Chapter 7 — Atoms and the periodic table
Read the atom before reading the table
The periodic table arranges elements by atomic number, the number of protons in each atom's nucleus. That rule is more reliable than guessing from mass or appearance. Hydrogen has one proton; an atom with six protons is carbon regardless of how many neutrons it contains. A neutral atom has as many electrons as protons. An ion has gained or lost electrons, but its atomic number remains the same. A positive sodium ion, for example, still belongs to sodium. If it had a different number of protons, it would be a different element.
Mass number counts protons plus neutrons in one specified atom. It is an integer. For elements with naturally occurring isotope mixtures, the relative atomic mass commonly printed on a periodic table reflects their weighted mean atomic mass relative to the atomic mass constant. It is dimensionless and often not an integer. Some radioactive elements instead have a bracketed isotope mass number displayed; read the table key. Do not substitute that tabulated average for the mass number of a particular isotope. For the common school example chlorine-35, the 35 identifies the selected isotope's mass number, while chlorine's atomic number 17 tells us it has 17 protons. Its neutron count is 35−17=18. A neutral chlorine-35 atom has 17 electrons; a chloride ion with a charge of 1− has 18. It has gained an electron, not a neutron.
For police-exam counting questions, make a three-column scratch table labelled proton, neutron and electron. First set protons equal to Z. Next subtract Z from A to get neutrons. Last interpret charge: a positive charge means electrons were removed; a negative charge means electrons were added. The order prevents a common error in which a student subtracts charge from the neutron count. Charge in an ordinary ion question tells you about electrons unless a nuclear process is explicitly described.
Electron arrangement as a useful school model
At school level, electrons are often grouped into shells around the nucleus. A simple arrangement for the first few elements helps explain why their chemistry repeats. For sodium (Z=11), the familiar shell count is 2, 8, 1. For chlorine (Z=17), it is 2, 8, 7. Sodium readily forms a 1+ ion by losing one outer electron; chlorine can form a 1− ion by gaining one. The shell picture predicts many introductory bonding patterns, but it is a model. It does not describe every electronic detail of heavier elements or every oxidation state.
The electrons most relevant to ordinary bonding in this simplified account are called valence electrons. In the main groups, elements in a column often share a valence-electron pattern and show related chemical behaviour. This is why a periodic table is a predictive tool, not simply an alphabetic list. Fluorine and chlorine are both halogens; their atoms each have seven outer electrons in the introductory model. That does not mean every compound they make has identical properties. Chemical behaviour depends on the complete bonding situation and conditions.
The term valency is useful at this level but must be applied carefully. It describes combining capacity in an elementary formula exercise; it is not a universal fixed charge printed permanently on every atom. Oxygen commonly has valency two in familiar school-level compounds, yet an oxygen atom is not always present as a free ion with 2− charge. When a question asks for the formula formed from ions, use the stated ion charges and demand total charge zero. When it asks for a covalent molecule, think of shared bonds. The two routes should not be mixed merely because both yield a formula.
Periods and groups
Horizontal rows are periods; vertical columns are groups. Atomic number increases as one moves through the modern table in its ordering. Elements in the same period do not have the same outer electron count: the outer occupancy generally changes across the row. Elements within a main-group column tend to have related outer arrangements. A question that describes “same row, similar outer configuration” has swapped the roles of period and group. Mark the table's direction with a quick arrow before choosing an option.
Metals occupy much of the left and centre of the table; nonmetals are concentrated toward the upper right; metalloids lie near a stepped boundary in common school diagrams. These are broad patterns rather than a rule that every element in a large region has identical physical properties. Hydrogen appears above the alkali-metal column in many layouts because of its one electron, but it is a nonmetal with distinctive chemistry. Treat its placement as an instructive exception to any oversimplified “column equals identical substance” slogan.
Group 1 alkali metals readily lose one outer electron in the simple model. Group 17 halogens tend to gain one electron in common ionic examples. Group 18 noble gases have filled outer shells in the introductory pattern and are comparatively unreactive under ordinary conditions. “Inert in every circumstance” is too absolute: chemistry has identified compounds of some heavier noble gases. In an exam stem, distinguish the useful general comparison from an unqualified claim. A safe statement is that noble gases are much less reactive than many neighbouring elements under ordinary conditions.
