Metallic - non-metallic character trends: Exercises

Before we start, four tools we will lean on every time — stated in plain words so nothing is used before it is built:

Definition The levers (plain words)

==Effective nuclear charge $Z_{eff}$ == — the net positive pull a valence (outermost) electron actually feels. Written $Z_{eff} = Z - S$ : take all the protons $Z$ , subtract the "screen" $S$ from inner electrons that block part of that pull. Bigger $Z_{eff}$ = tighter grip. See Effective Nuclear Charge (Zeff).
==Atomic radius $r$ == — how far the outer electron sits from the nucleus. Farther = looser grip, because pull weakens with distance. See Atomic and Ionic Radii.
Ionization energy — the energy needed to rip off an electron. We write it $I E_{1}$ , where the ==subscript $1$ means the first ionization energy== — the energy to remove the single easiest (outermost) electron. $I E_{2}$ would be the energy to remove the next one after that, and so on; every number on this page is $I E_{1}$ unless stated. Low $I E_{1}$ ⇔ outer electron leaves easily ⇔ more metallic. See Ionization Energy trends.
==Electron affinity $E A$ == — the energy released when a neutral atom gains an electron. Convention: atoms that love gaining electrons release a lot of energy, so their $E A$ is a large negative number (energy out). Because the sign only encodes "energy leaves," we compare magnitudes $∣ E A ∣$ : a big $∣ E A ∣$ = a strong pull to grab an electron = strongly non-metallic. See Electron Affinity.

Formula Why

$I E_{1} \propto Z_{eff} / r^{2}$ — the derivation This is the single relation the whole page rests on, so we build it, never just assert it.

Coulomb's law (the force between two charges): a nucleus of net charge $+ Z_{eff} e$ pulls an electron of charge $- e$ sitting a distance $r$ away with force $F = \frac{k Z _{eff} e ^{2}}{r ^{2}} .$ Here $k$ and $e$ are just fixed constants; what varies between atoms is $Z_{eff}$ and $r$ . So the pull grows with $Z_{eff}$ and dies off as $1/ r^{2}$ (twice as far ⇒ one-quarter the pull).
Energy $\approx$ force $\times$ distance. To free the electron you must drag it out against that pull, from its shell at radius $r$ to "infinity." A quick estimate of that work is the force at the shell times the distance scale $r$ : $I E_{1} \approx F \times r = \frac{k Z _{eff} e ^{2}}{r ^{2}} \times r = \frac{k Z _{eff} e ^{2}}{r} .$ (The full Bohr result carries an extra $Z_{eff}$ from the orbit-size relation, giving $I E_{1} \propto Z_{eff}^{2} / n^{2}$ — but the takeaway is identical: up with $Z_{eff}$ , down with $r$ .)
Keeping the plain force-picture, the ratio that decides "how hard is the grip" is $I E_{1} \propto \frac{Z _{eff}}{r ^{2}} (force form) or \frac{Z _{eff}}{r} (energy form) .$ Either way: bigger $Z_{eff}$ tightens the grip, bigger $r$ loosens it.

And metallic character is just the opposite of grip strength: $Metallic character \propto \frac{1}{I E _{1}} .$ Read it as: tighter grip (big $Z_{eff}$ , small $r$ ) → high $I E_{1}$ → low metallic character.

The compass figure below turns those two levers into a picture of the whole table — read it like this: it is a coarse periodic-table grid; the blue square (bottom-left) marks the most metallic corner (Cs, Fr), the red square (top-right) marks the champion non-metal (F). The yellow arrow along the top points right and is labelled "metallic DOWN" (across a period metallic character falls); the green arrow down the left points down and is labelled "metallic UP" (down a group it rises); the dashed white diagonal points from top-right toward bottom-left in the direction of increasing metallic character. Glance at it whenever you're unsure which way "more metallic" points.

Figure — Metallic - non-metallic character trends

Level 1 — Recognition

Goal: read the direction off the table.

Recall Solution L1.1

Same row means same outer shell, so we compare left vs right.

Moving right K → Ca adds one proton into the same shell. Same-shell electrons shield each other poorly, so $Z_{eff}$ rises.
Radius $r$ shrinks (tighter pull pulls the cloud in).
Tighter grip → higher $I E_{1}$ → harder to lose an electron.

So K is more metallic. ✔ Sanity check with real $I E_{1}$ : $I E_{1} (K) = 419 kJ/mol < I E_{1} (Ca) = 590 kJ/mol$ . Lower $I E_{1}$ ⇒ more metallic. ✅

Recall Solution L1.2

Down a group: each row adds a whole new shell, so $r$ jumps up while $Z_{eff}$ barely moves (inner shells shield well). Loose grip → more metallic.
To the left across a period: less $Z_{eff}$ , bigger $r$ than the right side → looser grip → more metallic.

