Classification of Elements Class 11 Chemistry NCERT Solutions

Step-by-step solution

1. 1The modern periodic law arranges elements in order of increasing atomic number, not atomic mass.
2. 2A period is a horizontal row with the same highest principal quantum number for the valence shell.
3. 3Elements in the same group have a similar valence-shell pattern, so they show related chemical behaviour and often comparable physical properties.
4. 4This arrangement therefore makes the study of elements and their compounds systematic rather than a collection of isolated facts.

Step-by-step solution

1. 1Mendeleev's periodic law treated physical and chemical properties as periodic functions of atomic weight, so he arranged known elements broadly in increasing atomic weight.
2. 2A strict atomic-weight order produced chemical inconsistencies, so Mendeleev placed some pairs according to strongly related properties rather than weight.
3. 3For example, modern standard atomic weights are approximately 127.60 for tellurium and 126.90 for iodine, yet Mendeleev placed Te before I because iodine closely resembles F, Cl and Br.
4. 4His arrangement was therefore a useful empirical rule, but the exceptional placements showed that atomic weight alone was not the fundamental organizing quantity.

Step-by-step solution

1. 1Mendeleev's law correlates an element's physical and chemical properties with its atomic weight.
2. 2The modern periodic law correlates those same properties with atomic number, which is the number of protons and does not depend on isotope mass.
3. 3Atomic number gives a unique, continuous order for all elements and explains the recurring electronic structures that produce the periodic patterns.

Step-by-step solution

1. 1A period is identified by the principal quantum number n of the newly starting valence shell; for the sixth period, n = 6.
2. 2According to the Aufbau order, the subshells filled during this period are 6s, 4f, 5d and 6p in that energy order.
3. 3Count the orbitals in these subshells:
4. 4
   $$N_{\mathrm{orbital}}=1+7+5+3=16$$
5. 5Each orbital can contain two electrons of opposite spin under the Pauli exclusion principle:
6. 6
   $$N_{e,\mathrm{capacity}}=2(16)=32\mathrm{electrons}$$
7. 7Each successive atomic number adds one electron, so a capacity of 32 electrons corresponds to 32 elements in the period.

Step-by-step solution

1. 1The seventh period contains the elements from francium, Z = 87, to oganesson, Z = 118, so Z = 114 lies in period 7.
2. 2The p-block of period 7 starts with nihonium, Z = 113, corresponding to group 13; the next element, Z = 114, therefore belongs to group 14.
3. 3Its distinguishing outer-shell pattern is 7s²7p⁴, which supplies four valence electrons when the filled 7s pair is included.

Step-by-step solution

1. 1The third period runs from sodium, Z = 11, to argon, Z = 18.
2. 2Argon is the group 18 member at the end of the period, so the preceding group 17 member is one atomic number lower.
3. 3The period-3 group-17 element is chlorine.
4. 4
   $$Z=18-1=17$$

Step-by-step solution

1. 1The actinide elements associated with the Lawrence Berkeley Laboratory are lawrencium, with Z = 103, and berkelium, with Z = 97.
2. 2Seaborg's research group proposed seaborgium for element 106, so seaborgium has Z = 106.
3. 3These names also reflect the close historical connection of the Berkeley and Seaborg teams with transuranium-element research.

Step-by-step solution

1. 1Elements in the same group generally possess the same outer-shell electron configuration or the same number of valence electrons.
2. 2Their atoms therefore form bonds and undergo reactions in similar ways; for example, alkali metals readily lose one electron and halogens readily gain one.
3. 3Comparable valence-electron interactions also produce related physical properties such as similar crystal structures and melting behaviour within many main-group families.

Step-by-step solution

1. 1Atomic radius is an operational estimate of atomic size. For a metal it is the metallic radius, commonly half the internuclear distance between neighbouring atoms in the metal lattice.
2. 2The exercise uses a 256 pm separation between neighbouring copper atoms as its illustrative metallic-lattice value:
3. 3
   $$r_{\mathrm{Cu}}=\frac{256\mathrm{pm}}{2}=128\mathrm{pm}$$
4. 4For a non-metal, covalent radius is half the internuclear distance in a homonuclear single bond. The exercise uses the 198 pm Cl–Cl distance:
5. 5
   $$r_{\mathrm{Cl}}=\frac{198\mathrm{pm}}{2}=99\mathrm{pm}$$
6. 6Ionic radius is the effective radius of a cation or anion, inferred from interionic distances in an ionic solid with a stated radius convention.
7. 7The exercise's illustrative values compare Na⁺ at 95 pm with Na at 186 pm and F⁻ at 136 pm with F at 64 pm, illustrating contraction of a cation and expansion of an anion.

