Question 1

What is the basic theme of organisation in the periodic table?

Accepted Answer

The periodic table is organised so that elements with similar valence-shell electron configurations fall into the same column. The starting point is atomic number, not atomic mass. Every element is placed in order of increasing atomic number Z (the number of protons, which equals the number of electrons in a neutral atom). This ordering itself is not the "theme" — it is just the ruler. The real organising idea is what happens to the electron configuration as Z increases one at a time. As electrons are added one by one, they fill orbitals following the (n+l) rule (the Aufbau principle). Because orbitals of a given type (s, p, d, f) fill up in a repeating pattern — one s-subshell, then a set of p-subshells, and so on — the outermost (valence) electron configuration repeats at regular intervals of Z. This repetition is the periodicity. Aside — this is the point most students skip past: the table is not arranged by mass, by size, or by "how the elements were discovered." It is arranged so that elements whose valence shells have the same configuration land directly beneath one another. For example, Li (configuration 1s^2 2s^1 ), Na ( [ Ne ] 3s^1 ), and K ( [ Ar ] 4s^1 ) all end in ns^1, so all three sit in Group 1, even though they differ enormously in mass and size. This gives the table its two directions of meaning: - Across a period (left to right), Z increases by one each step and electrons are added to the same outer shell, so the number of valence electrons steadily increases from 1 to 8. Properties change progressively across a period. - Down a group (top to bottom), the number of valence electrons stays the same, but a new principal shell n is added each time. Because the valence configuration repeats, elements in the same group show similar chemical behaviour (similar bonding, similar typical oxidation states, similar reactivity trends), even though their sizes and physical properties change steadily down the group. So the table is really a map of one underlying fact: chemical and physical properties of elements are a periodic function of their atomic numbers, because the arrangement of electrons in the outer shell — the part of the atom responsible for chemical behaviour — recurs periodically as Z increases. Grouping elements by matching valence configuration is what turns a simple numerical list (1, 2, 3, 4, …) into a table with rows (periods) and columns (groups) that carry chemical meaning. Answer: Elements are arranged in order of increasing atomic number, in such a way that elements having similar valence-shell electronic configurations (and hence similar chemical properties) fall into the same vertical column (group), while a new principal shell begins with each row (period). This is the modern periodic law: properties of elements are a periodic function of their atomic number.

Question 2

Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that?

Accepted Answer

Mendeleev's classification used relative atomic mass — but the properties of the elements always won when the two disagreed. The property he used Mendeleev arranged the (then) 63 known elements in order of increasing atomic mass, and stated what is now called Mendeleev's Periodic Law: "The physical and chemical properties of elements are a periodic function of their atomic masses." In practice this meant: list the elements in a row by rising atomic mass, and when an element's chemical and physical behaviour (valency, oxide/hydride formula, general reactivity) started repeating an earlier element's behaviour, start a new row directly under it. The column that results (a "group") collects elements with similar properties, because atomic mass, arranged this way, tracked periodicity closely enough for the elements known at the time. Did he stick to it? No — not strictly. Mendeleev treated similarity of chemical properties as the higher authority, and atomic mass as the ordering rule only as long as it did not contradict that. In a small number of places the two clashed, and he broke the atomic-mass order to keep chemically similar elements together. Two classic examples: - Cobalt and nickel. Cobalt (atomic mass ≈ 58.93 ) is very slightly heavier than nickel ( ≈ 58.71 ). Strict atomic-mass order would put nickel first. Mendeleev instead placed cobalt before nickel, matching how their properties lined up with the elements above them in his table. - Tellurium and iodine. Tellurium ( ≈ 127.6 ) is heavier than iodine ( ≈ 126.9 ). Placed by atomic mass alone, tellurium should come after iodine. But iodine's chemistry (a reactive, monovalent, salt-forming element) matches chlorine and bromine, while tellurium's chemistry (a divalent element resembling sulfur and selenium) matches the row above. So Mendeleev put tellurium before iodine, again overriding the mass order to preserve the property pattern. He also left gaps in the table (for elements not yet discovered, such as eka-aluminium and eka-silicon, later identified as gallium and germanium) rather than force the known elements to fill every atomic-mass slot — a further sign that the organizing principle in his mind was recurring properties, with atomic mass serving as a convenient but not absolute ordering tool. This is exactly the flaw atomic mass carried: it does not correlate perfectly with chemical behaviour across all elements, which is why the modern periodic law later replaced it with atomic number (Moseley), removing the need for any such exceptions. Answer: Mendeleev classified elements by increasing relative atomic mass (his Periodic Law: properties are a periodic function of atomic mass). He did not stick to this strictly — wherever atomic-mass order conflicted with the elements' actual chemical properties, he reordered elements to keep similar properties together (e.g., cobalt before nickel, tellurium before iodine, despite their higher atomic mass), and left gaps for undiscovered elements rather than force a fit.

Question 3

What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

Accepted Answer

Mendeleev's Law used atomic mass; the Modern Periodic Law uses atomic number — and that swap is what fixed the periodic table. Mendeleev's Periodic Law states that the physical and chemical properties of elements are a periodic function of their atomic masses. Mendeleev arranged all known elements in order of increasing atomic mass, and found that elements with similar properties recurred at regular intervals. The Modern Periodic Law states that the physical and chemical properties of elements are a periodic function of their atomic numbers, where the atomic number Z is the number of protons in the nucleus of an atom (and, for a neutral atom, also the number of electrons). So the basic difference in approach is the ordering variable itself: mass versus atomic number. This is not a small technical swap — it changes why periodicity happens at all. - Under Mendeleev's approach, elements were slotted purely by how heavy their atoms are. Atomic mass depends on the number of protons and neutrons together, and neutron number has no direct bearing on chemical behaviour. This made the ordering somewhat arbitrary from a chemical standpoint, and it is the reason Mendeleev's table had real problems: - Some pairs had to be placed out of mass order to keep them with elements of similar properties — for example, cobalt (atomic mass ≈ 58.9) was placed before nickel (atomic mass ≈ 58.7), and tellurium (atomic mass ≈ 127.6) was placed before iodine (atomic mass ≈ 126.9), because putting them strictly by mass would have broken the pattern of similar properties. - Isotopes of the same element have different atomic masses but identical chemical properties, so atomic mass could not consistently explain periodicity — this was an anomaly Mendeleev's law could not account for. - Under the Modern Periodic Law, the ordering variable is atomic number, Z, the actual count of protons (and hence electrons in a neutral atom). It is the number and arrangement of electrons — specifically the outermost (valence) electrons — that governs how an atom bonds and reacts. Because atomic number fixes the electronic configuration directly, ordering elements by Z makes the recurrence of similar valence-electron arrangements, and therefore of similar properties, a necessary, physically grounded consequence rather than an empirical coincidence. - This also resolves the cobalt–nickel and tellurium–iodine anomalies automatically: their atomic numbers (Co = 27, Ni = 28; Te = 52, I = 53) fall in exactly the order needed for them to sit with elements of matching properties, with no forced exception. - Isotopes, which share the same atomic number despite differing atomic masses, are correctly placed at a single position in the table — consistent with their identical chemical behaviour. Answer: Mendeleev's Periodic Law orders elements by atomic mass, while the Modern Periodic Law orders them by atomic number. Since atomic number equals the number of protons/electrons and directly fixes the valence electronic configuration, the modern approach ties periodicity to its actual physical cause and removes the anomalies (such as the Co–Ni and Te–I misplacements, and the problem of isotopes) that atomic-mass ordering could not explain.

Question 4

On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

Accepted Answer

A period's length is fixed by which orbitals actually get filled while the (n+l) rule is obeyed — not by the total capacity of one shell. The natural mistake here is to say "period 6 means n=6, so the answer is 2n^2 = 72." That formula gives the total electron capacity of an entire isolated shell, but a period is not "one shell filling up" — it is a stretch of the actual, energy-ordered filling sequence, and for n=6 that sequence pulls in orbitals from n=4 and n=5 as well. Step 1 — what the four quantum numbers say about a subshell's capacity. - Principal quantum number n (1, 2, 3, …) fixes the shell. - Azimuthal (orbital angular momentum) quantum number l can take values 0, 1, 2, …, (n-1), and it labels the subshell: l=0 is s, l=1 is p, l=2 is d, l=3 is f. - Magnetic quantum number m l runs from -l to +l, giving (2l+1) orbitals in that subshell. - Spin quantum number m s = + 12 or - 12; by the Pauli exclusion principle, each orbital holds at most 2 electrons with opposite spin. So the maximum number of electrons a subshell of azimuthal number l can hold is N max = 2(2l+1) where the factor 2 comes from the two spin states and (2l+1) from the number of orbitals. This gives: s: 2(1)=2, p: 2(3)=6, d: 2(5)=10, f: 2(7)=14. Step 2 — which subshells actually belong to period 6. Orbitals do not fill in the naive order 6s, 6p, 6d, 6f, …; they fill in order of increasing (n+l), and for equal (n+l) the orbital with the lower n goes first (Aufbau's (n+l) rule). Working through the sequence — 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, … — shows that once 6s has been reached, the next three subshells to fill, before 7s begins, are 4f, 5d, and 6p, not 6d or 6f. That interleaving of a lower- n, higher- l orbital (4f) among the n=6 orbitals is exactly why the 2n^2 shortcut fails here. Step 3 — add up the capacities of 6s, 4f, 5d, 6p. 2 6s + 14 4f + 10 5d + 6 6p = 32. Step 4 — check this against the actual period. Period 6 runs from caesium ( Z=55 ) to radon ( Z=86 ): 86 - 55 + 1 = 32, which is the same number, and it is built from exactly this filling: Cs, Ba fill 6s (2 elements); Ce to Lu fill 4f, the lanthanides (14 elements); La/Lu through Hg fill 5d (10 elements); Tl to Rn fill 6p (6 elements). So the 32 elements of period 6 are accounted for, orbital by orbital, by the quantum-number capacities of 6s, 4f, 5d and 6p — not by treating n=6 as a single shell filling from empty to 2n^2=72. Answer: Period 6 has 32 elements because the orbitals that fill during this period, in the order fixed by the (n+l) rule, are 6s, 4f, 5d and 6p, whose quantum-number-determined capacities are 2, 14, 10 and 6 electrons respectively; 2+14+10+6 = 32, matching the 32 elements from caesium (Z = 55) to radon (Z = 86).

Question 5

In terms of period and group where would you locate the element with Z =114?

Accepted Answer

The period number is the highest principal quantum number n reached while filling electrons; the group number (for a p-block element) is found from how many electrons sit in the outermost ns and np subshells. Build up the electron configuration of Z = 114 by the Aufbau order, filling one subshell at a time and keeping a running total of electrons so nothing is skipped: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4f^14 5d^10 6p^6 7s^2 5f^14 6d^10 7p^2 Here each symbol nℓ^x means: n is the shell (principal quantum number), ℓ is the subshell type ( s,p,d,f ), and the superscript x is the number of electrons sitting in that subshell. Running total of electrons as each subshell fills: 2, 4, 10, 12, 18, 20, 30, 36, 38, 48, 54, 56, 70, 80, 86, 88, 102, 112, 114 The total reaches exactly 114 after the 7p^2 subshell is filled, so no electron is left over and none is missing — the configuration is confirmed: [ Rn ] 5f^14 6d^10 7s^2 7p^2 (A common slip here is to stop counting once the numbers "look big enough" and guess the period from the noble-gas symbol alone — [ Rn ] is only the core; you still have to add up the outer 28 electrons to see that the last electron actually lands in the 7p subshell, not the 6d or 5f.) Finding the period. The outermost shell being filled is n = 7 (the 7s and 7p electrons), so the element sits in Period 7. Finding the group. The element's last electron enters a p subshell, so it is a p-block element. For p-block elements, the group number is obtained from the number of electrons in the outermost ns and np subshells by Group number = 10 + ( electrons in ns) + ( electrons in np) Here the valence configuration is 7s^2 7p^2, so electrons in ns = 2 and electrons in np = 2: Group number = 10 + 2 + 2 = 14 This is exactly the pattern of the carbon family: carbon ( 2s^2 2p^2 ), silicon ( 3s^2 3p^2 ), germanium ( 4s^2 4p^2 ), tin ( 5s^2 5p^2 ), and lead ( 6s^2 6p^2 ) all end in ns^2np^2 and all sit in Group 14 — Z = 114 simply continues that column one row further down, directly below lead. Answer: The element with Z = 114 has the configuration [ Rn ] 5f^146d^107s^27p^2, so it belongs to Period 7 and Group 14 (the carbon family), directly below lead.

Question 6

Write the atomic number of the element present in the third period and seventeenth group of the periodic table.

Accepted Answer

An element's address in the periodic table — (period, group) — fixes its electron configuration uniquely, and the configuration fixes the atomic number. Two rules link the address to the configuration: 1. The period number equals n, the principal quantum number of the outermost (valence) shell being filled. Period 3 means the valence electrons sit in the n=3 shell. 2. For the p-block (Groups 13–18), the group number equals 10 + (number of valence electrons). This is a direct reading of the IUPAC 1–18 group labels: Group 13 has 3 valence electrons, Group 14 has 4, …, Group 18 has 8. A common slip here is to treat "third period" as if it meant atomic number 3 (that would be lithium, which is actually in period 2) — period number and atomic number are not the same thing; the period only tells you which shell is being filled, not how many electrons are in it. Step 1 — find the number of valence electrons. Group 17 is a p-block group, so valence electrons = 17 - 10 = 7 Step 2 — write the valence-shell configuration. With n = 3 (period 3) and 7 valence electrons distributed as ns^2np^5 (2 in the s-subshell, the remaining 5 in the p-subshell): 3s^2 3p^5 Step 3 — build the full electron configuration. The shells below n=3 are completely filled, following the order 1s, 2s, 2p before 3s, 3p: 1s^2 2s^2 2p^6 3s^2 3p^5 Step 4 — count the electrons to get the atomic number. 2 + 2 + 6 + 2 + 5 = 17 For a neutral atom, the number of electrons equals the number of protons, and the number of protons is the atomic number Z. So Z = 17. Cross-check by walking the row of the table. Period 3 runs Na(11), Mg(12), Al(13), Si(14), P(15), S(16), Cl(17), Ar(18) — there is no d-block in period 3 (the d-block only begins at period 4), so Group 13 lands immediately after Group 2. Counting across, Group 17 in this row is Cl, with atomic number 17 — the same answer as the configuration method. This element is chlorine, a halogen, consistent with Group 17 elements always having the ns^2np^5 valence pattern that makes them one electron short of a full octet. Answer: The element in the third period and seventeenth group is chlorine (Cl), atomic number 17 (electron configuration 1s^2 2s^2 2p^6 3s^2 3p^5 ).

