CLASS 11 · CHEMISTRY · CHAPTER 3

Classification of Elements and Periodicity in Properties

Simple answers to ten worked chapter problems and all 40 end-of-chapter questions on periodic classification, configurations and element trends.

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Worked Questions from the Chapter

Textbook pages 82–95

Question Problem 3.1

What temporary IUPAC name and symbol are assigned to element 120?

Steps only

Digit roots are 1=un, 2=bi and 0=nil. The temporary name is unbinilium and the symbol is Ubn.

Answer

Digit roots are 1=un, 2=bi and 0=nil. The temporary name is unbinilium and the symbol is Ubn.

Question Problem 3.2

Why does period 5 contain 18 elements?

Steps only

The 5s, 4d and 5p subshells fill in this period. Their capacities are 2+10+6=18 electrons, so there are 18 elements.

Answer

The 5s, 4d and 5p subshells fill in this period. Their capacities are 2+10+6=18 electrons, so there are 18 elements.

Question Problem 3.3

Place predicted elements Z=117 and Z=120 in groups and give their expected outer configurations.

Steps only

Z=117 is a group 17 halogen, ending in 7p⁵. Z=120 is a group 2 element, ending in 8s².

Answer

Z=117 is a group 17 halogen, ending in 7p⁵. Z=120 is a group 2 element, ending in 8s².

Question Problem 3.4

Arrange Si, Be, Mg, Na and P in increasing metallic character.

Steps only

P < Si < Be < Mg < Na. Metallic character decreases across a period and increases down a group.

Answer

P < Si < Be < Mg < Na. Metallic character decreases across a period and increases down a group.

Question Problem 3.5

Among Mg, Mg²⁺, Al and Al³⁺, identify the largest and smallest.

Steps only

Largest: Mg atom. Smallest: Al³⁺. Atoms are larger than their cations; Al³⁺ is smaller than Mg²⁺ because it has greater nuclear charge with the same electron count.

Answer

Largest: Mg atom. Smallest: Al³⁺. Atoms are larger than their cations; Al³⁺ is smaller than Mg²⁺ because it has greater nuclear charge with the same electron count.

Question Problem 3.6

Given first ionisation enthalpies Na=496, Mg=737 and Si=786 kJ mol⁻¹, is Al closer to 575 or 760 kJ mol⁻¹? Explain.

Steps only

Closer to 575 kJ mol⁻¹. Aluminium’s removed electron is in 3p, which is higher in energy and shielded by 3s electrons, so it is easier to remove than magnesium’s 3s electron.

Answer

Closer to 575 kJ mol⁻¹. Aluminium’s removed electron is in 3p, which is higher in energy and shielded by 3s electrons, so it is easier to remove than magnesium’s 3s electron.

Question Problem 3.7

Among P, S, Cl and F, which has the most negative and least negative electron-gain enthalpy? Explain.

Steps only

Cl is most negative; P is least negative. Chlorine gains an electron into a larger 3p shell with less crowding than fluorine’s compact 2p shell; phosphorus has a stable half-filled 3p³ arrangement.

Answer

Question Problem 3.8

Predict formulas for compounds formed by (a) Si and Br (b) Al and S.

Steps only

(a) SiBr₄: Si valence 4 and Br valence 1. (b) Al₂S₃: Al valence 3 and S valence 2.

Answer

(a) SiBr₄: Si valence 4 and Br valence 1. (b) Al₂S₃: Al valence 3 and S valence 2.

Question Problem 3.9

Are aluminium’s oxidation state and covalency the same in [AlCl(H₂O)₅]²⁺?

Steps only

No. Al has oxidation state +3, while it forms six coordinate bonds (one Cl and five H₂O), so its covalency/coordination number here is 6.

Answer

No. Al has oxidation state +3, while it forms six coordinate bonds (one Cl and five H₂O), so its covalency/coordination number here is 6.

Question Problem 3.10

Use reactions with water to show Na₂O is basic and Cl₂O₇ is acidic.

Steps only

Na₂O + H₂O → 2NaOH (base). Cl₂O₇ + H₂O → 2HClO₄ (acid).

Answer

Na₂O + H₂O → 2NaOH (base). Cl₂O₇ + H₂O → 2HClO₄ (acid).

Chapter 3 Exercises

Textbook pages 96–100

Question 3.1

Exercise 3.1: What is the basic organising idea of the periodic table?

Steps only

Atomic number determines electron count and configuration; recurring valence configurations create periodic trends.

Answer

Elements are arranged by increasing atomic number. Their repeating electron configurations place elements with similar properties in the same groups.

Question 3.2

Exercise 3.2: Which property did Mendeleev use to classify elements, and did he follow it strictly?

Steps only

Tellurium and iodine are a known example: he placed them by chemical similarity.

