Periodic Trends in Chemical Properties — Valence, Oxides & Anomalous Behaviour

Chemical Properties Inherit Periodicity

Every periodic property we've studied so far — atomic radius, IE, $\Delta_{\text{eg}}H$, electronegativity — was a physical property. But since chemistry is the behaviour of electrons, these physical trends inevitably force chemical properties to be periodic too.

The three drivers of chemical periodicity

1. Valence electrons control bonding. Elements in the same group have the same valence-electron count, so they form compounds with similar stoichiometry.
2. Effective nuclear charge decides how tightly those electrons are held. High $Z_{\text{eff}}$ = hard to ionise, good at pulling bonding electrons (non-metallic behaviour). Low $Z_{\text{eff}}$ = easy to ionise, good at donating electrons (metallic behaviour).
3. Orbital availability. The second period has no $d$-orbitals, which limits valence to 8 electrons (octet rule). From period 3 onward, expanded valence shells become possible in many compounds such as PF$_{5}$ and SF$_{6}$.

What chemical properties to expect

Because of these drivers, moving across a period or down a group we see systematic changes in:

• Valence (towards H and towards O).
• Oxidation states available.
• Acidic vs basic character of oxides and hydrides.
• Metallic vs non-metallic character.
• Typical reactions (reducing/oxidising strength, reactivity with water/air, etc.).
• Physical state, melting/boiling points, conductivity.

This section knits all of these together under one roof.

Valence — The Simplest Chemical Property

Valence with respect to hydrogen

The hydrogen valence of an element is the number of hydrogen atoms bonded to one of its atoms in the "normal" hydride. For period-3 elements:

Element	Hydride	H-valence
Na	NaH	1
Mg	MgH$_{2}$	2
Al	AlH$_{3}$	3
Si	SiH$_{4}$	4
P	PH$_{3}$	3
S	H$_{2}$S	2
Cl	HCl	1

The H-valence rises from 1 to 4 (in Groups 1, 2, 13, 14) and then falls back from 3 to 1 (in Groups 15, 16, 17). The "peak at Group 14" is a classic pattern.

Quick rule: for Group $n$ (where $n$ = 1, 2, 13, 14, 15, 16, 17),

$$\text{H-valence} = \begin{cases} n & \text{for } n = 1, 2 \\ n - 10 & \text{for } n = 13, 14 \\ 18 - n & \text{for } n = 15, 16, 17 \end{cases}$$

Valence with respect to oxygen

The oxygen valence of an element is the effective number of oxygen atoms needed (counting each O = 2 bonds). For the period-3 elements' highest oxides:

Element	Highest oxide	O-valence
Na	Na$_{2}$O	1
Mg	MgO	2
Al	Al$_{2}$O$_{3}$	3
Si	SiO$_{2}$	4
P	P$_{2}$O$_{5}$	5
S	SO$_{3}$	6
Cl	Cl$_{2}$O$_{7}$	7

The O-valence climbs monotonically from 1 (Group 1) to 7 (Group 17). For main-group elements in their highest oxidation state:

$$\text{O-valence} = \begin{cases} n & \text{for } n = 1, 2 \\ n - 10 & \text{for } n = 13, 14, 15, 16, 17 \end{cases}$$

[Board favourite] — The classic question "what is the formula of the highest oxide of element X in Group 15?" — answer: $X_{2}O_{5}$, because O-valence = 5.

Why two different trends?

For H: an atom in Group $n$ can either donate electrons to hydrogen (H-valence equals the electron-donating capacity, Groups 1–4) or accept electrons from hydrogen (H-valence equals 8 minus $n$, Groups 5–7 because they need $(8-n)$ more to complete octet).

For O: an element generally forms oxides in which oxygen is assigned oxidation state $-2$, so the O-valence simply reflects the element's highest valence in its oxide.

