Arrange the halogens F2, Cl2, Br2, I2, in order of – Class 11 Chemistry

NCERT Exemplar · Q2

Q. Arrange the halogens F2, Cl2, Br2, I2, in order of their increasing reactivity with alkanes.

(i) I2 < Br2 < Cl2 < F2

(ii) Br2 < Cl2 < F2 < I2

(iii) F2 < Cl2 < Br2 < I2

(iv) Br2 < I2 < Cl2 < F2

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Concept understanding — Periodic Trend Metallic Character

What "Metallic Character" Actually Means

Imagine you have a piece of copper wire and a lump of charcoal. The copper is shiny, you can hammer it into a thin sheet, and it conducts electricity. The charcoal is dull, brittle, and does not conduct electricity well. That difference — the set of properties that make a metal "metallic" — is what we call metallic character.

Metallic character is not a single number you can measure directly. It is a qualitative trend that describes how strongly an element behaves like a metal. The more metallic an element is, the more it shows these traits:

Shiny (lustrous) appearance
High electrical and thermal conductivity
Malleability (can be hammered into sheets) and ductility (can be drawn into wires)
Tendency to lose electrons and form positive ions (cations)

The last point is the key chemical reason behind the trend. Metals are electron-losers. Non-metals are electron-gainers.

The Periodic Trend: The Precise Statement

Important

Metallic character decreases from left to right across a period, and increases from top to bottom down a group.

Let's break that into two parts.

Across a Period (Left to Right)

Take Period 3: Na → Mg → Al → Si → P → S → Cl → Ar.

Sodium (Na) is a highly reactive metal — it loses its one valence electron very easily. As you move right, the elements become less willing to lose electrons. Magnesium loses two electrons but holds them a bit tighter. Aluminium still behaves like a metal but is less reactive. Silicon is a metalloid — it has some metallic and some non-metallic properties. Phosphorus, sulfur, chlorine, and argon are clearly non-metals.

Why? The nuclear charge (number of protons) increases across the period, pulling the electrons in tighter. The valence electrons are held more strongly, so the atom is less willing to give them away. Losing electrons becomes harder → metallic character decreases.

Down a Group (Top to Bottom)

Take Group 1: Li → Na → K → Rb → Cs → Fr.

Lithium is a metal, but it is relatively hard and has a high melting point for a metal. Caesium is so metallic that it melts in your hand and explodes on contact with water. The metallic character increases dramatically as you go down.

Why? The atomic radius increases down the group. The valence electron is farther from the nucleus and is shielded by more inner electron shells. The nucleus holds it much more loosely. Losing that electron becomes very easy → metallic character increases.

Note

The same logic applies to all groups. Even in Group 14, carbon (top) is a non-metal, silicon and germanium are metalloids, and tin and lead (bottom) are metals. The trend is consistent.

The One Reason Behind Both Trends

Both trends come down to a single idea: how easily an atom can lose an electron.

Direction	Change in electron loss ease	Effect on metallic character
Left → Right	Harder (higher ionization energy)	Decreases
Top → Bottom	Easier (lower ionization energy)	Increases

Tip

If you ever forget the trend, remember: Metals are electron-losers. The easier it is to lose an electron, the more metallic the element. Ionization energy (the energy needed to remove an electron) is your best friend here — lower ionization energy = higher metallic character.

A Quick Visual Summary

Period	Left side	Middle	Right side
2	Li (metal)	Be (metal)	B (metalloid), C, N, O, F, Ne (non-metals)
3	Na (metal)	Mg, Al (metals)	Si (metalloid), P, S, Cl, Ar (non-metals)
4	K (metal)	...	...

Down any group, metallic character increases. Across any period, it decreases.

Common Mistake to Avoid

Watch out

Do not confuse metallic character with reactivity. While they often go together, they are not the same. For example, fluorine is the most reactive non-metal but has almost zero metallic character. Gold is a very unreactive metal but has high metallic character. Metallic character is about the type of properties, not how violently the element reacts.

Final Takeaway

Metallic character is a measure of how "metal-like" an element is. It depends on how easily the atom loses electrons. Across a period, electrons are held tighter → metallic character falls. Down a group, electrons are held looser → metallic character rises. This trend is one of the most reliable patterns in the periodic table, and it explains why metals are on the left and bottom, while non-metals are on the right and top.

