How does the metallic and non metallic character vary on moving from left to right in a period?

NCERT Exemplar · Q44

Q. How does the metallic and non metallic character vary on moving from left to right in a period?

Andhra Pradesh BieapShort· 2mImportance★★★★★est

86% · 94/109 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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

|--------|-----------|--------|------------| …

Start your 14-day free trial to unlock the full solution →

Metallic character decreases and non-metallic character increases as we move from left to right across a period. This is because the effective nuclear charge increases, pulling the valence electrons closer and making it harder to lose them (metallic property) but easier to gain them (non-metallic property).

The key to understanding this trend lies in one simple idea: electronegativity and ionization energy increase across a period. Let me explain why.

Across a period, electrons are added to the same principal quantum shell (same energy level). At the same time, protons are added to the nucleus. The shielding effect from inner electrons remains roughly constant (since no new inner shells are added), so the effective nuclear charge — the net positive charge felt by the valence electrons — increases steadily.

This stronger pull means:

The atomic radius decreases (electrons are pulled in tighter).
It becomes harder to remove an electron (ionization energy increases).
It becomes easier to gain an electron (electron affinity becomes more negative, and electronegativity increases).

Now, metallic character is essentially the tendency to lose electrons and form positive ions. Non-metallic character is the tendency to gain electrons and form negative ions. So as we go right, losing electrons gets harder and gaining them gets easier — hence the shift from metallic to non-metallic.

Let's walk through a concrete example: Period 3.

Start with sodium (Na) — Group 1. It has one valence electron, very low ionization energy, and readily loses that electron to form Na⁺. It's a classic metal: shiny, malleable, good conductor.
Move to magnesium (Mg) — Group 2. Two valence electrons, slightly higher ionization energy. Still metallic, but less reactive than sodium.
Aluminium (Al) — Group 13. Three valence electrons. It shows some metallic properties (conducts electricity) but its oxide is amphoteric — a hint of non-metallic behaviour creeping in.
Silicon (Si) — Group 14. A metalloid. It looks metallic but behaves chemically like a non-metal. Its ionization energy is high enough that it doesn't readily form cations. …

Start your 14-day free trial to unlock the full solution →

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: $B<C<N<F$ …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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: $Ga\,(30) < In\,(157) < Al\,(660)$ . …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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: $Na>Mg>Al>Si$ . …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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). …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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). …
Start your 14-day free trial to see the full solution →

←Previous questionQ43 — Explain the following: (a) Electronegativity of elements increase on m…→Next questionQ45 — The radius of Na^+ cation is less than that of Na atom. Give reason.

Q.How does the metallic and non metallic character vary on moving from left to right in a period?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days