom, and the result is then reduced to the smallest whole‑number ratio.(a) Lithium and oxygen. Li is group $\displaystyle 1$, so \(\displaystyle \text{Li}^{+}\). O is group $\displaystyle 16$, so \(\displaystyle \text{O}^{2-}\). Criss‑crossing the charges gives \(\displaystyle \text{Li}_2\text{O}_1\), written \(\displaystyle \text{Li}_2\text{O}\).(b) Magnesium and nitrogen. Mg is group $\displaystyle 2$, so \(\displaystyle \text{Mg}^{2+}\). N is group $\displaystyle 15$, so \(\displaystyle \text{N}^{3-}\). Criss‑crossing gives \(\displaystyle \text{Mg}_3\text{N}_2\) — already in lowest terms, since $\displaystyle 3$ and $\displaystyle 2$ share no common factor.(c) Aluminium and iodine. Al is group $\displaystyle 13$, so \(\displaystyle \text{Al}^{3+}\). I is group $\displaystyle 17$, so \(\displaystyle \text{I}^{-}\). Criss‑crossing gives \(\displaystyle \text{AlI}_3\).(d) Silicon and oxygen. Si is group 14. Against a highly electronegative partner like oxygen, silicon loses all four valence electrons rather than gaining four, giving \(\displaystyle \text{Si}^{4+}\). With \(\displaystyle \text{O}^{2-}\), criss‑crossing gives \(\displaystyle \text{Si}_2\text{O}_4\), which reduces to \(\displaystyle \text{SiO}_2\).(e) Phosphorus and fluorine — the step where a plain group-number reading goes wrong. Reading straight off group $\displaystyle 15$ would suggest phosphorus only ever takes the \(\displaystyle -3\)/\(\displaystyle +3\) route, giving \(\displaystyle \text{PF}_3\). But phosphorus sits in period $\displaystyle 3$, so it has empty \(\displaystyle 3d\) orbitals available and is not boxed in by the strict octet the way nitrogen (period $\displaystyle 2$, no accessible d‑orbitals) is. That lets phosphorus reach its full group oxidation state of \(\displaystyle +5\). Fluorine is exactly the partner that pushes it there: it is the smallest, most electronegative halogen, small enough to pack five atoms around one phosphorus and electronegative enough to pull out and stabilize that higher \(\displaystyle +5\) state. (This is also why the parallel compound with the bulkier iodine stops at \(\displaystyle \text{PI}_3\).) So the stable compound formed is \(\displaystyle \text{PF}_5\), phosphorus pentafluoride, not \(\displaystyle \text{PF}_3\).(f) Element $\displaystyle 71$ and fluorine. Element $\displaystyle 71$ is lutetium, Lu, with configuration \(\displaystyle [\text{Xe}]\,4f^{14}\,5d^{1}\,6s^{2}\). Its $\displaystyle 14$ f‑electrons form a filled, chemically inert \(\displaystyle 4f^{14}\) core, so lutetium behaves like a group‑$\displaystyle 3$ element: it loses only the two \(\displaystyle 6s\) electrons and the one \(\displaystyle 5d\) electron, giving the single stable oxidation state \(\displaystyle +3\), i.e. \(\displaystyle \text{Lu}^{3+}\). Criss‑crossing with \(\displaystyle \text{F}^{-}\) gives \(\displaystyle \text{LuF}_3\).Answer: (a) \(\displaystyle \text{Li}_2\text{O}\) (b) \(\displaystyle \text{Mg}_3\text{N}_2\) (c) \(\displaystyle \text{AlI}_3\) (d) \(\displaystyle \text{SiO}_2\) (e) \(\displaystyle \text{PF}_5\) (f) \(\displaystyle \text{LuF}_3\)

A period is a horizontal row of the periodic table, and moving across it means adding electrons to the same shell — the same value of \(\displaystyle n \).The modern periodic table is arranged on the basis of electronic configuration, following the Aufbau principle (electrons fill orbitals in order of increasing energy) and building on Bohr's idea of shells labelled by the principal quantum number \(\displaystyle n = 1, 2, 3, \dots \).Go through what the row number actually tracks: In each case, the period number is the value of \(\displaystyle n \) for the outermost (valence) shell. That is the entire logic of why a new row starts: a new principal shell has begun.Now check why the other options fail: So the row (period) is fixed by the principal quantum number of the outermost shell, while the block (s, p, d, f) is fixed by the azimuthal quantum number of the subshell being filled.Answer: (c) principal quantum number — the period number equals the principal quantum number \(\displaystyle n \) of the outermost shell being filled.

