Classification of Elements and Periodicity in Properties Class 11: Periodic Table and Trends

What Classification of Elements and Periodicity in Properties is about

Weightage: This chapter is Unit III, Classification of Elements and Periodicity in Properties, which the syllabus gives 6 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit III Classification of Elements and Periodicity in Properties

Key concepts

Why classify elements, and early attempts

NCERT section 3.1 Why do We Need to Classify Elements?; 3.2 Genesis of Periodic Classification

As more elements were discovered (63 by 1865, 118 today), chemists needed a way to organise their properties so that one could learn families instead of individual elements.

Döbereiner's triads (1817-1829): groups of three elements with similar properties in which the atomic mass of the middle one is roughly the average of the other two (Li, Na, K; Ca, Sr, Ba; Cl, Br, I). It worked for only a few elements.

Chancourtois (1862) arranged elements by increasing atomic weight on a cylinder; his work was not noticed.

Newlands' law of octaves (1865): when elements are arranged by increasing atomic mass, every eighth element has properties similar to the first, like the eighth note of a musical scale. It worked only up to calcium.

Lothar Meyer plotted physical properties such as atomic volume against atomic weight and found a periodically repeating pattern.

These attempts showed that properties repeat, but none gave a complete, predictive arrangement.

Worked example

The atomic masses of calcium and barium are 40 and 137. Use Döbereiner's idea to estimate the atomic mass of strontium, and compare it with the actual value of 87.6. Why was the idea abandoned? (2 marks)

Try it first, then show the solutionSolution

1. Average of the two outer members = (40 + 137) ÷ 2 = 88.5. (1)
2. This is close to 87.6, so Ca, Sr, Ba form a triad. The idea was abandoned because only a few such triads could be found among the known elements. (1)

Answer: Estimated 88.5, close to 87.6; triads covered too few elements to be a general law.

Mendeleev's periodic law and table

NCERT section 3.2 Genesis of Periodic Classification (Mendeleev)

Mendeleev's periodic law (1869): the properties of elements are a periodic function of their atomic weights.

He placed the elements in a grid, ordered by increasing atomic weight, so that elements with similar properties fell in the same column. His achievements:

• He used a wide range of physical and chemical properties, especially the formulas of oxides and hydrides, to group elements.
• He did not follow atomic weight blindly: where properties demanded it, he placed a heavier element before a lighter one (for example iodine after tellurium) to keep families together.
• He left gaps for undiscovered elements and predicted their properties. Eka-aluminium and eka-silicon were later found as gallium (1871 prediction, discovered 1875) and germanium (1886), with properties very close to his predictions.
• He corrected the atomic weights of some elements.

Limitations: the position of hydrogen was uncertain; isotopes, which have different atomic masses, did not get separate places; and the reversals of atomic weight order (Ar–K, Co–Ni, Te–I) could not be explained until atomic numbers were understood.

Worked example

Mendeleev predicted for eka-silicon an atomic weight of 72 and an oxide of formula EO₂. Germanium, discovered later, has atomic weight 72.6 and forms GeO₂. What does this show about Mendeleev's table? Give one limitation of his table. (2 marks)

Try it first, then show the solutionSolution

1. The close match between predicted and actual properties shows that his table captured a real periodic pattern and could be used to predict undiscovered elements. (1)
2. Limitation: it was based on atomic weight, so isotopes had no separate positions (or: the anomalous pairs Te–I and Co–Ni had to be reversed). (1)

Answer: It proved the predictive power of the periodic table; a limitation is the lack of places for isotopes (or atomic weight reversals).

Modern periodic law and the long form of the periodic table

NCERT section 3.3 Modern Periodic Law and the Present Form of the Periodic Table

In 1913 Moseley found that the square root of the frequency of X-rays emitted by elements plotted against atomic number (not atomic mass) gave a straight line. Atomic number is therefore the fundamental property.

Modern periodic law: the properties of elements, both physical and chemical, repeat periodically with their atomic numbers.

