Classification of Elements and Periodicity in Properties | ISC Class 11 Chemistry Notes

This note covers periodic classification, Mendeleev’s and modern periodic laws, groups and periods, electronic configurations, element nomenclature, the four blocks, atomic and ionic sizes, ionisation enthalpy, electron gain enthalpy, electronegativity, valency, second-period anomalies, diagonal relationships and the acidic or basic character of oxides.

Why are elements classified, and what did Mendeleev achieve?

What does classification make possible?

Classification organises elements into families with related properties. It reduces the need to study each element and its compounds separately, reveals patterns in known behaviour and helps predict properties of elements that have not yet been studied.

Periodicity means the recurrence of similar properties at intervals when elements are arranged systematically. Early classifications used atomic weight, also called relative atomic mass: the mass of an atom compared with one-twelfth of the mass of a carbon-12 atom.

Dobereiner grouped certain elements into triads, or sets of three with similar properties. The middle element had an atomic weight about halfway between those of the other two. This relationship seemed to work only for a few elements.

Newlands arranged elements by increasing atomic weight and proposed his Law of Octaves: every eighth element resembled the first. The relationship seemed to be true only for elements up to calcium. Lothar Meyer also recognised recurring patterns in physical properties.

How did Mendeleev use properties as well as mass?

Definition: Mendeleev’s periodic law states that the properties of elements are a periodic function of their atomic weights.

Mendeleev arranged elements in rows and columns, bringing those with similar properties together. He used the formulas and properties of their compounds as evidence. A chemical formula identifies the elements in a substance and the relative numbers of their atoms.

He sometimes departed from strict atomic-weight order. Iodine had a lower atomic weight than tellurium, yet he placed iodine with fluorine, chlorine and bromine because their properties were similar. Thus, chemical resemblance could take priority over the mass sequence.

His table left gaps for undiscovered elements. Eka-aluminium and eka-silicon were predictions of the elements later identified as gallium and germanium. Successful predictions were a major advantage: the table could guide discovery as well as organise existing knowledge.

What were the limitations?

Atomic-weight reversals showed that mass alone could not provide a consistent ordering principle. Hydrogen’s resemblance to more than one family also made its position difficult. The scheme did not explain why properties recur, which required an understanding of atomic structure and electron arrangement.

Isotopes are atoms of the same element with the same atomic number, meaning the same number of protons in the nucleus, but different masses. Protons are positively charged particles; the nucleus is the atom’s central region. A classification based strictly on atomic weight cannot give these atoms one unambiguous position on the basis of mass alone. Classification by atomic number resolves this difficulty.

How does the modern periodic table connect atomic number with electron arrangement?

What is the modern periodic law?

The atomic number, Z, is the number of protons, positively charged particles, in an atom’s nucleus. A neutral atom contains the same number of negatively charged electrons. Moseley’s work established atomic number as a more fundamental basis of classification than atomic mass.

Definition: The physical and chemical properties of elements are periodic functions of their atomic numbers. This is the modern periodic law.

An electronic configuration describes how electrons are distributed among atomic orbitals. An atomic orbital is a mathematical function describing an electron’s state and spatial probability distribution in an atom. The principal quantum number, n, identifies its main energy level or shell.

The outermost shell is the valence shell, and its electrons are valence electrons. Similar outer electronic configurations recur as atomic number increases. This recurrence explains why elements in the same family show similar chemical behaviour.

How are periods and groups arranged?

The long form has seven horizontal periods and eighteen vertical groups. Groups are numbered 1 to 18. The period number corresponds to the highest principal quantum number occupied in the atom’s ground state, its lowest-energy electronic arrangement.

Electrons occupy subshells, subdivisions of shells, labelled s, p, d and f. These contain one, three, five and seven orbitals respectively. Each orbital accommodates at most two electrons. In a configuration such as 2s², 2 identifies the shell, s the subshell and the superscript its electron count.

The seventh period involves 7s, 5f, 6d and 7p filling. Period lengths follow the orbitals being filled, not just the maximum capacity of the outermost shell. For example, filling inner 3d orbitals between 4s and 4p contributes to the length of the fourth period.

Derivation: Why does the fifth period contain eighteen elements?

1. The subshells filled across the fifth period are 5s, 4d and 5p. They contain one, five and three orbitals respectively.
2. Adding these gives the number of available orbitals: $N = 1 + 5 + 3 = 9$.
3. Each orbital accommodates two electrons, so the number of successive elements is $P = 2N = 18$.

Result: The fifth period contains eighteen elements because its nine available orbitals accommodate eighteen electrons.