The periodic trend in metallic character across periods and down groups is a core topic from the NCERT Class 11 Chemistry chapter on Classification of Elements and Periodicity, and "metallic and non-metallic character periodic trend" is a widely searched revision query for CBSE boards and JEE Main/NEET. This trend, linked closely to ionization energy, is a frequent basis for "periodic properties important questions" in competitive chemistry papers.

Halogen reactivity with alkanes follows the ease of homolytic bond cleavage and the exothermicity of the overall halogenation reaction. Fluorine is explosively reactive, chlorine reacts readily, bromine requires heat/light, and iodine is essentially unreactive. The order is I₂ < Br₂ < Cl₂ < F₂.

Why Halogens React Differently with Alkanes

Halogenation of alkanes is a free-radical substitution reaction. The reactivity of a halogen depends on two competing factors:

Bond dissociation energy – how easily the X–X bond breaks to form radicals
Overall thermodynamics – whether the net reaction releases or absorbs energy

The reaction proceeds through initiation (X₂ → 2X·), propagation (X· + R–H → R· + HX, then R· + X₂ → R–X + X·), and termination steps. The key is that even though fluorine has the strongest F–F bond, the formation of the very strong H–F and C–F bonds makes fluorination overwhelmingly exothermic. In contrast, iodination is endothermic overall, making it thermodynamically unfavourable.

Step-by-Step Analysis

Fluorine (F₂) – The F–F bond energy is relatively high (~158 kJ/mol), but the H–F bond formed is exceptionally strong (~570 kJ/mol). The overall reaction is so exothermic (ΔH ≈ −430 kJ/mol for CH₄ + F₂) that it is violent and uncontrollable, often leading to combustion rather than clean substitution. Fluorine reacts explosively with alkanes even in the dark and at low temperatures.
Chlorine (Cl₂) – The Cl–Cl bond (~243 kJ/mol) is weaker than F–F, and while H–Cl (~432 kJ/mol) is not as strong as H–F, chlorination is still exothermic (ΔH ≈ −100 kJ/mol). Chlorine reacts readily with alkanes in the presence of UV light or heat, giving good yields of chloroalkanes. This is the standard laboratory halogenation.
Bromine (Br₂) – The Br–Br bond (~193 kJ/mol) is weaker still, but H–Br (~366 kJ/mol) is also weaker. Bromination is only slightly exothermic or nearly thermoneutral (ΔH ≈ −30 kJ/mol). Bromine reacts with alkanes, but requires heat or strong light and proceeds more slowly than chlorination. It is selective, preferring tertiary > secondary > primary hydrogens.
Iodine (I₂) – The I–I bond is the weakest (~151 kJ/mol), but H–I is also very weak (~298 kJ/mol). The overall iodination reaction is endothermic (ΔH ≈ +50 kJ/mol), meaning it is thermodynamically unfavourable. Iodine does not react with alkanes under normal conditions; the reaction is reversible and the equilibrium lies far to the left.

Watch out

A common mistake is to assume that weaker X–X bonds automatically mean higher reactivity. While bond strength matters for initiation, the overall enthalpy change determines whether the reaction proceeds. Iodine's weak I–I bond cannot compensate for the weak H–I and C–I bonds formed.

Reactivity Order

Putting it all together, the increasing order of reactivity is:

\text{I}_2 < \text{Br}_2 < \text{Cl}_2 < \text{F}_2

Iodine is essentially unreactive, bromine reacts sluggishly, chlorine reacts well under standard conditions, and fluorine reacts explosively.

✓Final answer

The correct option is (i) : I₂ < Br₂ < Cl₂ < F₂.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$B > C > Si > N > F$
›Reveal solutionSolution
This tests periodic trends in non-metallic character (electronegativity) across a period and down a group. The correct order is $F>N>C>B>Si$ .