Trends without false absolutes
Moving across a period, atomic size often decreases in an introductory comparison because nuclear charge grows while added electrons enter the same broad shell. Moving down a group, atomic size generally increases as additional occupied shells appear. The precise definition of atomic radius matters in advanced chemistry, so the simple arrows belong to school-level trends rather than exact measurements for every atom. If a question asks for the larger of lithium and sodium atoms in the same group, sodium is the expected introductory answer. If it offers an obscure transition-metal exception, do not extrapolate the simple rule without more information.
Metallic character generally decreases from left to right across a period and increases down many groups. This follows a broad tendency for atoms on the left to lose electrons more readily than common nonmetals on the right. Reactivity itself cannot be represented by one universal arrow across all elements. For instance, highly reactive metals and highly reactive nonmetals occur at different ends of the table. Always ask “reactivity of which family and by which reaction?” before accepting a single direction.
An element's position helps predict a likely simple ion and a possible formula. Sodium in Group 1 often gives Na⁺; chlorine in Group 17 often gives Cl⁻. One of each makes a neutral NaCl formula unit. Magnesium commonly gives Mg²⁺ and chloride is Cl⁻, so two chloride ions balance one magnesium ion: MgCl₂. This is charge balancing, not “crossing numbers” as a magic operation. Check the final formula: (+2)+2(−1)=0. For aluminium oxide, the simple ions Al³⁺ and O²⁻ balance as 2(+3)+3(−2)=0, giving Al₂O₃.
What periodic classification can and cannot tell you
The periodic table is powerful because chemically relevant electron patterns recur as Z increases. It supports predictions about likely bonding and broad trends. It does not, by itself, tell you whether a proposed reaction happens rapidly, safely or at all under specified conditions. A metal's group and an acid's name alone are insufficient for a real laboratory procedure. In this book, reaction equations are conceptual exam examples rather than directions to mix substances.
The table also contains element symbols that must be read exactly. Na is sodium, K is potassium, Fe is iron, Cu is copper and Ag is silver. The first letter is capitalised; a second letter is lower case. CO means carbon monoxide, a compound formula containing carbon and oxygen, whereas Co is the symbol for cobalt. This typographic distinction can decide an MCQ. Memorise a short list of frequently tested traditional symbols by using them in formulas, not by copying an entire table without context.
Natural isotopic abundance can change the listed average atomic mass without changing the definition of the element. A hypothetical element with two isotopes of masses 10 and 11, present at 20% and 80%, has a simplified weighted average of (10×0.20)+(11×0.80)=10.8. Neither individual atom has mass number 10.8. The number is an average over the mixture. This example uses rounded isotope masses only to teach the arithmetic; actual tabulated relative masses use more precise data.
Build an element profile from its electron count
For a neutral atom, total electron count equals atomic number. A shell arrangement of 2,8,7 therefore describes an atom with 17 electrons and 17 protons. It has three occupied shells in this elementary model, placing it in Period 3, and seven outer-shell electrons, placing it in Group 17 among the main-group elements. It can gain one electron to form the familiar chloride ion. The ion's arrangement becomes 2,8,8, but the element does not move to Group 18: the periodic-table position still follows its proton count of 17. This is an important distinction between an ion's electron arrangement and an element's identity.
Compare neutral atoms with arrangements 2,8,1 and 2,8,2. They occupy the same period because both have three occupied shells, but they are in different groups. In introductory ionic examples, the first commonly loses one electron and the second two. Now compare 2,1 with 2,8,1. These have different numbers of occupied shells and lie in different periods, but both have one outer electron and belong to Group 1. Asking which pair shares a group is therefore different from asking which pair shares a period. Count shells and outer electrons separately rather than trying to infer both from the total alone.
The shell-capacity relation 2n² gives a maximum capacity for a shell with principal number n: 2, 8 and 18 for the first three shells. It is not a simple instruction to fill every shell to its maximum before any electron enters the next shell in every atom. For example, the common school arrangements of potassium and calcium are 2,8,8,1 and 2,8,8,2. Their fourth-shell occupancy begins before the third shell reaches 18. More detailed subshell energy ordering explains this. For elementary exam questions, use the stated arrangement and the familiar first-20-element model; do not force all atoms into a rigid 2,8,18 sequence.