The long form of the periodic table:

• 7 periods (horizontal rows). The period number equals the highest principal quantum number n of the element. Periods contain 2, 8, 8, 18, 18, 32 and 32 elements.
• 18 groups (vertical columns), numbered 1 to 18 as recommended by IUPAC. Elements in a group have the same outer electronic configuration and so similar properties.
• The 14 elements after lanthanum (lanthanoids) and after actinium (actinoids) are placed separately at the bottom to keep the table compact.

Atomic number removed Mendeleev's anomalies: argon (Z = 18) correctly comes before potassium (Z = 19), even though it is heavier.

Worked example

Argon (atomic mass 39.9) is placed before potassium (atomic mass 39.1). Explain this using the modern periodic law, and state how many elements the fourth period contains and why. (3 marks)

Try it first, then show the solutionSolution

1. The modern periodic law arranges elements by atomic number. Argon has Z = 18 and potassium Z = 19, so argon comes first even though it is heavier. (1)
2. Argon is a noble gas like neon and krypton, and potassium is an alkali metal like sodium; ordering by Z puts each in its correct family. (1)
3. Period 4 fills 4s, 3d and 4p: 1 + 5 + 3 = 9 orbitals, holding 18 electrons, so it has 18 elements. (1)

Answer: Order is by atomic number (18 before 19); period 4 has 18 elements because 9 orbitals (4s, 3d, 4p) are filled.

Nomenclature of elements with atomic number above 100

NCERT section 3.4 Nomenclature of Elements with Atomic Numbers > 100

Newly made heavy elements often exist for only a fraction of a second, and there may be disputes over who discovered them. Until IUPAC approves a permanent name, each one gets a systematic temporary name made from its atomic number.

Roots for the digits:
0 nil, 1 un, 2 bi, 3 tri, 4 quad, 5 pent, 6 hex, 7 sept, 8 oct, 9 enn.

Rules:
1. Join the roots for the digits in order.
2. Add the ending -ium.
3. The symbol is the first letters of the three roots, with the first letter capitalised.
4. Drop a final 'n' of enn before nil (enn + nil → ennil) and the 'i' of bi or tri before -ium (bi + ium → bium, tri + ium → trium).

Examples: Z = 101 is unnilunium (Unu; now mendelevium, Md); Z = 104 is unnilquadium (Unq; now rutherfordium); Z = 120 is unbinilium (Ubn).

All elements up to Z = 118 now have permanent names, such as nihonium (113), moscovium (115), tennessine (117) and oganesson (118).

Worked example

Give the IUPAC temporary names and symbols for the elements with atomic numbers (a) 109 (b) 112 (c) 119. (3 marks)

Try it first, then show the solutionSolution

1. (a) 1-0-9: un + nil + enn + ium = unnilennium, symbol Une. (1)
2. (b) 1-1-2: un + un + bi + ium; drop 'i' of bi: ununbium, symbol Uub. (1)
3. (c) 1-1-9: un + un + enn + ium = ununennium, symbol Uue. (1)

Answer: (a) unnilennium, Une (b) ununbium, Uub (c) ununennium, Uue

Electronic configuration, periods, groups and the s, p, d, f blocks

NCERT section 3.5 Electronic Configurations of Elements and the Periodic Table; 3.6 Types of Elements: s-, p-, d-, f- Blocks (3.6.1 to 3.6.5)

The position of an element follows from its electronic configuration:

• Period = the value of n for the outermost (valence) shell.
• In each period, the number of elements = 2 × the number of orbitals filled in that period: period 1 (1s) 2; periods 2 and 3 (ns, np) 8; periods 4 and 5 (ns, (n−1)d, np) 18; periods 6 and 7 (ns, (n−2)f, (n−1)d, np) 32.

