📌 Snapshot
- Periodic trends in radius, ionisation enthalpy and electronegativity, explained with solved CUET examples.
- How Mendeleev's gaps predicted germanium, why Moseley changed the law, and how the long form table is built.
- Why each trend holds, with the exceptions CUET loves: Be and B, N and O, and Cl against F.
- Block positions, IUPAC names above Z = 100, diagonal relationships and the acidic to basic oxide trend.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 3.
Could you predict an element nobody had seen?
In 1869 Dmitri Mendeleev arranged the known elements in a table, and where a pattern broke he left a gap and said an element would be found there. He even gave its properties. For the gap under silicon he predicted an atomic weight near 72, a density of 5.5 g cm⁻³ and an oxide of formula EO₂. In 1886 germanium was discovered with atomic weight 72.6, density 5.36 and the oxide GeO₂. A table that predicts the future is a powerful tool, and this chapter shows why it works. The answer is the electron configurations you met in Structure of Atom. Elements repeat their properties because they repeat their outer electron arrangement.
Why classify, and the first attempts
By 1800 only 31 elements were known, by 1865 there were 63, and today 118 have been named. Studying each element and its compounds one by one is hopeless, so chemists looked for patterns. Every attempt used atomic weight, because atoms of different elements were first compared by mass.
| Scientist | Idea | Why it fell short |
|---|---|---|
| Döbereiner (early 1800s) | Triads: in a group of three similar elements, the middle one has an atomic weight near the mean of the other two | Only a few triads existed |
| De Chancourtois (1862) | Elements listed by atomic weight on a spiral around a cylinder, and similar ones lined up | Received little attention |
| Newlands (1865) | Law of Octaves: every eighth element resembles the first, like musical notes | Worked only up to calcium |
| Mendeleev and Lothar Meyer (1869) | Properties are a periodic function of atomic weight | Several elements out of order, and no explanation of why |
Check a triad. Lithium (7), sodium (23) and potassium (39): the mean of the outer two is $\dfrac{7 + 39}{2} = 23$, exactly sodium's weight. Calcium, strontium, barium (40, 88, 137) give $\dfrac{40 + 137}{2} = 88.5$, and chlorine, bromine, iodine (35.5, 80, 127) give $81.3$, both close.
Mendeleev's table: gaps that predicted the future
Mendeleev put elements in order of increasing atomic weight but grouped them by chemical similarity, and when the two clashed, similarity won. Iodine (126.9) is lighter than tellurium (127.6), yet he put iodine after tellurium so that it sat in the same column as chlorine and bromine. He also left gaps, naming the missing elements eka-aluminium, eka-silicon and eka-boron (later found as gallium, germanium and scandium). The predictions were remarkably accurate.
| Property | Eka-aluminium (predicted) | Gallium (found, 1875) | Eka-silicon (predicted) | Germanium (found, 1886) |
|---|---|---|---|---|
| Atomic weight | 68 | 70 | 72 | 72.6 |
| Density (g cm⁻³) | 5.9 | 5.94 | 5.5 | 5.36 |
| Oxide | E₂O₃ | Ga₂O₃ | EO₂ | GeO₂ |
| Chloride | ECl₃ | GaCl₃ | ECl₄ | GeCl₄ |
The table had real defects, though. In order of atomic weight, potassium should come before argon, nickel before cobalt and iodine before tellurium, but their properties demanded the opposite order. Hydrogen had to be placed in two groups. There was no place for isotopes, which have different weights but the same chemistry, and no way to fit the lanthanoids. The cause was unknown until atomic structure was understood.
Moseley and the Modern Periodic Law
In 1913 Henry Moseley bombarded different metals with electrons and measured the frequency $\nu$ of the characteristic X-rays they gave off. A plot of $\sqrt{\nu}$ against atomic number $Z$ was a straight line, which means the relation is
$$\sqrt{\nu} = a\,(Z - b)$$
for constants $a$ and $b$ specific to the type of X-ray line. Plotted against atomic weight, the same data scatter. So atomic number, the number of protons, is the real ordering quantity. This gives the Modern Periodic Law: the physical and chemical properties of elements are periodic functions of their atomic numbers. Mendeleev's anomalies vanish at once, because Ar ($Z = 18$) does come before K ($Z = 19$).