Down a group

Going down a group the maximum valence usually stays the same (same group = same valence), but the stability of the highest oxidation state often DECREASES due to the inert-pair effect:

• Pb$^{4+}$ is less stable than Pb$^{2+}$
• Bi$^{5+}$ is less stable than Bi$^{3+}$
• Tl$^{3+}$ is less stable than Tl$^{+}$

The $ns^{2}$ pair becomes inert (reluctant to ionise) in heavier elements due to poor shielding by the intervening $d$ and $f$ electrons.

[JEE Main tip] — "PbO$_{2}$ is a strong oxidising agent" — this follows from the inert-pair effect, since Pb$^{4+}$ wants to revert to Pb$^{2+}$.

Oxide Character — Basic, Amphoteric, Acidic

The acidic-or-basic character of an oxide shifts systematically:

Across a period (Na → Cl)

• Groups 1, 2 (Na, Mg): strongly basic oxides. Na$_{2}$O + H$_{2}$O → 2 NaOH.
• Groups 13, 14 (Al, Si): transitional. Al$_{2}$O$_{3}$ is amphoteric (reacts with both acid and base); SiO$_{2}$ is weakly acidic.
• Groups 15, 16, 17 (P, S, Cl): acidic oxides. SO$_{3}$ + H$_{2}$O → H$_{2}$SO$_{4}$; Cl$_{2}$O$_{7}$ + H$_{2}$O → 2 HClO$_{4}$.

Down a group

Basic character generally increases down a group (because atoms become more metallic). Example: in Group 14 —

• CO$_{2}$: acidic
• SiO$_{2}$: weakly acidic
• GeO$_{2}$: weakly acidic / amphoteric
• SnO$_{2}$: amphoteric
• PbO$_{2}$: amphoteric / weakly acidic (inert-pair quirk)

And lower oxidation-state oxides (SnO, PbO) are basic.

Why the trend

An oxide $M_{x}O_{y}$ can react in two ways in water:

• If the M–O bond is highly ionic (M has low EN, like Na), the oxide releases O$^{2-}$ character, which leads to OH$^{-}$ formation in water and therefore basic solution.
• If the M–O bond is covalent (M has high EN, like S), the oxide behaves as an acidic anhydride and forms an oxoacid in water.

So: low-EN metal → ionic M-O → basic oxide. High-EN non-metal → covalent M-O → acidic oxide.

Amphoteric case study: Al$_{2}$O$_{3}$

Aluminium has intermediate EN (1.61). Its oxide reacts with both:

As a base (reacts with acid): $$Al_{2}O_{3} + 6\,HCl \;\longrightarrow\; 2\,AlCl_{3} + 3\,H_{2}O$$

As an acid (reacts with base): $$Al_{2}O_{3} + 2\,NaOH + 3\,H_{2}O \;\longrightarrow\; 2\,Na[Al(OH)_{4}]$$

This dual behaviour makes Al$_{2}$O$_{3}$ the quintessential amphoteric oxide.

Hydride character tracks electronegativity too

• H bonded to electropositive metal: hydride ion H$^{-}$ present; the compound is ionic hydride (NaH, CaH$_{2}$).
• H bonded to electronegative non-metal: H carries $\delta^{+}$; the compound is acidic hydride (HF, HCl, H$_{2}$S, NH$_{3}$ to varying degrees).
• H bonded to metalloid: covalent hydride (SiH$_{4}$, BH$_{3}$).

[NEET favourite] — Classification of NaH vs HF, or of Al$_{2}$O$_{3}$ vs SO$_{3}$, comes up almost every year. Remember: low-EN partner = basic/ionic; high-EN partner = acidic/covalent.

Metallic ↔ Non-metallic Character

Across a period: metallic character DECREASES

Moving left → right, elements go from metals (Na, Mg) to metalloids (B, Si, Ge) to non-metals (C, N, O, S, Cl). This reflects:

• Increasing IE (harder to lose electrons)
• Increasing EN (easier to gain electrons)
• Increasing non-metallic behaviour

The zig-zag "staircase" line separates metals (left) from non-metals (right), with metalloids (B, Si, Ge, As, Sb, Te) sitting along the boundary.