The block depends on the subshell receiving the last electron:

• s-block (groups 1 and 2): ns¹ and ns². Reactive metals with low ionisation enthalpies (alkali and alkaline earth metals). They lose 1 or 2 electrons to form +1 or +2 ions.
• p-block (groups 13 to 18): ns² np¹ to ns² np⁶. Includes metals, non-metals and metalloids. Group 18 noble gases have full shells and are very unreactive; halogens (17) and chalcogens (16) have highly negative electron gain enthalpies.
• d-block (groups 3 to 12): (n−1)d¹⁻¹⁰ ns⁰⁻². Transition elements: all metals, often coloured ions, variable oxidation states, paramagnetism, many used as catalysts. Zn, Cd and Hg (d¹⁰ ns²) do not show most of these typical properties.
• f-block: (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns². Inner transition elements: lanthanoids (Ce to Lu) and actinoids (Th to Lr). Actinoids are radioactive; those beyond uranium are made artificially (transuranium elements).

Group number for s-block = number of valence electrons; for p-block = 10 + number of valence electrons; for d-block = number of (n−1)d + ns electrons.

Metals, non-metals and metalloids: metals (over 78% of known elements) lie on the left; non-metals on the top right. Metallic character increases down a group and decreases across a period. Elements along the zig-zag line (Si, Ge, As, Sb, Te) show properties of both and are called metalloids or semi-metals.

Worked example

An element has the configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁵. (a) Give its period, group and block. (b) Is it a metal or non-metal? (c) Predict the formula of its compound with sodium. (3 marks)

Try it first, then show the solutionSolution

1. (a) Highest n = 4, so period 4. Last electron in 4p, so p-block; valence electrons = 2 + 5 = 7, group = 10 + 7 = 17. (1)
2. (b) Group 17 is the halogens: a non-metal (this is bromine, Z = 35). (1)
3. (c) It gains one electron to complete its octet, forming X⁻; with Na⁺ the formula is NaX (NaBr). (1)

Answer: (a) period 4, group 17, p-block (b) non-metal (bromine) (c) NaBr

Atomic and ionic radius

NCERT section 3.7.1 Trends in Physical Properties (a) Atomic Radius, (b) Ionic Radius

An atom has no sharp boundary, so its size is estimated from the distances between atoms in compounds or solids.

• Covalent radius: half the distance between the nuclei of two atoms joined by a single covalent bond (for example, Cl₂ bond length 198 pm, so the radius of Cl is 99 pm).
• Metallic radius: half the distance between neighbouring atoms in a metal crystal (Cu: 256 pm ÷ 2 = 128 pm).

Trends:

• Across a period (left to right) the radius decreases. Electrons enter the same shell, but the nuclear charge increases, so the effective nuclear charge rises and pulls the electrons closer. Noble gases are left out because their radii are measured as van der Waals radii, which are larger.
• Down a group the radius increases, because a new shell is added each period; the extra inner electrons shield the outer electrons from the growing nuclear charge.

Ionic radius:

• A cation is smaller than its parent atom: it has fewer electrons for the same nuclear charge, and often loses its whole outer shell (Na 186 pm, Na⁺ about 100 pm).
• An anion is larger than its parent atom: the extra electron(s) increase repulsion and lower the effective nuclear charge per electron (F 64 pm, F⁻ 136 pm).
• Isoelectronic species have the same number of electrons. Among them, the greater the nuclear charge, the smaller the radius: O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺ (all 10 electrons).

Worked example

Which is larger in each pair? Give a reason. (a) Fe²⁺ or Fe³⁺ (b) O²⁻ or F⁻ (2 marks)

Try it first, then show the solutionSolution

1. (a) Both have 26 protons. Fe²⁺ has 24 electrons and Fe³⁺ only 23, so in Fe³⁺ the same nuclear charge pulls fewer electrons and they are held closer: Fe²⁺ is larger. (1)
2. (b) O²⁻ and F⁻ are isoelectronic (10 electrons each). F has 9 protons against 8 for O, so F⁻ pulls its electrons in more strongly: O²⁻ is larger. (1)