Worked example 1. The fifth period fills 5s, 4d and 5p subshells, containing one, five and three orbitals respectively. Each orbital holds at most two electrons. Explain its length.

Answer: Formula: $N = N_s + N_d + N_p$ and $P = 2N$, where N counts available orbitals and P counts successive elements. Substitute: $N = 1 + 5 + 3 = 9$, so $P = 2 \times 9 = 18$. The period therefore contains eighteen elements.

How are systematic names constructed for elements with atomic number above 100?

What do the numerical roots represent?

The International Union of Pure and Applied Chemistry (IUPAC) provides a systematic naming method for use until an element’s discovery is established and its name officially recognised. The temporary name is derived directly from the digits of its atomic number.

Write the digits in order, replace each by its numerical root and combine the roots with the ending ium. The temporary symbol contains three letters based on the root abbreviations. Its first letter is a capital and the remaining letters are lower case.

How does a temporary name differ from an official name?

Worked example 2. Construct the systematic name and symbol for atomic number 120, using the roots un for 1, bi for 2 and nil for 0.

Answer: The ordered roots are un, bi and nil. Their combination with ium gives unbinilium; the root initials give the symbol Ubn.

A systematic name should not be confused with the subsequently approved permanent name. For example, element 104 has the temporary systematic name unnilquadium and symbol Unq, whereas its official name is rutherfordium and its official symbol is Rf.

The atomic number identifies the same element through both naming stages. Learning the root method therefore means understanding a naming procedure, rather than replacing the approved permanent names of elements with their temporary names.

How do electronic configurations define the four blocks?

What distinguishes the main-group elements?

An element’s block depends on the type of subshell being filled. The s-block and p-block together form the representative or main-group elements. Similar valence configurations within a group explain their related properties.

Group 1 elements are alkali metals; Group 2 elements are alkaline earth metals. They readily lose their outermost electrons to form positive ions, charged atoms produced here by electron loss. Their compounds, except those of lithium and beryllium, are predominantly ionic, involving attraction between oppositely charged ions.

Group 17 elements are halogens, and Group 18 elements are noble gases. Noble gases have completely filled valence shells and very low chemical reactivity. Helium has configuration 1s², yet its filled shell and properties justify its placement in Group 18.

Hydrogen is a special case. Its single s electron suggests a relationship with Group 1. Its ability to gain an electron and attain a filled shell also resembles the behaviour of halogens. One resemblance does not remove the other.

What distinguishes transition and inner-transition elements?

The d-block elements occupy the centre of the table and are metals. They mostly form coloured ions and can exhibit variable oxidation states, meaning different assigned atomic charges in different compounds. Zinc, cadmium and mercury do not show most characteristic transition-element properties.

The f-block contains the lanthanoids and actinoids. These are inner-transition elements, shown separately to preserve a convenient table structure. Within each series their properties are quite similar. Actinoid elements are radioactive, meaning their nuclei undergo spontaneous change.

Metalloids, such as silicon and germanium, have properties characteristic of both metals and non-metals. This classification concerns behaviour, whereas classification into blocks concerns electron filling. The two methods describe different aspects of the same periodic table.

How are atomic radii measured, and why do they change periodically?

What does atomic size mean?

The electron cloud has no sharp boundary, so an atom cannot be measured like a solid ball. Atomic radius is estimated from distances between atoms in a combined state. The kind of distance used must be specified before comparing radii.

A covalent bond involves sharing electrons. The covalent radius is half the distance between the nuclei of two identical atoms joined by a single covalent bond. The metallic radius is half the distance between adjacent metal cores in a metallic crystal.

Atomic distances are commonly expressed in picometres (pm), where 1 pm = 10⁻¹² m and m denotes metre. A van der Waals radius describes non-bonded atomic size; it must not be directly equated with a covalent radius.

Worked example 3. The internuclear distance, meaning the distance between nuclei, in a chlorine molecule is 198 pm. Find the covalent radius of chlorine.

Answer: For two identical singly bonded atoms, $r = d/2$, where d is the internuclear distance. Hence $r = \frac{198}{2}\,\mathrm{pm} = 99\,\mathrm{pm}$. In metres, $r = 99 \times 10^{-12}\,\mathrm{m} = \text{0.000000000099 m}$.

Worked example 4. Adjacent copper atoms in solid copper have an internuclear separation of 256 pm. Find the metallic radius.

Answer: The metallic radius is half the separation of adjacent metal cores: $r = \frac{256}{2}\,\mathrm{pm} = 128\,\mathrm{pm}$. In metres, $r = 128 \times 10^{-12}\,\mathrm{m} = \text{0.000000000128 m}$.

Why does size generally decrease across a period?