Down a group: metallic character INCREASES

Going top → bottom, atomic size increases, IE decreases, outer electrons are less tightly held — so elements become more metallic. In Group 14:

• C (non-metal) → Si (metalloid) → Ge (metalloid) → Sn (metal) → Pb (metal)

Reactivity with water

• Group 1 metals: react vigorously with cold water to give H$_{2}$ and MOH. Reactivity increases down the group (Li slow, Cs explosive). $$2\,Na + 2\,H_{2}O \;\longrightarrow\; 2\,NaOH + H_{2}\!\uparrow$$
• Group 2 metals: slower. Mg reacts with hot water/steam; Ca with cold water.
• Transition/post-transition metals: mostly do not react with water (Fe slowly rusts; Al forms protective oxide).

Reactivity with air/oxygen

• Group 1 metals form peroxides or superoxides: Na$_{2}$O$_{2}$, KO$_{2}$.
• Group 2 metals form normal oxides that are basic.
• Halogens generally do not combine directly with oxygen easily; many halogen oxides are unstable. Fluorine is special because it forms oxygen fluorides such as OF$_{2}$, where oxygen is the positive partner.

Standard reduction/oxidation potentials

• Metals have negative $E^{\circ}$ (they are reducing agents), and more negative down Group 1 (Li is the strongest reducing agent in aqueous solution due to hydration energy, even though Cs loses its electron most easily in the gas phase).
• Non-metals have positive $E^{\circ}$ (they are oxidising agents), and less positive down the group. F is the strongest oxidising element.

[JEE/NEET classic trap] — In Group 1, the gas-phase trend for IE (Li highest, Cs lowest) suggests Cs is the best reducing agent. But in aqueous solution, Li is the best because its small size gives enormous hydration energy. Learn to distinguish between "ease of ionisation in gas phase" and "reducing power in solution".

Amphoteric metals

A handful of metals form amphoteric oxides: Al, Zn, Sn, Pb, Be (alkaline earth diagonal). These are the pre-requisite for many PYQ questions about "dissolves in both HCl and NaOH".

Connection to EN

Pauling EN is the cleanest single predictor:

• $\chi < 1.8$: metal (usually)
• $\chi = 1.8$ to 2.1: metalloid
• $\chi > 2.1$: non-metal

Anomalous Behaviour of Second-Period Elements

The first member of each $p$-block group (Li, Be, B, C, N, O, F) differs noticeably from the rest of its group. Three causes:

1. Small atomic size: leads to very high charge density, strong polarisation of neighbouring electrons (Fajan's rules).
2. High electronegativity: makes bonds more polar and more covalent in compounds that would be ionic for the heavier group members.
3. Absence of $d$-orbitals in the valence shell: strict octet limit, no expanded valence.

Element-by-element snapshot

Li vs Na — LiCl shows much more covalent character than NaCl and is soluble in alcohol. LiNO$_{3}$ decomposes to Li$_{2}$O, unlike NaNO$_{3}$ which gives NaNO$_{2}$. Li resembles Mg (diagonal relationship): both form normal oxides, nitrides with N$_{2}$, and carbonates that decompose to oxides.

Be vs Mg — BeCl$_{2}$ is a polymeric covalent solid, whereas MgCl$_{2}$ is predominantly ionic. Be(OH)$_{2}$ is amphoteric; Mg(OH)$_{2}$ is purely basic. Be resembles Al (diagonal).

B vs Al — B is a metalloid; Al is a metal. B$_{2}$H$_{6}$ is electron-deficient (3-centre 2-electron bonds); AlH$_{3}$ is polymeric. Neutral boron compounds are commonly tricoordinate and electron-deficient, although tetra-coordinate borates such as [B(OH)$_{4}$]$^{-}$ are also known. Aluminium more readily forms 4- or 6-coordinate compounds.