How the long form of the table is built
The long form has 7 periods (rows) and 18 groups (columns, numbered 1 to 18 under the IUPAC system). The period number is the highest principal quantum number $n$ occupied. The number of elements in a period comes straight from the filling order: count the orbitals filled between one noble gas and the next, and double.
| Period | Orbitals filled | Elements | Starts with, ends with |
|---|---|---|---|
| 1 | 1s | 2 | H, He |
| 2 | 2s, 2p | 8 | Li, Ne |
| 3 | 3s, 3p | 8 | Na, Ar |
| 4 | 4s, 3d, 4p | 18 | K, Kr |
| 5 | 5s, 4d, 5p | 18 | Rb, Xe |
| 6 | 6s, 4f, 5d, 6p | 32 | Cs, Rn |
| 7 | 7s, 5f, 6d, 7p | 32 | Fr, Og |
Period 4, for example, has $1 + 5 + 3 = 9$ orbitals, and 9 orbitals hold 18 electrons, which is why the fourth period has 18 elements. The 14 lanthanoids (Ce to Lu) and 14 actinoids (Th to Lr) belong to periods 6 and 7 but are printed in two rows at the bottom to keep the table narrow.
To place an element, write its configuration, and then use these rules.
| Block | Group number | Outer configuration |
|---|---|---|
| s | Number of valence electrons (1 or 2) | $ns^{1-2}$ |
| p | 10 + number of valence electrons ($ns$ and $np$) | $ns^2np^{1-6}$ |
| d | Number of electrons in $(n-1)d$ and $ns$ | $(n-1)d^{1-10}ns^{0-2}$ |
| f | Lanthanoids and actinoids (not given a group number) | $(n-2)f^{1-14}(n-1)d^{0-1}ns^2$ |
Worked example: find the position. Locate the elements with $Z$ = 35, 26, 56, 82 and 29.
| $Z$ | Configuration | Period | Group | Block |
|---|---|---|---|---|
| 35 (Br) | [Ar] 3d¹⁰ 4s² 4p⁵ | 4 | $10 + 2 + 5 = 17$ | p |
| 26 (Fe) | [Ar] 3d⁶ 4s² | 4 | $6 + 2 = 8$ | d |
| 56 (Ba) | [Xe] 6s² | 6 | 2 | s |
| 82 (Pb) | [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p² | 6 | $10 + 2 + 2 = 14$ | p |
| 29 (Cu) | [Ar] 3d¹⁰ 4s¹ | 4 | $10 + 1 = 11$ | d |
The common wrong step: giving copper group 1 because it ends in 4s¹. The 3d electrons count towards the group number, so Cu is in group 11.
Beyond the textbook. For the exam: hydrogen ($1s^1$) is placed at the top of group 1, and helium ($1s^2$), although its configuration is s block, is placed in group 18 with the other noble gases because its shell is completely filled. In reality: hydrogen fits no group well. It gains an electron like the halogens and loses one like the alkali metals, yet it is a non metal. Some tables place it above the halogens and others show it on its own. Why it matters: it shows that a periodic table is a convenient arrangement of repeating properties, and a few elements will never fit perfectly.
Naming the elements above Z = 100
New heavy elements are made in laboratories and exist for fractions of a second. While discovery is disputed, IUPAC gives each a temporary systematic name built from its atomic number, using a root for each digit and ending in "ium".
| Digit | Root | Symbol letter |
|---|---|---|
| 0 | nil | n |
| 1 | un | u |
| 2 | bi | b |
| 3 | tri | t |
| 4 | quad | q |
| 5 | pent | p |
| 6 | hex | h |
| 7 | sept | s |
| 8 | oct | o |
| 9 | enn | e |
Worked example: systematic names. Name the elements with $Z$ = 120, 119 and 102, and give their three letter symbols.
| $Z$ | Digits | Roots | Name | Symbol |
|---|---|---|---|---|
| 120 | 1, 2, 0 | un + bi + nil + ium | Unbinilium | Ubn |
| 119 | 1, 1, 9 | un + un + enn + ium | Ununennium | Uue |
| 102 | 1, 0, 2 | un + nil + bi + ium | Unnilbium | Unb |
The common wrong step: dropping the "nil" for zero, which turns 120 into "Unbium". When a root ending in "i" (bi, tri) meets "ium", only one "i" is kept, as in Unnilbium.