Concept and Intuition

Non-metallic character is closely tied to electronegativity: elements become more non-metallic as you move left to right across a period (increasing nuclear charge pulls electrons in more strongly, with a similar shielding), and less non-metallic as you move down a group (the valence shell gets farther from the nucleus, so it holds electrons less tightly). Here B, C, N, F all sit in Period 2, increasing in non-metallic character left to right, while Si sits directly below C in Period 3 — so despite being to the 'right' of B's period-2 position in some naive sense, going down a group weakens non-metallic character enough that Si actually becomes less non-metallic than boron.

Step-by-Step Solution

Within Period 2 (B, C, N, F), non-metallic character increases left to right due to increasing effective nuclear charge:

$B<C<N<F$

Si is directly below C in Group 14. Moving down a group, non-metallic character decreases because the outer electrons are farther from the nucleus and more shielded, so Si is markedly less non-metallic than C — in fact even less than B, since silicon's electronegativity ( $\approx1.90$ ) is lower than boron's ( $\approx2.04$ ).

Combining both trends, the overall order from most to least non-metallic is:

$F>N>C>B>Si$
Common Mistakes

Placing Si between C and B, or above B, by only counting periods and forgetting that its position in Period 3 (one row down) drops its non-metallic character below even boron's.

Confusing 'non-metallic character' trend direction with 'atomic size' or 'metallic character' trends, which run oppositely.

✓Final answer
$F > N > C > B > Si$

ANSWER: C
AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
$S > Se > O$
›Reveal solutionSolution
This tests periodic trends across four properties; the flawed set is (D), because metallic character increases down a group, so K (below Na) must be more metallic than Na, and alkaline-earth Ba is less metallic than either alkali metal.

Concept and Intuition

Periodic trends: electron gain enthalpy (magnitude) is generally highest for elements just before a stable configuration but oxygen's small size causes strong electron-electron repulsion upon adding an electron, making its EGE magnitude anomalously lower than sulfur's. Electronegativity increases up a group and across a period (F > O > Cl holds since F and O are in period 2, both above Cl, and F > O within period 2). Metallic radius increases down a group and decreases across a period. Metallic character (nature) increases down a group and decreases across a period (left to right), and alkali metals (Group 1) are generally more metallic than alkaline earth metals (Group 2) of comparable periods due to lower ionization energy.

Step-by-Step Solution

(A) Electron gain enthalpy magnitude: O(≈141) < Se(≈195) < S(≈200) kJ/mol, so S > Se > O holds — correctly arranged.

(B) Electronegativity (Pauling): F(4.0) > O(3.5) > Cl(3.0) — correctly arranged.

(C) Metallic radius: Na(186 pm) > Li(152 pm) > Al(143 pm) — correctly arranged.

(D) Metallic nature: K lies directly below Na in Group 1, so K is more metallic than Na (metallic character increases down a group). Ba, an alkaline-earth metal, is less metallic than the alkali metals. So the correct order is K > Na > Ba, not Na > K > Ba as given — this set is incorrectly arranged.

Common Mistakes

Assuming electron gain enthalpy trends strictly follow electronegativity trends (oxygen is a classic exception).

Overlooking that metallic character comparisons across different groups (alkali vs. alkaline earth) still follow the general periodic trend of decreasing metallic character across a period/increasing ionization energy.

✓Final answer
$The correct option is (D) — Na > K > Ba (Metallic nature) is the incorrectly arranged set.$

ANSWER: D
AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$\mathrm{Ga < In < Al}$
›Reveal solutionSolution
Gallium has an anomalously low melting point (~30°C, it melts in your palm) while aluminium's is very high (~660°C); indium sits in between. So the increasing order is $Ga < In < Al$ .

Concept and Intuition

Group 13 melting points do not fall in a simple monotonic trend down the group. Aluminium has a fairly high melting point typical of a light metal with strong metallic bonding. Gallium is the famous outlier — its unusual crystal structure (made of $Ga_2$ dimeric units rather than simple close packing) gives it a melting point barely above room temperature. Indium's melting point (~157°C) is higher than gallium's but well below aluminium's.

Step-by-Step Solution

Recall/estimate melting points: $Al \approx 660^\circ C$ , $Ga \approx 30^\circ C$ , $In \approx 157^\circ C$ .

Rank from lowest to highest: $Ga\,(30) < In\,(157) < Al\,(660)$ .