C vs Si — Carbon shows catenation on an unmatched scale (long C–C–C chains → organic chemistry); Si forms mostly Si–O frameworks (silicates). C forms strong multiple bonds (C=C, C≡C); Si rarely does because $3p$ orbitals are too diffuse for effective $\pi$ overlap.

N vs P — N$_{2}$ is an exceptionally unreactive diatomic because of its triple bond (bond energy ~945 kJ/mol). P is reactive and exists as P$_{4}$ tetrahedra. NF$_{3}$ is stable; NCl$_{5}$ does not exist because N has no available valence-shell expansion beyond the octet. PF$_{5}$ and PCl$_{5}$ are stable.

O vs S — O is a diatomic gas (O$_{2}$, bond order 2); S is a solid with S$_{8}$ rings. O has maximum covalency of 2 (octet limit); S can expand to 6 (SF$_{6}$). O–O $\pi$-bonds are strong; S=S is weak.

F vs Cl — F is the most electronegative element but has a less negative $\Delta_{\text{eg}}H$ than Cl (compact-shell repulsion). F–F bond is unusually weak (~159 kJ/mol vs Cl–Cl's 242 kJ/mol) due to lone-pair repulsion in the tiny $2p$ shell. F shows only the –1 oxidation state; Cl can be +1, +3, +5, +7 in oxoacids.

The three diagonal relationships

Due to similar size, charge density, and related periodic effects, the following pairs behave alike:

• Li–Mg
• Be–Al
• B–Si

This is the diagonal relationship, a JEE/NEET classic. See solved examples 10 (in Section 10) and 6 (in this section) for worked comparisons.

[JEE Advanced favourite] — A question that asks "explain why $PCl_{5}$ exists but $NCl_{5}$ does not" is tagged to the absence of period-2 valence-shell expansion — a canonical anomalous-behaviour question.

Cheat-Sheet & Quick Decision Rules

The twin master arrows

$$\boxed{\;\text{Across period (left → right): metallic character} \downarrow,\; \text{non-metallic character} \uparrow,\; \text{oxide becomes more acidic}\;}$$

$$\boxed{\;\text{Down a group: metallic character} \uparrow,\; \text{non-metallic character} \downarrow,\; \text{oxide becomes more basic}\;}$$

Quick lookup: predict the oxide's nature

Given an element, you can usually tell its oxide's nature from a glance at the periodic table:

Region	Typical oxide
Left (Groups 1, 2)	Strongly basic (Na$_{2}$O, MgO)
Middle (Al, Zn, Sn, Pb, Be)	Amphoteric
Right (C-block and beyond)	Acidic (CO$_{2}$, SO$_{3}$, Cl$_{2}$O$_{7}$)
Lower-right non-metals	Very strongly acidic

Five-test rule for "Why is X anomalous?" questions

Run through these five checks whenever an anomaly question appears:

1. Is it a second-period element? (Li, Be, B, C, N, O, F) → size + EN + no d-orbitals.
2. Does the compound involve a maximum valence beyond 4 (e.g., NCl$_{5}$)? → no valence-shell expansion in period 2 → anomaly.
3. Is there catenation (long chains of the same element)? → C is special.
4. Is there strong $\pi$-bonding in a double/triple bond? → C=C, N≡N, O=O favoured; Si, P, S much less so.
5. Is the compound unexpectedly covalent (like BeCl$_{2}$ or LiCl)? → Fajan's rules + small cation + high polarising power.