Step-by-step solution

1. 1Across a period, the general trend is a decrease in atomic radius from left to right.
2. 2In the same period, electrons enter the same valence shell while nuclear charge increases, so effective nuclear charge rises and pulls the electrons inward more strongly.
3. 3Down a group, the general trend is an increase in atomic radius from top to bottom.
4. 4The principal quantum number increases and additional inner shells shield the valence electrons, placing the outer electrons farther from the nucleus.

Step-by-step solution

1. 1Isoelectronic species contain the same number of electrons. Calculate the electron count by subtracting positive charge or adding gained electrons:
2. 2
   $$N_e=Z-(\text{positive charge})+(\text{electrons gained})$$
3. 3(i) F⁻ has 9 + 1 = 10 electrons; Ne is isoelectronic with it.
4. 4(ii) Ar has 18 electrons; Cl⁻ is isoelectronic with it.
5. 5(iii) Mg²⁺ has 12 − 2 = 10 electrons; F⁻ is isoelectronic with it.
6. 6(iv) Rb⁺ has 37 − 1 = 36 electrons; Br⁻ is isoelectronic with it.

Step-by-step solution

1. 1Count electrons for each ion: N³⁻ has 7 + 3 = 10, O²⁻ has 8 + 2 = 10, F⁻ has 9 + 1 = 10, Na⁺ has 11 − 1 = 10, Mg²⁺ has 12 − 2 = 10 and Al³⁺ has 13 − 3 = 10.
2. 2All six species are therefore isoelectronic.
3. 3For species with the same electron count, the ion with the smaller nuclear charge has weaker attraction for the electrons and is larger.
4. 4The nuclear charges increase in the order N, O, F, Na, Mg, Al, so their ionic radii increase in the reverse order:
5. 5
   $$\mathrm{A}{\mathrm{l}}^{3+}<\mathrm{M}{\mathrm{g}}^{2+}<\mathrm{N}{\mathrm{a}}^{+}<{\mathrm{F}}^{-}<{\mathrm{O}}^{2-}<{\mathrm{N}}^{3-}$$

Step-by-step solution

1. 1Formation of a cation removes one or more electrons while nuclear charge remains unchanged.
2. 2The remaining electrons experience a greater effective nuclear charge per electron, electron–electron repulsion decreases and the electron cloud contracts; hence cation radius is smaller.
3. 3Formation of an anion adds one or more electrons without changing nuclear charge.
4. 4The added electron increases electron–electron repulsion and lowers the effective attraction per outer electron, so the electron cloud expands; hence anion radius is larger.

Step-by-step solution

1. 1'Isolated gaseous atom' places the atom in the gas phase, where atoms are widely separated and intermolecular forces are negligible compared with those in liquids and solids.
2. 2The term 'ground state' selects the electronic arrangement of lowest energy, so the starting atomic configuration is clearly specified.
3. 3Without these conditions, the measured enthalpy would also include interactions with neighbours or excitation to another state, making values difficult to compare.
4. 4For ionization enthalpy, the energy change is measured from the isolated ground-state atom to a gaseous ion plus an electron at very large separation; electron gain enthalpy uses the analogous reverse process with an incoming electron.

Step-by-step solution

1. 1Ionization takes the electron from the bound ground state to a zero-energy state at infinite separation, so the required energy is the magnitude of the given negative energy.
2. 2
   $$\Delta E_{\mathrm{H}}=0-(-2.18\times 10^{-18}\mathrm{J})=2.18\times 10^{-18}\mathrm{J}\mathrm{ato}{\mathrm{m}}^{-1}$$
3. 3One mole contains 6.022 × 10²³ atoms, so multiply the energy per atom by Avogadro's number:
4. 4
   $${\Delta }_{\mathrm{i}}H=(2.18\times 10^{-18}\mathrm{J}\mathrm{ato}{\mathrm{m}}^{-1})(6.022\times 10^{23}\mathrm{atoms}\mathrm{mo}{\mathrm{l}}^{-1})$$
5. 5
   $${\Delta }_{\mathrm{i}}H=1.3128\times 10^6\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}\approx 1.31\times 10^6\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$$

Step-by-step solution

1. 1For Be the configuration is 1s²2s², while for B it is 1s²2s²2p¹.
2. 2A 2s electron is closer to the nucleus and more penetrating than a 2p electron. In addition, Be has a filled 2s subshell, so its first electron is more tightly held than the first 2p electron of B.
3. 3For N, the three 2p electrons occupy separate 2p orbitals with parallel spins. In O, the fourth 2p electron must pair with one of them, increasing repulsion; removing this paired electron therefore needs less energy than removing an unpaired 2p electron from N.
4. 4From O to F, nuclear charge increases while the added electron remains in the compact 2p shell. The increase in effective nuclear attraction outweighs the added electron–electron repulsion, so F has the higher first ionization enthalpy.