Answer

He mainly used atomic mass and chemical properties. He usually followed increasing mass, but did not follow it strictly when properties required a different order.

Question 3.3

Exercise 3.3: How does Mendeleev’s Periodic Law differ from the Modern Periodic Law?

Steps only

Atomic number, unlike mass, gives the proton count and electronic configuration that controls properties.

Answer

Mendeleev’s law relates periodic properties to atomic mass. The modern law relates them to atomic number.

Question 3.4

Exercise 3.4: Use quantum numbers to show why period 6 contains 32 elements.

Steps only

Their capacities are 2(2l+1): s=2, f=14, d=10 and p=6.

Answer

The sixth period fills 6s, 4f, 5d and 6p subshells: 2+14+10+6=32 elements.

Question 3.5

Exercise 3.5: Locate element Z=114 by period, group and block.

Steps only

Its outer configuration ends in 7s²7p²: n=7 gives period 7 and four valence electrons place it in group 14.

Answer

Period 7, group 14, p-block (flerovium).

Question 3.6

Exercise 3.6: Give the atomic number of the period-3, group-17 element.

Steps only

The third-period halogen is chlorine.

Answer

Chlorine, atomic number 17.

Question 3.7

Exercise 3.7: Which element names honour (i) Lawrence Berkeley Laboratory and (ii) Seaborg’s group?

Steps only

Lawrencium honours the laboratory’s namesake Ernest Lawrence; seaborgium honours Glenn Seaborg.

Answer

(i) Lawrence Berkeley Laboratory: lawrencium (Lr).
(ii) Seaborg’s group: seaborgium (Sg).

Question 3.8

Exercise 3.8: Why do elements in one group have similar physical and chemical properties?

Steps only

Chemical behaviour is mostly controlled by valence electrons.

Answer

They have similar outer-shell electron configurations and the same number of valence electrons, so they tend to form similar ions and bonds.

Question 3.9

Exercise 3.9: What do atomic radius and ionic radius mean?

Steps only

Atoms and ions have no sharp outer edge, so radii are defined by measurable neighbour distances.

Answer

Atomic radius is commonly half the distance between nuclei of identical bonded atoms. Ionic radius is the effective size of an ion in a crystal, inferred from distances between neighbouring ions.

Question 3.10

Exercise 3.10: How does atomic radius change across a period and down a group? Explain.

Steps only

More pull from the nucleus shrinks atoms across; added shells enlarge them down a group.

Answer

Across a period, atomic radius generally decreases because nuclear charge rises while electrons enter the same shell. Down a group, it increases because new electron shells are added.

Question 3.11

Exercise 3.11: Define isoelectronic species and give one matching species for F⁻, Ar, Mg²⁺ and Rb⁺.

Steps only

Count electrons for each species, then name a neutral atom or ion with the same count.

Answer

Isoelectronic species have equal electron counts. Examples: F⁻: Ne; Ar: K⁺; Mg²⁺: Ne; Rb⁺: Kr.

Question 3.12

Exercise 3.12: For N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺ and Al³⁺, state the shared feature and order their ionic radii.

Steps only

For equal electron counts, more protons pull the same electron cloud more strongly and make the ion smaller.

Answer

(a) All have 10 electrons. (b) Increasing radius: Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻ < N³⁻.

Question 3.13

Exercise 3.13: Why are cations smaller and anions larger than their parent atoms?

Steps only

Both changes alter the balance between nuclear attraction and electron repulsion.

Answer

A cation is smaller because it loses electron(s), often losing an outer shell, and the remaining electrons feel stronger nuclear pull. An anion is larger because added electrons increase electron–electron repulsion.

Question 3.14

Exercise 3.14: Why do definitions of ionisation enthalpy and electron-gain enthalpy specify an isolated gaseous ground-state atom?

Steps only

These conditions make energy values comparable between elements.

Answer

“Isolated gaseous atom” means one atom separated from neighbours, so bonding effects are absent. “Ground state” means its electrons are in the lowest-energy arrangement.

Question 3.15

Exercise 3.15: Given hydrogen ground-state energy −2.18×10⁻¹⁸ J per atom, calculate its ionisation enthalpy in J mol⁻¹.

Steps only

Ionising an atom needs the opposite of its negative ground-state energy; multiply per-atom energy by Avogadro’s constant.

Answer

E per mole=2.18×10⁻¹⁸×6.022×10²³=1.31×10⁶ J mol⁻¹ (1312 kJ mol⁻¹).

Question 3.16

Exercise 3.16: Explain (i) Be having higher ionisation enthalpy than B and (ii) O having lower ionisation enthalpy than N and F.

Steps only

Subshell stability and electron pairing cause small exceptions to the general rise across a period.