A subshell's width is fixed by \(\displaystyle 2(2l+1) \): count the orbitals, double it for spin — that number is also how many columns wide its block is.Here \(\displaystyle l\) is the azimuthal (subshell) quantum number, the number of orbitals in a subshell is \(\displaystyle (2l+1)\), and each orbital holds at most $\displaystyle 2$ electrons (opposite spins, Pauli exclusion) — so the maximum electron capacity of a subshell is \[2(2l+1) \]Apply this subshell by subshell, since each period is built by filling one subshell across the periodic table: The trap in statement (b) is treating "$\displaystyle 8$" as some kind of natural cap on a subshell (perhaps borrowed from the octet rule for s+p electrons together). A d-subshell has $\displaystyle 5$ orbitals, not $\displaystyle 4$, so its true capacity is \(\displaystyle 2 \times 5 = 10\), and the d-block correspondingly spans $\displaystyle 10$ columns, not 8. So (b) is wrong on both the stated number of columns and the stated reason.Checking the other statements against this same rule: Only (b) fails the test, since the d-subshell's real capacity is $\displaystyle 10$ electrons ($\displaystyle 5$ orbitals \(\displaystyle \times\) $\displaystyle 2$), which makes the d-block $\displaystyle 10$ columns wide (groups $\displaystyle 3$–$\displaystyle 12$), not 8.Answer: Statement (b) is incorrect — a d-subshell has $\displaystyle 5$ orbitals, so its maximum capacity is \(\displaystyle 2(2\cdot2+1) = 10\) electrons, not $\displaystyle 8$, and the d-block is accordingly $\displaystyle 10$ columns wide (groups $\displaystyle 3$–$\displaystyle 12$).

Chemistry depends on how strongly the nucleus holds the valence electrons — and nuclear mass has nothing to do with that pull.Go through each factor and ask: does it change the force the valence electrons feel, or the shell they sit in?(a) Valence principal quantum number \(\displaystyle n\) This number fixes which shell is the valence shell. A larger \(\displaystyle n\) means the valence electrons are farther from the nucleus and more loosely held (lower ionisation enthalpy, larger radius). Changing \(\displaystyle n\) changes the whole chemistry of the element — think of the jump from Li (\(\displaystyle n=2\)) to Na (\(\displaystyle n=3\)): very different reactivity. So \(\displaystyle n\) does affect the valence shell.(b) Nuclear charge \(\displaystyle Z\) \(\displaystyle Z\) is the number of protons, i.e. the raw positive charge pulling on every electron. A bigger \(\displaystyle Z\) pulls the valence shell in more tightly (this is exactly why atomic radius shrinks and ionisation enthalpy rises across a period). So \(\displaystyle Z\) does affect the valence shell.(d) Number of core electrons The core (inner-shell) electrons sit between the nucleus and the valence shell and screen — partly cancel — the nuclear charge felt by the valence electrons. This screening is captured in the idea of effective nuclear charge, \[Z_{\text{eff}} = Z - \sigma, \] where \(\displaystyle Z\) is the actual nuclear charge and \(\displaystyle \sigma\) is the shielding contributed by the core electrons. More core electrons means more shielding, so the valence electrons feel a smaller pull. So the number of core electrons does affect the valence shell.(c) Nuclear mass Nuclear mass comes from the number of protons and neutrons. Neutrons carry no charge, so adding or removing neutrons changes the mass number but leaves \(\displaystyle Z\), the electron configuration, and the shielding \(\displaystyle \sigma\) completely unchanged. This is exactly why isotopes of an element — same \(\displaystyle Z\), different mass — are chemically identical: \(\displaystyle ^{1}_{1}\text{H}\), \(\displaystyle ^{2}_{1}\text{H}\) (deuterium), and \(\displaystyle ^{3}_{1}\text{H}\) (tritium) all react the same way, because chemistry runs on electron behaviour, not on how heavy the nucleus is. So nuclear mass is the one factor that does not touch the valence shell.The aside people miss: it is easy to lump "nuclear charge" and "nuclear mass" together as one "nucleus" property, but for chemistry only the charge (and its shielding) matters — the mass is irrelevant to how electrons are held.Answer: (c) Nuclear mass does not affect the valence shell — chemical behaviour depends on the principal quantum number of the valence shell, the nuclear charge, and the shielding by core electrons (i.e., \(\displaystyle Z_{\text{eff}} = Z - \sigma\)), none of which involve the mass of the nucleus. This is also why isotopes of an element, which differ only in nuclear mass, show identical chemical properties.