Answer: (a) Fe²⁺ (fewer electrons in Fe³⁺ for the same nuclear charge) (b) O²⁻ (lower nuclear charge for the same 10 electrons)

Ionisation enthalpy

NCERT section 3.7.1 (c) Ionization Enthalpy

Ionisation enthalpy (${\Delta }_{i}H$) is the energy needed to remove an electron from an isolated gaseous atom in its ground state: $X(g)\to X^{+}(g)+e^{-}$. It is always positive (energy is absorbed) and is measured in kJ mol⁻¹. 'Isolated gaseous' means no neighbours influence the atom; 'ground state' means the electron is removed from the most stable arrangement.

The second ionisation enthalpy (removing an electron from X⁺) is always higher than the first, because the electron is pulled away from a positive ion.

Trends:

• Across a period: generally increases, because effective nuclear charge increases and the atom gets smaller. Alkali metals have the lowest and noble gases the highest values in each period.
• Down a group: decreases, because the outer electron is farther from the nucleus and inner electrons shield it (shielding or screening).

Two exceptions in period 2 and 3:

• B (801) < Be (899) kJ mol⁻¹, and Al (577) < Mg (737): the electron removed from B or Al is a 2p or 3p electron, which is shielded by the inner s electrons and does not penetrate as close to the nucleus as an s electron.
• O (1314) < N (1402): nitrogen's 2p³ is half-filled with three parallel spins; in oxygen two of the four 2p electrons share an orbital, and their mutual repulsion makes one easier to remove.

Worked example

The first ionisation enthalpies of C, N and F are 1086, 1402 and 1681 kJ mol⁻¹. Would the value for oxygen be closer to 1250 or 1500 kJ mol⁻¹? Justify. (2 marks)

Try it first, then show the solutionSolution

1. The general trend across period 2 would put O between N and F, near 1500 kJ mol⁻¹. (½)
2. But N is 2p³ with three singly occupied orbitals, while in O (2p⁴) two electrons share one 2p orbital; their repulsion makes one of them easier to remove. (1)
3. So O lies below N: closer to 1250 kJ mol⁻¹ (actual value 1314). (½)

Answer: Closer to 1250 kJ mol⁻¹: the paired 2p electron of oxygen is easier to remove than an electron from nitrogen's half-filled 2p³.

Electron gain enthalpy and electronegativity

NCERT section 3.7.1 (d) Electron Gain Enthalpy, (e) Electronegativity

Electron gain enthalpy (${\Delta }_{eg}H$) is the enthalpy change when an electron is added to an isolated gaseous atom: $X(g)+e^{-}\to X^{-}(g)$. It may be negative (energy released) or positive (energy needed).

• Halogens have very negative values: one electron completes their noble gas configuration.
• Noble gases have large positive values: the added electron must go into a new, higher shell.
• Across a period: becomes more negative (higher effective nuclear charge).
• Down a group: becomes less negative (larger atoms).
• Exception: O and F have less negative values than S and Cl. Because O and F are very small, an added electron enters the crowded n = 2 shell and feels strong repulsion from the electrons already there. In the larger n = 3 shell of S and Cl, there is more room. So Cl (−349) is more negative than F (−328 kJ mol⁻¹), and chlorine has the most negative electron gain enthalpy of all elements.

Electronegativity is the tendency of an atom in a chemical compound to attract the shared electron pair towards itself. It is not measurable, has no unit, and is given on scales such as Pauling's (F = 4.0, the highest; O 3.5; N and Cl 3.0).

• It increases across a period and decreases down a group.
• Non-metals have high electronegativity; metals low. So electronegativity is linked to non-metallic character.

Do not confuse the two: electron gain enthalpy is for an isolated atom and is an energy; electronegativity is for a bonded atom and is a relative number.