Ten high-yield facts for MCQ speed

1. Most EN element: F (Pauling 3.98).
2. Most negative $\Delta_{\text{eg}}H$: Cl ($-349$ kJ/mol).
3. Highest first IE: He (2372 kJ/mol).
4. Largest atom among stable elements: Cs.
5. Most metallic: Cs/Fr.
6. Most acidic oxide in Period 3: Cl$_{2}$O$_{7}$.
7. Strongly basic oxides include Na$_{2}$O and K$_{2}$O.
8. Amphoteric champion: Al$_{2}$O$_{3}$.
9. Period-2 elements are restricted to octet-type behaviour in simple compounds.
10. Strongest diagonal resemblance: Li ↔ Mg (and Be ↔ Al).

Five traps to avoid

1. "F has the most negative $\Delta_{\text{eg}}H$ because it's most EN." WRONG — Cl has more negative $\Delta_{\text{eg}}H$.
2. "All period-3 oxides are acidic." WRONG — Na$_{2}$O and MgO are basic.
3. "Metallic character increases across a period." WRONG — it decreases.
4. "BeCl$_{2}$ is a typical ionic salt like CaCl$_{2}$." WRONG — BeCl$_{2}$ is covalent/polymeric.
5. "N forms $NCl_{5}$ analogous to $PCl_{5}$." WRONG — period-2 nitrogen cannot exceed the octet in such compounds.

Solved Examples

Example 1: Valence of an unknown

Element X is in Group 15. What is the formula of its hydride and its highest oxide?

Solution:

Group 15: H-valence $= 18 - 15 = 3$. O-valence $= 15 - 10 = 5$.

• Hydride formula: $XH_{3}$ (e.g., NH$_{3}$, PH$_{3}$, AsH$_{3}$).
• Highest oxide formula: $X_{2}O_{5}$ (e.g., N$_{2}$O$_{5}$, P$_{2}$O$_{5}$, As$_{2}$O$_{5}$).

Takeaway: H-valence peaks at Group 14 (4) and falls to 1 at Group 17; O-valence climbs monotonically to 7 at Group 17.

Example 2: Classify the oxide

Classify the following oxides as acidic, basic, or amphoteric: Na$_{2}$O, Al$_{2}$O$_{3}$, SO$_{3}$, MgO, Cl$_{2}$O$_{7}$.

Solution:

Oxide	Central atom EN	Nature
Na$_{2}$O	Na ($\chi = 0.93$)	Strongly basic
MgO	Mg ($\chi = 1.31$)	Basic
Al$_{2}$O$_{3}$	Al ($\chi = 1.61$)	Amphoteric
SO$_{3}$	S ($\chi = 2.58$)	Acidic
Cl$_{2}$O$_{7}$	Cl ($\chi = 3.16$)	Strongly acidic

Takeaway: rising central-atom EN progressively shifts oxide character from basic → amphoteric → acidic.

Example 3: Amphoteric oxide equations

Write balanced equations showing that Al$_{2}$O$_{3}$ is amphoteric.

Solution:

Acts as a base (reacts with acid):

$$Al_{2}O_{3}(s) + 6\,HCl(aq) \;\longrightarrow\; 2\,AlCl_{3}(aq) + 3\,H_{2}O(l)$$

Acts as an acid (reacts with base):

$$Al_{2}O_{3}(s) + 2\,NaOH(aq) + 3\,H_{2}O(l) \;\longrightarrow\; 2\,Na[Al(OH)_{4}](aq)$$

This shows that Al$_{2}$O$_{3}$ reacts with both acids and bases, which is the defining feature of an amphoteric oxide.

Takeaway: an oxide is amphoteric only if it reacts with BOTH strong acid and strong base. Al$_{2}$O$_{3}$, ZnO, SnO, PbO, BeO fit this bill.

Example 4: Why doesn't NCl$_{5}$ exist?

NCl$_{3}$ is a known compound but NCl$_{5}$ does not exist. Why, when PCl$_{5}$ is stable?

Solution:

Both N and P are Group 15 elements, so both have 5 valence electrons.

• N has only $2s$ and $2p$ valence orbitals and cannot accommodate more than 8 electrons around itself in such compounds. So only NCl$_{3}$ (with 3 bonds + 1 lone pair) is accessible.
• P, being a period-3 element, can form hypervalent compounds such as PCl$_{5}$ and PF$_{5}$.