Nuclear charge is the positive charge of the nucleus. Shielding, also called screening, is the reduction of the nucleus’s attraction for outer electrons by intervening electrons. The net positive charge experienced by an electron after shielding is the effective nuclear charge.

Across a period, electrons enter the same valence shell while nuclear charge increases. Inner-electron shielding does not increase enough to compensate. The stronger effective attraction draws the outer electrons closer, so atomic size generally decreases from left to right.

Why does size increase down a group?

Down a group, the outermost electrons occupy shells with higher principal quantum numbers. Their distance from the nucleus increases, while filled inner levels shield them. The resulting increase in radius occurs despite the increase in nuclear charge.

The horizontal axis shows atomic number and the vertical axis shows atomic radius in pm. The plotted line descends through labelled points from lithium to fluorine.

See Fig. 3.4(a) in your NCERT textbook

Note: Noble-gas non-bonded radii are not directly comparable with the covalent radii of neighbouring elements. Compare radii defined in the same way before drawing a conclusion about periodic trends.

How do ionic radii and isoelectronic comparisons work?

What changes when an atom gains or loses electrons?

An ion is a charged atom or group of atoms. Losing electrons produces a positively charged cation; gaining electrons produces a negatively charged anion. An ionic radius is estimated from distances between ions in ionic crystals.

A cation is smaller than its parent atom. Its electron count falls while the nuclear charge remains the same. An anion is larger than its parent atom because added electrons increase electron-electron repulsion and reduce the effective attraction experienced by each electron.

A superscript positive or negative sign identifies an ion’s charge. Thus Na⁺ has lost one electron, whereas F⁻ has gained one. In magnesium ion Mg²⁺, 2+ means two positive elementary charges; an elementary charge is the magnitude of a proton’s charge.

How is nuclear charge isolated as a factor?

Isoelectronic species, in these atomic and ionic comparisons, have the same number of electrons. The oxide ion O²⁻, fluoride ion F⁻, sodium ion Na⁺ and magnesium ion Mg²⁺ each contain ten electrons, but their nuclear charges differ.

Within such a series, increasing nuclear charge pulls the same electron population more strongly and decreases radius. The order of decreasing size is O²⁻, F⁻, Na⁺, Mg²⁺. This reasoning applies to an isoelectronic comparison, not indiscriminately to all positive and negative ions.

Worked example 5. Compare magnesium, Mg, and aluminium, Al, which occur successively in Period 3, and their ions Mg²⁺ and Al³⁺. Their atomic numbers are 12 and 13 respectively. Identify the largest and smallest species.

Answer: Mg is largest because atomic size decreases across the period and cations are smaller than their parent atoms. Both ions have ten electrons, but Al³⁺ has the greater nuclear charge. Therefore Al³⁺ is smallest.

What controls ionisation enthalpy and its exceptions?

What process does ionisation enthalpy describe?

Enthalpy is an energy quantity whose change measures heat absorbed or released at constant pressure.

First ionisation enthalpy is the energy required to remove an electron from an isolated gaseous atom in its ground state. It measures the difficulty of electron removal, and is commonly expressed in kilojoules per mole, written kJ mol⁻¹.

A joule is a unit of energy; kJ denotes kilojoule and mol denotes mole, the unit of amount of substance. The superscript −1 means “per mole”. The symbol ΔᵢH denotes ionisation enthalpy, with Δ indicating change and H denoting enthalpy.

For a general element represented by X, the first process is X(g) → X⁺(g) + e⁻. Here g means gaseous state, e⁻ means an electron and the arrow means “forms”. Isolation excludes attraction to neighbouring atoms or ions.

The second process is X⁺(g) → X²⁺(g) + e⁻. It removes an electron from the gaseous singly charged cation. Energy is always required for these removals, so ionisation enthalpies are positive; successive ionisation enthalpies increase.

Derivation: How is hydrogen’s molar ionisation enthalpy calculated?

The ground-state electron energy in hydrogen is $-2.18 \times 10^{-18}\,\mathrm{J}$. Let E denote this initial energy and $E_f$ the energy of the separated electron. Let $N_A$ denote the number of atoms per mole.

1. Choose the separated electron’s energy as zero: $E_f = 0$.
2. The energy required for one atom is final energy minus initial energy: $\Delta E = E_f - E$. Subtracting a negative energy gives a positive ionisation energy.
3. Multiply the energy for one atom by the number of atoms per mole: $\Delta_i H = N_A\Delta E$. The units combine as $\mathrm{mol^{-1}}\times\mathrm{J}=\mathrm{J\,mol^{-1}}$.