Worked example

Arrange O, S, F and Cl in order of increasingly negative electron gain enthalpy, and explain why O and F come below S and Cl. (3 marks)

Try it first, then show the solutionSolution

1. Across a period the value becomes more negative, so O is less negative than F and S less negative than Cl. (1)
2. Down a group the value normally becomes less negative, but O and F are exceptions: their added electron enters the small, crowded n = 2 shell and is repelled by the electrons already there, so O is less negative than S and F less negative than Cl. (1)
3. Values: O −141, S −200, F −328, Cl −349 kJ mol⁻¹, giving O < S < F < Cl. (1)

Answer: O < S < F < Cl (Cl most negative); O and F are held back by electron–electron repulsion in their small 2p subshells.

Periodic trends in chemical properties: valence, anomalous behaviour and reactivity

NCERT section 3.7.2 Periodic Trends in Chemical Properties; 3.7.3 Periodic Trends and Chemical Reactivity

Valence is the combining capacity of an element. For s- and p-block elements, valence = number of valence electrons, or 8 minus that number. So the formulas of compounds can be predicted:

• Group 1 elements form oxides M₂O and hydrides MH; group 2 forms MO and MH₂; group 13 forms M₂O₃; group 14 forms MO₂ and MH₄; group 15 forms M₂O₅ (oxide) and MH₃; group 16 forms MH₂; group 17 forms HX.
• Oxidation state and covalency can differ: in [AlCl(H₂O)₅]²⁺ aluminium has oxidation state +3 but covalency 6 (six bonds).

Anomalous properties of second-period elements: Li, Be, B, C, N, O and F differ from the heavier members of their groups because they have (i) small size, (ii) high electronegativity and high ionisation enthalpy, and (iii) no d orbitals in the valence shell, so their maximum covalency is 4 (Be and B form [BeF₄]²⁻ and [BF₄]⁻, while Al forms [AlF₆]³⁻). They can also form pπ–pπ multiple bonds (C=C, C≡C, N≡N, C=O) more easily. Diagonal relationships also appear: Li resembles Mg, and Be resembles Al.

Reactivity:

• Metallic character and reducing power are highest at the far left (lowest ionisation enthalpy); non-metallic character and oxidising power at the far right (highest electron gain enthalpy, apart from the noble gases).
• Oxides: elements on the left form basic oxides ($N{a}_{2}O+H_{2}O\to 2NaOH$); elements on the right form acidic oxides ($C{l}_2{O}_7+H_{2}O\to 2HCl{O}_4$); those in the middle form amphoteric oxides (Al₂O₃, As₂O₃) or neutral oxides (CO, NO, N₂O).

Worked example

Show with equations that aluminium oxide is amphoteric. (2 marks)

Try it first, then show the solutionSolution

1. With an acid it acts as a base: $A{l}_2{O}_3+6HCl\to 2AlC{l}_3+3H_{2}O$. (1)
2. With an alkali it acts as an acid: $A{l}_2{O}_3+6NaOH+3H_{2}O\to 2N{a}_3[Al(OH{)}_6]$. (1)

Answer: Al₂O₃ reacts with both HCl and NaOH to give salts, so it is amphoteric.

Chemistry in real life: Classification of Elements and Periodicity in Properties

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Atomic Radius, Electronegativity and Chemical Periodicity
• Ionisation Enthalpy and Electron Gain Enthalpy

Classification of Elements and Periodicity in Properties formula sheet

Elements in a period

$\text{elements}=2\times \text{orbitals filled in that period}$

Periods: 2, 8, 8, 18, 18, 32, 32.

Group number from configuration

$\text{s-block: }g=\text{valence }{e}^{-};\text{p-block: }g=10+\text{valence }{e}^{-};\text{d-block: }g=(n-1)d+ns{e}^{-}$

Valence electrons for p-block = ns + np electrons.

IUPAC roots

$0nil,1un,2bi,3tri,4quad,5pent,6hex,7sept,8oct,9enn$

Add -ium; symbol is the three initials.