Step-by-step solution

1. 1Neutral Na has the outer configuration 3s¹, whereas Mg has 3s²; Mg is smaller and has a higher effective nuclear charge, so removing an electron initially requires more energy.
2. 2Therefore, the first ionization enthalpy follows ΔᵢH₁(Na) < ΔᵢH₁(Mg).
3. 3After the first ionization, Na⁺ has the stable neon configuration 1s²2s²2p⁶, so its second ionization removes a core electron.
4. 4Mg⁺ retains the valence configuration 3s¹, so its second ionization removes a relatively weakly held 3s electron rather than a core electron.
5. 5The much greater energy needed to remove the second electron from Na⁺ reverses the first-ionization order.

Step-by-step solution

1. 1The principal quantum number increases down a group, placing the valence electrons in an outer shell that is farther from the nucleus and therefore larger in atomic radius.
2. 2The number of inner-shell electrons also increases, and these core electrons shield the valence electrons from the nuclear charge more effectively.
3. 3Although nuclear charge increases as atomic number rises, the combined increase in distance and shielding dominates for the outer electron in the main-group trend.
4. 4The valence electron is consequently less strongly attracted and requires less energy for removal.

Step-by-step solution

1. 1The listed group-13 values are the exercise's tabulated first ionization enthalpies, all in kJ mol⁻¹.
2. 2From B to Al, the expected decrease is very large because size and shielding increase when the new 3s valence shell begins.
3. 3From Al to Ga, the value rises slightly from 577 to 579 kJ mol⁻¹ because the inserted 3d electrons shield poorly, so the Ga valence electron experiences greater effective nuclear charge than a simple size trend predicts.
4. 4From Ga to In, the value falls from 579 to 558 kJ mol⁻¹ as the larger 5s valence shell is less strongly held, consistent with the normal trend.
5. 5From In to Tl, it rises from 558 to 589 kJ mol⁻¹ because poor shielding by inserted 4f and 5d electrons, together with lanthanide contraction, keeps the Tl valence electron relatively strongly bound.

Step-by-step solution

1. 1(i) O and F are in the same period. F has a smaller radius and a greater effective nuclear charge, so it attracts the incoming electron more strongly and also needs only one electron to complete a valence octet.
2. 2The compact 2p shell of F causes some repulsion, but the increase in nuclear attraction dominates; consequently F has the more negative electron gain enthalpy.
3. 3(ii) F and Cl are in the same group. Cl is larger, and its incoming electron enters the less compact n = 3 shell, where electron–electron repulsion is lower.
4. 4The repulsion outweighs the benefit of F's smaller size, producing the familiar exception that Cl has a more negative electron gain enthalpy than F.

Step-by-step solution

1. 1Adding the first electron to gaseous oxygen forms O⁻ with a filled 2p⁵ subshell and releases energy, so the first electron gain enthalpy is negative.
2. 2
   $${\mathrm{O}}_{(\mathrm{g})}+{\mathrm{e}}^{-}⟶{\mathrm{O}}_{(\mathrm{g})}^{-}{\mathrm{\Delta }}_{\mathrm{eg}}{\mathrm{H}}_1<0$$
3. 3The small O⁻ ion has concentrated negative charge and strongly repels a second incoming electron.
4. 4The electron also must enter the compact 2p shell and pair with an electron, creating additional repulsion even though an octet is completed.
5. 5Energy must therefore be supplied for the second process:
6. 6
   $${\mathrm{O}}_{(\mathrm{g})}^{-}+{\mathrm{e}}^{-}⟶{\mathrm{O}}_{(\mathrm{g})}^{2-}{\mathrm{\Delta }}_{\mathrm{eg}}{\mathrm{H}}_2>0$$

Step-by-step solution

1. 1Electron gain enthalpy is an absolute molar enthalpy change for adding an electron to an isolated gaseous atom in a specified state.
2. 2Electronegativity instead describes an atom's ability to attract a shared electron pair when that atom is already bonded in a chemical compound.
3. 3Electron gain enthalpy is a measurable thermochemical quantity, whereas electronegativity is an assigned comparative scale that depends on the bonded environment.