Answer: down a group, and to the left (i.e. it decreases going right) across a period.

Level 2 — Application

Goal: apply the $Z_{eff}$ / $r$ logic to a fresh pair, then verify with numbers.

Recall Solution L2.1

Across Period 3, left → right, the naive rule says metallic character falls: Na > Mg > Al. But look at the $I E_{1}$ numbers — Al's $I E_{1}$ (578) is lower than Mg's (738)! That means Al loses an electron more easily than Mg.

Why the dip: Mg is $[Ne] 3 s^{2}$ (a filled, stable $3 s$ subshell). Al is $[Ne] 3 s^{2} 3 p^{1}$ — its lone $3 p$ electron is higher in energy and a bit shielded by the $3 s$ pair, so it pops off easily. This is a small local wobble in the Ionization Energy trends.
Lower $I E_{1}$ ⇒ more metallic, so on the $I E_{1}$ evidence: Na (496) > Al (578) > Mg (738).

Order: Na > Al > Mg. ✔ (The broad trend still holds Na most metallic; the Mg/Al swap is the twist.)

Recall Solution L2.2

Oxide chemistry is a test strip for character (see Acidic Basic Amphoteric Oxides):

Metallic atom → basic oxide. Na is far left → very metallic → $Na_{2} O$ is basic: $Na_{2} O + H_{2} O \to 2 NaOH$
Non-metallic atom → acidic oxide. S is far right → non-metallic → $SO_{3}$ is acidic: $SO_{3} + H_{2} O \to H_{2} SO_{4}$

So $Na_{2} O$ basic, $SO_{3}$ acidic. ✔

Level 3 — Analysis

Goal: pull apart competing effects (period vs group pulling opposite ways).

Recall Solution L3.1

Set up the tug-of-war explicitly:

Group pull (Al is one period down): would make Al more metallic — bigger $r$ .
Period pull (Al is 12 columns right): would make Al less metallic — much bigger $Z_{eff}$ .

Which wins? Use the referee: $I E_{1}$ . Li = 520 vs Al = 578. Al's $I E_{1}$ is higher, so Al holds its electron tighter → Li is more metallic. ✔ Interpretation: moving 12 columns right stacks up so much $Z_{eff}$ that one extra shell can't undo it. The period effect wins this diagonal.

Recall Solution L3.2

The oxide nature simply reads out the metal→non-metal slide:

Na (very metallic) → $Na_{2} O$ basic (reacts with acids).
Al (borderline) → $Al_{2} O_{3}$ amphoteric = it reacts both with acids and with bases, because Al sits right on the metal/non-metal boundary. Example both ways: $Al_{2} O_{3} + 6 HCl \to 2 AlCl_{3} + 3 H_{2} O (acts as base)$ $Al_{2} O_{3} + 2 NaOH + 3 H_{2} O \to 2 NaAl(OH)_{4} (acts as acid)$
S (non-metallic) → $SO_{3}$ acidic.

As metallic character falls left→right, oxides march basic → amphoteric → acidic. ✔ (Acidic Basic Amphoteric Oxides)

Level 4 — Synthesis

Goal: weave several trends into one prediction.

Recall Solution L4.1

Cs is bottom-left, F is top-right — the two champions on opposite ends of the single metallic ↔ non-metallic axis. Chain all four levers:

Radius: Cs has a huge $r$ (six shells); F has a tiny $r$ (two shells). Far electron = loose grip.
$Z_{eff}$ : F's few inner electrons shield poorly, so its valence electrons feel a strong net pull; Cs's outer $6 s$ electron is buried behind many shells → weak net pull.
$I E_{1}$ : the pull ranking shows up directly — $I E_{1} (Cs) = 376 ≪ I E_{1} (F) = 1681$ . Cs sheds an electron with almost nothing; F clings hard.
$E N$ (electronegativity, the pull inside a bond, see Electronegativity trends): $E N (Cs) = 0.79$ (barely pulls shared electrons) vs $E N (F) = 3.98$ (the highest of all elements). Also mirrors Electron Affinity: F loves gaining an electron (large $∣ E A ∣$ ).

Every lever agrees: Cs is overwhelmingly more metallic; F is the strongest non-metal. ✔ Quantify the gap: $I E_{1} (F) / I E_{1} (Cs) = 1681/376 \approx 4.47$ — F's electron is held about 4.5× harder.

Recall Solution L4.2

Recall $∣ E A ∣$ is the size of the energy released on gaining an electron — big $∣ E A ∣$ = strong hunger for an extra electron.