Result: $\Delta_i H = -N_A E$ when the separated electron has zero energy. This converts the energy needed for one atom into the energy needed for one mole of atoms.

Worked example 8. The ground-state electron energy of a hydrogen atom is $-2.18 \times 10^{-18}\,\mathrm{J}$. Calculate its ionisation enthalpy in joules per mole, using $N_A = 6.02214076 \times 10^{23}\,\mathrm{mol^{-1}}$.

Answer: Formula: $\Delta E = E_f - E$ and $\Delta_i H = N_A\Delta E$. Substitute: $\Delta E = 0 - (-2.18 \times 10^{-18}) = 2.18 \times 10^{-18}\,\mathrm{J}$.

Then $\Delta_i H = (6.02214076 \times 10^{23})(2.18 \times 10^{-18}) = 1312826.68568\,\mathrm{J\,mol^{-1}}$. To three significant figures, $\Delta_i H = \text{1310000 J}\,\mathrm{mol^{-1}}$, or $1.31 \times 10^6\,\mathrm{J\,mol^{-1}}$. The positive sign means that energy must be supplied.

Why does the general trend occur?

First ionisation enthalpy generally increases across a period. Increasing nuclear charge outweighs the small additional shielding within the same shell. The outer electrons become more tightly held, making removal more difficult.

It generally decreases down a group. The outer electron lies farther from the nucleus and experiences greater shielding by inner electrons. These effects outweigh the increased nuclear charge, so less energy is needed for removal.

Noble gases have high first ionisation enthalpies associated with stable closed shells. Alkali metals have low values associated with easy loss of their outer electron. These recurring maxima and minima connect electron configuration with periodic reactivity.

Why are boron and oxygen exceptions?

Boron’s first ionisation enthalpy is slightly less than beryllium’s. Beryllium loses a 2s electron, but boron loses a 2p electron. Penetration describes how closely an electron’s distribution approaches the nucleus; a 2s electron penetrates more than a 2p electron.

The boron 2p electron is more shielded and easier to remove. Thus, the higher atomic number of boron does not by itself determine the result. The subshell occupied by the departing electron must also be considered.

Oxygen has a smaller first ionisation enthalpy than nitrogen. Nitrogen’s three 2p electrons occupy separate orbitals. Oxygen has four 2p electrons, so two share one orbital. Repulsion within this pair makes removal of the fourth electron easier.

Atomic number is on the horizontal axis and first ionisation enthalpy in kJ mol⁻¹ on the vertical axis. The line generally rises from lithium to neon, with downward steps at boron and oxygen.

See Fig. 3.6(a) in your NCERT textbook

Worked example 6. First ionisation enthalpies of sodium, magnesium and silicon are 496, 737 and 786 kJ mol⁻¹ respectively. Aluminium lies between magnesium and silicon. Its removed electron is 3p, shielded by 3s electrons. Is its value nearer 575 or 760 kJ mol⁻¹?

Answer: It is nearer 575 kJ mol⁻¹, equivalent to $575\,\mathrm{kJ\,mol^{-1}} = \text{575000 J}\,\mathrm{mol^{-1}}$. The 3p electron is easier to remove, so aluminium’s value is lower than magnesium’s rather than intermediate between those of magnesium and silicon.

How does electron gain enthalpy vary, and why is chlorine exceptional?

What do the sign and definition mean?

Electron gain enthalpy, written ΔₑgH, is the enthalpy change when an electron is added to a neutral gaseous atom to form a negative ion. Its defining process is X(g) + e⁻ → X⁻(g).

An exothermic process releases energy and has a negative enthalpy change; an endothermic process absorbs energy and has a positive enthalpy change. Electron addition may be either, depending on the element. Electron gain enthalpy is therefore not invariably negative.

Halogens have highly negative electron gain enthalpies because electron addition produces a stable noble-gas configuration. Noble gases have large positive values because the added electron must enter a higher principal energy level, producing a very unstable electronic configuration.

What are the trend and important departures?

As a general rule, electron gain enthalpy becomes more negative across a period. Increased effective nuclear charge favours electron addition. It generally becomes less negative down a group, where the added electron lies farther from the nucleus.

The variation is less systematic than for ionisation enthalpy. In particular, oxygen and fluorine have less negative values than sulphur and chlorine respectively. Their small second-shell orbitals crowd the incoming electron into a region with substantial electron-electron repulsion.

In sulphur and chlorine, the electron enters the larger third-shell region. Repulsion is much less, so addition releases more energy than in oxygen and fluorine respectively. Small size can therefore oppose, as well as favour, electron addition.