Covalent radius

$r_{cov}=\frac{\text{bond length of }X-X}{2}$

Metallic radius is half the distance between neighbouring metal atoms.

Ionisation enthalpy

$X(g)\to X^{+}(g)+e^{-},{\Delta }_{i}H>0$

${\Delta }_i{H}_2>{\Delta }_i{H}_1$ always.

Electron gain enthalpy

$X(g)+e^{-}\to X^{-}(g),{\Delta }_{eg}H$

Negative for halogens, positive for noble gases.

Energy per atom from molar value

$E_{atom}=\frac{\Delta H({\text{J mol}}^{-1})}{N_A}$

Useful in numericals linking Bohr energies to ionisation enthalpy.

Classification of Elements and Periodicity in Properties: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Classification of Elements and Periodicity in Properties

The question patterns to expect, and the method that works for each.

Tricks to remember Classification of Elements and Periodicity in Properties

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Triad average

  In Döbereiner's triads, middle mass ≈ (first + last) ÷ 2. Li 7, Na 23, K 39: (7 + 39) ÷ 2 = 23.

  Use it when: Questions on early classification.

• Uncle Bob's Tricky Quiz Pages Have Some Odd Endings Now

  un 1, bi 2, tri 3, quad 4, pent 5, hex 6, sept 7, oct 8, enn 9, nil 0.

  Use it when: IUPAC names for Z > 100.

• Radius: shrinks Right, grows Down

  Across a period radius shrinks; down a group it grows. Ionisation enthalpy, electron gain (more negative) and electronegativity do the opposite.

  Use it when: Every trend question.

• Bees and Boron, Magnets and Aluminium; Nitrogen beats Oxygen

  Be > B, Mg > Al and N > O in first ionisation enthalpy: s beats p, half-full beats crowded.

  Use it when: Exception questions.

• Chlorine wins the electron

  Most negative electron gain enthalpy: Cl, not F (F is too small and crowded).

  Use it when: MCQs and assertion-reason items.

Classification of Elements and Periodicity in Properties: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Classification of Elements and Periodicity in Properties

The slips examiners see most often, and what to write instead.

• ✗Mistake: Writing that F has the most negative electron gain enthalpy

  Students extend the trend to the top of the group.

  Wrong working

  Most negative ΔegH: F > Cl

  Right working

  Cl (−349) is more negative than F (−328 kJ mol⁻¹)

  ✓Do this: Remember the small-size, high-repulsion exception for F and O.

  Usually costs: 1 mark.

• ✗Mistake: Confusing electronegativity with electron gain enthalpy

  Both are about attracting electrons.

  ✓Do this: Electron gain enthalpy: isolated gaseous atom, measured in kJ mol⁻¹. Electronegativity: atom in a bond, no unit.

  Usually costs: 1-2 marks in a 'distinguish' question.

• ✗Mistake: Ignoring nuclear charge for isoelectronic species

  Students think same electrons means same size.

  Wrong working

  Na⁺ and F⁻ have the same size because both have 10 electrons.

  Right working

  Na⁺ (Z = 11) is smaller than F⁻ (Z = 9): more protons pull the same electrons closer.

  ✓Do this: For isoelectronic species, order by Z: higher Z, smaller ion.

  Usually costs: 1 mark.

• ✗Mistake: Wrong group number for p-block elements

  Students give the number of valence electrons.

  Wrong working

  Configuration [Ne] 3s² 3p³ is in group 5.

  Right working

  Group = 10 + 5 = 15.

  ✓Do this: For p-block add 10 to the number of valence electrons.

  Usually costs: ½-1 mark.

• ✗Mistake: Mis-spelling IUPAC names

  Students forget the elision rules.

  Wrong working

  112 = ununbiium; 190 = unennnilium

  Right working

  112 = ununbium; 190 = unennilium

  ✓Do this: Drop the 'i' of bi/tri before -ium and the final 'n' of enn before nil.

  Usually costs: ½ mark each.

