Valence from electronic configuration
The valence of an element is its combining capacity, the number of bonds its atom forms or the number of hydrogen atoms it can combine with. NCERT states the rule simply: for representative elements the valence is usually equal to the number of electrons in the outermost orbitals, or equal to eight minus that number, whichever is appropriate. Because the outer configuration is what repeats down a group, valence is itself a periodic property.
The NIOS treatment makes the cut-off explicit. When an atom has four or fewer valence electrons it tends to share or lose all of them, so the valence equals the number of valence electrons. When it has more than four, it is easier to complete the octet by gaining the deficit, so the valence equals eight minus the count.
| Outer electrons | Rule applied | Valence | Representative groups |
|---|---|---|---|
| 1 | = number of e⁻ | 1 | Group 1 (Li, Na, K) |
| 2 | = number of e⁻ | 2 | Group 2 (Be, Mg, Ca) |
| 3 | = number of e⁻ | 3 | Group 13 (B, Al) |
| 4 | = number of e⁻ | 4 | Group 14 (C, Si) |
| 5 | = 8 − e⁻ | 3 | Group 15 (N, P) |
| 6 | = 8 − e⁻ | 2 | Group 16 (O, S) |
| 7 | = 8 − e⁻ | 1 | Group 17 (F, Cl) |
| 8 | = 8 − e⁻ | 0 | Group 18 (Ne, Ar) |
NIOS adds the important caveat that elements of the third and higher periods can show higher valencies than the octet rule predicts, because empty d orbitals let them accommodate more than eight electrons in the valence shell. This is precisely why sulphur reaches a valence of 6 in $\ce{SO3}$ and chlorine reaches 7 in $\ce{Cl2O7}$, while their second-period analogues oxygen and fluorine cannot.
Periodicity of valence across a period
Walk across a period and the valence does not simply rise; it rises and then falls, depending on whether you measure it against hydrogen or against oxygen. Hydrogen forms one bond, so the valence with respect to hydrogen follows the "lower of the two" rule and traces the pattern 1, 2, 3, 4, 3, 2, 1, 0. Oxygen forms two bonds and pulls electrons the other way, so the valence with respect to oxygen climbs steadily to a maximum equal to the group number.
| Period-3 element | Outer config | Hydride | Valence (H) | Highest oxide | Valence (O) |
|---|---|---|---|---|---|
| Na | 3s¹ | $\ce{NaH}$ | 1 | $\ce{Na2O}$ | 1 |
| Mg | 3s² | $\ce{MgH2}$ | 2 | $\ce{MgO}$ | 2 |
| Al | 3s²3p¹ | $\ce{AlH3}$ | 3 | $\ce{Al2O3}$ | 3 |
| Si | 3s²3p² | $\ce{SiH4}$ | 4 | $\ce{SiO2}$ | 4 |
| P | 3s²3p³ | $\ce{PH3}$ | 3 | $\ce{P4O10}$ | 5 |
| S | 3s²3p⁴ | $\ce{H2S}$ | 2 | $\ce{SO3}$ | 6 |
| Cl | 3s²3p⁵ | $\ce{HCl}$ | 1 | $\ce{Cl2O7}$ | 7 |
The contrast is the heart of the topic. Against hydrogen the valence is symmetric, peaking at silicon in the centre; against oxygen it grows monotonically because the heavier p-block elements can expand their octet and use all their valence electrons in bonding to the electronegative oxygen. NCERT's Table 3.9 lists the corresponding formulae of hydrides and oxides that encode exactly these valences.
Using only the periodic table, predict the formula of the compound between (a) silicon and bromine, (b) aluminium and sulphur.
(a) Silicon is a group-14 element with valence 4; bromine is a halogen with valence 1. Combining 4 with 1 gives $\ce{SiBr4}$.
(b) Aluminium (group 13) has valence 3; sulphur (group 16) has valence 2. Cross-multiplying the valences gives $\ce{Al2S3}$. This is NCERT Problem 3.8 worked from group numbers alone.
Valence vs oxidation state
The term oxidation state is now used in place of valence in most contexts, but the two are not identical. Valence counts the bonds and is an unsigned whole number. Oxidation state is the charge an atom would carry if every shared pair were handed entirely to the more electronegative atom, so it carries a sign.
NCERT illustrates the distinction with two oxygen compounds. The electronegativity order is $\ce{F} > \ce{O} > \ce{Na}$. In $\ce{OF2}$ fluorine is the most electronegative, so it is assigned $-1$ each, and oxygen, sharing with two fluorines, takes oxidation state $+2$. In $\ce{Na2O}$ oxygen is now the more electronegative partner and accepts an electron from each sodium, taking oxidation state $-2$, while each sodium is $+1$.
The oxidation state of an element in a compound is the charge acquired by its atom on the basis of an electronegativity-based assignment of every bonding pair to the more electronegative atom.