Compare atoms and ions without changing identity
In a simple comparison of an atom with its common ion, a cation is generally smaller than its neutral parent atom and an anion larger. Removing electrons can reduce electron–electron repulsion and may remove the outer occupied shell; adding electrons increases repulsion within the occupied arrangement. Thus the sodium ion is smaller than the sodium atom in the usual comparison, and the chloride ion larger than the chlorine atom. The ion's size has changed while proton number has stayed fixed. A claim that a smaller ion must be a different element confuses size with identity.
Species with the same electron count are called isoelectronic. Na⁺, Mg²⁺ and Ne each have 10 electrons, but they have 11, 12 and 10 protons respectively. They are not the same element and do not have the same charge. For this simple isoelectronic comparison, stronger nuclear attraction generally makes the species with more protons smaller, so Mg²⁺ is smaller than Na⁺. The reasoning should be applied to a suitable comparable series, not treated as a universal size rule for unrelated species. First establish equal electron count; only then compare nuclear charge.
Try an example with a neutral atom having Z=16 and a 2− ion of that element. The neutral atom has 16 electrons; the ion has 18. An atom with Z=18 also has 18 electrons when neutral, but its nucleus has two more protons than the ion with Z=16. Equal electron counts alone do not imply equal chemical identities. The periodic table makes that distinction explicit by ordering nuclei through atomic number rather than sorting species by whatever electron count they happen to have after ion formation.
Distinguish three ideas about electrons
Ionisation energy concerns removing an electron from an isolated gaseous species. Electronegativity concerns an atom's tendency to attract shared electrons in a chemical bond. Electron affinity concerns the energy change associated with adding an electron to an isolated gaseous atom, with care needed about sign conventions. These are related ideas, but they are not synonyms. An item about a polar bond usually needs electronegativity; an item about the energy needed to form a gaseous positive ion needs ionisation energy. Reading the process in the stem is more reliable than remembering that all three terms involve electrons.
Across many main-group periods, first ionisation energy broadly increases, while down a group it broadly decreases. These trends have exceptions related to electronic structure. Electronegativity also commonly increases across a period and decreases down a group, with fluorine conventionally the most electronegative element on the familiar Pauling scale. That statement does not automatically mean fluorine has the most favourable isolated-atom electron affinity under every convention; chlorine's electron-addition energetics provide a familiar caution. Use each term for its defined process. An exam option that substitutes one term for another should be judged by the definition, not by a broad trend arrow.
Read a formula from the group pattern
Suppose X is a main-group element that forms X²⁺ and Y forms Y⁻. A neutral formula needs one X for two Y, giving XY₂. If X³⁺ combines with Y²⁻, the smallest total positive and negative charge is six units: two X ions and three Y ions give X₂Y₃. The lowest whole-number ratio matters. X₄Y₆ has the same overall charge balance but is not the simplest formula-unit ratio. After writing a formula, check both charge and notation rather than treating the crossed numbers as an answer that needs no verification.
For polyatomic ions, the group must stay intact in formula writing. A magnesium ion Mg²⁺ and hydroxide OH⁻ form Mg(OH)₂ because two hydroxide groups are needed. One formula unit then contains one magnesium, two oxygen and two hydrogen atoms. The subscript outside the parentheses multiplies every atom in the group. This mechanical counting step connects the periodic-table prediction of common ionic charge with the equation-balancing work in the next chapter.
Retrieval exercise with worked checks
Use these four prompts without viewing the answers. Which period contains a neutral atom with arrangement 2,8,3? How many electrons are in an ion with Z=13 and charge 3+? Are two species with equal mass number necessarily isotopes? What formula balances a 2+ cation with a 3− anion? The answers are Period 3, 10 electrons, no, and X₃Y₂ for cation X and anion Y. The third answer follows because isotopes require equal atomic number, whereas equal mass number with different atomic number describes isobars. For the last answer, 3(+2)+2(−3)=0 verifies neutrality.
Worked route: identify the species
Suppose species X has Z=12, A=24 and a 2+ charge. It has 12 protons, 12 neutrons and 10 electrons. Its element identity follows Z=12: magnesium. An isotope with A=25 would keep 12 protons and have 13 neutrons. An atom of a different element with A=24 could have the same mass number but a different Z; that pair would be isobars. A question may give all three descriptions and ask which has the same chemical identity. Match atomic number first.