Answer

(i) Be is higher than B because Be has a filled 2s subshell; B loses a higher-energy 2p electron more easily.
(ii) O is lower than N because one O 2p orbital contains a paired pair whose repulsion eases removal; N has a stable half-filled 2p³ subshell. F is higher because its greater nuclear charge holds electrons more strongly.

Question 3.17

Exercise 3.17: Why is sodium’s first ionisation enthalpy below magnesium’s, but sodium’s second above magnesium’s?

Steps only

Compare the configurations: Na⁺ is closed-shell, while Mg⁺ still has one 3s electron.

Answer

Na’s first ionisation is lower because Na loses its single 3s electron easily. After that, Na⁺ has a stable neon configuration, so its second ionisation is very high. Mg’s second ionisation removes its remaining 3s electron and is lower than Na’s second.

Question 3.18

Exercise 3.18: Why does main-group ionisation enthalpy generally decrease down a group?

Steps only

The added nuclear charge is outweighed by added shells and shielding.

Answer

Atomic radius and shielding increase down the group, so the outer electron is farther from and less strongly attracted to the nucleus. It takes less energy to remove.

Question 3.19

Exercise 3.19: Explain the group-13 first ionisation enthalpy values B 801, Al 577, Ga 579, In 558, Tl 589 kJ mol⁻¹.

Steps only

Poor shielding by inner d/f electrons creates the small upward deviations.

Answer

Values generally fall B→Al→In, but Ga is slightly above Al because filled 3d electrons shield poorly. Tl is above In because filled f and d electrons shield poorly, increasing effective nuclear attraction.

Question 3.20

Exercise 3.20: Which has more negative electron-gain enthalpy: (i) O or F (ii) F or Cl?

Steps only

Adding an electron is more favourable for halogens; chlorine’s larger 3p shell has less crowding than fluorine’s compact 2p shell.

Answer

(i) F is more negative than O.
(ii) Cl is more negative than F.

Question 3.21

Exercise 3.21: Is oxygen’s second electron-gain enthalpy positive or negative? Explain.

Steps only

The incoming electron is repelled by the already negative ion.

Answer

It is positive (endothermic). Adding the first electron releases energy, but adding a second electron to O⁻ requires overcoming repulsion between the negative ion and incoming electron.

Question 3.22

Exercise 3.22: Distinguish electron-gain enthalpy from electronegativity.

Steps only

One is an energy change for a free atom; the other is a relative bonding property.

Answer

Electron-gain enthalpy is the energy change when an isolated gaseous atom gains an electron. Electronegativity is an atom’s relative pull on shared electrons while bonded.

Question 3.23

Exercise 3.23: Is nitrogen’s Pauling electronegativity exactly 3.0 in every compound? Explain.

Steps only

Electronegativity is relative and depends on the chemical environment.

Answer

The Pauling value 3.0 is a useful average scale value, not an identical fixed value in every compound. Effective electronegativity can vary with bonding and oxidation state.

Question 3.24

Exercise 3.24: Explain how an atom’s radius changes when it (a) gains (b) loses an electron.

Steps only

The electron cloud expands on gaining electrons and contracts on losing them.

Answer

(a) Gaining electron(s) makes the radius larger due to increased electron repulsion. (b) Losing electron(s) makes it smaller due to reduced repulsion and stronger attraction per remaining electron.

Question 3.25

Exercise 3.25: Would two isotopes of one element have the same first ionisation enthalpy? Explain.

Steps only

Ionisation mainly depends on electron–nucleus attraction, which is nearly unchanged.

Answer

They are almost the same. Isotopes have the same nuclear charge and electron configuration; their different neutron counts have only a very small effect.

Question 3.26

Exercise 3.26: Give major differences between metals and non-metals.

Steps only

These are broad trends; some elements and allotropes are exceptions.

Answer

Metals usually lose electrons, conduct heat and electricity, and are malleable. Non-metals generally gain/share electrons and are poor conductors; many are brittle as solids.

Question 3.27

Exercise 3.27: Use the periodic table to identify (a) an element with five outer-subshell electrons (b) one likely to lose two (c) one likely to gain two (d) a group containing gases, a liquid and a metal.

Steps only

Use valence-electron counts and common group trends.

Answer

(a) Five outer p electrons: Nitrogen (group 15).
(b) Tends to lose two electrons: Magnesium (group 2).
(c) Tends to gain two electrons: Oxygen (group 16).
(d) Gases, a liquid and a metal: Group 17 includes gaseous fluorine and chlorine, liquid bromine, and the predicted metallic element tennessine.

Question 3.28

Exercise 3.28: Explain why group-1 reactivity increases down the group while group-17 reactivity decreases.

Steps only

Metals react by losing an electron; halogens react by gaining one.

Answer

Group 1 reactivity rises down the group because the outer electron is farther away and easier to remove. Group 17 reactivity falls down the group because larger atoms attract an added electron less strongly.