After the discoveries were confirmed, permanent names were approved.
| $Z$ | Name (symbol) | $Z$ | Name (symbol) | $Z$ | Name (symbol) |
|---|---|---|---|---|---|
| 101 | Mendelevium (Md) | 107 | Bohrium (Bh) | 113 | Nihonium (Nh) |
| 102 | Nobelium (No) | 108 | Hassium (Hs) | 114 | Flerovium (Fl) |
| 103 | Lawrencium (Lr) | 109 | Meitnerium (Mt) | 115 | Moscovium (Mc) |
| 104 | Rutherfordium (Rf) | 110 | Darmstadtium (Ds) | 116 | Livermorium (Lv) |
| 105 | Dubnium (Db) | 111 | Roentgenium (Rg) | 117 | Tennessine (Ts) |
| 106 | Seaborgium (Sg) | 112 | Copernicium (Cn) | 118 | Oganesson (Og) |
The four blocks and the metal to non metal divide
Where an element sits depends on which subshell receives its last electron, and the block tells you how it behaves.
| Block | Groups | Members | Typical behaviour |
|---|---|---|---|
| s | 1 and 2 | Alkali metals (group 1), alkaline earth metals (group 2) | Soft reactive metals with low ionisation enthalpy forming 1+ and 2+ ions, mostly ionic compounds, coloured flames |
| p | 13 to 18 | Boron, carbon and nitrogen families, chalcogens (16), halogens (17), noble gases (18) | Mix of metals, metalloids and non metals. Groups 16 and 17 have very negative electron gain enthalpy |
| d | 3 to 12 | Transition elements | Coloured ions, variable oxidation states, paramagnetism, catalysts |
| f | Two bottom rows | Lanthanoids and actinoids | All metals. Actinoids are radioactive, and elements beyond uranium (transuranium) are man made |
More than 78% of elements are metals, found on the left and centre. They are usually solid with high melting and boiling points, and are malleable, ductile and good conductors. Non metals sit at the top right, are often gases or low melting solids, and are poor conductors. A zig zag line separates them, and along it lie the metalloids (Si, Ge, As, Sb, Te) with intermediate properties. Metallic character falls from left to right across a period and rises down a group.
Beyond the textbook. For the exam: group 12 elements Zn, Cd and Hg are d block elements. In reality: the IUPAC definition of a transition element is one with an incomplete d subshell in the atom or in a common ion. Zn, Cd and Hg have full $d^{10}$ shells in the atom and in their usual +2 ions, so they do not behave as typical transition metals. Why it matters: you will meet this again in the d and f block chapter, where colour and magnetism need partly filled d orbitals.
The master idea: effective nuclear charge
Every periodic trend rests on one question: how strongly does the nucleus hold the outermost electron? The answer is the effective nuclear charge, the pull of the nucleus after the inner electrons have shielded part of it.
$$Z_{\text{eff}} = Z - S$$
where $S$ measures the shielding. Across a period, the electron added goes into the same shell, where electrons shield one another poorly, so $Z_{\text{eff}}$ rises and the atom shrinks. Down a group, a new shell is added, which is far from the nucleus and well shielded by all the inner shells, so the outer electron is held more loosely despite the higher $Z$. With these two facts you can predict almost every trend below.