$X$ : low $I E_{1}$ (500) means its electron leaves cheaply → $X$ is metallic → forms a cation $X^{+}$ . Its tiny $∣ E A ∣$ (50) shows it barely wants extra electrons.
$Y$ : high $I E_{1}$ (1300) means $Y$ keeps its own electrons, and large $∣ E A ∣$ (330) means it grabs more → $Y$ is non-metallic → forms an anion $Y^{-}$ .
In a compound, the metallic partner donates: $X$ becomes $X^{+}$ , $Y$ becomes $Y^{-}$ , giving ionic $X^{+} Y^{-}$ . ✔ Rule extracted: electron flows from low- $I E_{1}$ (metallic) to high- $∣ E A ∣$ (non-metallic).

Level 5 — Mastery

Goal: handle the subtle, exception-laden, "why does the rule bend" cases.

Recall Solution L5.1

By the naive rule Fr (one shell below Cs) should win. But two subtleties intervene:

Relativistic contraction: Fr's $7 s$ electron moves fast enough that relativistic effects pull the $7 s$ orbital slightly inward, tightening the grip a little — partly cancelling the extra-shell loosening.
Data scarcity: Fr is intensely radioactive and rare, so its $I E_{1}$ is estimated, not cleanly measured.

Result: measured $I E_{1} (Fr) \approx 393$ kJ/mol is slightly higher than $I E_{1} (Cs) \approx 376$ kJ/mol, implying Cs is the practical champion of metallic character. ✔ Takeaway: trends are guides, not laws — always let $I E_{1}$ arbitrate the close calls.

Recall Solution L5.2

(a) Why the dip: Configurations — P is $[Ne] 3 s^{2} 3 p^{3}$ (each $3 p$ orbital singly filled, a stable half-filled set). S is $[Ne] 3 s^{2} 3 p^{4}$ — its fourth $3 p$ electron must pair up in an already-occupied orbital, and electron–electron repulsion in that pair makes it easier to remove. So S's outermost electron leaves more cheaply → $I E_{1} (S) < I E_{1} (P)$ . (b) Does S become more metallic? On this single measure, yes — locally S loses that electron more easily than P, so at this micro-level S looks a hair more metallic. But the broad Period-3 trend still climbs to Cl (1251), so both P and S remain clearly non-metals. The dip is a real, explainable wobble — not experimental error. ✔

Recall Solution L5.3

Litmus reports acid/base, and oxide acidity mirrors character:

Red → blue means the solution is basic → came from a basic oxide → from a metallic element (like Na near the left, giving NaOH).
Blue → red means acidic → acidic oxide → from a non-metallic element (like S near the right, giving $H_{2} SO_{4}$ ).

So the base-forming oxide's element is more metallic and sits farther left; the acid-forming oxide's element is more non-metallic and sits farther right. ✔ This is the whole chapter compressed into one bench test: one drop of litmus tells you which side of the period the parent element lives on, because oxide acid/base nature is the visible fingerprint of the metallic ↔ non-metallic axis.

Wrap-up recall

Recall One-line answers (cover them)

Two levers controlling metallic character? ::: $Z_{eff}$ and radius $r$ . What does the subscript in $I E_{1}$ mean? ::: The first ionization energy — removing the single easiest (outermost) electron. Low first ionization energy means the atom is more...? ::: Metallic. Direction of maximum metallic character on the table? ::: Bottom-left (Cs region). Why is $I E_{1} (Al) < I E_{1} (Mg)$ ? ::: Al's lone $3 p$ electron is higher-energy and shielded by the filled $3 s^{2}$ , so it leaves easily. Why is $I E_{1} (S) < I E_{1} (P)$ ? ::: S must pair a $3 p$ electron; pair repulsion makes removal easier (P has a stable half-filled $3 p^{3}$ ). Why do we compare $∣ E A ∣$ rather than $E A$ itself? ::: $E A$ is negative by convention (energy released on gaining an electron); the magnitude measures how strongly the atom grabs one. A basic oxide implies its element is...? ::: Metallic (left side). Is Fr guaranteed more metallic than Cs? ::: No — relativistic contraction + sparse data make Cs the practical champion.

Connections

Effective Nuclear Charge (Zeff) — the pull behind every prediction here.
Ionization Energy trends — our numerical referee ( $I E_{1}$ dips at Al and S).
Electronegativity trends — the bonding-side mirror of non-metallic character.
Atomic and Ionic Radii — the $r$ lever.
Acidic Basic Amphoteric Oxides — the oxide test strip (L2, L3, L5).
Electron Affinity — the anion-forming drive (L4).