More negative means a lower algebraic value and a greater energy release for electron addition. Chlorine’s −349 kJ mol⁻¹ is more negative than fluorine’s −328 kJ mol⁻¹. Avoid the ambiguous statement that one simply has “higher” electron gain enthalpy.

Note: Electron gain enthalpy describes addition to an isolated gaseous atom. Electronegativity describes attraction for shared electrons in a chemical compound. The most electronegative element need not have the most negative electron gain enthalpy.

Worked example 9. Which member of each pair has the more negative electron gain enthalpy: O or F, and F or Cl? Their values are −141, −328 and −349 kJ mol⁻¹ respectively.

Answer: Fluorine is more negative than oxygen: $-328 < -141$. Chlorine is more negative than fluorine: $-349 < -328$. Their selected values in joules per mole are $\Delta_{eg}H(\mathrm{F}) = \text{-328000 J}\,\mathrm{mol^{-1}}$ and $\Delta_{eg}H(\mathrm{Cl}) = \text{-349000 J}\,\mathrm{mol^{-1}}$.

Across Period 2, increasing effective nuclear charge favours electron addition. Chlorine’s larger third-shell region reduces repulsion compared with fluorine’s compact second shell, making chlorine the exception to the usual down-group trend.

How does electronegativity relate to metallic and non-metallic character?

What is being attracted?

Electronegativity is a qualitative measure of an atom’s ability, in a chemical compound, to attract shared electrons towards itself. It is not a directly measurable quantity like ionisation enthalpy, although numerical scales allow comparisons.

The widely used Pauling scale assigns fluorine a value of 4.0. Electronegativity is not constant for an element in every compound; its value depends on the atom to which it is bonded. It describes bonding behaviour rather than electron capture by an isolated atom.

Electronegativity generally increases across a period and decreases down a group. Smaller size and stronger attraction between the nucleus and outer electrons favour attraction of shared electrons. Increasing atomic radius down a group has the opposite effect.

How does the trend connect with chemical character?

Metallic character is associated with electron loss, while non-metallic character is associated with electron gain. Increasing electronegativity across a period accompanies increasing non-metallic character. Down a group, declining electronegativity accompanies increasing metallic character.

Metals are usually solids at room temperature and usually have high melting and boiling points. They conduct heat and electricity and are malleable, meaning they can be hammered into sheets, and ductile, meaning they can be drawn into wires.

Non-metals are usually solids or gases at room temperature. Most non-metallic solids are brittle and are neither malleable nor ductile. The transition between metallic and non-metallic behaviour is not abrupt: metalloids display characteristics of both.

A periodic-table outline is surrounded by arrows. Atomic-radius arrows point left and down; electronegativity arrows point right and up. Slanting arrows across the outline mark metallic and non-metallic character in opposite directions.

See Fig. 3.7 in your NCERT textbook

How do valency and oxidation state express chemical periodicity?

How can valence electrons suggest combining capacity?

Valency, or valence, expresses an element’s combining capacity. For representative elements, it is usually, though not necessarily, equal to the number of outermost electrons or to eight minus that number. The qualification matters because many elements exhibit variable valence.

The recurrence of similar valence-shell configurations produces related compound formulas down a group. Hydrogen and oxygen compounds are useful comparisons because the same partner element is retained while the other element changes.

For example, the Group 1 oxides lithium oxide, sodium oxide and potassium oxide have formulas Li₂O, Na₂O and K₂O. The subscript 2 indicates two metal atoms for each oxygen atom. Their repeated formula pattern reflects similar combining behaviour.

How are formulas predicted from supplied valencies?

Worked example 7. Silicon has valency 4 and bromine has valency 1. Aluminium has valency 3 and sulphur has valency 2. Predict the compounds formed by each pair.

Answer: One silicon atom combines with four bromine atoms, giving SiBr₄. Two aluminium atoms supply a total combining capacity of six, matching three sulphur atoms, giving Al₂S₃.

Oxidation state assigns an atom a charge on the basis of relative electronegativity in a compound. It includes a positive or negative sign, unlike the simple unsigned combining-capacity description of valency.

In oxygen difluoride, OF₂, fluorine is more electronegative than oxygen. Each fluorine has oxidation state −1 and oxygen has +2. In sodium oxide, Na₂O, oxygen is more electronegative than sodium: oxygen has −2 and each sodium has +1.

Thus, the same element can have different oxidation states in different compounds. Oxygen’s combining capacity in these examples is two, but its assigned oxidation state changes sign. A formula must therefore be interpreted with the identities and electronegativities of its atoms in mind.

Why are second-period elements anomalous, and what is a diagonal relationship?

Why does the first member of a group differ?