Now suppose an unknown neutral main-group atom has the school shell count 2, 8, 2. Its electron total is 12, so Z=12. The two outer electrons indicate a common 2+ ionic form in simple compounds. Pair it with a stated oxide ion O²⁻: one of each balances, so the formula is MgO. Do not state that every magnesium compound is formed only by literal transfer of two electrons; the simple model is enough for this exam-level prediction, while real bonding can be more nuanced.
Recall and review
Without looking back, explain why atomic number identifies an element while mass number distinguishes isotopes. Calculate protons, neutrons and electrons for a neutral atom and a charged ion. Point to a period and a group on a blank table. Describe the common shell arrangement and typical simple ion of sodium, chlorine and magnesium. Predict the formulas MgCl₂ and Al₂O₃ by balancing charge. Finally, explain why a trend arrow is a broad comparison rather than proof that every pair of elements follows it. Attempt the chapter practice set, then use the explanations to revisit any uncertain concepts.
Figure 7. Original chapter diagram.
Practice 43 questions
Choose one answer for each item. The answer key and explanations follow this chapter set.
1. The maximum number of electrons that can be accommodated in a shell with principal quantum number n is given by:
A. n² B. 2n² C. 2n D. n + 2
2. The rule that no two electrons in the same atom can have identical values of all four quantum numbers is known as:
A. Hund's rule of maximum multiplicity B. Aufbau principle C. Pauli exclusion principle D. Heisenberg's uncertainty principle
3. Which scientist proposed the 'plum pudding' model of the atom, in which negatively charged electrons are embedded in a sphere of positive charge?
A. J.J. Thomson B. Ernest Rutherford C. Niels Bohr D. John Dalton
4. How many vertical columns (groups) are there in the modern long form of the periodic table?
A. 7 B. 16 C. 18 D. 8
5. Which element is used to cross-link rubber molecules in the process of vulcanisation?
A. Phosphorus B. Chlorine C. Nitrogen D. Sulphur
6. Which halogen is present in traditional tincture preparations used as skin antiseptics?
A. Iodine B. Fluorine C. Bromine D. Astatine
7. In the traditional school-level shorthand for bleaching powder, which formula is commonly written?
A. CaCl₂ B. CaOCl₂ C. CaCO₃ D. CaSO₄
8. Which scientist established that the elements in the modern periodic table should be arranged in order of increasing atomic number rather than atomic mass?
A. Dmitri Mendeleev B. John Newlands C. Johann Döbereiner D. Henry Moseley
9. Which is the only non-metal that exists as a liquid at room temperature?
A. Bromine B. Mercury C. Iodine D. Chlorine
10. Arrange the following scientific milestones chronologically: 1. Discovery of the neutron; 2. Bohr’s model of quantised orbits; 3. Discovery of the electron; 4. Gold-foil scattering results and Rutherford’s nuclear interpretation.
A. 3-4-2-1 B. 3-2-4-1 C. 4-3-2-1 D. 3-4-1-2
11. The subshells 3d, 4p and 5s all have the same value of (n + l) = 5. In which order are they filled according to the Aufbau (n + l) rule?