In summary: N cannot expand its octet in this case; P can form 5-coordinate compounds.

Takeaway: period-2 elements are octet-limited in ordinary compounds, which is why NCl$_{5}$ does not exist while PCl$_{5}$ does.

Example 5: Predict reactivity with water

Which of Na, Mg, Al, Si reacts most vigorously with water? Justify.

Solution:

Reactivity with water depends on metallic character and ease of electron loss.

• Na: reacts violently with cold water → NaOH + H$_{2}$.
• Mg: reacts slowly with cold water, vigorously with hot water/steam.
• Al: reacts very slowly because of a protective Al$_{2}$O$_{3}$ coating.
• Si: does not react with water under ordinary conditions.

Sodium is therefore the most reactive:

$$2\,Na(s) + 2\,H_{2}O(l) \;\longrightarrow\; 2\,NaOH(aq) + H_{2}(g)\!\uparrow$$

Takeaway: reactivity with water decreases left → right in a period (metallic to non-metallic) and generally increases down Groups 1 and 2.

Example 6: Diagonal relationship

Explain why Li shows more similarity to Mg than to Na, giving three common chemical points of similarity.

Solution:

Reasons:

• Similar ionic radii: Li$^{+}$ (76 pm) vs Mg$^{2+}$ (72 pm) — close; Na$^{+}$ (102 pm) is much larger.
• Similar charge density and polarising power: both Li$^{+}$ and Mg$^{2+}$ are small ions with strong polarising effect.
• Their positions in the periodic table produce a diagonal balance of periodic trends, giving broadly similar chemical behaviour.

Three common points of similarity:

1. Reaction with N$_{2}$: Both form nitrides (Li$_{3}$N, Mg$_{3}$N$_{2}$) when heated in N$_{2}$. Other alkali metals do not.
2. Thermal decomposition of carbonate: Li$_{2}$CO$_{3}$ → Li$_{2}$O + CO$_{2}$ on heating; MgCO$_{3}$ → MgO + CO$_{2}$. Na$_{2}$CO$_{3}$ is thermally stable.
3. Normal oxide formation: Li$_{2}$O and MgO are the primary products from burning in air; Na gives peroxide (Na$_{2}$O$_{2}$), K gives superoxide.

Takeaway: diagonal similarity arises from a combination of size, charge density, and related periodic effects, not from electronegativity closeness alone.

Example 7: Acidic character of hydrides

Arrange HF, HCl, HBr, HI in order of increasing acidic strength in water.

Solution:

Acidic strength in water depends more on bond-dissociation energy than on electronegativity alone.

• H–F: bond energy 569 kJ/mol (very strong)
• H–Cl: 431 kJ/mol
• H–Br: 366 kJ/mol
• H–I: 299 kJ/mol (weakest)

The weaker the H–X bond, the more readily H$^{+}$ is released in water, hence more acidic. Therefore:

$$HF < HCl < HBr < HI$$

HF is only a weak acid in water despite being the most polar; HI is a strong acid.

Takeaway: across the halogens, acidic strength tracks bond energy (weaker bond → stronger acid), not electronegativity.

Example 8: Inert-pair effect

Why is Pb$^{2+}$ more stable than Pb$^{4+}$, while C(IV) is more stable than C(II)?

Solution:

Carbon has configuration $2s^{2}\,2p^{2}$. Both $s$ and $p$ electrons are valence electrons and readily participate in bonding, so the +4 state is normal.

Lead has configuration $[Xe]\,4f^{14}\,5d^{10}\,6s^{2}\,6p^{2}$. The $6s^{2}$ pair experiences poor shielding by the intervening $4f^{14}$ and $5d^{10}$ electrons, so it is held strongly and tends to remain non-bonding. Lead therefore prefers the +2 state, in which only the $6p^{2}$ electrons are used.