Match to the options: this is exactly option (A).

Common Mistakes

Assuming melting point falls monotonically down Group 13 the way, say, reactivity of alkali metals rises — gallium's anomaly breaks any such assumption.

Confusing gallium's low melting point with a boiling point property — gallium actually has a very high boiling point, a separate anomaly.

✓Final answer
$Ga < In < Al$

ANSWER: A
AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
$\mathrm{C,\ Si,\ Ge,\ Sn,\ Pb}$
›Reveal solutionSolution
In Group 14 (C, Si, Ge, Sn, Pb), metallic character increases down the group; only Sn and Pb are classified as metals.

Concept and Intuition

Metallic character increases down a group because atomic size increases and ionization energy decreases, making it progressively easier for atoms to lose electrons and behave like metals (delocalized electron sea, malleability, conductivity as a "true" metal). Carbon at the top of Group 14 is a clear non-metal (as diamond/graphite); silicon and germanium in the middle are metalloids (semiconducting behavior, intermediate properties); tin and lead at the bottom are genuine metals (malleable, good conductors, form basic/amphoteric oxides typical of metals).

Step-by-Step Solution

List Group 14 elements in order of increasing atomic number: C, Si, Ge, Sn, Pb.

Apply the down-the-group trend: metallic character increases from top to bottom.

Classify: C = non-metal; Si, Ge = metalloids; Sn, Pb = metals.

The elements with metallic nature are therefore Sn and Pb.

Common Mistakes

Classifying Ge as a metal (it is usually classified as a metalloid, though it shows some borderline metallic character — the standard classification used in these papers places only Sn and Pb as metals).

Forgetting that metallic character trends increase down a group (opposite of the across-a-period trend).

✓Final answer
$The correct option is (D) — Sn, Pb.$

ANSWER: D
AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
$\mathrm{Si > Al > Na > Hg}$
›Reveal solutionSolution
Metallic character falls across a period; for Na, Mg, Al, Si the correct decreasing order is $Na>Mg>Al>Si$ .

Concept and Intuition

Metallic character reflects how easily an element loses electrons to form a cation. Moving left to right across a period, nuclear charge increases while the number of shells stays the same, so atoms hold their outer electrons more tightly and become less willing to lose them — metallic character therefore decreases across a period.

Step-by-Step Solution

Identify the elements Na, Mg, Al, Si as consecutive period-3 elements (groups 1, 2, 13, 14).

Across period 3, effective nuclear charge rises steadily from Na to Si.

Ionisation enthalpy therefore rises from Na to Si, meaning it becomes progressively harder to remove an electron.

Metallic (electropositive) character is highest for Na and lowest for Si among these four: $Na>Mg>Al>Si$ .

Common Mistakes

Confusing metallic character trend with atomic size trend direction (both decrease left to right, easy to mix up with group trends).

Including Hg (a transition/post-transition metal from a different part of the periodic table) in the same simple periodic trend as an option distractor.

✓Final answer
$Na > Mg > Al > Si$

ANSWER: B
AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$Q. The correct order of the metallic character of the elements Be, Al, Na, K is (A) K > Na > Al > Be (B) K > Al > Na > Be (C) Al > K > Na > Be (D) Na > K > Be > Al$
›Reveal solutionSolution
Metallic character increases down a group and decreases across a period (left to right); ranking these four elements by their group/period positions gives the order directly.

Concept and Intuition

Metallic character reflects how easily an element loses electrons (low ionization energy, electropositive nature). It increases as you go down a group (larger atoms, outer electrons less tightly held) and decreases as you go across a period left to right (increasing nuclear charge holds electrons tighter).

Step-by-Step Solution

K is in period 4, group 1 — the most metallic of the four (largest atomic size among alkali metals shown, lowest ionization energy).

Na is in period 3, group 1 — more metallic than any period-3 element to its right, but less metallic than K (which is one period further down the same group).

Al is in period 3, group 13 — less metallic than Na (same period, further right), but still a genuine metal, more metallic than Be.

Be is in period 2, group 2 — smallest atom of the four, highest ionization energy among them, so the least metallic (Be is famously anomalous/covalent-leaning among alkaline earths).