A. 5s → 4p → 3d B. 3d → 4p → 5s C. 4p → 3d → 5s D. 3d → 5s → 4p
12. According to the Aufbau order of filling, which subshell is filled immediately after 4s?
A. 4p B. 3d C. 5s D. 4d
13. J.J. Thomson's cathode-ray experiments enabled him to measure which property of the electron?
A. Its charge-to-mass ratio (e/m) B. Its charge alone C. Its spin quantum number D. Its mass alone
14. Which of the following is the correct condensed electronic configuration of iron (Z = 26)?
A. [Ar] 3d⁸ B. [Ar] 3d⁵ 4s² C. [Ar] 3d⁷ 4s² D. [Ar] 3d⁶ 4s²
15. While filling orbitals of equal energy, electrons first occupy each orbital singly with parallel spins before any pairing occurs. This is:
A. Aufbau principle B. (n + l) rule C. Hund's rule D. Pauli exclusion principle
16. Match the scientist (List-I) with the atomic model/idea (List-II):
List-I: (a) John Dalton (b) J.J. Thomson (c) Ernest Rutherford (d) Niels Bohr
List-II: (i) Nuclear model with mostly empty space (ii) Quantised energy levels (iii) Indivisible atoms (iv) Plum pudding model
A. a-iv, b-iii, c-i, d-ii B. a-iii, b-iv, c-i, d-ii C. a-iii, b-iv, c-ii, d-i D. a-iii, b-i, c-iv, d-ii
17. Match the subshell in List-I with the number of orbitals it contains in List-II.
List-I: (A) s subshell (B) p subshell (C) d subshell (D) f subshell
List-II: (i) 5 (ii) 1 (iii) 7 (iv) 3
A. A-ii, B-i, C-iv, D-iii B. A-ii, B-iv, C-i, D-iii C. A-iv, B-ii, C-i, D-iii D. A-ii, B-iv, C-iii, D-i
18. A proton is approximately how many times heavier than an electron?
A. 18 B. 18,360 C. 1836 D. 183
19. Which of the following third-period elements is a noble gas?
A. Magnesium B. Chlorine C. Argon D. Silicon
20. How many unpaired electrons are present in a ground-state nitrogen atom (Z = 7)?
A. 5 B. 1 C. 2 D. 3
21. A neutral sodium atom has electron distribution 2,8,1. After it loses one electron to form Na⁺, what is its electron distribution?
A. 2,8 B. 2,8,1 C. 2,7 D. 2,8,2
22. A hypothetical element has two isotopes of mass numbers 35 and 37, present in 75% and 25% of its atoms respectively. What approximate weighted mean mass number would you calculate?
A. 35.0 B. 35.5 C. 36.0 D. 37.0
23. Using Bohr's expression Eₙ = −13.6/n² eV, what is the energy of the electron in the second orbit (n = 2) of a hydrogen atom?
A. −1.51 eV B. −6.8 eV C. −13.6 eV D. −3.4 eV
24. Using Eₙ = −13.6/n² eV for hydrogen, how much energy is emitted as a photon when the electron falls from n = 3 to n = 2 (the red H-alpha line)?
A. 3.40 eV B. 1.51 eV C. 12.09 eV D. 1.89 eV
25. A neutral chlorine atom has 17 protons and 17 electrons. After it gains one electron, how many protons and electrons does the resulting chloride ion have?
A. 18 protons and 17 electrons B. 17 protons and 17 electrons C. 18 protons and 18 electrons D. 17 protons and 18 electrons
26. According to Bohr's model, how much energy is required to remove the electron from a hydrogen atom in its ground state (its ionisation energy)?
A. 13.6 eV B. 3.4 eV C. 1.51 eV D. 10.2 eV
27. The prominent visible lines of the Balmer series in the hydrogen spectrum are found in which region?
A. Ultraviolet B. Visible C. Infrared D. Microwave
28. Which observation was difficult for Rutherford’s classical nuclear model of the atom to explain?
A. That most alpha particles cross thin foil B. That a few alpha particles turn through large angles C. That the atom has a concentrated positive nucleus D. That atoms remain stable and emit discrete spectral lines
29. What is the shape of a p orbital?
A. Cloverleaf-shaped B. Ring-shaped C. Dumbbell-shaped D. Spherical
30. Which quantum number describes the spatial orientation of an orbital?
A. Magnetic quantum number, mₗ B. Principal quantum number, n C. Azimuthal quantum number, l D. Spin quantum number, mₛ
31. Which quantum number determines the shape of an orbital (s, p, d or f)?
A. Principal quantum number, n B. Azimuthal (angular momentum) quantum number, l C. Magnetic quantum number, mₗ D. Spin quantum number, mₛ
32. Consider the following statements about Bohr's model:
Statement I: It accurately explains the spectra of multi-electron atoms such as lithium and oxygen.
Statement II: It states that an electron in a permitted orbit does not radiate energy.
Which of the above is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
33. Consider the following statements:
Statement I: The atomic number of an element equals the number of protons in its nucleus.
Statement II: The mass number of an atom equals the sum of its protons and electrons.
Which of the above is/are correct?