| Property | Across a period (left to right) | Down a group |
|---|---|---|
| Atomic radius | Decreases | Increases |
| Ionisation enthalpy | Increases (with two exceptions) | Decreases |
| Electron gain enthalpy | Becomes more negative (with exceptions) | Becomes less negative (with the F exception) |
| Electronegativity | Increases | Decreases |
| Metallic character | Decreases | Increases |
Atomic radius
An atom has no sharp edge, so its size is defined from distances between nuclei, and the definition depends on how it is bonded.
| Type | Definition | Example |
|---|---|---|
| Covalent radius | Half the distance between the nuclei of two like atoms joined by a single covalent bond | Cl–Cl is 198 pm, so $r = 99$ pm |
| Metallic radius | Half the distance between the cores in a metallic crystal | Cu–Cu is 256 pm, so $r = 128$ pm |
| van der Waals radius | Half the distance between nuclei of two non bonded atoms of neighbouring molecules, for noble gases | Always larger than the covalent radius |
| Element (period 2) | Radius (pm) | IE₁ (kJ mol⁻¹) | Electronegativity |
|---|---|---|---|
| Li | 152 | 520 | 1.0 |
| Be | 111 | 899 | 1.5 |
| B | 88 | 801 | 2.0 |
| C | 77 | 1086 | 2.5 |
| N | 70 | 1402 | 3.0 |
| O | 66 | 1314 | 3.5 |
| F | 64 | 1681 | 4.0 |
| Ne | not defined | 2081 | not defined |
| Element (period 3) | Radius (pm) | IE₁ (kJ mol⁻¹) | Electronegativity |
|---|---|---|---|
| Na | 186 | 496 | 0.9 |
| Mg | 160 | 738 | 1.2 |
| Al | 143 | 578 | 1.5 |
| Si | 117 | 787 | 1.8 |
| P | 110 | 1012 | 2.1 |
| S | 104 | 1000 | 2.5 |
| Cl | 99 | 1251 | 3.0 |
| Element (group 1) | Radius (pm) | IE₁ (kJ mol⁻¹) | Electronegativity |
|---|---|---|---|
| Li | 152 | 520 | 1.0 |
| Na | 186 | 496 | 0.9 |
| K | 231 | 419 | 0.8 |
| Rb | 244 | 403 | 0.8 |
| Cs | 262 | 376 | 0.7 |
| Element (group 17) | Radius (pm) | Electron gain enthalpy (kJ mol⁻¹) | Electronegativity |
|---|---|---|---|
| F | 64 | −328 | 4.0 |
| Cl | 99 | −349 | 3.0 |
| Br | 114 | −325 | 2.8 |
| I | 133 | −295 | 2.5 |
| At | not listed | −270 | 2.2 |
These four tables hold most of the data you will need for the rest of the chapter. Radii fall steadily across a period and rise down a group. The covalent and metallic radii are listed together here, and the noble gas values (van der Waals) are left out because they are measured differently.
Worked example: radius from bond length. The Br–Br bond length is 228 pm. Estimate the covalent radius of bromine, and then the C–Br bond length (covalent radius of C is 77 pm).
| Step | What we do | Result |
|---|---|---|
| 1 | Radius is half the bond length | $r_{\text{Br}} = \dfrac{228}{2} = 114$ pm |
| 2 | In a heteronuclear bond the bond length is the sum of the radii | $77 + 114 = 191$ pm |
The common wrong step: comparing a noble gas radius with its neighbours without noticing it is a van der Waals radius. Neon looks bigger than fluorine only because it is measured differently.
Ionic radius
An ion's size depends on its charge, and the rule is simple. A cation is smaller than its atom because it has lost electrons (often a whole shell), while the same nuclear charge now pulls fewer electrons. An anion is larger than its atom because the extra electrons repel each other and $Z_{\text{eff}}$ per electron falls. Na is 186 pm but Na⁺ is about 100 pm. Cl is 99 pm but Cl⁻ is 181 pm. The more positive the charge, the smaller the cation, so Fe³⁺ is smaller than Fe²⁺.
Species with the same number of electrons are isoelectronic. Among these, size is controlled by nuclear charge alone, and more protons pull the same electrons closer.