The first members of Groups 1 and 2, lithium and beryllium, and of Groups 13 to 17, boron through fluorine, differ in many respects from later members of their groups. This is called anomalous behaviour: departure from a simple group resemblance.

The main causes are small atomic size, a large charge-to-radius ratio and high electronegativity. Charge-to-radius ratio compares ionic charge with ionic radius: a substantial charge concentrated in a small species produces a strong electrical influence on nearby electrons.

Lithium and beryllium form compounds with pronounced covalent character, meaning substantial electron sharing. Other alkali and alkaline earth metals predominantly form ionic compounds. Group membership therefore predicts similarities without requiring all members to behave identically.

How does bonding capacity differ?

Second-period atoms have four valence orbitals available, one 2s and three 2p. Their maximum covalency, meaning the number of shared electron pairs used in bonding, is four. Later members can show greater covalency.

For example, boron forms the ion [BF₄]⁻, while aluminium forms [AlF₆]³⁻. Square brackets enclose the whole ion; the charge outside belongs to the entire bracketed species. Four and six fluorine atoms respectively illustrate the difference in covalency.

The first p-block members also have a greater ability to form multiple bonds, which share more than one electron pair. Examples include carbon-carbon double and triple bonds and the nitrogen-nitrogen triple bond. Subsequent members of the same groups show less tendency towards this bonding.

Which similarities run diagonally?

A diagonal relationship is a resemblance between a second-period element and the third-period element placed one group to its right. Lithium resembles magnesium, and beryllium resembles aluminium. These similarities supplement the usual vertical group relationships.

To identify the direction, move one period down and one group right. The relationship concerns selected chemical properties. It does not mean that the two elements have identical electron configurations, the same group number or the same behaviour in every reaction.

How do periodic trends explain reactivity and the nature of oxides?

Why are both ends of a period reactive?

Chemical reactivity tends to be high among Group 1 metals, lower towards the middle of the table and high again among Group 17 non-metals. Low ionisation enthalpy favours electron loss at the metallic end; favourable electron gain favours reaction at the halogen end.

Among alkali metals, reactivity increases down the group. Among halogens, it decreases down the group. These examples show why “reactivity increases down every group” is not a sound general rule: different reactions depend on different tendencies to lose or gain electrons.

Worked example 10. Three metals labelled I, II and VI have first ionisation enthalpies of 520, 419 and 738 kJ mol⁻¹ respectively. Their second ionisation enthalpies are 7300, 3051 and 1451 kJ mol⁻¹. Which is most reactive, and which can form a stable binary halide MX₂?

Answer: Metal II is the most reactive because its first electron is easiest to remove: $\Delta_i H_1(\mathrm{II}) = \text{419000 J}\,\mathrm{mol^{-1}}$, the lowest of the three first ionisation enthalpies.

Metal VI can form MX₂. Its second ionisation enthalpy, $\Delta_i H_2(\mathrm{VI}) = \text{1451000 J}\,\mathrm{mol^{-1}}$, is much lower than those of I and II. Losing two electrons gives a divalent metal ion, which combines with two singly charged halide ions.

How do acidic, basic, amphoteric and neutral oxides differ?

An oxide is a compound of oxygen with another element. Across a period, oxide character changes broadly from basic towards acidic. Sodium oxide is a basic oxide; dichlorine heptoxide, Cl₂O₇, is an acidic oxide.

A basic oxide reacts with acids, whereas an acidic oxide reacts with bases. Sodium oxide reacts with water, H₂O, to give sodium hydroxide, NaOH, a strong base. Dichlorine heptoxide reacts with water to give perchloric acid, HClO₄, a strong acid.

Na₂O + H₂O → 2NaOH

Cl₂O₇ + H₂O → 2HClO₄

Here H₂O is water. A number before a formula gives the relative number of formula units or molecules involved. The coefficient 2 balances both reactions without changing the formulas of the products.

Amphoteric oxides behave as acids towards bases and as bases towards acids. Aluminium oxide, Al₂O₃, and arsenic oxide, As₂O₃, are examples. Neutral oxides have neither acidic nor basic properties.

Carbon monoxide, CO, nitrogen monoxide, NO, and nitrous oxide, N₂O, are neutral oxides. “Neutral” must not be confused with “amphoteric”: having neither acid-base behaviour differs from being able to show both depending on the reacting substance.