Question 3.29

Exercise 3.29: Give general outer electronic configurations of s-, p-, d- and f-block elements.

Steps only

The block letter identifies the subshell receiving the differentiating electron.

Answer

s: ns¹–²; p: ns²np¹–⁶; d: (n−1)d¹–¹⁰ns⁰–²; f: (n−2)f¹–¹⁴(n−1)d⁰–¹ns².

Question 3.30

Exercise 3.30: Find the periodic-table positions of (i) ns²np⁴,n=3 (ii) (n−1)d²ns²,n=4 (iii) (n−2)f⁷(n−1)d¹ns²,n=6.

Steps only

Highest n gives period. For p/s use valence count; for d use (n−1)d+ns; the f case is a lanthanoid.

Answer

(i) 3s²3p⁴: period 3, group 16, p-block (S).
(ii) 3d²4s²: period 4, group 4, d-block (Ti).
(iii) 4f⁷5d¹6s²: period 6, group 3, f-block (Gd).

Question 3.31

Exercise 3.31: Using the ionisation and electron-gain data for elements I–VI, identify (a) least reactive element (b) most reactive metal (c) most reactive non-metal (d) least reactive non-metal (e) metal forming stable MX₂ (f) metal forming a mainly covalent MX halide.

Steps only

Match low first ionisation with reactive metals, strongly negative electron gain with halogens, and the large second-ionisation jump with group 1.

Answer

(a) Least reactive element: V. It fits a noble gas.
(b) Most reactive metal: II. It has the lowest first ionisation enthalpy.
(c) Most reactive non-metal: III. It has the most negative electron-gain enthalpy.
(d) Least reactive non-metal: IV. Its electron-gain enthalpy is less negative than III’s.
(e) Stable MX₂-forming metal: VI. It can lose two electrons.
(f) Mainly covalent MX-forming metal: I. It forms a small, highly charged cation.

Question 3.32

Exercise 3.32: Predict stable binary formulae for (a) lithium/oxygen (b) magnesium/nitrogen (c) aluminium/iodine (d) silicon/oxygen (e) phosphorus/fluorine (f) element 71/fluorine.

Steps only

Balance the usual charges/valencies; element 71 is lutetium (Lu), which commonly forms +3 ions.

Answer

(a) Li₂O (b) Mg₃N₂ (c) AlI₃ (d) SiO₂ (e) PF₃ (common stable fluoride) (f) LuF₃.

Question 3.33

Exercise 3.33: In the modern periodic table, the period indicates (a) atomic number (b) atomic mass (c) principal quantum number (d) azimuthal quantum number.

Steps only

The period number gives the highest occupied principal shell in the ground-state atom.

Answer

Question 3.34

Exercise 3.34: Which statement is incorrect? (a) p-block has six columns (b) d-block has eight columns (c) block column count matches subshell capacity (d) block shows l of last-filled subshell.

Steps only

Five d orbitals hold two electrons each.

Answer

(b) is incorrect: the d-block has 10 columns because a d subshell holds 10 electrons, not 8.

Question 3.35

Exercise 3.35: Which factor does not affect the valence shell: valence n, nuclear charge, nuclear mass, or core-electron count?

Steps only

Valence configuration depends on nuclear charge and electron arrangement; isotopic mass usually has little chemical effect.

Answer

Question 3.36

Exercise 3.36: Which factor controls the sizes of isoelectronic F⁻, Ne and Na⁺: nuclear charge, valence n, outer-electron repulsion, or none?

Steps only

F⁻, Ne and Na⁺ have the same number of electrons; increasing proton count pulls them into a smaller radius.

Answer

(a) Nuclear charge.

Question 3.37

Exercise 3.37: Which ionisation-enthalpy statement is incorrect? Successive values rise; a large jump follows removal of valence electrons; core removal causes a large jump; lower-n electrons are easier to remove.

Steps only

The other statements describe successive ionisation and the large jump after valence electrons are removed.

Answer

(d) is incorrect. An electron in a lower-n orbital is generally more strongly held, so it is harder—not easier—to remove.

Question 3.38

Exercise 3.38: Choose the correct metallic-character order for B, Al, Mg and K.

Steps only

Metallic character increases down a group and toward the left; potassium is farthest down-left among these.

Answer

(d) K > Mg > Al > B.

Question 3.39

Exercise 3.39: Choose the correct non-metallic-character order for B, C, N, F and Si.

Steps only

Non-metallic character rises toward the upper right; within group 14, C is above Si.

Answer

Question 3.40

Exercise 3.40: Choose the correct oxidising reactivity order for F, Cl, O and N.

Steps only

Fluorine is the strongest oxidiser; oxygen is next, followed by chlorine and nitrogen for this comparison.

Answer

(b) F > O > Cl > N.