Step-by-step solution

1. 1Electronegativity is not an isolated-atom thermochemical constant; it describes the attraction of an atom for bonding electrons in a molecular environment.
2. 2Changing the bonded partner and oxidation environment can change the charge distribution on nitrogen and hence its effective electronegativity.
3. 3Therefore, assigning the same effective value to nitrogen in every compound ignores the chemical context of the bond.

Step-by-step solution

1. 1(a) On gaining an electron, proton number and nuclear charge stay constant while electron number rises.
2. 2The added electron increases electron–electron repulsion and reduces the effective nuclear attraction felt per outer electron, so the electron cloud expands and the radius increases.
3. 3(b) On losing an electron, proton number and nuclear charge stay constant while electron number falls.
4. 4Repulsion decreases and the remaining electrons experience a higher effective nuclear charge, so the electron cloud contracts and the radius decreases.

Step-by-step solution

1. 1Isotopes have the same atomic number and therefore the same number of protons.
2. 2For neutral atoms they also have the same number of electrons and the same ground-state electronic configuration, so nuclear attraction on the electron being removed is the same.
3. 3The differing nuclear masses do not materially change the electrostatic attraction responsible for chemical ionization enthalpy.
4. 4Small isotope-dependent nuclear-volume effects can exist in very precise physical measurements, but they are not part of the periodic trend expected in this exercise.

Step-by-step solution

1. 1(i) Metals have low ionization enthalpies and lose electrons to form cations; non-metals generally have high ionization enthalpies and gain electrons to form anions.
2. 2(ii) Metals have relatively high reducing power and less-negative electron gain enthalpies; non-metals commonly have more-negative electron gain enthalpies and oxidizing character.
3. 3(iii) Metals are electropositive and have low electronegativity, whereas non-metals are electronegative and attract bonding electrons more strongly.
4. 4(iv) Metal–non-metal combinations are usually ionic, while compounds formed mainly between non-metals are usually covalent.
5. 5(v) Metal oxides are generally basic or amphoteric, while non-metal oxides are generally acidic, though neutral oxides also occur.
6. 6(vi) Metals readily lose electrons and act as reducing agents; non-metals tend to gain electrons and act as oxidizing agents.
7. 7(vii) Metals are generally lustrous, malleable and ductile and have relatively high density and melting points; solid non-metals are generally brittle and have lower melting points.

Step-by-step solution

1. 1(a) Five electrons in the outer subshell follow the pattern ns²np⁵, so bromine is one valid example; the other halogens F, Cl, I and At also fit.
2. 2(b) Losing two electrons is favoured by the ns² configuration of group 2; magnesium is a suitable example.
3. 3(c) Gaining two electrons is favoured by ns²np⁴, the group 16 pattern; oxygen or sulfur is a suitable example.
4. 4(d) Group 17 is the intended family: F and Cl are gases, Br is a liquid, and I and At are solids at room temperature, while metallic character increases down the group and is expected to be appreciable in At.
5. 5Thus group 17 displays both the physical states and the trend from strongly non-metallic to increasingly metallic character described in the question.

Step-by-step solution

1. 1Group 1 atoms have one valence electron and tend to lose it. Atomic size and shielding increase down the group, so ionization enthalpy decreases and electron loss becomes easier; therefore reactivity increases.
2. 2Group 17 atoms need one electron for a complete valence octet. Down the group, increased size and shielding generally make electron-gain enthalpy less negative, so electron acceptance becomes less favourable.
3. 3Fluorine is the exception in electron-gain enthalpy because its very compact 2p shell produces strong incoming-electron repulsion, so Cl has the more negative value.
4. 4Fluorine nevertheless has the greatest oxidizing reactivity because F₂ has a low bond dissociation enthalpy and F has a very high effective attraction for electrons; its small size also gives strong interaction with many reductants.