Isoelectronic species have the same number of electrons, so size differences among them come down to how hard the nucleus is pulling on that fixed electron count.Check what F⁻, Ne, and Na⁺ have in common first, because that tells you which options can immediately be ruled out.Each species has $\displaystyle 10$ electrons: \[\text{F}^- : 9 + 1 = 10, \qquad \text{Ne} : 10, \qquad \text{Na}^+ : 11 - 1 = 10 \]In every case those $\displaystyle 10$ electrons are arranged as \(\displaystyle 1s^2\, 2s^2\, 2p^6\) — the electron configuration of neon. So the outermost (valence) shell is \(\displaystyle n = 2\) for all three species.Now go through the options with that fact in hand.(b) Valence principal quantum number — this is the same, \(\displaystyle n = 2\), for F⁻, Ne, and Na⁺. Since it does not change across the series, it cannot be the factor causing their sizes to differ. Ruled out.(c) Electron–electron interaction in the outer orbitals — with the same number of electrons ($\displaystyle 10$) sitting in the same set of orbitals (\(\displaystyle 2s\), \(\displaystyle 2p\)) in all three species, the mutual repulsion among electrons in the outer shell is essentially unchanged across the series. This is the detail people get wrong: it is tempting to think more shielding or repulsion is doing the work, but the electron count and their orbital arrangement are identical here, so this interaction stays constant and cannot explain a size difference. Ruled out.(d) None of the factors, because their size is the same — this is false as a matter of experimental fact. Isoelectronic species do not have equal radii; they shrink steadily as you add protons. Ruled out.(a) Nuclear charge (Z) — this is the one quantity that actually changes across the series: \[Z(\text{F}) = 9, \qquad Z(\text{Ne}) = 10, \qquad Z(\text{Na}) = 11 \]With the number of electrons held fixed at $\displaystyle 10$, a larger nuclear charge pulls that same electron cloud in more strongly (each electron feels a greater effective nuclear charge, since shielding by the other $\displaystyle 9$ electrons is nearly identical across the series). More pull on the same number of electrons means a smaller radius. So the radii fall in the order\[\text{F}^- > \text{Ne} > \text{Na}^+ \]matching the increase in \(\displaystyle Z\) exactly. This is the only factor among the four choices that varies from species to species, and it is the one that correctly predicts the observed shrinking trend.Answer: (a) nuclear charge (Z). All three species have the same $\displaystyle 10$ electrons in the same \(\displaystyle n=2\) shell, so options (b) and (c) are constant across the series and cannot be the cause; size is not equal, ruling out (d). What changes is Z ($\displaystyle 9$ for F⁻, $\displaystyle 10$ for Ne, $\displaystyle 11$ for Na⁺), and the greater pull on the same electron count is exactly why the radius decreases along F⁻ > Ne > Na⁺.

Successive ionization enthalpies always rise, and they jump hardest right when the core is reached — but "lower shell electrons are easier to remove" gets the shielding argument backwards.Go through each statement using what governs ionization enthalpy: nuclear charge felt by the electron being removed, and how many shells sit between that electron and the nucleus.(a) Ionization enthalpy increases for each successive electron. Once one electron is removed, the atom becomes a cation. The remaining electrons feel a larger effective nuclear charge per electron (same nuclear charge \(\displaystyle Z \), fewer electrons to share the pull and to shield each other), so each further electron is held more tightly. \(\displaystyle \Delta_i H_1 < \Delta_i H_2 < \Delta_i H_3 < \ldots \) always. This statement is correct.(b) The greatest increase in ionization enthalpy is on removing an electron from the core noble-gas configuration. As successive electrons are stripped, you first remove all the valence-shell electrons — these are shielded by the inner core and come off with gradually increasing (but comparable) energy. The moment you try to remove the next electron, you are pulling it from a completed, noble-gas-like inner shell that sits much closer to the nucleus and is far less shielded. That jump in energy is by far the largest of the whole sequence. This statement is correct.(c) The end of the valence electrons is marked by a big jump in ionization enthalpy. This is the same fact as (b), stated the other way round: the discontinuity in the successive-\(\displaystyle \Delta_i H \) values is exactly what tells you how many valence electrons the atom had (for example, sodium shows a huge jump between \(\displaystyle \Delta_i H_1 \) and \(\displaystyle \Delta_i H_2 \), confirming one valence electron). This statement is correct.(d) Removal of an electron from an orbital with lower \(\displaystyle n \) is easier than from one with higher \(\displaystyle n \). Here is the aside worth stopping on: this is the statement that reverses cause and effect. An orbital with a lower principal quantum number \(\displaystyle n \) is, on average, closer to the nucleus and has fewer inner shells shielding it from the nuclear charge \(\displaystyle Z \). It therefore feels a larger effective nuclear charge, \(\displaystyle Z_{\text{eff}} = Z - \sigma \) (where \(\displaystyle \sigma \) is the shielding constant contributed by electrons inside that orbital), and is held more tightly. An electron in a higher-\(\displaystyle n \) orbital is farther out, more shielded, and comes off more easily. So it is the higher-\(\displaystyle n \) electron that is easier to remove, not the lower-\(\displaystyle n \) one — statement (d) says the opposite of what actually happens. This statement is incorrect.Since (a), (b), and (c) all correctly describe how ionization enthalpy behaves, and (d) inverts the true relationship between \(\displaystyle n \) and ease of removal, (d) is the statement that does not hold.Answer: (d) is incorrect — electrons in orbitals of lower \(\displaystyle n\) are held closer to the nucleus with less shielding, so they are harder (not easier) to remove than electrons in orbitals of higher \(\displaystyle n\).