A. Only I B. Only II C. Both I and II D. Neither I nor II
34. Bohr's model accurately explains the spectra of single-electron (hydrogen-like) species. Which of the following is NOT such a species?
A. He (neutral helium atom) B. H C. He⁺ D. Li²⁺
35. For a hydrogen-like He⁺ ion, the Bohr ground-state energy is E₁ = −13.6 Z² eV, where Z = 2. What is its ground-state ionisation energy?
A. 13.6 eV B. 27.2 eV C. 54.4 eV D. 108.8 eV
36. Which statement correctly defines the first ionisation energy of an element?
A. Energy released when a gaseous atom gains one electron B. Energy needed to remove one electron from a gaseous neutral atom C. Energy needed to remove a proton from a nucleus D. Energy released when a neutral atom becomes an anion
37. Which of the following sets of quantum numbers (n, l, mₗ) is NOT permitted for an electron?
A. n = 2, l = 2, mₗ = 0 B. n = 3, l = 2, mₗ = −2 C. n = 2, l = 1, mₗ = +1 D. n = 1, l = 0, mₗ = 0
38. Among the following elements of the third period, which has the largest atomic radius?
A. Magnesium B. Silicon C. Chlorine D. Sodium
39. Arrange the following alkali metals in increasing order of reactivity with water:
1. Sodium 2. Rubidium 3. Lithium 4. Potassium
A. 3-4-1-2 B. 2-4-1-3 C. 1-3-4-2 D. 3-1-4-2
40. Arrange the following halogens in decreasing order of chemical reactivity:
1. Bromine 2. Fluorine 3. Iodine 4. Chlorine
A. 4-2-1-3 B. 3-1-4-2 C. 2-1-4-3 D. 2-4-1-3
41. Assertion (A): The first ionisation energy of oxygen is lower than that of nitrogen, even though oxygen lies to the right of nitrogen in period 2.
Reason (R): Nitrogen has a stable half-filled 2p³ configuration, whereas removing one electron from oxygen's 2p⁴ relieves pairing repulsion and leaves a half-filled 2p³ ion.
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
42. An element has the electronic configuration 2, 8, 3. To which period and group of the modern periodic table does it belong?
A. Period 3, Group 11 B. Period 3, Group 3 C. Period 2, Group 13 D. Period 3, Group 13
43. Which subshell is progressively filled across most of the lanthanoid series?
A. 5d B. 5f C. 4d D. 4f
Answers and explanations
1. B. A shell can hold at most 2n² electrons: K (n = 1) holds 2, L (n = 2) holds 8, M (n = 3) holds 18 and N (n = 4) holds 32. The expression n² gives the number of orbitals in the shell; each orbital holds two electrons.
2. C. Pauli's exclusion principle forbids two electrons in an atom from sharing all four quantum numbers, so an orbital can hold at most two electrons of opposite spin. Hund's rule governs single filling of equal-energy orbitals and the Aufbau principle gives the filling order.
3. A. J.J. Thomson, who discovered the electron in 1897, pictured the atom as a positive 'pudding' studded with electron 'plums'. Rutherford's gold-foil experiment replaced it with the nuclear model, Bohr added fixed orbits, and Dalton's atoms were indivisible.
4. C. The modern periodic table has 18 groups numbered 1 to 18 (IUPAC system) and 7 periods. Older tables used Roman numerals I–VIII with A and B sub-groups, which is where the figure 8 comes from.
5. D. In vulcanisation natural rubber is heated with sulphur, which forms cross-links between the polymer chains and makes the rubber harder, more elastic and less sticky. Sulphur is a Group 16 element also used to make sulphuric acid and fertilisers.
6. A. Traditional tincture of iodine contains iodine; iodine-containing preparations are used as skin antiseptics. Fluorine and bromine are not substitutes for this use, and astatine is a rare radioactive element. The question identifies the element rather than giving treatment instructions.
7. B. The conventional school-level notation is CaOCl₂ (calcium oxychloride). It is a shorthand for a chlorinated-lime product; commercial bleaching products and calcium hypochlorite preparations need not have that single exact composition. CaCl₂, CaCO₃ and CaSO₄ denote different compounds.