Step-by-step solution

1. 1In an s-block element, the differentiating electron enters an ns subshell:
2. 2
   $$n{s}^{1-2}$$
3. 3In a p-block element, the ns subshell is filled and the differentiating electron enters np:
4. 4
   $$n{s}^{2}n{p}^{1-6}$$
5. 5In a d-block element, the last electron enters the penultimate (n − 1)d subshell while the outer ns subshell is filled or nearly filled:
6. 6
   $$(n-1)d^{1-10}n{s}^{0-2}$$
7. 7In an f-block element, the differentiating electron enters (n − 2)f while ns² is filled and the penultimate d subshell has 0 or 1 electron:
8. 8
   $$(n-2)f^{1-14}(n-1)d^{0-1}n{s}^2$$
9. 9These patterns identify the block from the subshell receiving the last electron; the principal shell containing ns also gives the period number.

Step-by-step solution

1. 1(i) The valence shell has n = 3, so the period is 3; an np⁴ ending identifies a p-block element. With two s-block groups and ten d-block groups before it, the group number is 2 + 10 + 4 = 16, identifying sulfur.
2. 2(ii) The valence shell has n = 4, so the period is 4; the d² ending places the element in the d-block. Its group is 2 + 2 = 4, identifying titanium.
3. 3(iii) The valence shell has n = 6, so the period is 6; the f⁷ ending places the element in the f-block, conventionally shown in group 3.
4. 4Complete the configuration as [Xe]4f⁷5d¹6s² and calculate the atomic number:
5. 5
   $$Z=54+7+1+2=64$$
6. 6The element with Z = 64 is gadolinium.

Step-by-step solution

1. 1Identify the characteristic ionization patterns. I and II have low first values but large jumps to the second value, indicating one easily lost electron followed by removal from a noble-gas core; their data match Li and K, respectively.
2. 2VI has first and second ionization enthalpies of 738 and 1451 kJ mol⁻¹ with no core-sized jump between them, matching Mg. III and IV match F and I, while the very high first value and positive electron gain enthalpy of V match He.
3. 3(a) V is least reactive because its first ionization enthalpy of 2372 kJ mol⁻¹ is highest and electron addition is endothermic at +48 kJ mol⁻¹.
4. 4(b) II is most reactive as a metal because its first ionization enthalpy of 419 kJ mol⁻¹ is lowest and its electron gain tendency is weak.
5. 5(c) III is most reactive as a non-metal because it combines a high first ionization enthalpy with the most negative electron gain enthalpy, −328 kJ mol⁻¹.
6. 6(d) V is the least reactive non-metal in the listed set because its closed-shell configuration makes both electron loss and electron gain energetically unfavourable.
7. 7(e) VI forms MX₂: its comparatively low second ionization enthalpy of 1451 kJ mol⁻¹ allows two electrons to be removed, so it behaves as a Group 2 metal and gives a stable predominantly ionic halide.
8. 8(f) I fits MX: its huge jump from 520 to 7300 kJ mol⁻¹ identifies an alkali metal. Among the alkali metals, the small Li⁺ ion strongly polarizes a halide ion, so lithium halide such as LiF has substantial covalent character and the predominantly stable MX pattern is characteristic of Li.

Step-by-step solution

1. 1(a) Li forms Li⁺ and O forms O²⁻, so charge balance requires two Li⁺ ions for each oxide ion.
2. 2
   $$\mathrm{L}{\mathrm{i}}_2\mathrm{O}$$
3. 3(b) Mg forms Mg²⁺ and nitride is N³⁻; three Mg²⁺ ions balance two nitride ions.
4. 4
   $$\mathrm{M}{\mathrm{g}}_3{\mathrm{N}}_2$$
5. 5(c) Al forms Al³⁺ and iodide is I⁻, requiring a 1:3 ratio.
6. 6
   $$\mathrm{Al}{\mathrm{I}}_3$$
7. 7(d) Silicon commonly forms Si⁴⁺ and oxide is O²⁻, giving a 1:2 ratio.
8. 8
   $$\mathrm{Si}{\mathrm{O}}_2$$
9. 9(e) Fluorine is monovalent and phosphorus can show valencies 3 or 5, so stable binary fluorides include PF₃ and PF₅; PF₅ is the expected higher-valency product under excess fluorine.
10. 10
    $$\mathrm{P}{\mathrm{F}}_3\text{or}\mathrm{P}{\mathrm{F}}_5$$
11. 11(f) Z = 71 is lutetium, a lanthanide with the characteristic oxidation state +3; F has oxidation state −1.
12. 12
    $$\mathrm{Lu}{\mathrm{F}}_3$$

Step-by-step solution

1. 1Each new period begins when electrons start entering a shell with a new principal quantum number n.
2. 2The period number is therefore the principal quantum number of the outermost shell for the elements in that row, apart from details of individual configurations that do not change the row assignment.
3. 3Atomic number increases across a whole table rather than defining a single period, while azimuthal quantum number l identifies subshell type rather than period number.