Metallic character depends on how easily an atom loses its valence electron — it rises as you go down a group and falls as you go across a period, left to right.Losing an electron becomes easier when the electron is farther from the nucleus (larger atom, more shielding) and harder when the nuclear charge pulling on it is stronger. That is exactly what happens down a group and across a period, so the two trends drive metallic character in opposite ways depending on which direction you move.Locate each element first. Step $\displaystyle 1$ — compare Mg and Al (same period). Both are in Period 3. Moving left to right across a period, nuclear charge increases while shielding stays roughly the same, so the outer electron is held more tightly and metallic character falls. Mg (Group $\displaystyle 2$) sits to the left of Al (Group $\displaystyle 13$), so \[\text{Mg} > \text{Al} \]Step $\displaystyle 2$ — compare Al and B (same group). Both are in Group 13. Moving down a group, each successive element adds a new shell, so the valence electron sits farther from the nucleus and is more shielded — it comes off more easily, so metallic character rises. Al (Period $\displaystyle 3$) is below B (Period $\displaystyle 2$), so \[\text{Al} > \text{B} \] This is the step people get backwards: it is tempting to think a "bigger" nucleus (more protons) always holds electrons tighter, but within a group the extra shells outweigh the extra nuclear charge, so the lower element is the more metallic one, not the less metallic one.Step $\displaystyle 3$ — place K. K is an alkali metal, Group $\displaystyle 1$, and moreover it sits in Period $\displaystyle 4$ — one period below Mg and Al, and two below B. Group $\displaystyle 1$ elements are already the most metallic in any period they belong to (least nuclear pull relative to shielding among that period's elements), and going down a period only adds to that. So K outranks all three of Mg, Al, and B.Putting the three comparisons together: \[\text{K} > \text{Mg} > \text{Al} > \text{B} \]This is exactly the trend you'd read straight off the periodic table: K is far down-left (most metallic corner), B is far up-right relative to the other three (least metallic among them), and Mg, Al fall in between in period-$\displaystyle 3$ order.Answer: (d) K > Mg > Al > B — metallic character rises down a group and falls across a period, so K (Group $\displaystyle 1$, Period $\displaystyle 4$) is the most metallic and B (Group $\displaystyle 13$, Period $\displaystyle 2$) is the least; within Period $\displaystyle 3$, Mg outranks Al because Al sits farther right.