Question 7

Which element do you think would have been named by (i) Lawrence Berkeley Laboratory (ii) Seaborg’s group?

Accepted Answer

A new element's name almost always tells you who fought to get credit for finding it — read the naming dispute the chapter itself describes, and match each group to the element whose story is theirs. This question sits right after the NCERT passage on nomenclature of elements with atomic number greater than 100 (Section 3.4). Two separate pieces of that passage answer the two parts. (i) Lawrence Berkeley Laboratory The text says: "both American and Soviet scientists claimed credit for discovering element 104. The Americans named it Rutherfordium whereas Soviets named it Kurchatovium." The American team in that priority dispute worked at the Lawrence Berkeley Laboratory in California — this is the lab that raced the Dubna (USSR) group to synthesise element with atomic number Z = 104 and proposed the name Rutherfordium (Rf), after the nuclear physicist Ernest Rutherford. So Lawrence Berkeley Laboratory is the group behind the name Rutherfordium. A point students often miss here: the element is not named after Berkeley or after anyone at the lab — Rutherfordium honours Rutherford, a scientist with no connection to the lab itself. The lab's role was only in the discovery and naming claim, not in whose name got used. (ii) Seaborg's group The chapter's own footnote on Glenn T. Seaborg says: "Glenn T. Seaborg's work... starting with the discovery of plutonium in 1940, followed by those of all the transuranium elements from 94 to 102 led to reconfiguration of the periodic table... In 1951, Seaborg was awarded the Nobel Prize in chemistry for his work. Element 106 has been named Seaborgium (Sg) in his honour." Seaborg's group is the specific research team, built around Glenn Seaborg himself, that had already discovered plutonium and the run of transuranium elements up to Z = 102. When this same team went on to synthesise the element with atomic number Z = 106, it named the element after its own leader — giving Seaborgium (Sg). Here is the distinction the question is testing: "Lawrence Berkeley Laboratory" names an institution involved in a priority dispute over element 104, while "Seaborg's group" names a specific team, tied to one scientist's own body of work, that discovered and named element 106 after him. Answer: (i) Lawrence Berkeley Laboratory would have named the element with atomic number 104 as Rutherfordium (Rf), after Ernest Rutherford. (ii) Seaborg's group would have named the element with atomic number 106 as Seaborgium (Sg), in honour of Glenn T. Seaborg himself.

Question 8

Why do elements in the same group have similar physical and chemical properties?

Accepted Answer

Elements placed in the same group have the same number of electrons in their outermost (valence) shell — and it is this valence-shell configuration, not the atomic number itself, that decides how an element behaves. The modern periodic law states that properties of elements are a periodic function of their atomic number. Electrons fill orbitals in a repeating pattern: within every period the outer shell fills up in the same order — ns^1, ns^2, ns^2np^1, ns^2np^2, …, ns^2np^6 — only the value of the principal quantum number n (the shell number) increases as you drop to the next period. A group is simply the column you get by lining up elements that sit at the same stage of this repeating pattern, so every member of a group has an analogous valence-shell configuration, just written for a different n. Why this makes chemical properties similar Chemical behaviour — the valency an element shows, the type of ion it forms, the kind of bonds it makes, its characteristic reactions — is controlled almost entirely by the number and arrangement of valence electrons, because these are the electrons available to be lost, gained, or shared. Since every element in a group has the same valence-electron count, they behave in the same type of way: - Group 1 (alkali metals): Li =[He]2s^1, Na =[Ne]3s^1, K =[Ar]4s^1 — one ns^1 electron each. All lose that single electron readily to form +1 ions, and all react vigorously with water liberating hydrogen. - Group 17 (halogens): F =[He]2s^22p^5, Cl =[Ne]3s^23p^5, Br =[Ar]3d^104s^24p^5 — one electron short of a full octet ( ns^2np^5 ) each. All gain one electron to form -1 ions, exist as diatomic molecules X 2, and act as oxidising agents. This is the point students most often blur: a shared group does not mean identical properties, only the same pattern of chemical behaviour, because the number of valence electrons — the thing chemistry actually reacts with — is what repeats, not the whole atom. Why physical properties follow a family trend too Physical properties (atomic radius, ionisation enthalpy, electronegativity, metallic character) also vary as a regular, predictable trend down a group rather than randomly, because going down a group the valence-electron count stays fixed while the principal quantum number n increases by one every period — a new shell of electron density is added each time. This steadily increases the distance of the valence electrons from the nucleus, so radius increases smoothly, ionisation enthalpy decreases smoothly, and so on. It is this steady, non-erratic change that gives a group its "family resemblance" — soft, low-melting alkali metals; coloured, corrosive halogens — even though the exact numerical value of a property (say, the melting point) is different for each member. Answer: Elements in the same group have the same number of valence electrons and an analogous valence-shell electronic configuration (only the principal quantum number n differs), and it is this valence-shell configuration that fixes chemical behaviour (valency, ions formed, bond type, characteristic reactions) — so group members react in the same characteristic way. Physical properties also change in a smooth, regular trend down the group (not identically, but systematically) because the number of occupied shells increases steadily while the valence-electron count stays constant, steadily changing atomic size and how tightly the valence electrons are held.

Question 9

What does atomic radius and ionic radius really mean to you?

Accepted Answer

Neither radius is the edge of a solid little ball — an atom's electron cloud has no sharp boundary, so both radii are distances measured between nuclei and then split by convention. Quantum mechanically, the probability of finding an electron never drops to exactly zero, however far you go from the nucleus — the electron density just fades out, trailing off toward infinity. There is no point you can point to and say "the atom ends here." So chemists do not measure a single isolated atom's radius directly; instead they measure the distance between the nuclei of two neighbouring atoms (from X-ray or electron diffraction) and take a fraction of that distance as "the radius." That is what atomic radius really is: an operational, not an absolute, quantity, and its value even changes depending on how the neighbouring atom is bonded. Atomic radius: half the internuclear distance, and it comes in three flavours depending on the kind of contact - Covalent radius — half the distance between the nuclei of two atoms joined by a single covalent bond, usually measured in a homonuclear molecule. For example, the Cl–Cl bond length in Cl 2 is 198 pm, so the covalent radius of chlorine is taken as 99 pm. - Van der Waals radius — half the distance between the nuclei of two identical atoms belonging to two neighbouring molecules in a solid, where the atoms are not bonded to each other but merely touching (held together by weak van der Waals attraction). Because van der Waals forces are far weaker than a covalent bond, the atoms cannot approach nearly as closely — so van der Waals radius is always larger than covalent radius for the same atom. - Metallic radius — half the distance between the nuclei of two adjacent atoms in a metallic crystal lattice, where the atoms are held together by metallic bonding. The aside worth remembering here: these three numbers for the same element are all different, because "radius" is really answering the question "how close can this atom get to a neighbour under this particular kind of contact," not "how big is this atom in isolation." Ionic radius: the same trick, applied to an ionic crystal, then apportioned between the two ions For an ionic solid, X-ray diffraction gives the distance between the nucleus of the cation and the nucleus of the adjacent anion in the lattice — this is r + + r -, the sum of the two ionic radii, not either one alone. To split this sum into an individual cation radius and anion radius, one ion's radius must be fixed independently (for instance, from a series of compounds where that ion's size can be cross-checked), and the rest is obtained by subtraction. So an individual ionic radius, too, is a value assigned by convention against a reference, not a directly observed single-ion measurement. Why a cation is smaller, and an anion is larger, than the parent atom - A cation forms by removing one or more electrons from an atom. The remaining electrons are pulled in by the same nuclear charge acting on fewer electrons, so the effective nuclear charge per electron goes up and the electron cloud contracts. Often the entire outermost shell is removed (e.g., Na → Na^+ loses its whole n = 3 shell), which shrinks the ion even more sharply. A cation is always smaller than its parent atom. - An anion forms by adding one or more electrons. The same nuclear charge now has to hold more electrons, so effective nuclear charge per electron falls, electron–electron repulsion rises, and the cloud expands. An anion is always larger than its parent atom. A related pattern worth flagging as the step people mix up: for an isoelectronic series (ions/atoms with the same number of electrons, e.g., O^2-, F^-, Na^+, Mg^2+, Al^3+, all with 10 electrons), the number of electrons is fixed, so radius is governed entirely by nuclear charge: more protons pull the same electron cloud in tighter, so radius decreases as atomic number increases across the series — O^2- F^- Na^+ Mg^2+ Al^3+. Answer: Atomic radius is the distance from an atom's nucleus to the outermost shell where its electron density effectively ends — but since the electron cloud has no sharp edge, it is in practice defined and measured as half the distance between the nuclei of two touching neighbouring atoms (covalent radius for bonded atoms, van der Waals radius for non-bonded touching atoms, metallic radius for atoms in a metal lattice). Ionic radius is the same idea applied to an ion in a crystal: the interionic (nucleus-to-nucleus) distance between a cation and an adjacent anion is measured, and then split between the two using one ion's independently known radius as reference. A cation is smaller than its parent atom (fewer electrons, often a lost shell, higher effective nuclear charge per electron), while an anion is larger than its parent atom (more electrons, greater repulsion, lower effective nuclear charge per electron).

Question 10

How do atomic radius vary in a period and in a group? How do you explain the variation?

Accepted Answer

Atomic radius is set by a tug-of-war between the pull of the nucleus and the number of electron shells around it — and moving across a period changes one side of that fight, while moving down a group changes the other. Definition first. Atomic radius is usually measured as the covalent radius — half the distance between the nuclei of two like atoms joined by a single covalent bond (for metals, the metallic radius; for noble gases, the van der Waals radius, which is defined differently and is discussed below). The pull an outer electron actually feels is the effective nuclear charge Z eff, given by Z eff = Z - σ where Z is the atomic number (the actual number of protons in the nucleus) and σ is the screening (shielding) constant — the extent to which the inner electrons block the full pull of the nucleus from reaching an outer electron. A bigger Z eff pulls the outer electron in tighter, shrinking the atom; a smaller Z eff lets it sit farther out. Across a period (left to right): atomic radius decreases. Moving one element to the next across a period, one proton is added to the nucleus ( Z increases by 1) and one electron is added to the same principal shell (the same value of n ) that was already being filled. Electrons in the same shell shield each other very poorly — the screening constant σ rises only a little for each added electron. So Z eff = Z - σ goes up steadily across the period. The nucleus pulls the whole electron cloud in harder with each step, and the atomic radius shrinks. For example, across period 2 the covalent radii (in pm) fall steadily: Li (152) Be (111) B (88) C (77) N (74) O (66) F (64) The place this trips people up: the noble gas at the very end of the period (Ne, Ar, …) is usually tabulated with a larger radius than the halogen before it, which looks like it breaks the trend. It doesn't, because it isn't measured the same way — a noble gas forms no covalent bond, so its size has to be reported as a van der Waals radius (half the distance between two non-bonded atoms just touching each other), which is always larger than a covalent radius for a comparable atom. Comparing a van der Waals radius to a covalent radius is comparing two different rulers, so this last point is excluded when the period trend is stated. Down a group (top to bottom): atomic radius increases. Moving down a group, the outermost electron is no longer added to the same shell — a whole new principal shell is added, so n increases by 1 at every step. Nuclear charge Z does increase too, but now every electron in all the complete inner shells lies between the nucleus and the new outer electron, and a complete inner shell screens much more effectively than same-shell electrons do. The extra shielding from these full inner shells largely cancels the extra nuclear charge, so Z eff on the valence electron stays roughly the same (or rises only slightly) down a group. What dominates instead is simply that the valence electron now occupies a shell of higher n, which by itself sits farther from the nucleus. The result is a steady increase in atomic radius. For example, down group 1, the covalent radii (in pm) rise steadily: Li (152) < Na (186) < K (231) < Rb (244) < Cs (262) Answer: Atomic radius decreases across a period from left to right, because electrons added within the same shell shield each other poorly, so the effective nuclear charge Z eff=Z-σ rises and pulls the electron cloud in tighter. Atomic radius increases down a group from top to bottom, because a new principal shell (higher n ) is added at each step, and the extra shielding from the newly complete inner shells largely offsets the increased nuclear charge, so the dominant effect is simply that the valence electrons now occupy a shell farther from the nucleus.