Worked example: rank isoelectronic ions. Arrange O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺ in order of decreasing radius.
| Step | What we do | Result |
|---|---|---|
| 1 | Check the electron count | All have 10 electrons |
| 2 | List nuclear charge | $Z$ = 8, 9, 11, 12, 13 |
| 3 | More protons means a smaller ion | O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺ |
Pause and predict: Which is larger, Fe²⁺ or Fe³⁺?
Check your answer: Fe²⁺. Both have the same nuclear charge of 26 protons, but Fe³⁺ has one electron fewer, so the same pull acts on fewer electrons that repel each other less, and the ion shrinks. The more positive the charge on a cation of one element, the smaller it is.
Ionisation enthalpy
The first ionisation enthalpy is the energy needed to remove the most loosely held electron from an isolated gaseous atom. It is always positive, and it is the measure of how well an element can form a cation.
X(g) → X⁺(g) + e⁻ X⁺(g) → X²⁺(g) + e⁻ (second)
Removing each further electron is harder than the one before, because the electron is leaving an ion of higher positive charge: IE₃ > IE₂ > IE₁. A sudden huge jump shows that the next electron comes from an inner, tightly held shell.
Worked example: group from successive ionisation enthalpies. An element has successive ionisation enthalpies (kJ mol⁻¹) of 738, 1451, 7733 and 10543. Which group is it in?
| Step | What we do | Result |
|---|---|---|
| 1 | Look at the ratios between successive values | 1451/738 is 2.0, 7733/1451 is 5.3 |
| 2 | The big jump comes after the second electron | Two valence electrons, then a noble gas core |
| 3 | Two valence electrons | Group 2 (the data are for magnesium) |
Trend. Across a period, ionisation enthalpy rises because $Z_{\text{eff}}$ rises and the radius shrinks. The peaks are at noble gases (full shells) and the lows at alkali metals (one loosely held electron). Down a group it falls, because the outer electron is farther and better shielded. Group 1 shows this clearly, with IE₁ falling from 520 kJ mol⁻¹ for Li to 376 kJ mol⁻¹ for Cs.
Look again at the period 2 and period 3 columns of the data tables. Each period follows the rule except at two places. The full period 2 order is Li < B < Be < C < O < N < F < Ne. Two dips explain the exceptions. Boron's electron comes from a 2p orbital, which is farther out and shielded by the 2s electrons, so it leaves more easily than the 2s electron of beryllium. In oxygen, the fourth 2p electron must pair with another in the same orbital, and the repulsion helps it escape. Nitrogen's $2p^3$ is half filled, with three unpaired electrons, which is a stable arrangement.
Pause and predict: Why is the second ionisation enthalpy of sodium (4562 kJ mol⁻¹) so much higher than that of magnesium (1451 kJ mol⁻¹)?
Check your answer: Removing the second electron from Na⁺ means breaking into the stable $2p^6$ core. In Mg⁺ the second electron is still the outer 3s electron. A noble gas core is far more tightly held.
Electron gain enthalpy
Electron gain enthalpy is the enthalpy change when a gaseous atom takes up an electron.
X(g) + e⁻ → X⁻(g)
It is negative when energy is released. The more negative it is, the more readily the atom accepts an electron. Halogens are the most negative, because one electron completes their octet.
In the group 17 table above, chlorine, not fluorine, is the most negative. Fluorine is so small that its 2p subshell is crowded, and the incoming electron is repelled strongly by the electrons already there. In the larger 3p subshell of chlorine the repulsion is less. The same effect makes O (−141 kJ mol⁻¹) less negative than S (−200 kJ mol⁻¹). Noble gases have positive values (He +48, Ne +116, Ar +96 kJ mol⁻¹), because their shells are full and the new electron must go into the next shell. Nitrogen, with its stable half filled $2p^3$, has a value close to zero, much less negative than carbon (−122 kJ mol⁻¹).
The common wrong step: treating electron gain enthalpy and ionisation enthalpy as opposites that must follow the same order. They share trends but the exceptions are different.