This is the inert-pair effect, most pronounced in heavier $p$-block elements such as Tl, Pb, and Bi.

Takeaway: going down a $p$-block group, the lower oxidation state often becomes more stable than the group oxidation state because the $ns^{2}$ pair becomes inert. PbO$_{2}$ is a strong oxidising agent because Pb(IV) tends to revert to Pb(II).

Example 9: Reducing and oxidising strength

Why is fluorine the strongest oxidising agent in the periodic table?

Solution:

An oxidising agent gains electrons and gets reduced. Strength depends on:

1. Electron gain enthalpy / electron affinity.
2. Bond dissociation energy of X$_{2}$ (easier to break → easier to react).
3. Hydration energy of X$^{-}$ in aqueous solution.

For F:

• $\Delta_{\text{eg}}H(\text{F}) = -328$ kJ/mol (less negative than Cl but still strongly favourable).
• F–F bond is unusually weak (~159 kJ/mol) due to lone-pair repulsion in the small $2p$ shell.
• Hydration of F$^{-}$ is very large because F$^{-}$ is small and highly charge-dense.

The combination of easy bond breaking and very large hydration energy gives F$_{2}$ the highest standard reduction potential (+2.87 V). That makes F$_{2}$ the strongest oxidising agent.

Takeaway: oxidising strength in solution is not determined by electron gain enthalpy alone. Bond dissociation and hydration energy are also crucial.

Example 10: Formula prediction for oxoacids

Predict the formula of the highest-oxidation-state oxoacid of chlorine, sulfur, and phosphorus.

Solution:

Highest oxidation states:

• Cl (Group 17): +7 → HClO$_{4}$ (perchloric acid)
• S (Group 16): +6 → H$_{2}$SO$_{4}$ (sulfuric acid)
• P (Group 15): +5 → H$_{3}$PO$_{4}$ (phosphoric acid)

Takeaway: knowing the group and highest oxidation state is usually enough to predict the formula of the common highest oxoacid: HClO$_{4}$, H$_{2}$SO$_{4}$, H$_{3}$PO$_{4}$, H$_{4}$SiO$_{4}$.

Example 11: Reactivity of alkali metals

Arrange Li, Na, K, Rb, Cs in order of increasing reactivity with water.

Solution:

Down Group 1, atomic size increases and the outermost $s$ electron becomes progressively easier to lose (lower IE). This makes the metals more reactive toward water.

• Li: reacts gently with water.
• Na: faster, with a yellow flame sometimes visible.
• K: often ignites evolved H$_{2}$ (lilac flame).
• Rb: reacts even more violently.
• Cs: extremely explosive.

So reactivity with water increases as:

$$Li < Na < K < Rb < Cs$$

Subtle point: despite this, in terms of standard reduction potential in aqueous solution, Li is actually the most negative because of its very high hydration energy. So "Li is the strongest reducing agent in water" and "Cs is the most reactive with water" can both be true — they refer to different ideas.

Takeaway: for reactivity with water in the plain kinetic sense, go with the "down the group = more reactive" trend.

Example 12: Identify element from clues

An element X has IE$_{1}$ $\approx 786$ kJ/mol, $\Delta_{\text{eg}}H \approx -134$ kJ/mol, atomic radius $\approx 111$ pm, and forms an oxide XO$_{2}$ which is weakly acidic and polymeric. Identify X.

Solution:

• XO$_{2}$ formula → O-valence = 4 → Group 14.
• "Weakly acidic, polymeric oxide" strongly suggests SiO$_{2}$.
• Check the data: IE$_{1}$ of Si is about 786 kJ/mol, atomic radius about 111 pm, and $\Delta_{\text{eg}}H$ about $-134$ kJ/mol.

Best fit: X = Silicon (Si).

Takeaway: group + oxide formula is usually enough to identify the element; the other properties confirm it.