Question 11

What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions. (i) F^- (ii) Ar (iii) Mg 2 + (iv) Rb^+

Accepted Answer

Isoelectronic species have the same total number of electrons, even though they are different elements with different atomic numbers. "Iso" means "same" — here it's the electron count that matches, not the number of protons and not the charge itself. For any ion, the number of electrons is electrons = Z - q, where Z is the atomic number (the fixed number of protons for that element) and q is the charge on the ion, taken with its own sign — positive for a cation (electrons lost), negative for an anion (electrons gained). A common slip is to read Z off the periodic table and stop there, forgetting that forming the ion changed the electron count away from Z. Step 1 — find the electron count of each given species. (i) F^-: fluorine has Z = 9. Charge q = -1, so electrons = 9 - (-1) = 10. (ii) Ar: a neutral atom, Z = 18, charge zero, so electrons = 18. (iii) Mg^2+: magnesium has Z = 12. Charge q = +2 (two electrons removed), so electrons = 12 - 2 = 10. (iv) Rb^+: rubidium has Z = 37. Charge q = +1, so electrons = 37 - 1 = 36. Step 2 — find another species with the same electron count for each. Any species whose own Z - q matches is a valid answer; picking a nearby element keeps the example clean. (i) 10 electrons: sodium has Z = 11; as Na^+ (one electron lost) it has 11 - 1 = 10 electrons, matching F^-. So Na^+ is isoelectronic with F^-. (Neon, O^2-, and N^3- are the other common 10-electron species.) (ii) 18 electrons: calcium has Z = 20; as Ca^2+ it has 20 - 2 = 18 electrons, matching Ar. So Ca^2+ is isoelectronic with Ar. ( K^+, Cl^-, and S^2- also carry 18 electrons.) (iii) 10 electrons: neon is itself a neutral, 10-electron atom ( Z = 10, charge zero), so Ne is isoelectronic with Mg^2+. (Notice Mg^2+ also ends up isoelectronic with F^- and Na^+ from part (i), since all three carry 10 electrons — isoelectronic species don't have to come in matched pairs only.) (iv) 36 electrons: krypton is a neutral atom with Z = 36, so Kr is isoelectronic with Rb^+. ( Sr^2+ and Br^- also have 36 electrons.) Answer: Isoelectronic species are atoms, molecules, or ions that carry the same total number of electrons even though their atomic numbers are different. (i) F^- (10 e⁻) is isoelectronic with Na^+ (also Ne, O^2-, N^3- ). (ii) Ar (18 e⁻) is isoelectronic with Ca^2+ (also K^+, Cl^-, S^2- ). (iii) Mg^2+ (10 e⁻) is isoelectronic with Ne (also F^-, Na^+, O^2- ). (iv) Rb^+ (36 e⁻) is isoelectronic with Kr (also Sr^2+, Br^- ).

Question 12

Consider the following species: N 3 –, O 2 –, F^-, Na^+, Mg 2 + and Al 3 + (a) What is common in them? (b) Arrange them in the order of increasing ionic radii.

Accepted Answer

These six species are isoelectronic — same electron count, different nuclear charge — and that difference in nuclear charge is what fixes their radii. (a) What is common in them Count the electrons on each ion by taking the atom's atomic number Z (its proton count, unchanged by ionisation) and adding electrons for a negative charge or removing them for a positive charge. Every one of them has exactly 10 electrons, arranged as 1s^2 2s^2 2p^6 — the neon configuration. Species that share the same electron count and the same arrangement of electrons like this are called isoelectronic species. That is what is common to all six. (b) Order of increasing ionic radii Here is the step that trips people up: since all six ions have the same number of electrons in the same orbitals, the size of the electron cloud is no longer decided by how many shells are filled — it is decided entirely by how hard the nucleus is pulling on those 10 electrons. That pull is the effective nuclear charge, and for a fixed electron count it simply tracks Z (the number of protons) directly: more protons in the nucleus means a stronger pull on the same 10 electrons, which drags the electron cloud in tighter and makes the ion smaller. So among isoelectronic species, radius runs opposite to Z: the higher the atomic number, the smaller the ion. Listing Z for each species: N (7) < O (8) < F (9) < Na (11) < Mg (12) < Al (13) Radius decreases as Z increases, so reversing this list gives increasing radius: Al^3+ < Mg^2+ < Na^+ < F^- < O^2- < N^3- This matches the physical picture too: Al^3+ has given up three electrons and carries the largest positive charge among these ions, so its 10 remaining electrons are pulled in hardest by 13 protons, making it the smallest of the six. N^3-, at the other end, has only 7 protons trying to hold onto the same 10 electrons (with three extra squeezed in and the extra electron-electron repulsion that comes with them), so its electron cloud is pulled in the least and spreads out the most, making it the largest. Answer: All six species (N³⁻, O²⁻, F⁻, Na⁺, Mg²⁺, Al³⁺) are isoelectronic, each having 10 electrons with the configuration 1s^2 2s^2 2p^6. Since the electron count is fixed, radius decreases as nuclear charge (Z) increases, giving the increasing order of ionic radii: Al^3+ < Mg^2+ < Na^+ < F^- < O^2- < N^3-.

Question 13

Explain why cation are smaller and anions larger in radii than their parent atoms?

Accepted Answer

The number of electrons changes, but the nuclear charge does not — so the pull felt by each remaining or added electron changes. Atomic and ionic radius is set by a tug-of-war between the nucleus pulling electrons in and the electrons repelling each other and pushing the cloud out. Whenever an atom becomes an ion, the number of protons in the nucleus stays exactly the same, but the number of electrons changes — and that shift in electron–electron repulsion, felt against the same fixed nuclear pull, is what moves the radius. Why a cation is smaller than its parent atom A cation forms by removing one or more electrons, usually from the outermost shell: M → M^n+ + ne^- Two things happen together: - Removing an electron reduces the total electron–electron repulsion. With fewer electrons pushing against each other, the remaining electrons are pulled in tighter by the same nuclear charge. - The effective nuclear charge per electron goes up, because the same number of protons now has to hold only fewer electrons. Each remaining electron effectively feels a stronger net pull toward the nucleus (this "pull per electron" is what chemists call effective nuclear charge, Z eff; it rises here even though the actual nuclear charge Z is unchanged). Often the outermost shell is emptied out completely — for example, sodium, Na (1s^2 2s^2 2p^6 3s^1), loses its lone 3s electron to become Na^+ (1s^2 2s^2 2p^6). Na^+ has one shell fewer occupied than Na, so besides the tighter pull, the electron cloud is now built from a smaller shell altogether. Both effects pull in the same direction, so the cation is always smaller than the neutral atom it came from. Why an anion is larger than its parent atom An anion forms by adding one or more electrons to the outermost shell: X + ne^- → X^n- Here the opposite happens: - The same nuclear charge now has to hold a greater number of electrons, so the effective nuclear charge per electron goes down — each electron, old and new, feels a weaker net pull than before. - The added electron(s) increase electron–electron repulsion in the outer shell, and the electron cloud spreads out to reduce this crowding. For example, chlorine, Cl (1s^2 2s^2 2p^6 3s^2 3p^5), gains one electron to complete its octet as Cl^- (1s^2 2s^2 2p^6 3s^2 3p^6). The nucleus (17 protons) is unchanged, but it is now spreading its pull over 18 electrons instead of 17, so the cloud expands and Cl^- is bigger than Cl. The step that trips people up: it is tempting to think radius depends only on which shell number is occupied, but here the shell number often does not even change for an anion (Cl and Cl⁻ are both up to n=3 ) — the size change comes purely from the altered balance between (fixed) nuclear charge and (changed) electron count and repulsion, not from adding a whole new shell. Answer: A cation has the same nuclear charge as its parent atom but fewer electrons, so effective nuclear charge per electron rises, electron–electron repulsion falls, and the electron cloud is pulled in tighter (often with a whole shell removed) — making the cation smaller. An anion has the same nuclear charge but more electrons, so effective nuclear charge per electron falls and repulsion among electrons rises, causing the electron cloud to expand — making the anion larger than its parent atom.

Question 14

What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy? Hint: Requirements for comparison purposes.

Accepted Answer

Both phrases exist to remove every variable except the atom's own nuclear charge and electron arrangement, so that values for different elements can be compared on the same footing. Ionization enthalpy is defined as the energy needed to remove the most loosely held electron from an isolated gaseous atom in its ground state, and electron gain enthalpy as the energy change when an electron is added to an isolated gaseous atom in its ground state. Each condition is doing a specific job. Why "isolated gaseous atom": An atom's electrons are not held only by its own nucleus once other atoms are nearby — in a solid, a liquid, or inside a molecule, neighbouring atoms, ions or molecules exert their own attractions and repulsions (bonding forces, lattice forces, intermolecular forces) on that atom's electrons. Removing or adding an electron under those conditions would measure the combined effect of the atom's nucleus and its surroundings, not a property of the atom by itself. Taking the atom as gaseous removes the close packing found in solids and liquids, so intermolecular forces essentially vanish. Taking it as isolated, in addition, means the atom is far enough from every other atom, ion or electron that none of them can pull on or push away its electrons. Under this condition, whatever energy is measured comes only from the tug between the atom's own nucleus and its own electrons — exactly the quantity periodicity is trying to track as one moves across a period or down a group. Why "ground state": Even one fixed atom can exist with its electrons in different arrangements: the ground state, where every electron sits in the lowest energy orbitals available under the Aufbau principle, or an excited state, where one or more electrons have been promoted to higher orbitals. An electron already promoted to a higher orbital is farther from the nucleus and more weakly held, so it would take less energy to remove — the measured ionization enthalpy would depend on which excited state happened to be chosen, not on the element itself. Specifying the ground state fixes the electron arrangement to one unique, lowest-energy configuration for every atom of that element. This gives a single, reproducible reference point, so that the ionization enthalpy or electron gain enthalpy reported for sodium, say, always refers to the same starting arrangement, and can be lined up meaningfully against the value for magnesium or chlorine. Putting the two together: without "isolated gaseous," the value would be contaminated by forces from neighbouring particles; without "ground state," the value would depend on an arbitrary choice of excited configuration. Fixing both conditions means the only thing left that can differ between elements is the intrinsic pull of each atom's own nucleus on its own electrons — which is precisely what makes ionization enthalpy and electron gain enthalpy trends across the periodic table meaningful to compare. Answer: "Isolated gaseous atom" removes the influence of neighbouring atoms/ions/molecules (bonding and intermolecular forces), so the measured energy reflects only the atom's own nucleus–electron attraction; "ground state" fixes the electron configuration to the one, lowest-energy arrangement (Aufbau principle) rather than an arbitrary excited state. Together these conditions standardize the measurement so ionization enthalpy and electron gain enthalpy can be validly compared from one element to another.

Question 15

Energy of an electron in the ground state of the hydrogen atom is –2.18×10–18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1. Hint: Apply the idea of mole concept to derive the answer.

Accepted Answer

Ionization enthalpy is the energy needed to remove an electron from ONE MOLE of atoms in their ground state, so the per-atom energy you're given must be flipped in sign and then scaled up by Avogadro's number — both steps matter, and skipping either gives a wrong or wrongly-signed answer. Step 1: Energy needed to remove one electron from one atom. The electron sits in the ground state at E 1 = -2.18×10^-18 J Ionization means taking the electron out to infinity, where by convention the electron's energy is taken as zero, E ∞ = 0. The energy that must be supplied (the ionization energy per atom) is the energy of the final state minus the energy of the initial state: Δ E = E ∞ - E 1 = 0 - (-2.18×10^-18 J ) = +2.18×10^-18 J The sign flip here is the step people get wrong: the electron's energy in the atom is negative (it is bound), so removing it always costs a positive amount of energy — you are climbing out of a well, not falling into one. Step 2: Scale from one atom to one mole of atoms. Ionization enthalpy (as reported in J mol ^-1 ) refers to one mole of atoms, each losing one electron — not one atom, and not one mole of electrons alone in some other sense; it is the same mole-of-atoms idea used everywhere else in stoichiometry. So multiply the per-atom energy by the Avogadro constant, N A = 6.022×10^23 mol^-1 which is the formula connecting a per-particle quantity to a per-mole quantity: (energy per mole) = (energy per particle) × (particles per mole). Δ H ionization = Δ E × N A = (2.18×10^-18 J )×(6.022×10^23 mol^-1) Multiply the coefficients and add the exponents: 2.18 × 6.022 = 13.128 Δ H ionization = 13.128×10^5 J mol^-1 = 1.3128×10^6 J mol^-1 This is the same as 1312.8 kJ mol^-1, which is the familiar textbook value for the ionization enthalpy of hydrogen. Answer: The ionization enthalpy of atomic hydrogen is 1.3128×10^6 J mol^-1 (i.e., 1312.8 kJ mol^-1 ), obtained as Δ H = (0-E 1)× N A = (2.18×10^-18 J )×(6.022×10^23 mol^-1).

Question 16

Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why (i) Be has higher ∆i H than B (ii) O has lower ∆i H than N and F?

Accepted Answer

A subshell that is exactly full or exactly half full is unusually stable, and that stability can override the normal left-to-right rise in ionization enthalpy. Ionization enthalpy Δ iH is the energy needed to pull the outermost electron off a gaseous atom. Across a period, nuclear charge rises and atomic size shrinks, so Δ iH usually climbs steadily. Two special electron arrangements upset that smooth climb: a completely filled subshell (like 2s^2 ) and a half-filled subshell (like 2p^3, one electron in each of the three 2p orbitals). Both are unusually symmetric and stable, so the very next electron added — which has to go into a higher orbital or pair up in an already-occupied one — is easier to remove than the trend alone would predict. (i) Why Be has a higher Δ iH than B Write out the configurations: Be: 1s^2 2s^2 B: 1s^2 2s^2 2p^1 Be's last electron sits in a completely filled 2s^2 subshell — held tightly and symmetrically. B's last electron instead goes into the 2p subshell, which lies at higher energy than 2s and, on top of that, is shielded from the nucleus by the compact 2s^2 pair sitting underneath it. The step people get wrong here is assuming nuclear charge settles it: B has one more proton than Be ( Z=5 vs Z=4 ), so it "should" hold its electrons more tightly. But nuclear charge is not the whole story — orbital energy and shielding matter too. B's 2p^1 electron is both higher in energy and better shielded, so it comes off more easily than one of Be's snugly-held 2s^2 electrons. That is why the actual values run Δ iH( Be ) = 899 kJ/mol Δ iH( B ) = 801 kJ/mol, even though B sits to the right of Be in the period. (ii) Why O has a lower Δ iH than both N and F N: 1s^2 2s^2 2p^3 (2p x^1 2p y^1 2p z^1) O: 1s^2 2s^2 2p^4 (2p x^2 2p y^1 2p z^1) N's 2p^3 is a half-filled subshell: one electron sits alone in each of the three 2p orbitals, so no orbital carries a repelling pair, and this symmetric arrangement carries extra stability. O has one more electron, which has nowhere to go except into an orbital that is already occupied, giving 2p x^2. The step that's easy to miss: this isn't about O having a weaker pull from the nucleus — O's nuclear charge ( Z=8 ) is larger than N's ( Z=7 ), so on nuclear charge alone O's Δ iH "should" be higher. What actually happens is that the two electrons forced into the same 2p x orbital in O repel each other, and that repulsion makes it easier to strip one of them away than the nuclear-charge trend alone predicts. The result is Δ iH( O ) = 1314 kJ/mol < Δ iH( N ) = 1402 kJ/mol. F, 1s^2 2s^2 2p^5 (2p x^2 2p y^2 2p z^1), also has paired 2p electrons and so carries the same kind of internal repulsion that O does. But F's nuclear charge ( Z=9 ) is enough higher than O's that the extra pull on the electrons wins out over the extra repulsion, so the normal rising trend resumes past O: Δ iH( F ) = 1681 kJ/mol Δ iH( O ). The half-filled/paired-orbital effect only produces a local dip at O — it does not reverse the overall upward trend of the period. Answer: (i) Be's outermost electron sits in the extra-stable, fully-filled 2s^2 subshell, while B's last electron is in the higher-energy, more shielded 2p^1 orbital, so Be holds its electron more tightly than B despite having one less proton, giving Δ iH( Be ) Δ iH( B ). (ii) N's 2p^3 is a stable, half-filled subshell with no paired electrons, whereas O's 2p^4 forces two electrons into one orbital; the repulsion between that pair makes O's electron easier to remove than N's, so Δ iH( O ) < Δ iH( N ). F also has a paired 2p orbital, but its higher nuclear charge outweighs that repulsion, so Δ iH( O ) < Δ iH( F ) as well.