8. D. Henry Moseley's X-ray studies (1913) showed that atomic number, the nuclear charge, is the fundamental property, which fixed anomalies in Mendeleev's mass-based table (1869). Newlands proposed the law of octaves and Döbereiner the triads.
9. A. Bromine is a reddish-brown liquid non-metal at room temperature. Mercury is also a liquid but it is a metal, not a non-metal; iodine is a solid and chlorine a gas. These two liquids are a classic exam trap.
10. A. Thomson’s electron was identified in 1897; Geiger–Marsden scattering measurements were reported in 1909 and Rutherford’s nuclear interpretation in 1911; Bohr’s model followed in 1913; Chadwick found the neutron in 1932. Thus the order is 3-4-2-1.
11. B. When two subshells have equal (n + l), the one with the lower n is filled first. So 3d (n = 3) precedes 4p (n = 4), which precedes 5s (n = 5), matching the standard order …4s, 3d, 4p, 5s, 4d…
12. B. By the (n + l) rule, 4s has n + l = 4 and 3d has 5; among the n + l = 5 subshells (3d, 4p, 5s) the one with lower n fills first, so the order is 4s → 3d → 4p → 5s.
13. A. Thomson deflected cathode rays with electric and magnetic fields and obtained the electron's charge-to-mass ratio, about 1.76 × 10¹¹ C/kg. The charge itself was measured later by Millikan's oil-drop experiment, and the mass was then calculated from e and e/m.
14. D. Iron has 26 electrons: 18 in the argon core, then 4s² and 3d⁶ (18 + 2 + 6 = 26). [Ar] 3d⁵ 4s² is manganese (25) and [Ar] 3d⁷ 4s² is cobalt (27).
15. C. Hund's rule of maximum multiplicity says degenerate orbitals are singly occupied with parallel spins before pairing, which minimises electron repulsion — e.g. nitrogen's three 2p electrons are all unpaired.
16. B. Dalton (1808) – indivisible atoms; Thomson (1897 onwards) – plum pudding; Rutherford (gold foil) – nuclear model, atom mostly empty; Bohr (1913) – quantised energy levels. Hence a-iii, b-iv, c-i, d-ii.
17. B. The number of orbitals in a subshell is 2l + 1: s (l = 0) has 1, p (l = 1) has 3, d (l = 2) has 5 and f (l = 3) has 7. Since each orbital holds two electrons, these subshells hold at most 2, 6, 10 and 14 electrons. Hence A-ii, B-iv, C-i, D-iii.
18. C. Proton mass ≈ 1.67 × 10⁻²⁷ kg and electron mass ≈ 9.1 × 10⁻³¹ kg; the ratio is about 1836. This is why the electron's contribution to atomic mass is negligible.
19. C. Argon (Z = 18) closes the third period with the electron distribution 2,8,8 and belongs to Group 18, the noble gases. Magnesium is Group 2, silicon Group 14 and chlorine Group 17.
20. D. Nitrogen is 1s² 2s² 2p³. By Hund's rule the three 2p electrons occupy the three 2p orbitals singly with parallel spins, so there are 3 unpaired electrons.
21. A. Neutral sodium has 11 electrons. Losing one produces Na⁺ with 10 electrons: two in the first shell and eight in the second, or 2,8. Its 11 protons are unchanged; the ionic charge is +1 because one electron was lost.
22. B. Multiply each mass number by its fractional abundance: 35 × 0.75 + 37 × 0.25 = 26.25 + 9.25 = 35.5. This is an illustrative weighted mean of mass numbers, not a precision atomic-weight determination using isotopic masses.
23. D. E₂ = −13.6/2² = −13.6/4 = −3.4 eV. The common mistake −6.8 eV divides by n instead of n²; −13.6 eV is the n = 1 value and −1.51 eV is the n = 3 value.
24. D. E₃ = −13.6/9 = −1.51 eV and E₂ = −13.6/4 = −3.40 eV. Energy emitted = E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV, i.e. 13.6 × (1/4 − 1/9) = 13.6 × 5/36. 12.09 eV is the n = 3 → 1 jump; 3.40 and 1.51 eV are single-level energies.
25. D. Gaining an electron changes charge but not nuclear identity. A chloride ion retains 17 protons and has 18 electrons, giving net charge −1. An option with 18 protons would describe a different element.