Glossary

• Periodicity — Recurrence of similar properties at intervals when elements are arranged in increasing atomic number.
• Atomic number — Number of protons in an atom’s nucleus, equal to its electron count when neutral.
• Period — Horizontal row of the periodic table associated with the highest occupied principal energy level.
• Group — Vertical family of elements whose similar outer electronic configurations produce related chemical properties.
• Valence shell — Outermost occupied electron shell of an atom, important in determining its chemical behaviour.
• Shielding — Reduction of nuclear attraction experienced by outer electrons because of intervening inner electrons.
• Covalent radius — Half the internuclear distance between identical atoms joined by a single covalent bond.
• Isoelectronic species — Atoms or ions containing the same total number of electrons in the comparison being made.
• Ionisation enthalpy — Energy required to remove an electron from an isolated gaseous atom in its ground state.
• Electron gain enthalpy — Enthalpy change when an electron is added to a neutral gaseous atom to form an anion.
• Electronegativity — Qualitative measure of an atom’s ability in a compound to attract shared electrons towards itself.
• Oxidation state — Charge assigned to an atom in a compound by considering relative electronegativities of bonded atoms.
• Diagonal relationship — Similarity between certain second-period elements and third-period elements located one group to their right.
• Amphoteric oxide — Oxide that reacts as an acid with bases and as a base with acids.

Common errors and misconceptions

• Misconception: Modern classification still follows increasing atomic mass. Correct: Atomic number determines the sequence. Similar outer electron configurations explain the recurring properties.
• Misconception: Every radius in a table measures the same thing. Correct: Covalent, metallic and non-bonded radii use different definitions. Noble-gas radii require a suitable non-bonded comparison.
• Misconception: More protons always make an atom smaller. Correct: Down a group, additional occupied shells and shielding increase atomic radius despite greater nuclear charge.
• Misconception: First ionisation enthalpy increases without exceptions across a period. Correct: Boron is slightly below beryllium, and oxygen is below nitrogen, because subshell and pairing effects matter.
• Misconception: Fluorine must have the most negative electron gain enthalpy because it is most electronegative. Correct: Chlorine’s electron gain enthalpy is more negative; the two properties describe different processes.
• Misconception: Valency and oxidation state are interchangeable in every context. Correct: Oxidation state is signed and depends on electronegativity; simple valency expresses combining capacity.
• Misconception: Neutral and amphoteric oxides mean the same thing. Correct: Neutral oxides show neither acidic nor basic properties; amphoteric oxides show both according to the reactant.

Exam-style questions with model answers

Q1. State the basis of Mendeleev’s periodic law and of the modern periodic law. [2 marks]

1. Mendeleev’s law treats the properties of elements as periodic functions of their atomic weights.
2. The modern law treats their physical and chemical properties as periodic functions of their atomic numbers.

Q2. In a chlorine molecule, the distance between the two nuclei is 198 pm. Define covalent radius for identical atoms joined by a single bond and calculate chlorine’s radius. [2 marks]

1. The covalent radius is half the internuclear distance between two identical atoms joined by a single covalent bond.
2. Chlorine’s covalent radius is 198 pm ÷ 2 = 99 pm.

Q3. O²⁻, F⁻, Na⁺ and Mg²⁺ each contain ten electrons. Their nuclei contain 8, 9, 11 and 12 protons respectively. Define their electron-count relationship, arrange them in decreasing radius and explain the order. [3 marks]

1. The ions are isoelectronic because each has the same total number of electrons, namely ten, even though they represent different elements.
2. The decreasing-radius order is O²⁻, F⁻, Na⁺, Mg²⁺, from the largest ion to the smallest ion.
3. With electron count fixed, increasing nuclear charge attracts the electrons more strongly. The ion with twelve protons therefore has the smallest radius.

Q4. Beryllium loses a 2s electron, whereas boron loses a more shielded 2p electron. Nitrogen has three singly occupied 2p orbitals; oxygen has one electron pair and two singly occupied 2p orbitals. Explain the two departures from the general first-ionisation-enthalpy trend across this period. [4 marks]

1. First ionisation enthalpy generally increases across a period as increasing nuclear charge holds the outer electrons more tightly.
2. Boron’s value is slightly less than beryllium’s because the more shielded 2p electron is easier to remove than a 2s electron.
3. Oxygen’s value is smaller than nitrogen’s because repulsion between the two electrons sharing an orbital assists removal of one electron.
4. Nitrogen has no paired 2p electrons in the given arrangement, so removing one lacks this repulsion advantage; nuclear charge alone cannot explain both comparisons.