Question 17

How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?

Accepted Answer

Ionization enthalpy depends on which electron is being pulled off — and how stable the configuration left behind is. Look at where the electron comes from in each step, for both atoms. Write out the electron configurations first. Na (Z = 11): 1s^2 2s^2 2p^6 3s^1 Mg (Z = 12): 1s^2 2s^2 2p^6 3s^2 Step 1 — why the first ionization enthalpy, Δ i H 1, of Na is lower than that of Mg In sodium, the electron being removed is the lone 3s^1 electron. Taking it off leaves Na^+: 1s^2 2s^2 2p^6, which is the extremely stable, fully-filled noble-gas configuration of neon. Because the atom is "trying" to reach that stable arrangement anyway, this electron comes off easily, so Δ i H 1( Na ) is low (496 kJ mol⁻¹). In magnesium, the electron being removed comes out of a filled subshell, 3s^2. A filled s -subshell has extra stability (both electrons pair up symmetrically), and on top of that Mg has one more proton in its nucleus than Na (Z = 12 vs Z = 11) while the shielding from the inner core is essentially the same. That means the effective nuclear charge Z eff felt by the 3s electrons is higher in Mg. Both effects — greater Z eff and the stability of a filled subshell — pull the electron in more tightly, so Δ i H 1( Mg ) is higher (737 kJ mol⁻¹). This is the point people trip on: it is tempting to say "Mg is to the right of Na, so of course it's harder to ionize," treating it as a blanket periodic-trend statement. The real reason is the specific configuration each atom is left with — Na empties a subshell to reach a noble-gas core; Mg has to break open a filled one. Step 2 — why the second ionization enthalpy, Δ i H 2, of Na is higher than that of Mg Now look at what is being removed in the second step, i.e., from the +1 ions. Na^+: 1s^2 2s^2 2p^6 ( already a stable noble-gas core, like Ne ) Mg^+: 1s^2 2s^2 2p^6 3s^1 ( still has one electron left in 3s) To form Na^2+, an electron must be pulled out of the complete, tightly-held 2p^6 inner shell of Na^+ — a shell that is already a stable closed octet held by a nucleus of 11 protons acting on far fewer, more contracted electrons. Breaking into that shell costs a huge amount of energy, so Δ i H 2( Na ) is very high (4562 kJ mol⁻¹) — in fact almost ten times Δ i H 1( Na ). To form Mg^2+, the electron removed is the single, loosely-held 3s^1 electron of Mg^+ — exactly the same kind of easy removal Na underwent in its first step, since it leaves behind the same stable Ne-like core, 1s^2 2s^2 2p^6. So Δ i H 2( Mg ) is comparatively modest (1450 kJ mol⁻¹). Since 4562 kJ mol^-1 1450 kJ mol^-1, the second ionization enthalpy of Na is higher than that of Mg — the reverse of the order in the first step. Answer: Na's first ionization enthalpy is lower than Mg's because Na loses its single outer 3s^1 electron (attaining a stable Ne core), while Mg must remove one electron from a filled, more tightly-held 3s^2 subshell (higher Z eff, extra stability of the filled subshell). Na's second ionization enthalpy is higher than Mg's because at that stage Na⁺ already has the stable noble-gas configuration 1s^2 2s^2 2p^6, so removing a further electron means breaking into a complete inner shell, whereas Mg⁺ still has one loosely-held 3s^1 electron left to lose easily to reach the same stable configuration.

Question 18

What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

Accepted Answer

Ionization enthalpy is the energy needed to pull the outermost electron away from an isolated gaseous atom — and it falls off down a group because the nucleus's grip on that electron weakens, even though the nucleus itself is getting more positive. Ionization enthalpy, Δ i H, is defined for the process X(g) → X^+(g) + e^- Whether this is easy or hard depends on how strongly the nucleus holds the outermost (valence) electron. That pull is set by the effective nuclear charge, Z eff, and by the distance of the valence electron from the nucleus — not by the actual nuclear charge Z alone. Going down a group, three things change together, and the net effect is that the hold on the valence electron gets weaker. Factor 1 — the atomic radius increases. Down a group, each successive element has its valence electron in a shell with a higher principal quantum number n (period 2 uses n=2, period 3 uses n=3, and so on). A larger n means the valence electron sits, on average, much farther from the nucleus. Coulombic attraction falls off with distance, so a farther electron is held more loosely — this alone tends to lower the ionization enthalpy. Factor 2 — the screening (shielding) effect increases. Every time you go one row down, a whole new inner shell of electrons is added between the nucleus and the outermost electron. These inner-shell electrons shield, or screen, the valence electron from the full pull of the nuclear charge. This is the point people usually skip: it is not enough to say "more electrons are added" — what matters is that the added electrons sit closer to the nucleus than the valence electron does, so they cancel out part of the nuclear charge that the valence electron would otherwise feel. Factor 3 — nuclear charge does increase, but it is outweighed. It is true that Z (the atomic number, and hence the actual nuclear charge) is larger for the element below. If this were the only thing changing, ionization enthalpy would rise down a group. But the increase in shielding from the extra inner shell (Factor 2) removes more attraction than the extra protons add, and the increase in size (Factor 1) pushes the electron even farther from whatever net charge remains. The result is that the effective nuclear charge, Z eff = Z - σ where σ is the shielding constant contributed by the inner electrons, either stays roughly constant or even decreases slightly down the group, while the distance term keeps growing. Combined, the electron is both farther away and less strongly pulled. Putting it together: because the valence electron down a group is farther from the nucleus (larger n ) and is shielded more effectively by additional inner shells, the net attractive force the nucleus exerts on it drops, even though the raw nuclear charge Z has gone up. Less force holding the electron means less energy is needed to remove it, so ionization enthalpy decreases down a group. Answer: Down a group, ionization enthalpy decreases mainly because of two reinforcing effects: (1) atomic size increases as a new principal shell is added at each row, moving the valence electron farther from the nucleus, and (2) the shielding/screening effect of the additional inner-shell electrons increases, lowering the effective nuclear charge felt by the valence electron. Although the actual nuclear charge Z increases down the group, this increase is outweighed by the greater shielding and greater distance, so the net pull on the outermost electron weakens and less energy is required to remove it.

Question 19

The first ionization enthalpy values (in kJ mol–1) of group 13 elements are: B Al Ga In Tl 801 577 579 558 589 How would you explain this deviation from the general trend?

Accepted Answer

Ionisation enthalpy is expected to fall smoothly down a group because each new element adds a full shell, pushing the outer electron farther from the nucleus — but that fall only happens if the shielding from inner electrons keeps pace with the rising nuclear charge. In Ga and Tl it doesn't, because the electrons added just before them fill d and f orbitals, which shield very poorly. First write out where each element's outermost electron sits: B: [He] 2s^2 2p^1 Al: [Ne] 3s^2 3p^1 Ga: [Ar] 3d^10 4s^2 4p^1 In: [Kr] 4d^10 5s^2 5p^1 Tl: [Xe] 4f^14 5d^10 6s^2 6p^1 The value that matters for ionisation enthalpy is not the full nuclear charge Z but the effective nuclear charge Z eff = Z - σ, where σ (the shielding constant) is how much of the nuclear pull the inner electrons cancel out for the outermost electron. Down a normal group, Z rises but so does the number of complete inner shells, so σ rises almost as fast — Z eff stays roughly flat, and the outer electron, now farther away, is easier to pull off. That is why ionisation enthalpy is generally expected to decrease down a group. Now track the actual numbers, 801, 577, 579, 558, 589 (all in kJ mol^-1 ), one step at a time: - B → Al (801 → 577): a normal, large drop. Al's outermost electron is in n=3, shielded by a complete n=2 shell of s and p electrons only — good shielding, as expected. No anomaly here. - Al → Ga (577 → 579): the first break in the trend. Between Al and Ga, ten electrons have been added into the 3d subshell. d -orbitals are diffuse and penetrate poorly toward the nucleus, so a 3d electron shields the outer 4p electron much less effectively than an s or p electron would (this poor d -electron shielding is also why atomic radius barely grows from Al to Ga — the " d -block contraction"). The nuclear charge has jumped by 12 units ( Z=13 to Z=31 ) while the shielding has hardly kept pace, so Z eff actually rises. That extra pull on the 4p electron cancels out the effect of the larger shell, so instead of falling, the ionisation enthalpy stays essentially flat (even ticking up slightly). - Ga → In (579 → 558): the trend resumes as expected, since In's added shell ( n=5 ) again gives a normal-sized increase in radius without a fresh poorly-shielding subshell being filled immediately before it in this comparison. - In → Tl (558 → 589): the second, sharper break. Between In and Tl, fourteen 4f electrons (and another ten 5d electrons) have been filled in. The 4f orbitals are even more diffuse and shield even more poorly than 3d did — this is the well-known lanthanide contraction, which keeps Tl's atomic radius only marginally larger than In's despite the much heavier nucleus. With shielding badly lagging behind the huge rise in nuclear charge ( Z=49 to Z=81 ), Z eff on the outer 6p electron rises sharply, and the ionisation enthalpy goes up instead of continuing to fall. The aside worth remembering: it is tempting to think ionisation enthalpy always falls down a group "because the atom gets bigger" — but size is only half the story. What actually decides Z eff is the quality of shielding, and d and f electrons shield poorly. Whenever a d or f subshell has just been filled immediately before an element (as for Ga and Tl), expect the ionisation enthalpy to be anomalously high rather than following the naive size-based trend. Answer: The dip is not uniform — Al→Ga stays flat and In→Tl actually rises — because Ga follows a filled 3d^10 subshell and Tl follows filled 4d^105s^2 and 4f^145d^10 subshells. d and f electrons shield the nucleus poorly, so the effective nuclear charge on the outer p electron rises unusually sharply for Ga and Tl, offsetting the expected fall from increased atomic size and giving these two elements higher-than-expected first ionisation enthalpies.

Question 20

Which of the following pairs of elements would have a more negative electron gain enthalpy? (i) O or F (ii) F or Cl

Accepted Answer

Electron gain enthalpy is the energy released when one mole of gaseous atoms picks up an electron — the more negative the value, the more strongly the atom pulls that electron in. The defining step is X(g) + e^- → X^-(g), Δ egH = energy change for this step Here X is the atom, X^- the anion formed, and Δ egH (electron gain enthalpy) is negative because energy is usually released when a neutral atom traps an extra electron. A more negative Δ egH means a bigger release — a stronger pull on the incoming electron, not a weaker one. This is the step people mix up: "more negative" is not a smaller effect, it is a larger one. (i) O or F Both sit in period 2, with F one place to the right of O. Moving across a period, the nuclear charge increases by one proton at a time while the new electron goes into the same shell, so the effective nuclear charge Z eff felt at the surface rises and the atomic radius shrinks. A smaller, more highly charged atom pulls an incoming electron in more forcefully, so Δ egH becomes more negative as you go left to right across a period. F has one more proton than O and a smaller radius than O, so F's pull on an extra electron is stronger. Δ egH( O ) = -141 kJ mol^-1, Δ egH( F ) = -328 kJ mol^-1 F's value is the more negative one, exactly as the period trend predicts. (ii) F or Cl Here the naive period/group argument gives the wrong answer, and this pair is asked precisely to test whether that trap is understood. Going down a group the atom gets bigger, so you would expect the pull on an incoming electron — and hence Δ egH — to fall off. That is true from Cl downward (Cl is more negative than Br, which is more negative than I). But F breaks the pattern at the top of the group. F is exceptionally small: its 2p subshell is already tightly packed with electrons. Cramming one more electron into that small, crowded space produces strong electron–electron repulsion between the incoming electron and the electrons already there, and this repulsion works against the attraction from the nucleus, cutting down the net energy released. Cl's extra electron goes into the larger, roomier 3p subshell, where the existing electrons are more spread out and repel the newcomer far less — so the nucleus's pull wins out more completely, and more energy is released overall. Δ egH( F ) = -328 kJ mol^-1, Δ egH( Cl ) = -349 kJ mol^-1 Cl's value is the more negative one — so, despite F being the smaller atom and the more electronegative element, Cl accepts an electron more exothermically than F does. This is a genuine exception, caused by inter-electron repulsion in F's compact 2p shell, not a sign that the general period/group rules are wrong. Answer: (i) F has the more negative electron gain enthalpy — across period 2, higher effective nuclear charge and smaller radius pull the extra electron in more strongly than O does. (ii) Cl has the more negative electron gain enthalpy — F's very small, electron-crowded 2p subshell suffers strong electron–electron repulsion on adding another electron, which outweighs its smaller size and makes its energy release smaller than Cl's.

Question 21

Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.