NCERT exercise guide: Classification of Elements and Periodicity in Properties

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Classification of Elements and Periodicity in Properties

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The IUPAC name of the element with Z = 107 is:
   (a) unnilseptium (b) unnilheptium (c) ununseptium (d) unnilsevium

   Hint

   1-0-7.

   Answer

   (a) unnilseptium (Uns); its permanent name is bohrium.

2. Q2 · 1 mark · MCQExam level

   Which has the largest first ionisation enthalpy?
   (a) Li (b) Be (c) B (d) C

   Hint

   Across period 2, with the Be/B exception.

   Answer

   (d) C (1086 kJ mol⁻¹)

3. Q3 · 2 marks · Very short answerExam level

   What are isoelectronic species? Give two species isoelectronic with Ne. (2 marks)

   Hint

   Same number of electrons.

   Answer

   Atoms or ions with the same number of electrons (1). Examples: Na⁺, Mg²⁺, F⁻, O²⁻ (any two) (1).

4. Q4 · 2 marks · Very short answerHarder exam level

   An element lies in period 5 and group 15. Write its outer electronic configuration and atomic number, and name it. (2 marks)

   Hint

   Period 5 starts after krypton (Z = 36) and fills 5s, 4d, then 5p.

   Answer

   Configuration [Kr] 4d¹⁰ 5s² 5p³ (group 15 has 5 valence electrons) (1). Z = 36 + 2 + 10 + 3 = 51: antimony, Sb, a metalloid (1).

5. Q5 · 3 marks · Short answerHarder exam level

   Arrange in increasing order of the property shown: (a) F, Cl, Br, I (electron gain enthalpy, negative values) (b) Na, Mg, Al (first ionisation enthalpy) (c) B, C, N, O, F (electronegativity). (3 marks)

   Hint

   Check the exceptions.

   Answer

   (a) I < Br < F < Cl (Cl most negative) (1). (b) Na < Al < Mg (1). (c) B < C < N < O < F (1).

6. Q6 · 2 marks · Very short answerHarder exam level

   Explain why second-period elements cannot have a covalency greater than 4, while their third-period group members can. Give one example. (2 marks)

   Hint

   Which orbitals are in the n = 2 shell?

   Answer

   The n = 2 shell has only 2s and 2p orbitals (four orbitals), so at most four bonds (1). Third-period elements can use 3d orbitals: N forms NF₃ only but P forms PCl₅; B forms [BF₄]⁻ but Al forms [AlF₆]³⁻ (1).

7. Q7 · 3 marks · Short answerHarder exam level

   Suppose you are given oxides X₂O, YO₂ and ZO, where X, Y and Z are Na, S and Mg in some order. Match each oxide to its element, state whether it is acidic, basic or amphoteric, and write the reaction of the acidic one with water. (3 marks)

   Hint

   Use valence: 1 for Na, 2 for Mg, 4 for S in SO₂.

   Answer

   X₂O = Na₂O (basic); ZO = MgO (basic); YO₂ = SO₂ (acidic) (2). $S{O}_2+H_{2}O\to H_{2}S{O}_3$ (1).

8. Q8 · 3 marks · Short answerHarder exam level

   The first ionisation enthalpy of potassium is 419 kJ mol⁻¹. Calculate the minimum energy in joules needed to ionise one potassium atom, and the maximum wavelength of light that could do it. ($N_A$ = 6.022 × 10²³, $hc$ = 1.988 × 10⁻²⁵ J m) (3 marks)

   Hint

   Divide by N_A, then λ = hc/E.

   Answer

   Per atom = 419 000 ÷ 6.022 × 10²³ = 6.96 × 10⁻¹⁹ J (1½). λ = 1.988 × 10⁻²⁵ ÷ 6.96 × 10⁻¹⁹ = 2.86 × 10⁻⁷ m = 286 nm (ultraviolet) (1½).