Oxygen's valence is 2 in both molecules; only its oxidation state flips sign, from $+2$ in $\ce{OF2}$ to $-2$ in $\ce{Na2O}$. Note also that oxidation state need not equal covalency. In $\ce{[AlCl(H2O)5]^2+}$ the oxidation state of aluminium is $+3$ but its covalency (the number of bonds it forms) is 6, which is NCERT Problem 3.9.
Valence, oxidation state and covalency are three different numbers
Students collapse all three into one. Valence is unsigned bond count; oxidation state is a signed charge from electronegativity bookkeeping; covalency is the actual number of bonds. For Al in $\ce{[AlCl(H2O)5]^2+}$ these are 3, $+3$ and 6 respectively. They coincide for simple ionic binaries but diverge in coordination species and in compounds with electronegativity reversals such as $\ce{OF2}$.
Ask first which partner is more electronegative; that single decision sets the sign of the oxidation state.
Anomalous second-period elements
The first member of each main group, lithium and beryllium in groups 1 and 2 and boron through fluorine in groups 13 to 17, differs markedly from the heavier members of its own group. Lithium, unlike the other alkali metals, and beryllium, unlike the other alkaline earth metals, form compounds with pronounced covalent character, whereas the heavier members are predominantly ionic.
NCERT attributes this anomaly to three structural facts about the small first-row atoms: their small size, their large charge-to-radius ratio and their high electronegativity. Two consequences follow directly and both are heavily examined.
| Structural cause | Consequence | Illustration |
|---|---|---|
| Only four valence orbitals (2s, 2p); no 2d | Maximum covalency capped at 4 | Boron forms only $\ce{BF4^-}$; aluminium reaches $\ce{AlF6^3-}$ |
| Heavier members have nine valence orbitals (3s, 3p, 3d) | Can expand the octet beyond 4 | $\ce{SF6}$, $\ce{PCl5}$ exist; $\ce{NF5}$ does not |
| Small size favours close p-orbital overlap | Strong pπ–pπ multiple bonds | $\ce{C=C}$, $\ce{C#C}$, $\ce{N#N}$, $\ce{C=O}$, $\ce{C#N}$ |
The first member's ability to form $p\pi\text{–}p\pi$ multiple bonds, both to itself and to other second-period atoms, is what gives carbon and nitrogen their rich chemistry of double and triple bonds; their heavier congeners silicon and phosphorus largely cannot sustain such bonds and prefer single-bonded, often polymeric, structures instead.
The small size and high electronegativity that drive these anomalies trace back to one master trend. Revisit electronegativity across periods and groups to see why fluorine sits at the extreme.
The diagonal relationship
A direct corollary of second-period anomaly is the diagonal relationship: the first element of a group resembles, in many chemical respects, the element placed diagonally below and to its right in the next group and next period. NCERT states it explicitly, that lithium behaves more like magnesium and beryllium more like aluminium than like the rest of their own groups.
The reason is a cancellation of opposing trends. Moving left to right across a period, atomic size decreases and electronegativity rises; moving top to bottom down a group, size increases and electronegativity falls. Along a diagonal the two motions oppose one another, so the diagonal neighbours end up with comparable size, electronegativity and, most decisively, similar charge-to-radius ratios, which governs polarising power and hence chemical behaviour.
| Diagonal pair | Groups | Shared chemistry |
|---|---|---|
| Li ~ Mg | 1 and 2 | Both form covalent, hygroscopic compounds; nitrides $\ce{Li3N}$, $\ce{Mg3N2}$; carbonates decompose on heating; $\ce{LiOH}$ and $\ce{Mg(OH)2}$ are weak/sparingly soluble |
| Be ~ Al | 2 and 13 | Both give amphoteric oxides/hydroxides; covalent halides that are Lewis acids; passivated by conc. $\ce{HNO3}$ |
| B ~ Si | 13 and 14 | Both are metalloids forming covalent, acidic oxides ($\ce{B2O3}$, $\ce{SiO2}$); volatile, hydrolysable halides; semiconducting/network solids |
The horizontal (period) and vertical (group) trends pull size and electronegativity in opposite senses; their resultant runs along the dashed purple diagonals, leaving Li~Mg, Be~Al and B~Si with matching charge-to-radius ratios and so similar chemistry.
Memorise the three diagonal pairs, not a vague idea
Examiners ask for the partner directly: "Which element resembles beryllium?" The answer is aluminium, not boron and not magnesium. Lock in Li–Mg, Be–Al, B–Si as the down-and-right diagonal. Confusing the diagonal partner with the next group-member is the most common error here.
Diagonal partner = one group right and one period down. Be (group 2, period 2) → Al (group 13, period 3).
Reactivity and metallic character
All chemical behaviour is a manifestation of electronic configuration, so chemical reactivity tracks the same trends as ionisation enthalpy and electron gain enthalpy. Across a period atomic and ionic radii fall, ionisation enthalpy rises, and electron gain enthalpy becomes more negative.