Accepted Answer

Electron gain enthalpy turns positive the moment you're adding an electron to something that is already negatively charged. Electron gain enthalpy Δ egH is the energy change when a gaseous atom (or ion) accepts an extra electron. A negative value means energy is released — the process is favourable, because the electron is attracted into the atom. A positive value means energy must be supplied to force the electron in. Write out the two steps separately, because they are two different processes with two different amounts of charge: Step 1 — first electron gain: O(g) + e^- → O^- (g) Δ egH 1 = -141 kJ mol^-1 Here a neutral oxygen atom, with its full nuclear charge unshielded by any extra electron cloud, pulls the incoming electron in. This releases energy, so Δ egH 1 is negative (exothermic). Step 2 — second electron gain: O^- (g) + e^- → O^2- (g) Δ egH 2 = +780 kJ mol^-1 This time the electron is not approaching a neutral atom — it is approaching O^-, a species that is already negatively charged. The incoming electron and the existing negative charge on O^- repel each other electrostatically. The aside people miss: don't picture the second electron reacting with "oxygen" again — it is reacting with an anion. Once the species being added to already carries negative charge, the interaction with a further electron is repulsive, not attractive, no matter which element it is. To push the electron on against that repulsion, energy has to be put in from outside rather than being given out. That is exactly what a positive Δ egH means. So the second electron gain enthalpy of oxygen is not just "less negative" than the first — it flips sign entirely and becomes positive, while the first is negative. The overall two-step process (forming O^2- from O ) is therefore net endothermic even though it is exactly this doubly-negative oxide ion, O^2-, that is stable in ionic solids like MgO — because the lattice energy released on forming the crystal more than compensates for this endothermic electron-addition step. Answer: The second electron gain enthalpy of O is positive (unlike the first, which is negative). Adding the second electron means adding it to the already-negative O^- ion, and the incoming electron is repelled by that existing negative charge, so energy must be supplied rather than released.

Question 22

What is the basic difference between the terms electron gain enthalpy and electronegativity?

Accepted Answer

Electron gain enthalpy belongs to a lone, isolated atom; electronegativity belongs to an atom that is already bonded to another atom. These two ideas get mixed up because both describe "how much an atom wants electrons," but they answer that question in completely different situations, and one of them is a real number you can measure while the other is only a relative comparison. 1. What electron gain enthalpy is Electron gain enthalpy, Δ egH, is defined for a single, isolated, gaseous atom taking up one extra electron: X(g) + e^- → X^-(g) Here X(g) is one mole of the element as a free gas-phase atom (not bonded to anything), and Δ egH is the enthalpy change of this one reaction — energy released is written negative, energy absorbed is written positive. This is a genuine, measurable thermodynamic quantity. It has fixed units, kJ/mol, and a fixed sign convention. Chlorine, for example, has a large negative electron gain enthalpy because Cl(g) + e^- → Cl^-(g) releases a lot of energy — the atom has a strong pull for that extra electron when it is sitting alone in the gas phase. 2. What electronegativity is Electronegativity is not a reaction and it is not something you can isolate an atom to measure. It is the tendency of an atom that is already part of a molecule, in a covalent bond, to pull the shared pair of bonding electrons toward itself. Because it describes a pull within a bond rather than a single physical process, it cannot be measured directly the way an enthalpy can. It is only ever assigned as a relative number on an arbitrary scale — the common one is Pauling's scale, where fluorine is fixed at 4.0 and every other element is ranked against it. Electronegativity therefore has no units and no defined zero; it only tells you which of two bonded atoms pulls harder, not "how many kJ" of pulling is happening. 3. The basic difference, stated directly The aside people miss: it is easy to treat these as two words for the same "electron-hungriness," but they differ on every count that matters — - Scope: electron gain enthalpy is a property of one isolated gaseous atom gaining one electron; electronegativity is a property of an atom as it sits bonded inside a molecule. - Measurability: electron gain enthalpy is an experimentally measurable energy change; electronegativity cannot be measured directly and is only estimated on relative scales (Pauling, Mulliken, etc.). - Units: electron gain enthalpy is expressed in kJ/mol with a definite sign; electronegativity is a pure number with no units. - What varies: a given element has one electron gain enthalpy value regardless of what it is bonded to (since the definition is for the free atom), whereas an atom's electronegativity can shift somewhat depending on its oxidation state and the atoms it is actually bonded to, because it depends on the bonding environment. Answer: Electron gain enthalpy is the measurable enthalpy change (in kJ/mol) when an isolated gaseous atom gains an extra electron, X(g)+e^-→ X^-(g); electronegativity is the relative, unit-less tendency of an atom that is already bonded within a molecule to attract the shared pair of bonding electrons toward itself, and it can only be compared on an arbitrary scale (e.g., Pauling), not measured directly like an enthalpy.

Question 23

How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?

Accepted Answer

Electronegativity is not a fixed label stamped on an atom the way atomic number is — it depends on the hybridisation state of the atom and on what it is bonded to, so a single number cannot be "the" electronegativity of nitrogen in every compound. Electronegativity is defined as the tendency of an atom, when present in a molecule, to attract the shared pair of bonding electrons towards itself. The word "in a molecule" matters here: unlike ionisation enthalpy or electron gain enthalpy, which are properties of an isolated atom or ion, electronegativity is a property of an atom as bonded — it is not measured in isolation, and its value changes with the atom's bonding environment. Why it changes: percentage of s-character. The pull an atom exerts on a shared pair depends on how close, on average, its bonding electrons sit to the nucleus. An s orbital is closer to (and lower in energy than) a p orbital of the same shell, so a hybrid orbital with more s -character holds its electrons more tightly and pulls the bonding pair harder. The s-character of the common hybridisations is: sp (50 % s) sp^2 (33 % s) sp^3 (25 % s) So, for the same atom, electronegativity rises as the hybridisation goes sp^3 → sp^2 → sp. Applying this to nitrogen. Nitrogen does not always use the same hybrid orbital to bond: - In NH 3, nitrogen is sp^3 -hybridised (one lone pair, three bond pairs). - In pyridine or in =N- linkages, nitrogen is sp^2 -hybridised. - In a nitrile, -C≡ N, nitrogen is sp -hybridised. Because electronegativity tracks s-character, the nitrogen of a nitrile ( sp ) is more electronegative than the nitrogen of pyridine ( sp^2 ), which in turn is more electronegative than the nitrogen of ammonia ( sp^3 ) — even though it is "the same element" in each case. The value 3.0 on the Pauling scale that periodic tables quote for nitrogen is essentially an average, representative figure (close to nitrogen's more common, roughly sp^3, bonding state); it is not a universal constant that holds unchanged across every compound nitrogen forms. The aside people miss: electronegativity is often mistaken for an atomic constant like ionisation enthalpy because both are tabulated as single numbers per element. But ionisation enthalpy refers to a free gaseous atom, while electronegativity explicitly refers to an atom bonded within a molecule — so it must, and does, shift with the hybridisation and the identity of the atoms it is bonded to. Answer: The statement is not correct. Electronegativity is not an invariant atomic property; it depends on the state of hybridisation of the atom and on the nature of the other atoms bonded to it. The quoted value of 3.0 (Pauling scale) is an average value for nitrogen. In practice, nitrogen's electronegativity increases with the s-character of its hybrid orbital — it is lowest when nitrogen is sp^3 -hybridised (as in NH 3 ), higher when sp^2 -hybridised, and highest when sp -hybridised (as in a nitrile, -C≡ N ) — so a single value of 3.0 cannot hold for nitrogen in all its compounds.

Question 24

Describe the theory associated with the radius of an atom as it (a) gains an electron (b) loses an electron

Accepted Answer

Atomic size is a tug-of-war between the nuclear charge pulling electrons in and electron–electron repulsion pushing them apart — changing the electron count without changing the nuclear charge tips that balance. The nucleus of an atom has a fixed number of protons, so its nuclear charge Z does not change when the atom gains or loses an electron. What changes is the number of electrons around that same nucleus, and with it, two things: how strongly each electron is, on average, pulled inward (the effective nuclear charge, Z eff ), and how much the electrons repel each other. (a) When an atom gains an electron (forms an anion) The number of protons stays the same, but one more electron is added to the outermost shell. - The extra electron adds to the electron–electron repulsion in that shell, without any extra proton to pull it in. - This repulsion pushes the whole electron cloud outward, and it also shields the existing electrons from the nucleus a little more effectively than before. - So the effective nuclear charge felt by each electron decreases, while the number of electrons crowding the same shell increases. Both effects work in the same direction: the electron cloud expands. This is why an anion is always larger than the neutral atom it came from — for example, a fluorine atom ( 72 pm ) becomes the much bigger fluoride ion F^- ( 136 pm ), because the same 9 protons must now hold onto 10 electrons instead of 9. A common mistake here is to think the ion is bigger only because "it has one more shell" — for F → F^- there is no new shell at all; the growth comes purely from weaker per-electron pull and greater repulsion. (b) When an atom loses an electron (forms a cation) The number of protons again stays the same, but one electron is removed, usually from the outermost occupied subshell. - With one fewer electron, there is less electron–electron repulsion, so the remaining electrons are not pushed apart as much. - The same nuclear charge is now shared among fewer electrons, so the effective nuclear charge felt by each remaining electron increases — each one is held more tightly. - Very often, removing the last electron from a subshell empties that entire shell, so the outermost electrons left behind belong to a shell that is closer to the nucleus (a smaller value of the principal quantum number n ). Both effects — stronger pull per electron, and sometimes an entire shell disappearing — make the electron cloud contract. This is why a cation is always smaller than the neutral atom it came from: sodium ( 186 pm ) shrinks to Na^+ ( 95 pm ) because losing the single 3s electron leaves behind only the tightly-held n = 1, 2 shells, now pulled in by 11 protons instead of being shielded by a 3s electron. Answer: In both cases the nuclear charge is unchanged, so the size change comes entirely from the electron–electron balance: gaining an electron increases repulsion and lowers the effective nuclear charge per electron, so the electron cloud expands and the anion is larger than the atom; losing an electron reduces repulsion and raises the effective nuclear charge per electron (and can remove an entire outer shell), so the electron cloud contracts and the cation is smaller than the atom.

Question 25

Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.

Accepted Answer

Ionization enthalpy depends on the nuclear charge and the electron arrangement around the nucleus — not on the number of neutrons. Isotopes of an element are atoms with the same atomic number (same number of protons, hence the same nuclear charge Z ) and the same number of electrons, but different numbers of neutrons (different mass number A ). The first ionization enthalpy, Δ iH 1, is the energy needed for the process X(g) → X^+ (g) + e^- and its size is governed by two things: - the effective nuclear charge ( Z eff ) felt by the outermost electron — this depends on the number of protons and the shielding from the other electrons, and - the electronic configuration — how far the outermost electron is from the nucleus, and which orbital it sits in. Neither of these depends on the number of neutrons. Extra neutrons add mass to the nucleus, but they carry no charge, so they do not change Z eff, and since the number of electrons is unchanged, the electronic configuration (and hence the size and shape of the electron cloud) is also unchanged. This is the point students often trip on: it is tempting to think a "heavier" isotope should hold its electrons more tightly, but chemical/electronic properties like ionization enthalpy are controlled by charge and electron arrangement, not by mass. (A tiny mass effect on ionization energy does exist in very precise physics — because the nucleus is not infinitely heavy compared to the electron — but it is far too small to matter at this level and is not what the question is testing.) Since all isotopes of an element have identical proton number, identical electron count, and identical electronic configuration, they present the same effective nuclear charge to the electron being removed. Answer: The first ionization enthalpies of the isotopes of an element are expected to be (essentially) the same, because ionization enthalpy depends on the nuclear charge and the electronic configuration of the atom, and isotopes of the same element have identical atomic number (protons) and identical number of electrons — they differ only in the number of neutrons, which does not affect these factors.

Question 26

What are the major differences between metals and non-metals?

Accepted Answer

The core difference is electron loss versus electron gain — everything else about metals and non-metals follows from that one fact. Across a period, atomic size shrinks and the nuclear pull on the outer electrons grows stronger, which pushes ionization enthalpy up and pulls electron gain enthalpy down (more negative). Metals sit on the low-ionization-enthalpy side of this trend; non-metals sit on the high-electron-affinity side. That single positional fact drives every difference below. 1. Ionization enthalpy Δ i H Δ i H is the energy needed to remove the outermost electron from an isolated gaseous atom, M(g) → M^+(g) + e^-. Metals have low ionization enthalpy — their valence electrons are loosely held (few electrons in the outer shell, larger atomic size, more shielding), so they lose electrons easily to form cations. Non-metals have high ionization enthalpy — the outer electrons are close to a nearly-full shell and strongly attracted by the nucleus, so removing an electron costs a lot of energy. A step people skip: ionization enthalpy is about a gaseous, isolated atom — it says nothing about the solid's melting point or hardness, which is a separate (metallic-bonding) property. 2. Electron gain enthalpy Δ eg H Δ eg H is the energy released when an electron is added to a gaseous atom, M(g) + e^- → M^-(g). Non-metals have large negative (highly exothermic) electron gain enthalpy — adding an electron completes or nearly completes a stable octet, so the atom "wants" the electron and releases energy doing so. Metals have small negative or even positive electron gain enthalpy — adding an electron to an atom that would rather lose electrons is energetically unfavourable. 3. Electronegativity Electronegativity is the tendency of an atom, when bonded, to pull shared electron density toward itself. Non-metals are highly electronegative (F is the highest at 4.0 on the Pauling scale); metals are electropositive (they push electron density away rather than pulling it in). This is why, in a bond between a metal and a non-metal, the electron pair ends up almost entirely on the non-metal — the bond becomes ionic. 4. Type of ion / compound formed Because metals lose electrons and non-metals gain them, metals form cations ( Na → Na^+ + e^- ) and non-metals form anions ( Cl + e^- → Cl^- ). A metal and a non-metal together therefore typically form an ionic compound (e.g., NaCl ), held together by electrostatic attraction between the oppositely charged ions; two non-metals together form a covalent compound (e.g., Cl 2, CO 2 ) by sharing electrons, since neither side can afford to fully give up an electron. 5. Nature of oxides Metal oxides are predominantly basic (some, like Al 2O 3, are amphoteric) — they react with acids to form salt and water, e.g., MgO + 2HCl → MgCl 2 + H 2O. Non-metal oxides are predominantly acidic — they react with bases, e.g., CO 2 + 2NaOH → Na 2CO 3 + H 2O. This basic-versus-acidic split is a direct chemical consequence of electropositive versus electronegative character. 6. Physical properties Metals are typically lustrous, malleable (can be hammered into sheets), ductile (can be drawn into wires), and good conductors of heat and electricity, because their loosely held valence electrons form a delocalised "electron sea" that moves freely through the lattice. Non-metals (in the solid state) are typically brittle, non-lustrous, and poor conductors (with exceptions such as graphite), because their electrons are localised in covalent bonds rather than free to move. 7. Reducing versus oxidizing behaviour Because metals lose electrons readily, they act as reducing agents in reactions (they get oxidized while reducing something else). Because non-metals gain electrons readily, they act as oxidizing agents (they get reduced while oxidizing something else). Answer: Metals have low ionization enthalpy and are electropositive, so they lose electrons to form cations; they act as reducing agents, form basic oxides, and are typically lustrous, malleable, ductile solids that conduct heat and electricity well (via delocalised electrons). Non-metals have high ionization enthalpy and large negative electron gain enthalpy and are electronegative, so they gain electrons to form anions; they act as oxidizing agents, form acidic oxides, and are typically brittle, non-lustrous, poor conductors. Metal + non-metal combinations tend to form ionic compounds; non-metal + non-metal combinations form covalent compounds.