Electronegativity
Electronegativity is the tendency of an atom bonded to another to attract the shared electrons towards itself. Unlike ionisation enthalpy, it is a relative scale and cannot be measured directly. The most widely used is the Pauling scale, in which fluorine, the most electronegative element, is fixed at 4.0.
The data tables show it rising steadily across each period (Li 1.0 to F 4.0, Na 0.9 to Cl 3.0) and falling down a group (F 4.0 to At 2.2, Li 1.0 to Cs 0.7). Electronegativity is directly related to non metallic character and inversely to metallic character. The three similar looking terms are different quantities.
| Term | Isolated or bonded? | Measures |
|---|---|---|
| Ionisation enthalpy | Isolated gaseous atom | Energy to remove an electron |
| Electron gain enthalpy | Isolated gaseous atom | Energy change on gaining an electron |
| Electronegativity | Atom in a molecule | Pull on a shared pair |
Valence and oxidation state
Valence is the combining capacity of an element. For groups 1, 2, 13 and 14 it equals the number of valence electrons, and for groups 15, 16 and 17 it is $8 - \text{valence electrons}$, so it is 3, 2 and 1. Transition elements show variable valence. The oxidation state is the charge an atom would have if the shared electrons went to the more electronegative atom. In OF₂, fluorine is more electronegative, so F is −1 and O is +2. In Na₂O, Na is +1 and O is −2.
Second period anomalies and diagonal relationships
The first member of each group (Li, Be, B, C, N, O, F) differs from the rest of its group. The reasons are small size, high charge to radius ratio, high electronegativity and only four valence orbitals (2s and three 2p), with no d orbitals. This puts a ceiling on covalency.
| Period | Maximum covalency | Why | Examples |
|---|---|---|---|
| 2 | 4 | Only 2s and 2p are available | CH₄, NH₄⁺, BF₄⁻ |
| 3 and below | 6 or more | d orbitals can be used, so the octet can expand | PCl₅, SF₆, AlF₆³⁻ |
That is why phosphorus makes PCl₃ and PCl₅ while nitrogen makes only NCl₃. The first members also form strong multiple bonds ($p\pi$–$p\pi$) such as C=C, N≡N and C=O.
Moving one step down and one step right, elements in a diagonal pair behave alike, because the increase in charge is offset by the increase in size. The two key pairs are lithium and magnesium, and beryllium and aluminium.
| Pair | Shared behaviour | Example |
|---|---|---|
| Li and Mg | Form a normal oxide with oxygen. React directly with nitrogen | 6Li + N₂ → 2Li₃N and 3Mg + N₂ → Mg₃N₂ |
| Li and Mg | Nitrides give ammonia with water | Mg₃N₂ + 6H₂O → 3Mg(OH)₂ + 2NH₃ |
| Be and Al | Amphoteric oxides and hydroxides | BeO + 2NaOH → Na₂BeO₂ + H₂O |
| Be and Al | Form nitrides | 3Be + N₂ → Be₃N₂ |
Reactivity and the nature of oxides
Reactivity is highest at the two ends of a period: at the left, metals lose electrons easily (Na reacts with water, 2Na + 2H₂O → 2NaOH + H₂), and at the right, non metals gain them easily. Reactivity is lowest at the centre. The nature of the oxide changes steadily across period 3, from strongly basic to strongly acidic.