Combining: K > Na > Al > Be.

Common Mistakes

Placing Al above Na by only considering "more electrons" without accounting for the increased nuclear charge across the period.

Forgetting Be's well-known anomalous, weakly-metallic behaviour due to its very small size.

✓Final answer
$The correct option is (A) — K > Na > Al > Be.$

ANSWER: A
AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
$Q. In which of the following, the elements Al, B, Mg and K are correctly arranged in the increasing order of their metallic character? (A) B < Al < Mg < K (B) B < Mg < Al < K (C) Al < Mg < K < B (D) B < Mg < K < Al$
›Reveal solutionSolution
Metallic character increases down a group and decreases across a period; ranking B, Al, Mg, K by this trend gives B < Al < Mg < K.

Concept and Intuition

Metallic character reflects how easily an atom loses electrons (low ionization energy, low electronegativity). Down a group, atomic size increases and valence electrons are held less tightly, so metallic character increases. Across a period, nuclear charge increases and electrons are held more tightly, so metallic character decreases. Boron sits at the top of group 13 and is actually a metalloid (least metallic here); potassium sits in group 1 at the bottom of the group range considered, making it the most metallic.

Step-by-Step Solution

Locate each element: B (period 2, group 13, metalloid), Al (period 3, group 13, metal but the least metallic true metal here), Mg (period 3, group 2, more metallic than Al since group 2 is more metallic than group 13), K (period 4, group 1, alkali metal — most metallic of all).

Apply group trend: within group 13, Al (period 3) is more metallic than B (period 2).

Apply period trend at period 3: Mg (group 2) is more metallic than Al (group 13), since metallic character falls left to right — wait, group 2 is to the left of group 13, so Mg (further left) is more metallic than Al, consistent with the period trend.

K (group 1, period 4) beats all of them, being both further left (group 1) and lower (period 4).

Combine: B < Al < Mg < K.

Common Mistakes

Forgetting boron is anomalously non-metallic (a metalloid) despite being in group 13.

Mixing up the period trend direction (metallic character decreases, not increases, left to right).

✓Final answer
$The correct option is (A) — B < Al < Mg < K.$

ANSWER: A
AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$Q. In which of the following the elements are in correct order of their chemical reactivity in terms of oxidizing property? (A) F > O > Cl > N (B) O > F > Cl > N (C) Cl > F > O > N (D) F > Cl > N > O$
›Reveal solutionSolution
Electronegativity order F > O > Cl > N tracks their oxidizing strength/reactivity as well: fluorine is the most powerful oxidizer known, followed by oxygen, then chlorine, with nitrogen the weakest of the four (its inert $N\equiv N$ triple bond makes $N_2$ sluggish despite decent electronegativity).

Concept and Intuition

Oxidizing power (an element's tendency to pull electrons from another species, getting reduced itself) generally rises with electronegativity, since a more electronegative atom holds added electrons more strongly. Fluorine sits at the top of the entire periodic table for electronegativity, making $F_2$ the strongest common oxidizer; oxygen and chlorine follow, both strong oxidizers used throughout chemistry (combustion, bleaching, disinfection); nitrogen, despite decent electronegativity, exists as unreactive $N_2$ gas at ordinary conditions due to its very strong triple bond, making it the weakest oxidizer of the four in practical reactivity terms.

Step-by-Step Solution

List electronegativities (Pauling scale, approx.): F = 3.98, O = 3.44, Cl = 3.16, N = 3.04.

This directly orders them: F > O > Cl > N.

Cross-check with known chemical behavior: $F_2$ reacts with almost everything (most reactive/strongest oxidizer known); $O_2$ supports combustion and oxidizes most elements; $Cl_2$ is a strong oxidizer/bleaching agent but milder than $F_2$ or $O_2$ 's more vigorous reactions; $N_2$ is famously inert (used as an unreactive blanket gas) despite being right next to O and F in the periodic table, due to the strength of its triple bond.

So the reactivity/oxidizing-power order F > O > Cl > N is consistent both with electronegativity trends and observed chemical behavior.