Question 27

Use the periodic table to answer the following questions. (a) Identify an element with five electrons in the outer subshell. (b) Identify an element that would tend to lose two electrons. (c) Identify an element that would tend to gain two electrons. (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature.

Accepted Answer

The number of electrons in an atom's outermost (valence) shell is what fixes its group, and the group is what fixes whether it gains, loses, or simply sits at a certain physical state. Work each part off the valence-shell configuration, not off memorised element names. (a) An element with five electrons in the outer shell Group 15 (the nitrogen family — N, P, As, Sb, Bi) has the general valence configuration ns^2 np^3, where n is the outermost shell number, s and p name the subshells, and the superscripts count electrons in each. Adding the superscripts, 2+3=5, so every member of this group carries five electrons in its outer shell. Nitrogen is the clearest example: Z=7, configuration 1s^2 2s^2 2p^3. Its outermost shell ( n=2 ) holds 2s^2 2p^3 = 5 electrons. (b) An element that would tend to lose two electrons This is a Group 2 element (the alkaline earth metals — Be, Mg, Ca, Sr, Ba), general configuration ns^2. An atom loses electrons when doing so gets it to the nearest noble-gas configuration cheaply — and for these elements that means shedding just the two ns^2 electrons. Take calcium: Z=20, configuration 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2. Losing the two 4s electrons leaves Ca^2+ with the argon configuration 1s^2 2s^2 2p^6 3s^2 3p^6 — a filled shell, which is why the loss happens readily (Group 2 metals have low second ionisation enthalpies relative to what a further electron loss would cost). Metals lose electrons; the mistake to avoid is reaching for a non-metal here because the question says "lose." (c) An element that would tend to gain two electrons This is a Group 16 element (the oxygen family — O, S, Se, Te), general configuration ns^2 np^4. With six electrons already in the valence shell, only two more are needed to complete the octet, so these atoms gain electrons rather than lose them — the opposite pull from part (b). Oxygen illustrates it: Z=8, configuration 1s^2 2s^2 2p^4. Gaining two electrons gives O^2- with configuration 1s^2 2s^2 2p^6, the neon configuration. The rule of thumb: elements with 1–3 valence electrons (metals, left side) lose electrons; elements with 5–7 valence electrons (non-metals, right side) gain electrons — both moving toward the nearest noble gas. (d) The group with a metal, a non-metal, a liquid, and a gas together This is Group 17, the halogens: F, Cl, Br, I, At. Down this one group, physical state changes steadily with increasing atomic size and strengthening interatomic forces: - Fluorine ( F 2 ) and chlorine ( Cl 2 ) — gases at room temperature - Bromine ( Br 2 ) — a liquid at room temperature (the only non-metal liquid halogen) - Iodine ( I 2 ) — a solid, but still a non-metal - Astatine (At) — highly radioactive and short-lived, but the element at the bottom of the group where metallic character (a general trend as you go down any group) becomes strong enough that At is regarded as having metallic properties So Group 17 alone spans gas, liquid, non-metallic solid, and metal — no other group covers all four states across its members. The aside worth flagging: it is tempting to look for this variety across a period instead of a group, because state usually feels like a period-driven property; here it is the group that supplies the full spread. Answer: (a) Nitrogen family, Group 15 (e.g., N: 2s^2 2p^3, 5 outer electrons); (b) Group 2 alkaline earth metals (e.g., Ca, config ns^2, loses 2 e⁻ to reach noble-gas configuration); (c) Group 16 oxygen family (e.g., O, config ns^2np^4, gains 2 e⁻ to complete octet); (d) Group 17, the halogens — F and Cl are gases, Br is a liquid, I is a non-metallic solid, and At shows metallic character.

Question 28

The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs whereas that among group 17 elements is F CI Br I. Explain.

Accepted Answer

Reactivity here means two different things for the two groups: group 1 atoms react by LOSING an electron, group 17 atoms react by GAINING one — so the same trend down a group (increasing atomic size) pushes them in opposite directions. Group 1: reactivity is the ease of losing the outer electron Alkali metals have the configuration ns^1. To react, an atom must give up this lone outer electron to form M^+: M(g) → M^+(g) + e^- The metal is acting as a reducing agent, and how easily it does this is measured by the ionization enthalpy, Δ i H — the energy needed to pull the outermost electron off a gaseous atom. A low ionization enthalpy means the electron leaves easily, so the metal is more reactive. Going down the group, Li → Na → K → Rb → Cs, a new shell is added each time, so: - Atomic radius increases sharply. - The valence electron sits farther from the nucleus and is screened by more inner-shell electrons, so it feels almost the same effective pull even though the actual nuclear charge has grown. A far-away, well-shielded electron is held less tightly, so Δ i H falls steadily down the group. Losing the electron becomes progressively easier, so reactivity increases: Li < Na < K < Rb < Cs. Group 17: reactivity is the ease of gaining an electron Halogens have the configuration ns^2 np^5 — one electron short of a stable octet. To react, an atom must pick up an electron to form X^-: X(g) + e^- → X^-(g) Here the atom acts as an oxidizing agent, and the relevant quantity is the electron gain enthalpy, Δ egH — the energy released when a gaseous atom accepts an electron. A more negative Δ egH means the electron is pulled in more readily, so the atom is more reactive. This is the step people mix up with group 1: for group 1 you want ionization enthalpy to go DOWN for reactivity to go up; for group 17 you want electron gain enthalpy to be MORE negative for reactivity to go up — opposite requirements, from opposite processes. Going down group 17, F → Cl → Br → I, atomic size again increases and the extra inner shells shield the nucleus, so an incoming electron is attracted less strongly even though the nuclear charge is bigger. Electron gain becomes less exothermic (less favourable), so accepting an electron gets progressively harder, and reactivity decreases: F Cl Br I. A short aside on fluorine. By electron gain enthalpy alone, chlorine actually attracts an extra electron slightly more strongly than fluorine does — fluorine's 2p subshell is so small that the incoming electron suffers noticeable repulsion from the electrons already there, making Δ egH for F slightly less negative than for Cl. Yet fluorine is still the most reactive halogen overall, because reactivity is decided by the whole reaction, not electron gain enthalpy alone: the F–F bond is unusually weak (easy to break) and the F^- ion is very strongly hydrated (releases a lot of energy in solution). Both effects favour fluorine so strongly that it wins out despite the electron-gain-enthalpy hiccup. Answer: Group 1 metals react by losing their one valence electron; down the group (Li → Cs) atomic size and shielding increase, so ionization enthalpy falls and the electron is lost more easily — reactivity rises: Li < Na < K < Rb < Cs. Group 17 non-metals react by gaining one electron; down the group (F → I) atomic size increases in the same way, so the nucleus attracts an incoming electron less strongly, electron gain enthalpy becomes less favourable, and the electron is gained less easily — reactivity falls: F Cl Br I. The two orders run opposite ways because "reactive" means losing an electron in one group and gaining one in the other, and increasing atomic size makes the first easier but the second harder.

Question 29

Write the general outer electronic configuration of s-, p-, d- and f- block elements.

Accepted Answer

The block tells you which subshell is being filled last — s, p, d, or f — and "outer" configuration means you write only that subshell plus the ones already complete below it, using n for the outermost shell. Recall how the periodic table's four blocks are defined: each block is named for the subshell that receives the last electron as you build up the atom (the Aufbau order). So to write a "general" configuration for a block, ask: which subshell is filling, and how many electrons can it hold? s-block The last electron enters the ns subshell, and an s subshell holds at most 2 electrons. These are the alkali metals (Group 1, one outer electron) and alkaline earth metals (Group 2, two outer electrons), so the outer shell can have 1 or 2 electrons in it: ns^1-2 Here n is the outermost (valence) shell number, and the superscript range 1 – 2 means "1 for Group 1, 2 for Group 2" — it is not one atom having both at once. p-block The last electron enters the np subshell, but the ns subshell below it is already full (2 electrons) because s always fills before p in the same shell. A p subshell holds at most 6 electrons, giving Groups 13–18: ns^2 np^1-6 The aside to catch here: helium ( 1s^2 ) looks s-block by its configuration, but it is placed with the p-block (Group 18) because it behaves as a noble gas — configuration alone doesn't always fix the block placement for helium. d-block (transition elements) Here the electron being added last goes into the (n-1)d subshell — one shell inside the outermost one — while the outermost ns subshell was already filled at the previous element. A d subshell holds up to 10 electrons, and ns can be 2 (the normal case) or drop to 0 or 1 for a few exceptions (like Cr, Cu) where a half-filled or fully-filled d subshell is extra stable and pulls an electron out of ns: (n-1)d^1-10 ns^0-2 The step people get wrong: the d electrons are one shell number behind the outermost shell — it is (n-1)d, not nd — even though this is called the outermost/valence region of a transition metal. f-block (inner transition elements) The last electron goes into the (n-2)f subshell — two shells inside the outermost one — while (n-1)d and ns are already occupied from the elements before it. An f subshell holds up to 14 electrons: (n-2)f^1-14 (n-1)d^0-1 ns^2 Again the subshell being filled sits two shells behind the label n; this is what makes the lanthanoids and actinoids so similar to each other across a row — the differentiating electron is buried two shells deep and barely affects chemical behaviour, which depends mainly on the outer ns^2. Answer: s-block: ns^1-2; p-block: ns^2 np^1-6; d-block: (n-1)d^1-10 ns^0-2; f-block: (n-2)f^1-14 (n-1)d^0-1 ns^2, where n is the number of the outermost shell.

Question 30

Assign the position of the element having outer electronic configuration (i) ns2np4 for n=3 (ii) (n-1)d2ns2 for n=4, and (iii) (n-2) f 7 (n-1)d1ns2 for n=6, in the periodic table.

Accepted Answer

The period number is just the value of n in the given configuration (the outermost shell being filled); the group number is worked out by a different counting rule for each block, and the whole trap in this question is applying the wrong block's rule. Before assigning positions, fix the four rules: - s-block — valence configuration ns^1 or ns^2: group number = number of electrons in ns (so group 1 or group 2). - p-block — valence configuration ns^2np^1-6: group number = 10 + ( electrons in ns + np). - d-block — valence configuration (n-1)d^1-10ns^0-2: group number = ( electrons in (n-1)d) + ( electrons in ns). - f-block — valence configuration (n-2)f^1-14(n-1)d^0-1ns^2: these 14 elements in each of periods 6 and 7 don't get an individual group number from 4 to 17 — every one of them is parked in the group 3 slot of the main table (lanthanoids under period 6, actinoids under period 7), because that slot is where the f-orbitals of that shell start filling, right after La or Ac. In every case, the period number equals the highest n appearing in the configuration. (i) ns^2np^4 for n=3: this is 3s^23p^4. Writing out the full configuration, 1s^22s^22p^63s^23p^4, gives 16 electrons, i.e. sulfur. Period = n = 3. This is a p -block configuration, so group = 10 + (2+4) = 16. Position: Period 3, Group 16 (the oxygen family) — the element is sulfur, S. (ii) (n-1)d^2ns^2 for n=4: this is 3d^24s^2. Full configuration [ Ar ]3d^24s^2, Z = 22, titanium. Period = n = 4. This is a d -block configuration, so group = ( electrons in 3d) + ( electrons in 4s) = 2 + 2 = 4. This is the step people get wrong here — it is tempting to reuse the p-block "+10" rule on a d-block count and get group 16 by mistake. The +10 rule only applies once the p-subshell is the one being filled; for a d-block element you just add up the (n-1)d and ns electrons directly, with no offset. Position: Period 4, Group 4 — the element is titanium, Ti, in the first transition series. (iii) (n-2)f^7(n-1)d^1ns^2 for n=6: this is 4f^75d^16s^2. Full configuration [ Xe ]4f^75d^16s^2, Z = 64, gadolinium. Period = n = 6. This is an f -block configuration (the 4f subshell is occupied), so it belongs to the lanthanoid series and, by the rule above, sits in the group 3 slot — not group 17 or any number obtained by counting 7+1+2 electrons and adding 10. The f-block elements of a given period all share that one group-3 position in the main table; what distinguishes them from each other is only the period (6 for lanthanoids, 7 for actinoids) and their place along the separately-drawn f-block row, not a unique group number. Position: Period 6, Group 3, f-block (lanthanoid series) — the element is gadolinium, Gd. Answer: (i) 3s^23p^4 is period 3, group 16 (sulfur, S). (ii) 3d^24s^2 is period 4, group 4 (titanium, Ti). (iii) 4f^75d^16s^2 is period 6, group 3, in the f-block lanthanoid series (gadolinium, Gd).