Step-by-step solution

1. 1The number of orbitals in a subshell is 2l + 1, and each orbital can hold 2 electrons; hence the maximum capacity is 2(2l + 1).
2. 2For p, d and f subshells, l = 1, 2 and 3, so the maximum occupancies are 6, 10 and 14 electrons respectively.
3. 3A d subshell has 5 orbitals and holds 10 electrons, not 8, so the d-block has 10 columns, not 8.
4. 4The other statements agree with the correspondence between block width, subshell capacity and the l value of the differentiating subshell.

Step-by-step solution

1. 1The valence principal quantum number fixes the shell in which valence electrons are located.
2. 2Nuclear charge governs attraction for valence electrons, and the number of core electrons controls shielding experienced by the valence shell.
3. 3Valence-shell chemistry is governed mainly by charge and electron arrangement, not by the small isotope-dependent change in nuclear mass.
4. 4Hence nuclear mass is the factor that does not determine the valence shell or its chemistry in the context of this question.

Step-by-step solution

1. 1Each species has 10 electrons: F⁻ has 9 + 1, Ne has 10, and Na⁺ has 11 − 1 electrons.
2. 2Their nuclear charges are different, increasing from Z = 9 for F to Z = 10 for Ne and Z = 11 for Na.
3. 3For an isoelectronic set, greater nuclear charge contracts the electron cloud, so increasing radius follows decreasing nuclear charge.
4. 4
   $$r(\mathrm{N}{\mathrm{a}}^{+})<r(\mathrm{Ne})<r({\mathrm{F}}^{-})$$

Step-by-step solution

1. 1Successive ionization enthalpies increase because each electron is removed from a species with greater positive charge and the remaining electrons are more strongly bound.
2. 2A large jump occurs when the outer valence electrons have been removed and the next electron must come from a closed noble-gas core.
3. 3Electrons in a lower-n orbital are closer to the nucleus, experience weaker shielding relative to their charge and are more strongly attracted than electrons in a higher-n orbital.
4. 4Therefore, an electron with higher n is easier to remove, which makes statement (d) incorrect.

Step-by-step solution

1. 1Metallic character generally increases from right to left across a period because size decreases and effective nuclear charge increases toward the right.
2. 2It generally increases down a group because valence electrons occupy higher principal shells and are more shielded.
3. 3Mg lies to the left of Al in period 3, so Mg is more metallic than Al.
4. 4K lies to the left of Mg in period 4, so K is more metallic than Mg; B is to the right of and above Al, making B the least metallic of the four.
5. 5
   $$\text{metallic character: }\mathrm{K}>\mathrm{Mg}>\mathrm{Al}>\mathrm{B}$$

Step-by-step solution

1. 1Non-metallic character generally increases from left to right across a period because size decreases and attraction for bonding electrons increases.
2. 2Thus, within period 2 the order is F > N > C > B.
3. 3Non-metallic character generally decreases down a group because atomic size and shielding increase; hence C is more non-metallic than its congener Si.
4. 4Si is also less non-metallic than B in this set, consistent with its lower period position and greater metallic character; the combined order is F > N > C > B > Si.
5. 5
   $$\text{non-metallic character: }\mathrm{F}>\mathrm{N}>\mathrm{C}>\mathrm{B}>\mathrm{Si}$$

Step-by-step solution

1. 1Oxidizing character is the ability to accept electrons and is supported by high effective nuclear attraction and a stable configuration after electron gain.
2. 2Across period 2, the general increase in electronegativity and electron-accepting tendency gives F > O > N; nitrogen is especially reluctant to add an electron because its 2p³ subshell is already half filled and stable.
3. 3Down group 17, oxidizing strength decreases because atoms become larger and more shielded, so F is a stronger oxidizing element than Cl even though Cl has the more negative electron gain enthalpy.
4. 4The electron-gain-enthalpy exception does not reverse the halogen order because F₂ also has a low bond dissociation enthalpy; its F atoms are small and attract transferred electrons strongly.
5. 5For the cross-period comparison required here, O is more electronegative and smaller than Cl and is the stronger oxidizing element, while N remains the weakest of the four.
6. 6
   $$\text{oxidizing character: }\mathrm{F}>\mathrm{O}>\mathrm{Cl}>\mathrm{N}$$