Non-metallic character means how strongly an atom pulls on electrons — and the number that measures this directly is electronegativity, not just a mental picture of "left-right, top-bottom."Start with the two standard periodic trends: Sorting B, C, N, F first. All four sit in Period $\displaystyle 2$, in increasing order of atomic number: \(\displaystyle \text{B}(Z=5) \), \(\displaystyle \text{C}(Z=6) \), \(\displaystyle \text{N}(Z=7) \), \(\displaystyle \text{F}(Z=9) \). Moving left to right across this period, non-metallic character only goes up, so \[\text{B} < \text{C} < \text{N} < \text{F}. \]Placing Si. Silicon sits directly below carbon, in the same group (Group $\displaystyle 14$: C, Si, Ge, Sn, Pb). Going down that group, non-metallic character falls, so \[\text{Si} < \text{C}. \]The step everyone trips on: where does Si sit relative to B? Si and B are neither in the same period nor the same group — B is Period $\displaystyle 2$/Group $\displaystyle 13$, Si is Period $\displaystyle 3$/Group 14. The two trends above are each defined within a period or within a group; they simply don't tell you how to compare two elements that share neither. You cannot "eyeball" this one — you need the actual electronegativity values.On the Pauling scale (the standard periodic-properties table for this chapter): \[\text{F} = 4.0,\quad \text{N} = 3.0,\quad \text{C} = 2.5,\quad \text{B} = 2.0,\quad \text{Si} = 1.8. \]Reading these off from highest (most non-metallic) to lowest: \[\text{F}(4.0) > \text{N}(3.0) > \text{C}(2.5) > \text{B}(2.0) > \text{Si}(1.8). \]This is consistent with everything derived from trends alone (F > N > C, and C > Si), and it resolves the one comparison the trends couldn't: boron, sitting higher up the table with a smaller atomic radius, holds its electrons slightly more tightly than silicon does, so \(\displaystyle \text{B} > \text{Si} \) in non-metallic character — even though Si often "looks" similar to B because both are metalloids. Similarity in chemical behaviour (a diagonal relationship) is not the same as being equal in electronegativity.So the full order is \[\text{F} > \text{N} > \text{C} > \text{B} > \text{Si}. \]Answer: (c) F > N > C > B > Si

Oxidizing power means how strongly an atom pulls in an extra electron and gets reduced itself — and that strength does not simply track "smaller atom, more to the right." It has to be built up piece by piece from electronegativity trends plus one exception that trips up nitrogen.Step $\displaystyle 1$ — Compare N, O, F (same period, Period $\displaystyle 2$). Moving left to right across a period, the nuclear charge increases while the electrons added go into the same shell (same shielding), so the pull on any extra electron gets stronger. Electronegativity therefore rises steadily: \[N < O < F \] So among these three, fluorine is the strongest oxidizer and nitrogen the weakest — no surprise yet.Step $\displaystyle 2$ — Bring in chlorine (same group as F, one period down). Down a group, an extra shell of electrons is added, atomic radius grows, and the extra shielding weakens the nucleus's grip on an incoming electron. So chlorine's electronegativity is well below fluorine's: \[Cl < F \] On the Pauling scale the numbers run F \(\displaystyle \approx 4.0\), O \(\displaystyle \approx 3.5\), Cl \(\displaystyle \approx 3.0\), N \(\displaystyle \approx 3.0\) — Cl and N look almost tied on this number alone. Deciding their order needs a second idea, not just the electronegativity value.Step $\displaystyle 3$ — Why nitrogen falls behind chlorine even though it's the smaller, more "electronegative-looking" atom. Electronegativity is a tendency; what actually decides oxidizing power is whether the atom accepts the extra electron easily (its electron gain enthalpy), and here nitrogen has an unusual problem. Its ground-state configuration is \[N: \; [He]\,2s^2\,2p^3 \] a subshell that is exactly half-filled — a genuinely stable arrangement (each of the three 2p orbitals holds one electron, no pairing, minimum repulsion). Forcing one more electron in destroys that stability, so nitrogen's electron gain enthalpy is close to zero (even slightly unfavourable), far less negative than a typical Period-$\displaystyle 2$ non-metal in that position. On top of that, elemental nitrogen exists as \(\displaystyle N_2\), held together by a very strong \(\displaystyle N \equiv N\) triple bond (about $\displaystyle 941$ kJ/mol) that must first be broken before nitrogen can act as an oxidizing agent at all. Both effects make \(\displaystyle N_2\) a poor oxidizer in practice — weaker than chlorine, even though chlorine sits a full period further down the table.Fluorine, by contrast, has the opposite bonus: the \(\displaystyle F-F\) bond in \(\displaystyle F_2\) is unusually weak (the two small F atoms crowd their lone pairs together), so \(\displaystyle F_2\) breaks apart easily and reacts with almost everything — reinforcing, not fighting, its already-highest electronegativity.Step $\displaystyle 4$ — Put it together. \[F > O > Cl > N \] Fluorine is the strongest oxidizer (highest electronegativity, weak F–F bond to boot); oxygen is next (high electronegativity, strong electron affinity); chlorine, though a period lower, still readily accepts an electron and outperforms nitrogen; nitrogen, despite its position in Period $\displaystyle 2$, is held back by its stable half-filled \(\displaystyle 2p^3\) configuration and the tough-to-break \(\displaystyle N \equiv N\) bond.This matches option (b).Answer: (b) F > O > Cl > N