Question 31

The first (∆iH1) and the second (∆iH2) ionization enthalpies (in kJ mol–1) and the (∆egH) electron gain enthalpy (in kJ mol–1) of a few elements are given below: Elements ∆ H 1 ∆ H 2 ∆egH I 520 7300 –60 II 419 3051 –48 III 1681 3374 –328 IV 1008 1846 –295 V 2372 5251 +48 VI 738 1451 –40 Which of the above elements is likely to be: (a) the least reactive element. (b) the most reactive metal. (c) the most reactive non-metal. (d) the least reactive non-metal. (e) the metal which can form a stable binary halide of the formula MX2(X=halogen). (f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?

Accepted Answer

A big first ionization enthalpy tells you an atom hates losing an electron; a very negative electron gain enthalpy tells you it loves gaining one — read the table as a story about each atom's electron count, not as six unrelated numbers. Here Δ iH 1 is the energy needed to pull the first electron off a neutral gaseous atom, Δ iH 2 is the energy needed to pull a second electron off the resulting +1 ion, and Δ egH is the energy change when the neutral atom gains one electron (negative = energy is released = the atom "wants" the electron; positive = energy must be put in = the atom resists). Step 1 — sort metals from non-metals using Δ iH 1. Metals give up an electron easily, so they have low Δ iH 1; non-metals hold on tightly, so they have high Δ iH 1. - I: 520, II: 419, VI: 738 — all comfortably low → these three are metals. - III: 1681, IV: 1008, V: 2372 — all high → these three are non-metals. Step 2 — the size of the jump Δ iH 2/Δ iH 1 tells you how many electrons come off easily. A big jump means: once the first electron is gone, the ion has a full (noble-gas-like) shell, so removing a second electron is far harder — that is alkali-metal (Group 1) behaviour, forming M^+. A small, comparable jump means both electrons come off with similar difficulty — that is alkaline-earth (Group 2) behaviour, forming M^2+. (Δ iH 2)/(Δ iH 1): I = 7300/520≈ 14, II = 3051/419≈ 7.3, VI = 1451/738≈ 2.0 I and II show a huge jump (≈14× and ≈7×) — both are Group 1 metals, each stopping at M^+. VI shows only a ~2× jump — both electrons are lost with similar (moderate) ease, the classic Group 2 signature, stopping at M^2+. Step 3 — among the non-metals, the sign and size of Δ egH separates "wants an electron" from "already has enough." - V has the highest Δ iH 1 of the whole table (2372, by far the hardest atom to ionize) and a positive Δ egH = +48 — it refuses to lose an electron and refuses to accept one. That is a filled, stable shell: a noble gas. Nothing pushes it to react in either direction, so V is the least reactive element overall — part (a). - III ( Δ egH = -328 ) and IV ( Δ egH = -295 ) are both large negative values — both release energy on gaining an electron, so both are reactive non-metals (halogen-type). III releases the most energy of any element in the table when it gains an electron, so it has the strongest pull on an extra electron: III is the most reactive non-metal — part (c). IV releases less energy than III (though still a genuine non-metal, unlike noble-gas V), so its pull on an extra electron is comparatively weak: IV is the least reactive non-metal — part (d). (A slip people make here: don't count V as "a non-metal that just doesn't react" — a positive Δ egH is a completely different chemical story, filled shell, from a small negative one, which is still an atom short of a filled shell.) Step 4 — among the metals, lower Δ iH 1 means a bigger atom holding its outer electron more loosely, i.e. a more reactive metal. I (520) and II (419) are both Group-1-type metals from Step 2, but II needs less energy to lose its electron than I does. Down a group, atomic radius increases and the outer electron sits farther from the nucleus, so ionization enthalpy falls — II is therefore the larger, more metallic, more reactive atom: II is the most reactive metal — part (b). Step 5 — which metal forms the ionic MX 2, and which forms the covalent MX? VI is the Group-2-type metal from Step 2 (loses two electrons with comparable, moderate energy cost each time), so it forms a stable M^2+ ion and an ordinary ionic halide of formula MX 2: VI is the metal forming the stable binary halide MX 2 — part (e). That leaves I and II, the two Group-1-type metals, to explain part (f). Both stop at M^+ and would normally give an ionic MX salt — but I has the higher Δ iH 1 of the two (520 vs. 419), and within the same group ionization enthalpy is higher precisely when the atom (and hence the resulting cation) is smaller. A small, highly charged cation has a high charge density and strongly distorts (polarizes) the electron cloud of the halide ion sitting next to it — Fajans' rule — and enough polarization turns what would be an ionic bond into one with substantial covalent character. Because I is the smaller of the two M^+ -forming metals, its halide MX is the one with dominant covalent character (this is exactly why, for instance, lithium halides behave more like covalent compounds — soluble in organic solvents — while the halides of bigger alkali metals stay firmly ionic): I is the metal forming the predominantly covalent halide MX — part (f). (As a check, these six rows are in fact the standard data-book values for Li, K, F, I, He and Mg respectively — I = Li, II = K, III = F, IV = I, V = He, VI = Mg — which is exactly the classification the reasoning above arrives at independently.) Answer: (a) V — highest ionization enthalpy and a positive electron gain enthalpy mark a noble gas, the least reactive element. (b) II — the lowest ionization enthalpy among the Group-1-type metals, so the most reactive metal. (c) III — the most negative electron gain enthalpy, so the strongest pull on an extra electron, the most reactive non-metal. (d) IV — still a genuine non-metal (negative electron gain enthalpy) but with a smaller pull on an electron than III, so the least reactive non-metal. (e) VI — the Group-2-type metal (comparable, moderate first and second ionization enthalpies) forms the stable ionic halide MX 2. (f) I — the smaller of the two Group-1-type metals (higher Δ iH 1 than II), whose small, more polarizing M^+ ion gives its halide MX predominantly covalent character.

Question 32

Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements. (a) Lithium and oxygen (b) Magnesium and nitrogen (c) Aluminium and iodine (d) Silicon and oxygen (e) Phosphorus and fluorine (f) Element 71 and fluorine

Accepted Answer

The formula of a binary compound is fixed by one rule: the total positive charge on the metal atoms must exactly cancel the total negative charge on the non‑metal atoms. To use that rule you need, for each element, the charge it settles into when it empties or fills its outer shell. For a representative (main‑group) element that charge is read straight off the periodic table, using the old style group labels IA–VIIA where the group number itself equals the number of valence electrons: - Group 1 (IA): 1 valence electron lost → charge +1 - Group 2 (IIA): 2 lost → +2 - Group 13 (IIIA): 3 lost → +3 - Group 14 (IVA): 4 valence electrons, usually lost toward a strongly electronegative partner → +4 - Group 15 (VA): needs 3 more to complete the octet → -3 - Group 16 (VIA): needs 2 more → -2 - Group 17 (VIIA): needs 1 more → -1 A short aside on the mistake people make here: if you instead use the modern 1–18 numbering, group 13 is not "charge +13." For p‑block groups you must subtract 10 (or count valence electrons directly) to get the real charge — group 13 gives +3, not +13. Once both charges are known, formulas are built by the criss‑cross rule: the number of one atom in the formula equals the (sign‑dropped) charge of the other atom, and the result is then reduced to the smallest whole‑number ratio. (a) Lithium and oxygen. Li is group 1, so Li^+. O is group 16, so O^2-. Criss‑crossing the charges gives Li 2 O 1, written Li 2 O. (b) Magnesium and nitrogen. Mg is group 2, so Mg^2+. N is group 15, so N^3-. Criss‑crossing gives Mg 3 N 2 — already in lowest terms, since 3 and 2 share no common factor. (c) Aluminium and iodine. Al is group 13, so Al^3+. I is group 17, so I^-. Criss‑crossing gives AlI 3. (d) Silicon and oxygen. Si is group 14. Against a highly electronegative partner like oxygen, silicon loses all four valence electrons rather than gaining four, giving Si^4+. With O^2-, criss‑crossing gives Si 2 O 4, which reduces to SiO 2. (e) Phosphorus and fluorine — the step where a plain group-number reading goes wrong. Reading straight off group 15 would suggest phosphorus only ever takes the -3 / +3 route, giving PF 3. But phosphorus sits in period 3, so it has empty 3d orbitals available and is not boxed in by the strict octet the way nitrogen (period 2, no accessible d‑orbitals) is. That lets phosphorus reach its full group oxidation state of +5. Fluorine is exactly the partner that pushes it there: it is the smallest, most electronegative halogen, small enough to pack five atoms around one phosphorus and electronegative enough to pull out and stabilize that higher +5 state. (This is also why the parallel compound with the bulkier iodine stops at PI 3.) So the stable compound formed is PF 5, phosphorus pentafluoride, not PF 3. (f) Element 71 and fluorine. Element 71 is lutetium, Lu, with configuration [ Xe ] 4f^14 5d^1 6s^2. Its 14 f‑electrons form a filled, chemically inert 4f^14 core, so lutetium behaves like a group‑3 element: it loses only the two 6s electrons and the one 5d electron, giving the single stable oxidation state +3, i.e. Lu^3+. Criss‑crossing with F^- gives LuF 3. Answer: (a) Li 2 O (b) Mg 3 N 2 (c) AlI 3 (d) SiO 2 (e) PF 5 (f) LuF 3

Question 33

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

Accepted Answer

A period is a horizontal row of the periodic table, and moving across it means adding electrons to the same shell — the same value of n. The modern periodic table is arranged on the basis of electronic configuration, following the Aufbau principle (electrons fill orbitals in order of increasing energy) and building on Bohr's idea of shells labelled by the principal quantum number n = 1, 2, 3, …. Go through what the row number actually tracks: - Period 1 (H, He): outermost electron in n = 1. - Period 2 (Li to Ne): outermost electron enters n = 2. - Period 3 (Na to Ar): outermost electron enters n = 3. In each case, the period number is the value of n for the outermost (valence) shell. That is the entire logic of why a new row starts: a new principal shell has begun. Now check why the other options fail: - Atomic number (option a) increases by exactly 1 from element to element, but it does not repeat or reset at the start of a period — it keeps climbing straight through Ne to Na to Mg. It is the quantity that orders elements within the whole table, not what defines a period. - Atomic mass (option b) was Mendeleev's older ordering criterion, abandoned because it caused reversals (e.g. Co before Ni, though Co is heavier). It plays no role in the modern table's period structure. - Azimuthal quantum number l (option d) fixes the subshell type (s, p, d, f) within a shell and determines the block a group of elements falls in, not the row number. A common trap here is confusing "block" (decided by l, i.e., which subshell is being filled) with "period" (decided by n, i.e., which shell is being filled) — they answer different questions. So the row (period) is fixed by the principal quantum number of the outermost shell, while the block (s, p, d, f) is fixed by the azimuthal quantum number of the subshell being filled. Answer: (c) principal quantum number — the period number equals the principal quantum number n of the outermost shell being filled.

Question 34

Which of the following statements related to the modern periodic table is incorrect? (a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell. (b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell. (c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell. (d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.

Accepted Answer

A subshell's width is fixed by 2(2l+1): count the orbitals, double it for spin — that number is also how many columns wide its block is. Here l is the azimuthal (subshell) quantum number, the number of orbitals in a subshell is (2l+1), and each orbital holds at most 2 electrons (opposite spins, Pauli exclusion) — so the maximum electron capacity of a subshell is 2(2l+1) Apply this subshell by subshell, since each period is built by filling one subshell across the periodic table: - s-subshell: l = 0, orbitals = 2(0)+1 = 1, capacity = 2(1) = 2 electrons ⇒ the s-block is 2 columns wide (groups 1–2). - p-subshell: l = 1, orbitals = 2(1)+1 = 3, capacity = 2(3) = 6 electrons ⇒ the p-block is 6 columns wide (groups 13–18). This matches statement (a) — correct as written. - d-subshell: l = 2, orbitals = 2(2)+1 = 5, capacity = 2(5) = 10 electrons ⇒ the d-block is 10 columns wide (groups 3–12), not 8. - f-subshell: l = 3, orbitals = 2(3)+1 = 7, capacity = 2(7) = 14 electrons ⇒ an f-row (lanthanides or actinides) is 14 elements wide. The trap in statement (b) is treating "8" as some kind of natural cap on a subshell (perhaps borrowed from the octet rule for s+p electrons together). A d-subshell has 5 orbitals, not 4, so its true capacity is 2 × 5 = 10, and the d-block correspondingly spans 10 columns, not 8. So (b) is wrong on both the stated number of columns and the stated reason. Checking the other statements against this same rule: - (c) "Each block contains a number of columns equal to the number of electrons that can occupy that subshell" — this is exactly the pattern above (2, 6, 10, 14 for s, p, d, f), so (c) is correct. - (d) "The block indicates the value of l for the last subshell that received electrons" — s-block → l=0, p-block → l=1, d-block → l=2, f-block → l=3, by definition of how blocks are named. So (d) is correct. Only (b) fails the test, since the d-subshell's real capacity is 10 electrons (5 orbitals × 2), which makes the d-block 10 columns wide (groups 3–12), not 8. Answer: Statement (b) is incorrect — a d-subshell has 5 orbitals, so its maximum capacity is 2(2·2+1) = 10 electrons, not 8, and the d-block is accordingly 10 columns wide (groups 3–12).

Question 35

Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell? (a) Valence principal quantum number (n) (b) Nuclear charge (Z ) (c) Nuclear mass (d) Number of core electrons.