| Oxide | Nature |
|---|---|
| Na₂O | Basic |
| MgO | Basic |
| Al₂O₃ | Amphoteric |
| SiO₂ | Acidic |
| P₄O₁₀ | Acidic |
| SO₃ | Acidic |
| Cl₂O₇ | Strongly acidic |
Basic oxide with water: Na₂O + H₂O → 2NaOH. Acidic oxide with water: Cl₂O₇ + H₂O → 2HClO₄. An amphoteric oxide reacts with both acid and base:
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O and Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄]
Neutral oxides, such as CO, NO and N₂O, react with neither. Metallic character grows down a group, so oxides become more basic as you descend.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "The Modern Periodic Law is based on atomic mass." | Mendeleev's law was | Moseley showed the basis is atomic number |
| "Fluorine has the most negative electron gain enthalpy." | It is the most electronegative | Chlorine (−349) beats fluorine (−328) because F is so small |
| "IE₁ of boron is higher than that of beryllium." | Boron has more protons | Be > B, since the 2s electron is more tightly held than the 2p electron |
| "IE₁ of oxygen is higher than that of nitrogen." | Oxygen has more protons | N > O, because the half filled 2p³ of nitrogen is stable |
| "Noble gases have negative electron gain enthalpy." | They are next to the halogens | Their values are positive, because the electron must enter a new shell |
| "A cation is larger than its parent atom." | The nucleus is the same | A cation is smaller. An anion is larger |
| "Lanthanoids belong to the d block." | They sit under the d block | The 4f subshell is filling, so they are f block |
| "Helium is a p block element." | It is placed with the p block noble gases | Its configuration is $1s^2$, so electronically it is s block, placed in group 18 |
Trend and reaction sheet
| Quantity | Rule or value | When to use |
|---|---|---|
| Group of a p block element | $10 + (\text{electrons in } ns \text{ and } np)$ | Locating an element from its configuration |
| Group of a d block element | Electrons in $(n-1)d$ plus $ns$ | Same |
| Period | Highest $n$ occupied | Same |
| Covalent radius | $r = \tfrac{1}{2}d(\text{X–X})$ | Radius from a bond length |
| Isoelectronic radius | More protons means smaller | O²⁻ > F⁻ > Na⁺ > Mg²⁺ > Al³⁺ |
| Period 2 ionisation order | Li < B < Be < C < O < N < F < Ne | Ranking IE₁ |
| Halogen electron gain | Cl > F > Br > I | Ranking electron gain enthalpy |
| Valence | Groups 15 to 17: $8 - \text{valence electrons}$ | Predicting formulas |
| Oxide trend | Na₂O (basic) to Cl₂O₇ (acidic) | Nature of an oxide |
60-second revision
| Topic | Lock this in |
|---|---|
| Early classifications | Döbereiner triads → Newlands octaves → Mendeleev (atomic weight) → Moseley (atomic number) |
| Mendeleev's gaps | Eka-aluminium = Ga, eka-silicon = Ge, eka-boron = Sc. Oxides E₂O₃ and EO₂ |
| Table layout | 7 periods, 18 groups. Periods hold 2, 8, 8, 18, 18, 32, 32 |
| Position | Period = highest $n$. Group (p block) = 10 + valence electrons |
| IUPAC roots | nil, un, bi, tri, quad, pent, hex, sept, oct, enn. Z = 120 is Unbinilium |
| Radius | Falls across a period, rises down a group. Cation < atom < anion |
| Ionisation enthalpy | Rises across, falls down. Exceptions: Be > B, N > O |
| Electron gain | Cl > F. Positive for noble gases |
| Electronegativity | Pauling, F = 4.0. Rises across, falls down |
| Period 2 anomaly | Maximum covalency 4. Diagonal pairs Li–Mg and Be–Al |
| Oxides | Basic → amphoteric → acidic across the period |
Where this chapter connects
- Before this: Structure of Atom gave the electron configurations and filling rules that decide every position in the table.
- Next: Chemical Bonding and Molecular Structure uses ionisation enthalpy, electron gain enthalpy and electronegativity to decide whether atoms form ionic or covalent bonds.
- Class XII link: The d and f Block Elements explains variable oxidation states, colour and the lanthanoid trends that begin here.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. Assertion (A): Mendeleev placed iodine after tellurium even though iodine has the lower atomic weight. Reason (R): He grouped elements by similarity of chemical properties, even where this broke the order of atomic weight.
▸ Show answer & explanation
Answer: A
Tellurium (127.6) is heavier than iodine (126.9), yet iodine resembles chlorine and bromine, so Mendeleev put it in the halogen column. The reason states exactly why he did this. Options B, C and D each deny a true statement.
Q2. Which of the following correctly matches Mendeleev's prediction for eka-silicon with the properties of germanium?