26. A. The ground-state (n = 1) energy of hydrogen is −13.6 eV and the ionised level (n = ∞) is 0 eV, so 13.6 eV must be supplied. 3.4 eV and 1.51 eV are the magnitudes of the n = 2 and n = 3 levels.
27. B. Balmer transitions end at n = 2. The well-known H-alpha, H-beta and other prominent lines are visible (H-alpha is red, near 656 nm). At high n, Balmer lines approach a series limit near the ultraviolet boundary, so saying every Balmer wavelength is strictly visible would be too broad. Lyman transitions end at n = 1 and lie in the ultraviolet.
28. D. The classical nuclear picture could account for alpha scattering and the concentrated nucleus. It did not explain why an orbiting charged electron would not radiate continuously and collapse inward, or why hydrogen emits discrete lines. Bohr introduced stationary energy levels to address these points for hydrogen-like atoms.
29. C. A p orbital (l = 1) is dumbbell-shaped with two lobes along one axis. s orbitals (l = 0) are spherical and most d orbitals (l = 2) have a cloverleaf shape.
30. A. The magnetic quantum number mₗ (values −l to +l) gives the orientation of an orbital in space; for l = 1 it takes three values, giving three p orbitals.
31. B. The azimuthal quantum number l identifies the subshell and hence the orbital shape (l = 0 s, 1 p, 2 d, 3 f). n gives the shell/energy level, mₗ the orientation and mₛ the spin.
32. B. Statement II is a Bohr postulate: electrons in stationary orbits do not radiate. Statement I is false: the model fails for multi-electron atoms because it ignores electron–electron repulsion.
33. A. Statement I is correct: Z = number of protons. Statement II is wrong: mass number A = protons + neutrons (nucleons); electrons contribute almost nothing to mass and are not counted.
34. A. A neutral helium atom has two electrons, so electron–electron repulsion makes Bohr's model fail for it. H (1 e⁻), He⁺ (2 − 1 = 1 e⁻) and Li²⁺ (3 − 2 = 1 e⁻) each have only one electron.
35. C. For He⁺, Z = 2, so E₁ = −13.6 × 2² = −54.4 eV. Ionising from the ground state to a free electron at 0 eV requires +54.4 eV. Multiplying by Z instead of Z² gives the 27.2 eV distractor.
36. B. The first ionisation energy is the energy required to remove one electron from a gaseous neutral atom to form a gaseous singly positive ion. Electron gain is related to electron affinity, whereas removing a proton changes the nucleus and is a different process.
37. A. The azimuthal quantum number can take only values 0 to (n − 1), so for n = 2, l can be 0 or 1 — l = 2 is impossible. In the other sets l < n and mₗ lies between −l and +l (3d, 2p and 1s electrons respectively), so all are allowed.
38. D. Across a period the nuclear charge increases while electrons enter the same shell, so atoms shrink from left to right. Sodium, at the start of period 3, is therefore the largest of these; chlorine near the end is the smallest.
39. D. Reactivity of alkali metals increases down Group 1 (Li < Na < K < Rb) because the single valence electron is farther from the nucleus and more easily lost. So the order is 3-1-4-2.
40. D. The standard reactivity/oxidising-strength order for these elemental halogens is F₂ > Cl₂ > Br₂ > I₂, giving 2-4-1-3. Fluorine is the strongest oxidant of the group. Do not justify that fact by claiming fluorine has the most exothermic atomic electron affinity: chlorine’s electron affinity is more exothermic, while the overall reaction energetics still favour fluorine.
41. A. Both are true and R explains A. Nitrogen's first ionisation energy (about 1402 kJ/mol) exceeds oxygen's (about 1314 kJ/mol) because N has a stable half-filled 2p³, while O loses a paired 2p electron and becomes 2p³. This is the Group 16 < Group 15 exception to the rising trend.
42. D. Three occupied shells place it in period 3. With 3 valence electrons in a p-block position, the group is 10 + 3 = 13. The element is aluminium (Z = 13).
43. D. The characteristic filling across most lanthanoids is of the 4f subshell. The end members and individual electron configurations have exceptions, so saying every differentiating electron from La to Lu enters 4f is too absolute. The analogous characteristic subshell of the actinoid series is 5f.