9. Q9 · 1 mark · MCQHarder exam level

   Which of these is the correct order of size?
   (a) Al³⁺ < Mg²⁺ < Na⁺ < F⁻ (b) F⁻ < Na⁺ < Mg²⁺ < Al³⁺ (c) Na⁺ < Al³⁺ < F⁻ < Mg²⁺ (d) Mg²⁺ < Al³⁺ < F⁻ < Na⁺

   Hint

   Isoelectronic: highest Z is smallest.

   Answer

   (a)

10. Q10 · 3 marks · Short answerHOTS / competency

    Mendeleev placed tellurium (127.6) before iodine (126.9) though it is heavier. (a) What property did he use to justify this? (b) How does the modern periodic law remove the problem? (c) Name one other pair that shows the same reversal. (3 marks)

    Hint

    Te has Z = 52 and I has Z = 53.

    Answer

    (a) Chemical similarity: I resembles Cl and Br (halogens), Te resembles S and Se (1). (b) Ordering by atomic number puts Te (52) before I (53) naturally (1). (c) Ar–K or Co–Ni (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    Suppose element E has these successive ionisation enthalpies (kJ mol⁻¹): 738, 1451, 7733, 10 540. (a) To which group does E belong? (b) Write the formula of its chloride and oxide. (c) Would its oxide react with water to give an acid or a base? (d) If E is in period 3, identify it and explain why its first ionisation enthalpy is higher than that of the next element. (5 marks)

    Hint

    Find the big jump.

    Answer

    (a) The big jump is after the second electron, so two valence electrons: group 2 (1). (b) ECl₂ and EO (1). (c) A base (metal oxide), e.g. E(OH)₂ (1). (d) Mg; its next element Al loses a 3p electron, which is better shielded and less tightly held than Mg's 3s electron from a filled 3s² subshell (2).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): Electronegativity of fluorine is higher than chlorine, but electron gain enthalpy of chlorine is more negative than fluorine.
    Reason (R): Electronegativity refers to an atom in a bond, while electron gain enthalpy refers to an isolated gaseous atom, and fluorine's small size causes strong repulsion for an incoming electron.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    Two different properties.

    Answer

    (a) Both are true and R explains the apparent contradiction.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State the modern periodic law.Show answer

   The physical and chemical properties of elements are periodic functions of their atomic numbers.

2. 2.Whose work showed atomic number is more fundamental than atomic mass?Show answer

   Henry Moseley (X-ray frequencies, 1913).

3. 3.How many periods and groups are there in the long form?Show answer

   7 periods and 18 groups.

4. 4.What did Mendeleev call the element later named gallium?Show answer

   Eka-aluminium.

5. 5.Give the general configuration of d-block elements.Show answer

   (n−1)d¹⁻¹⁰ ns⁰⁻².

6. 6.Name two metalloids.Show answer

   Any two of Si, Ge, As, Sb, Te.

7. 7.Which element has the highest electronegativity?Show answer

   Fluorine (4.0 on Pauling's scale).

8. 8.Why is a cation smaller than its atom?Show answer

   It has fewer electrons for the same nuclear charge (and often loses a whole shell), so the electrons are pulled closer.

9. 9.What type of oxide is Cl₂O₇?Show answer

   Acidic; with water it gives perchloric acid, HClO₄.

10. 10.Name two diagonal pairs.Show answer

    Li and Mg; Be and Al.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state Mendeleev's and the modern periodic law and give the key difference?
• ☐Can you explain why periods have 2, 8, 8, 18, 18, 32, 32 elements?
• ☐Can you name and give the symbol for element 116?
• ☐Can you find the period, group and block of Z = 29 and Z = 34?
• ☐Can you order isoelectronic ions by size?
• ☐Can you explain why Mg has a higher ΔᵢH₁ than Al and N higher than O?
• ☐Can you explain why Cl has a more negative ΔegH than F?
• ☐Can you show that Al₂O₃ is amphoteric with equations?

Classification of Elements and Periodicity in Properties: frequently asked questions