▸ Show answer & explanation
Answer: A
Eka-silicon was predicted with atomic weight 72, oxide EO₂ and chloride ECl₄, and germanium has 72.6, GeO₂ and GeCl₄. Options B and C use the formulas predicted for eka-aluminium (E₂O₃, ECl₃), and D uses its atomic weight, 68.
Q3. The systematic IUPAC name and symbol of the element with atomic number 119 are
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Answer: A
The digits 1, 1, 9 give the roots un, un and enn, so the name is Ununennium and the symbol is Uue. Option B corresponds to 109 (digits 1, 0, 9), C to 120 and D to 118.
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Q4. Match the elements in List I with their blocks in List II. List I: (i) Mg (Z = 12) (ii) Se (Z = 34) (iii) Zn (Z = 30) (iv) Th (Z = 90). List II: (p) s block (q) p block (r) d block (s) f block.
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Answer: A
Mg is 3s², so it is s block. Se is 4s² 4p⁴, in group 16, so p block. Zn is 3d¹⁰ 4s², in group 12, so d block. Thorium is an actinoid, so it is f block. The other options swap blocks for at least two of these elements.
Q5. The correct order of increasing first ionisation enthalpy among Be, B, N and O is
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Answer: B
The values are B 801, Be 899, O 1314 and N 1402 kJ mol⁻¹. Boron's 2p electron is easier to remove than beryllium's 2s electron, and nitrogen's half filled 2p³ holds on to its electrons more strongly than oxygen. Options A, C and D each reverse one of these two exceptions.
Q6. Which statement about electron gain enthalpy is correct?
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Answer: C
Chlorine (−349 kJ mol⁻¹) is more negative than fluorine (−328 kJ mol⁻¹) because of the strong electron repulsion in the small 2p subshell of F. Noble gases have positive values, so A is false. B is the reverse of the facts, and in D the values become less negative from chlorine to iodine.
Q7. Assertion (A): Lithium resembles magnesium more than it resembles the other alkali metals. Reason (R): Lithium and magnesium have similar atomic sizes and similar electronegativities.
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Answer: A
Li (152 pm, electronegativity 1.0) and Mg (160 pm, 1.2) are close in size and electronegativity, which produces the diagonal relationship, for example both form nitrides directly with nitrogen. So both statements are true and R explains A.
Q8. Among the isoelectronic species K⁺, Ca²⁺, Cl⁻ and S²⁻, the one with the largest radius is
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Answer: D
All four have 18 electrons, so the nuclear charge decides the size: S²⁻ ($Z$ = 16) has the fewest protons and is the largest, then Cl⁻ (17), K⁺ (19) and Ca²⁺ (20). Option C is the next largest, and A and B have the stronger pull of more protons.
Q9. Which pair of oxides is amphoteric?
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Answer: A
Al₂O₃ and BeO react with both acids and bases, an example of the Be–Al diagonal relationship. Na₂O is basic, CO and NO are neutral, and SO₃ and Cl₂O₇ are acidic.
Q10. An element has successive ionisation enthalpies (in kJ mol⁻¹) of 738, 1451, 7733 and 10543. The element belongs to
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Answer: B
There is a sharp jump between the second and third ionisation enthalpies, which marks the start of a noble gas core after two valence electrons are removed. Two valence electrons place the element in Group 2 (it is magnesium). A group 1 element would jump after IE₁, group 13 after IE₃ and group 14 after IE₄.
Q11. Phosphorus forms both PCl₃ and PCl₅, but nitrogen forms only NCl₃. The best explanation is that
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Answer: B
Nitrogen has only 2s and 2p orbitals, so it can form at most four bonds. Phosphorus can use 3d orbitals and expand its octet. Physical state (A) is irrelevant. Nitrogen does show +5 in compounds such as HNO₃, so C is false. D is a true fact about the two elements, but it does not explain the difference in covalency.
Q12. The correct order of increasing electronegativity is
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Answer: A
Electronegativity rises across period 3: Na 0.9, Si 1.8, P 2.1, Cl 3.0. Options B and C misplace silicon or sodium, and D is the exact reverse.
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