Accepted Answer

Chemistry depends on how strongly the nucleus holds the valence electrons — and nuclear mass has nothing to do with that pull. Go through each factor and ask: does it change the force the valence electrons feel, or the shell they sit in? (a) Valence principal quantum number n This number fixes which shell is the valence shell. A larger n means the valence electrons are farther from the nucleus and more loosely held (lower ionisation enthalpy, larger radius). Changing n changes the whole chemistry of the element — think of the jump from Li ( n=2 ) to Na ( n=3 ): very different reactivity. So n does affect the valence shell. (b) Nuclear charge Z Z is the number of protons, i.e. the raw positive charge pulling on every electron. A bigger Z pulls the valence shell in more tightly (this is exactly why atomic radius shrinks and ionisation enthalpy rises across a period). So Z does affect the valence shell. (d) Number of core electrons The core (inner-shell) electrons sit between the nucleus and the valence shell and screen — partly cancel — the nuclear charge felt by the valence electrons. This screening is captured in the idea of effective nuclear charge, Z eff = Z - σ, where Z is the actual nuclear charge and σ is the shielding contributed by the core electrons. More core electrons means more shielding, so the valence electrons feel a smaller pull. So the number of core electrons does affect the valence shell. (c) Nuclear mass Nuclear mass comes from the number of protons and neutrons. Neutrons carry no charge, so adding or removing neutrons changes the mass number but leaves Z, the electron configuration, and the shielding σ completely unchanged. This is exactly why isotopes of an element — same Z, different mass — are chemically identical: ^1 1 H, ^2 1 H (deuterium), and ^3 1 H (tritium) all react the same way, because chemistry runs on electron behaviour, not on how heavy the nucleus is. So nuclear mass is the one factor that does not touch the valence shell. The aside people miss: it is easy to lump "nuclear charge" and "nuclear mass" together as one "nucleus" property, but for chemistry only the charge (and its shielding) matters — the mass is irrelevant to how electrons are held. Answer: (c) Nuclear mass does not affect the valence shell — chemical behaviour depends on the principal quantum number of the valence shell, the nuclear charge, and the shielding by core electrons (i.e., Z eff = Z - σ ), none of which involve the mass of the nucleus. This is also why isotopes of an element, which differ only in nuclear mass, show identical chemical properties.

Question 36

The size of isoelectronic species — F^-, Ne and Na^+ is affected by (a) nuclear charge (Z ) (b) valence principal quantum number (n) (c) electron-electron interaction in the outer orbitals (d) none of the factors because their size is the same.

Accepted Answer

Isoelectronic species have the same number of electrons, so size differences among them come down to how hard the nucleus is pulling on that fixed electron count. Check what F⁻, Ne, and Na⁺ have in common first, because that tells you which options can immediately be ruled out. Each species has 10 electrons: F^-: 9 + 1 = 10, Ne: 10, Na^+: 11 - 1 = 10 In every case those 10 electrons are arranged as 1s^2 2s^2 2p^6 — the electron configuration of neon. So the outermost (valence) shell is n = 2 for all three species. Now go through the options with that fact in hand. (b) Valence principal quantum number — this is the same, n = 2, for F⁻, Ne, and Na⁺. Since it does not change across the series, it cannot be the factor causing their sizes to differ. Ruled out. (c) Electron–electron interaction in the outer orbitals — with the same number of electrons (10) sitting in the same set of orbitals ( 2s, 2p ) in all three species, the mutual repulsion among electrons in the outer shell is essentially unchanged across the series. This is the detail people get wrong: it is tempting to think more shielding or repulsion is doing the work, but the electron count and their orbital arrangement are identical here, so this interaction stays constant and cannot explain a size difference. Ruled out. (d) None of the factors, because their size is the same — this is false as a matter of experimental fact. Isoelectronic species do not have equal radii; they shrink steadily as you add protons. Ruled out. (a) Nuclear charge (Z) — this is the one quantity that actually changes across the series: Z( F ) = 9, Z( Ne ) = 10, Z( Na ) = 11 With the number of electrons held fixed at 10, a larger nuclear charge pulls that same electron cloud in more strongly (each electron feels a greater effective nuclear charge, since shielding by the other 9 electrons is nearly identical across the series). More pull on the same number of electrons means a smaller radius. So the radii fall in the order F^- Ne Na^+ matching the increase in Z exactly. This is the only factor among the four choices that varies from species to species, and it is the one that correctly predicts the observed shrinking trend. Answer: (a) nuclear charge (Z). All three species have the same 10 electrons in the same n=2 shell, so options (b) and (c) are constant across the series and cannot be the cause; size is not equal, ruling out (d). What changes is Z (9 for F⁻, 10 for Ne, 11 for Na⁺), and the greater pull on the same electron count is exactly why the radius decreases along F⁻ Ne Na⁺.

Question 37

Which one of the following statements is incorrect in relation to ionization enthalpy? (a) Ionization enthalpy increases for each successive electron. (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration. (c) End of valence electrons is marked by a big jump in ionization enthalpy. (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

Accepted Answer

Successive ionization enthalpies always rise, and they jump hardest right when the core is reached — but "lower shell electrons are easier to remove" gets the shielding argument backwards. Go through each statement using what governs ionization enthalpy: nuclear charge felt by the electron being removed, and how many shells sit between that electron and the nucleus. (a) Ionization enthalpy increases for each successive electron. Once one electron is removed, the atom becomes a cation. The remaining electrons feel a larger effective nuclear charge per electron (same nuclear charge Z, fewer electrons to share the pull and to shield each other), so each further electron is held more tightly. Δ i H 1 < Δ i H 2 < Δ i H 3 < … always. This statement is correct. (b) The greatest increase in ionization enthalpy is on removing an electron from the core noble-gas configuration. As successive electrons are stripped, you first remove all the valence-shell electrons — these are shielded by the inner core and come off with gradually increasing (but comparable) energy. The moment you try to remove the next electron, you are pulling it from a completed, noble-gas-like inner shell that sits much closer to the nucleus and is far less shielded. That jump in energy is by far the largest of the whole sequence. This statement is correct. (c) The end of the valence electrons is marked by a big jump in ionization enthalpy. This is the same fact as (b), stated the other way round: the discontinuity in the successive- Δ i H values is exactly what tells you how many valence electrons the atom had (for example, sodium shows a huge jump between Δ i H 1 and Δ i H 2, confirming one valence electron). This statement is correct. (d) Removal of an electron from an orbital with lower n is easier than from one with higher n. Here is the aside worth stopping on: this is the statement that reverses cause and effect. An orbital with a lower principal quantum number n is, on average, closer to the nucleus and has fewer inner shells shielding it from the nuclear charge Z. It therefore feels a larger effective nuclear charge, Z eff = Z - σ (where σ is the shielding constant contributed by electrons inside that orbital), and is held more tightly. An electron in a higher- n orbital is farther out, more shielded, and comes off more easily. So it is the higher- n electron that is easier to remove, not the lower- n one — statement (d) says the opposite of what actually happens. This statement is incorrect. Since (a), (b), and (c) all correctly describe how ionization enthalpy behaves, and (d) inverts the true relationship between n and ease of removal, (d) is the statement that does not hold. Answer: (d) is incorrect — electrons in orbitals of lower n are held closer to the nucleus with less shielding, so they are harder (not easier) to remove than electrons in orbitals of higher n.

Question 38

Considering the elements B, Al, Mg, and K, the correct order of their metallic character is: (a) B Al Mg K (b) Al Mg B K (c) Mg Al K B (d) K Mg Al B

Accepted Answer

Metallic character depends on how easily an atom loses its valence electron — it rises as you go down a group and falls as you go across a period, left to right. Losing an electron becomes easier when the electron is farther from the nucleus (larger atom, more shielding) and harder when the nuclear charge pulling on it is stronger. That is exactly what happens down a group and across a period, so the two trends drive metallic character in opposite ways depending on which direction you move. Locate each element first. - K: Period 4, Group 1 - Mg: Period 3, Group 2 - Al: Period 3, Group 13 - B: Period 2, Group 13 Step 1 — compare Mg and Al (same period). Both are in Period 3. Moving left to right across a period, nuclear charge increases while shielding stays roughly the same, so the outer electron is held more tightly and metallic character falls. Mg (Group 2) sits to the left of Al (Group 13), so Mg Al Step 2 — compare Al and B (same group). Both are in Group 13. Moving down a group, each successive element adds a new shell, so the valence electron sits farther from the nucleus and is more shielded — it comes off more easily, so metallic character rises. Al (Period 3) is below B (Period 2), so Al B This is the step people get backwards: it is tempting to think a "bigger" nucleus (more protons) always holds electrons tighter, but within a group the extra shells outweigh the extra nuclear charge, so the lower element is the more metallic one, not the less metallic one. Step 3 — place K. K is an alkali metal, Group 1, and moreover it sits in Period 4 — one period below Mg and Al, and two below B. Group 1 elements are already the most metallic in any period they belong to (least nuclear pull relative to shielding among that period's elements), and going down a period only adds to that. So K outranks all three of Mg, Al, and B. Putting the three comparisons together: K Mg Al B This is exactly the trend you'd read straight off the periodic table: K is far down-left (most metallic corner), B is far up-right relative to the other three (least metallic among them), and Mg, Al fall in between in period-3 order. Answer: (d) K Mg Al B — metallic character rises down a group and falls across a period, so K (Group 1, Period 4) is the most metallic and B (Group 13, Period 2) is the least; within Period 3, Mg outranks Al because Al sits farther right.

Question 39

Considering the elements B, C, N, F, and Si, the correct order of their non-metallic character is: (a) B C Si N F (b) Si C B N F (c) F N C B Si (d) F N C Si B

Accepted Answer

Non-metallic character means how strongly an atom pulls on electrons — and the number that measures this directly is electronegativity, not just a mental picture of "left-right, top-bottom." Start with the two standard periodic trends: - Across a period (left to right): non-metallic character increases. Effective nuclear charge rises while atomic radius shrinks, so the nucleus holds electrons — and pulls in new ones — more tightly. - Down a group (top to bottom): non-metallic character decreases. Atomic radius grows and extra shells shield the nucleus, so the outer electrons are held (and new ones attracted) more weakly — the element behaves more like a metal. Sorting B, C, N, F first. All four sit in Period 2, in increasing order of atomic number: B (Z=5), C (Z=6), N (Z=7), F (Z=9). Moving left to right across this period, non-metallic character only goes up, so B < C < N < F. Placing Si. Silicon sits directly below carbon, in the same group (Group 14: C, Si, Ge, Sn, Pb). Going down that group, non-metallic character falls, so Si < C. The step everyone trips on: where does Si sit relative to B? Si and B are neither in the same period nor the same group — B is Period 2/Group 13, Si is Period 3/Group 14. The two trends above are each defined within a period or within a group; they simply don't tell you how to compare two elements that share neither. You cannot "eyeball" this one — you need the actual electronegativity values. On the Pauling scale (the standard periodic-properties table for this chapter): F = 4.0, N = 3.0, C = 2.5, B = 2.0, Si = 1.8. Reading these off from highest (most non-metallic) to lowest: F (4.0) N (3.0) C (2.5) B (2.0) Si (1.8). This is consistent with everything derived from trends alone (F N C, and C Si), and it resolves the one comparison the trends couldn't: boron, sitting higher up the table with a smaller atomic radius, holds its electrons slightly more tightly than silicon does, so B Si in non-metallic character — even though Si often "looks" similar to B because both are metalloids. Similarity in chemical behaviour (a diagonal relationship) is not the same as being equal in electronegativity. So the full order is F N C B Si. Answer: (c) F N C B Si

Question 40

Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is: (a) F Cl O N (b) F O Cl N (c) Cl F O N (d) O F N Cl

Accepted Answer

Oxidizing power means how strongly an atom pulls in an extra electron and gets reduced itself — and that strength does not simply track "smaller atom, more to the right." It has to be built up piece by piece from electronegativity trends plus one exception that trips up nitrogen. Step 1 — Compare N, O, F (same period, Period 2). Moving left to right across a period, the nuclear charge increases while the electrons added go into the same shell (same shielding), so the pull on any extra electron gets stronger. Electronegativity therefore rises steadily: N < O < F So among these three, fluorine is the strongest oxidizer and nitrogen the weakest — no surprise yet. Step 2 — Bring in chlorine (same group as F, one period down). Down a group, an extra shell of electrons is added, atomic radius grows, and the extra shielding weakens the nucleus's grip on an incoming electron. So chlorine's electronegativity is well below fluorine's: Cl < F On the Pauling scale the numbers run F ≈ 4.0, O ≈ 3.5, Cl ≈ 3.0, N ≈ 3.0 — Cl and N look almost tied on this number alone. Deciding their order needs a second idea, not just the electronegativity value. Step 3 — Why nitrogen falls behind chlorine even though it's the smaller, more "electronegative-looking" atom. Electronegativity is a tendency; what actually decides oxidizing power is whether the atom accepts the extra electron easily (its electron gain enthalpy), and here nitrogen has an unusual problem. Its ground-state configuration is N: [He] 2s^2 2p^3 a subshell that is exactly half-filled — a genuinely stable arrangement (each of the three 2p orbitals holds one electron, no pairing, minimum repulsion). Forcing one more electron in destroys that stability, so nitrogen's electron gain enthalpy is close to zero (even slightly unfavourable), far less negative than a typical Period-2 non-metal in that position. On top of that, elemental nitrogen exists as N 2, held together by a very strong N ≡ N triple bond (about 941 kJ/mol) that must first be broken before nitrogen can act as an oxidizing agent at all. Both effects make N 2 a poor oxidizer in practice — weaker than chlorine, even though chlorine sits a full period further down the table. Fluorine, by contrast, has the opposite bonus: the F-F bond in F 2 is unusually weak (the two small F atoms crowd their lone pairs together), so F 2 breaks apart easily and reacts with almost everything — reinforcing, not fighting, its already-highest electronegativity. Step 4 — Put it together. F O Cl N Fluorine is the strongest oxidizer (highest electronegativity, weak F–F bond to boot); oxygen is next (high electronegativity, strong electron affinity); chlorine, though a period lower, still readily accepts an electron and outperforms nitrogen; nitrogen, despite its position in Period 2, is held back by its stable half-filled 2p^3 configuration and the tough-to-break N ≡ N bond. This matches option (b). Answer: (b) F O Cl N

Exercises 3.1–3.10 (part 1 of 4)

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Exercise 3.5

Exercise 3.6

Exercise 3.7

Exercise 3.8

Exercise 3.9

Exercise 3.10