NCERT draws the conclusion neatly: reactivity is highest at the two extremes of a period and lowest in the centre. The extreme-left element has the smallest ionisation enthalpy, so it loses an electron easily to form a cation; the extreme-right halogen has the most negative electron gain enthalpy, so it gains an electron easily to form an anion. Both extremes are reactive, the left through electron loss and metallic reducing behaviour, the right through electron gain and non-metallic oxidising behaviour. Noble gases, with filled shells and positive electron gain enthalpies, are essentially inert.
Left to right across period 3, metallic (electropositive) character falls and non-metallic (electronegative) character rises, with metalloids such as silicon straddling the centre. Down a group the trend reverses, metallic character increasing with atomic number.
The metallic character is therefore highest at the far left and decreases across the period, while the non-metallic character increases. Down a group the radii grow, ionisation enthalpies fall, and metallic character increases with atomic number, the mirror image of the period trend, for main-group elements. NCERT notes that the transition elements show a reversed trend, explained by their small change in size and intermediate ionisation enthalpies.
Nature of oxides across a period
The cleanest chemical reading of metallic-to-non-metallic character is the acid-base nature of the oxides, and this is the single most examined idea in the chapter. The chemical reactivity of an element is best shown by its reaction with oxygen, and the oxide formed by the extreme-left element is the most basic, while that formed by the extreme-right element is the most acidic.
| Oxide | Nature | Reaction with water / behaviour |
|---|---|---|
| $\ce{Na2O}$ | Strongly basic | $\ce{Na2O + H2O -> 2NaOH}$ (strong base) |
| $\ce{MgO}$ | Basic | Reacts with acids to give salts |
| $\ce{Al2O3}$ | Amphoteric | Acts as base toward acids and as acid toward bases |
| $\ce{As2O3}$ | Amphoteric | Reacts with both acids and alkalis |
| $\ce{CO}$, $\ce{NO}$, $\ce{N2O}$ | Neutral | No acidic or basic character |
| $\ce{Cl2O7}$ | Strongly acidic | $\ce{Cl2O7 + H2O -> 2HClO4}$ (strong acid) |
Two definitions anchor the centre of the row. Amphoteric oxides behave as acidic with bases and as basic with acids, the textbook example being $\ce{Al2O3}$. Neutral oxides such as $\ce{CO}$, $\ce{NO}$ and $\ce{N2O}$ have no acidic or basic properties at all. The acid-base reactions of the two extremes can be confirmed qualitatively with litmus.
Show by reaction with water that $\ce{Na2O}$ is basic and $\ce{Cl2O7}$ is acidic. (NCERT Problem 3.10)
$\ce{Na2O}$, the oxide of an extreme-left metal, dissolves to form a strong base: $\ce{Na2O + H2O -> 2NaOH}$. $\ce{Cl2O7}$, the oxide of an extreme-right non-metal in its highest oxidation state, dissolves to form a strong acid: $\ce{Cl2O7 + H2O -> 2HClO4}$. Litmus confirms blue and red respectively.
Oxide nature: don't force a strict monotonic order
The trend is basic → amphoteric → acidic across a period, but the centre is not always neatly "amphoteric only". $\ce{CO}$ and $\ce{NO}$ are neutral, not amphoteric, even though they sit in the central p-block. In the classic NEET matching item the safe assignments are $\ce{BaO}$ basic, $\ce{CO}$ neutral, $\ce{Al2O3}$ amphoteric and $\ce{Cl2O7}$ acidic. Match each oxide on its own merits rather than assuming every central oxide is amphoteric.
Higher oxidation state of the same element makes its oxide more acidic: $\ce{MnO}$ basic, $\ce{Mn2O7}$ acidic.
Five things to carry into the exam
- Valence rule: for representative elements, valence = number of outer electrons (if ≤ 4) or 8 − that number (if > 4); higher periods can exceed the octet using d orbitals.
- Two trends in one period: valence vs hydrogen runs 1,2,3,4,3,2,1,0; valence vs oxygen rises to a maximum equal to the group number ($\ce{Cl2O7}$, valence 7).
- Oxidation state ≠ valence ≠ covalency: oxidation state carries a sign from electronegativity; oxygen is $+2$ in $\ce{OF2}$ but $-2$ in $\ce{Na2O}$.
- Second-period anomaly & diagonal pairs: small size, no d orbital, max covalency 4, strong $p\pi\text{–}p\pi$ bonds; remember Li~Mg, Be~Al, B~Si.
- Across a period: metallic character falls, non-metallic rises, reactivity peaks at both ends, and oxides shift basic ($\ce{Na2O}$) → amphoteric ($\ce{Al2O3}$) → acidic ($\ce{Cl2O7}$).