Q5. Electron gain enthalpies of O, S, F and Cl are −141, −200, −328 and −349 kJ mol⁻¹ respectively. O and F receive the added electron in the smaller second shell; S and Cl receive it in the larger third shell. Define electron gain enthalpy, interpret its sign, identify the most negative value and explain both pairwise exceptions to the usual down-group trend. [5 marks]

1. Electron gain enthalpy is the enthalpy change when an electron is added to a neutral gaseous atom, producing a negative ion.
2. All four values are negative, so electron addition releases energy in each case. A more negative value represents a greater release of energy.
3. Chlorine has the most negative value, −349 kJ mol⁻¹. It releases more energy on electron addition than fluorine, whose value is −328 kJ mol⁻¹.
4. Sulphur’s value is more negative than oxygen’s because the incoming electron occupies a larger region and experiences less electron-electron repulsion.
5. Chlorine similarly has less crowding than fluorine. These two comparisons depart from the general expectation that electron gain enthalpy becomes less negative down a group.

Q6. Use the supplied facts to explain four features of classification: Group 1 has outer configuration ns¹; helium has a filled 1s² shell and properties of Group 18 noble gases; d-block elements fill inner d orbitals; f-block elements fill inner f orbitals. [4 marks]

1. Group 1 elements belong to the s-block because their characteristic outer electron occupies an s subshell, as shown by ns¹.
2. Helium has an s configuration, but its completely filled shell and noble-gas properties justify its position in Group 18.
3. The d-block classification follows the filling of inner d orbitals; it is therefore based on electron arrangement.
4. The f-block similarly follows inner f filling. Thus, distinguishing d-block from f-block elements requires identifying which type of inner subshell receives electrons.

Q7. Sodium oxide, Na₂O, reacts with water, H₂O, to form the strong base sodium hydroxide, NaOH. Dichlorine heptoxide, Cl₂O₇, reacts with water to form the strong acid perchloric acid, HClO₄. Write both balanced reactions and use the products to classify each oxide. [4 marks]

1. The first balanced reaction is Na₂O + H₂O → 2NaOH. Two sodium atoms on the left require two sodium hydroxide units on the right.
2. Sodium oxide is basic because reaction with water produces sodium hydroxide, which is a strong base.
3. The second balanced reaction is Cl₂O₇ + H₂O → 2HClO₄, conserving chlorine, hydrogen and oxygen atoms.
4. Dichlorine heptoxide is acidic because reaction with water produces perchloric acid, which is a strong acid.

Q8. Construct the temporary systematic name and symbol for Z = 120. The numerical roots for digits 1, 2 and 0 are un, bi and nil respectively; use the ending ium and root initials for the symbol. [2 marks]

1. Combining the roots in digit order with the ending gives the systematic name unbinilium.
2. The root initials give Ubn, with a capital first letter and lower-case remaining letters.

Key takeaways

• Modern classification uses atomic number, while recurring valence-shell electronic configurations explain why properties repeat periodically.
• Groups contain related elements, periods reflect successive electron filling, and blocks identify the subshell being filled.
• Atomic radius generally decreases across a period and increases down a group because nuclear attraction competes with shell addition and shielding.
• Cations are smaller and anions larger than their parent atoms; greater nuclear charge contracts an isoelectronic series.
• Ionisation enthalpy generally rises across a period, with subshell and electron-pairing effects explaining the boron and oxygen exceptions.
• Electron gain enthalpy generally becomes more negative across a period, but oxygen and fluorine have important crowding-related exceptions.
• Electronegativity concerns shared electrons, usually increases across a period, and accompanies increasing non-metallic character.
• Second-period anomalies, diagonal relationships and changing oxide character qualify simple predictions based on group position alone.

Test yourself

What does the atomic number count in a neutral atom?

It counts protons in the nucleus and equals the number of electrons in a neutral atom.

Why does helium belong with the noble gases despite its 1s² configuration?

Its valence shell is completely filled, giving properties characteristic of the noble gases in Group 18.

Why should noble-gas radii not be directly compared with neighbouring covalent radii?

Noble-gas radii are non-bonded values; a valid comparison requires radii defined using the same kind of atomic separation.

Why is the second ionisation enthalpy greater than the first?

The second electron must be removed from an already positively charged ion, which attracts the remaining electrons more strongly.

Does a negative electron gain enthalpy mean energy is required?

No. A negative value indicates release of energy when the electron is added to the gaseous atom.

Which diagonal pairs illustrate relationships involving lithium and beryllium?

Lithium resembles magnesium, while beryllium resembles aluminium, each located one period below and one group to the right.

Why is chlorine’s electron gain enthalpy more negative than fluorine’s?

The incoming electron enters chlorine’s larger third-shell region, where electron-electron repulsion is less than in fluorine’s compact second shell.

How does an amphoteric oxide differ from a neutral oxide?

An amphoteric oxide behaves as an acid or base depending on its partner; a neutral oxide has neither acidic nor basic properties.