Common Mistakes

Placing Cl above O based solely on standard reduction potentials in aqueous solution (where $Cl_2/Cl^-$ can appear competitive with $O_2$ ) — this question is asking about the general elemental reactivity/electronegativity-based trend, not aqueous half-cell potentials.

Overrating nitrogen's reactivity based on its electronegativity value alone, ignoring that $N_2$ 's triple bond makes it kinetically inert in most conditions.

✓Final answer
$The correct option is (A) — F > O > Cl > N.$

ANSWER: A
AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQ
$Q. Which among the following property increases, as we move down group-I of the periodic table? (A) Electro negativity (B) Ionic radius (C) Melting point (D) Ionization enthalpy$
›Reveal solutionSolution
Down Group 1, atomic/ionic size increases while electronegativity,

ionization enthalpy and melting point all decrease — so the property that

increases is ionic radius.

Concept and Intuition

Moving down any group, a new principal shell of electrons is added, and

despite increasing nuclear charge, the increased shielding and greater

principal quantum number dominate, so atomic and ionic radii increase steadily.

Because the outer electron is progressively farther from the nucleus (and more

shielded), the attraction on it weakens — so ionization enthalpy and

electronegativity both fall down the group. Metallic bonding also weakens with

increasing size, so melting points fall too (alkali metals get softer and

lower-melting down the group).

Step-by-Step Solution

Electronegativity of alkali metals: Li > Na > K > Rb > Cs — DECREASES down the group. So (A) is wrong.

Ionic radius: $Li^+ < Na^+ < K^+ < Rb^+ < Cs^+$ — INCREASES down the group. So (B) is correct.

Melting point: Li (180.5°C) > Na (97.8°C) > K (63.5°C) > Rb > Cs — DECREASES down the group. So (C) is wrong.

Ionization enthalpy: Li > Na > K > Rb > Cs — DECREASES down the group (easier to remove the outer electron as it gets farther from the nucleus). So (D) is wrong.

Common Mistakes

Confusing atomic/ionic SIZE trends (increase down the group) with ionization enthalpy/electronegativity trends (decrease down the group) — they move in opposite directions.

✓Final answer
$The correct option is (B) — Ionic radius.$

ANSWER: B
AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$Q. Alkali metals are powerful reducing agents because ________ (A) They are metals (B) They are monovalent (C) Their ionic radii are large (D) Their ionization energies are low$
›Reveal solutionSolution
A reducing agent works by losing electrons; alkali metals do this exceptionally easily because their ionization energies are the lowest among all elements, which is the direct reason they're such powerful reducing agents.

Concept and Intuition

A reducing agent reduces another species by itself being oxidized, i.e., by losing electrons. How easily a species loses an electron is governed by its ionization energy — the energy required to remove an electron. Alkali metals (Group 1) have exactly one electron in their outermost shell, shielded by all the inner shells and held only loosely by the nucleus (large atomic radius, poor effective nuclear charge on the valence electron), giving them the lowest ionization energies of any group in the periodic table. This is the direct physical reason they give up their electron so readily and hence act as powerful reducing agents.

Step-by-Step Solution

Reducing power = tendency to lose electrons and get oxidized.

The quantity that directly measures "how easily a species loses its electron" is ionization energy — lower ionization energy means the electron is lost more readily.

Alkali metals have the lowest ionization energies in their periods (single loosely-held valence electron, large atomic size, effective shielding by inner shells).

Therefore they lose their electron with minimal energy input, making them the strongest reducing agents among common metals — the cause-and-effect chain is: low ionization energy → easy electron loss → strong reducing power.

"They are metals" (A), "monovalent" (B), and "large ionic radii" (C) are all true but secondary/associated facts — none of them is the direct energetic reason for easy electron loss the way ionization energy is.

Common Mistakes

Picking "large ionic radii" (C) — radius influences ionization energy indirectly (larger radius → lower ionization energy), but the property that directly defines ease of electron loss is the ionization energy itself, not the radius.

✓Final answer
$The correct option is (D) — Their ionization energies are low.$

ANSWER: D

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Q.Arrange the halogens F2, Cl2, Br2, I2, in order of their increasing reactivity with alkanes.

Why Halogens React Differently with Alkanes

Step-by-Step Analysis

Reactivity Order

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