NCERT Solutions · Class 11 Chemistry Classification of Elements and Periodicity in Properties

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 themCount the electrons on each ion by taking the atom's atomic number \(\displaystyle Z\) (its proton count, unchanged by ionisation) and adding electrons for a negative charge or removing them for a positive charge.\[\begin{aligned} \text{N}^{3-} &: Z=7,\ \ 7+3=10\ \text{electrons}\\ \text{O}^{2-} &: Z=8,\ \ 8+2=10\ \text{electrons}\\ \text{F}^{-} &: Z=9,\ \ 9+1=10\ \text{electrons}\\ \text{Na}^{+} &: Z=11,\ 11-1=10\ \text{electrons}\\ \text{Mg}^{2+} &: Z=12,\ 12-2=10\ \text{electrons}\\ \text{Al}^{3+} &: Z=13,\ 13-3=10\ \text{electrons} \end{aligned} \]Every one of them has exactly $\displaystyle 10$ electrons, arranged as \(\displaystyle 1s^2\,2s^2\,2p^6\) — 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 radiiHere 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 $\displaystyle 10$ electrons. That pull is the effective nuclear charge, and for a fixed electron count it simply tracks \(\displaystyle Z\) (the number of protons) directly: more protons in the nucleus means a stronger pull on the same $\displaystyle 10$ electrons, which drags the electron cloud in tighter and makes the ion smaller.So among isoelectronic species, radius runs opposite to \(\displaystyle Z\): the higher the atomic number, the smaller the ion.Listing \(\displaystyle Z\) for each species:\[\text{N}(7) < \text{O}(8) < \text{F}(9) < \text{Na}(11) < \text{Mg}(12) < \text{Al}(13) \]Radius decreases as \(\displaystyle Z\) increases, so reversing this list gives increasing radius:\[\text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} < \text{O}^{2-} < \text{N}^{3-} \]This matches the physical picture too: \(\displaystyle \text{Al}^{3+}\) has given up three electrons and carries the largest positive charge among these ions, so its $\displaystyle 10$ remaining electrons are pulled in hardest by $\displaystyle 13$ protons, making it the smallest of the six. \(\displaystyle \text{N}^{3-}\), at the other end, has only $\displaystyle 7$ protons trying to hold onto the same $\displaystyle 10$ 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 $\displaystyle 10$ electrons with the configuration \(\displaystyle 1s^2\,2s^2\,2p^6\). Since the electron count is fixed, radius decreases as nuclear charge (Z) increases, giving the increasing order of ionic radii: \(\displaystyle \text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^{+} < \text{F}^{-} < \text{O}^{2-} < \text{N}^{3-}\).

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 atomA cation forms by removing one or more electrons, usually from the outermost shell: \[M \rightarrow M^{n+} + ne^- \] Two things happen together: Often the outermost shell is emptied out completely — for example, sodium, \(\displaystyle \text{Na} \; (1s^2 2s^2 2p^6 3s^1)\), loses its lone \(\displaystyle 3s\) electron to become \(\displaystyle \text{Na}^+ \; (1s^2 2s^2 2p^6)\). \(\displaystyle \text{Na}^+\) has one shell fewer occupied than \(\displaystyle \text{Na}\), 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 atomAn anion forms by adding one or more electrons to the outermost shell: \[X + ne^- \rightarrow X^{n-} \] Here the opposite happens: For example, chlorine, \(\displaystyle \text{Cl} \; (1s^2 2s^2 2p^6 3s^2 3p^5)\), gains one electron to complete its octet as \(\displaystyle \text{Cl}^- \; (1s^2 2s^2 2p^6 3s^2 3p^6)\). The nucleus ($\displaystyle 17$ protons) is unchanged, but it is now spreading its pull over $\displaystyle 18$ electrons instead of $\displaystyle 17$, so the cloud expands and \(\displaystyle \text{Cl}^-\) is bigger than \(\displaystyle \text{Cl}\).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 \(\displaystyle n=3\)) — 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.

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.

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 $\displaystyle 1$: Energy needed to remove one electron from one atom.The electron sits in the ground state at \[E_1 = -2.18\times10^{-18}\ \text{J} \] Ionization means taking the electron out to infinity, where by convention the electron's energy is taken as zero, \(\displaystyle E_\infty = 0\).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: \[\Delta E = E_\infty - E_1 = 0 - (-2.18\times10^{-18}\ \text{J}) = +2.18\times10^{-18}\ \text{J} \]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 $\displaystyle 2$: Scale from one atom to one mole of atoms.Ionization enthalpy (as reported in J mol\(\displaystyle ^{-1}\)) 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, \[N_A = 6.022\times10^{23}\ \text{mol}^{-1} \] which is the formula connecting a per-particle quantity to a per-mole quantity: (energy per mole) = (energy per particle) \(\displaystyle \times\) (particles per mole).\[\Delta H_{\text{ionization}} = \Delta E \times N_A = (2.18\times10^{-18}\ \text{J})\times(6.022\times10^{23}\ \text{mol}^{-1}) \]Multiply the coefficients and add the exponents: \[2.18 \times 6.022 = 13.128 \] \[\Delta H_{\text{ionization}} = 13.128\times10^{5}\ \text{J mol}^{-1} = 1.3128\times10^{6}\ \text{J mol}^{-1} \]This is the same as \(\displaystyle 1312.8\ \text{kJ mol}^{-1}\), which is the familiar textbook value for the ionization enthalpy of hydrogen.Answer: The ionization enthalpy of atomic hydrogen is \(\displaystyle 1.3128\times10^{6}\ \text{J mol}^{-1}\) (i.e., \(\displaystyle 1312.8\ \text{kJ mol}^{-1}\)), obtained as \(\displaystyle \Delta H = (0-E_1)\times N_A = (2.18\times10^{-18}\ \text{J})\times(6.022\times10^{23}\ \text{mol}^{-1})\).

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 \(\displaystyle \Delta_iH \) is the energy needed to pull the outermost electron off a gaseous atom. Across a period, nuclear charge rises and atomic size shrinks, so \(\displaystyle \Delta_iH \) usually climbs steadily. Two special electron arrangements upset that smooth climb: a completely filled subshell (like \(\displaystyle 2s^2 \)) and a half-filled subshell (like \(\displaystyle 2p^3 \), 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 \(\displaystyle \Delta_iH \) than BWrite out the configurations: \[\text{Be: } 1s^2\,2s^2 \qquad \text{B: } 1s^2\,2s^2\,2p^1 \]Be's last electron sits in a completely filled \(\displaystyle 2s^2 \) subshell — held tightly and symmetrically. B's last electron instead goes into the \(\displaystyle 2p \) subshell, which lies at higher energy than \(\displaystyle 2s \) and, on top of that, is shielded from the nucleus by the compact \(\displaystyle 2s^2 \) pair sitting underneath it.The step people get wrong here is assuming nuclear charge settles it: B has one more proton than Be ( \(\displaystyle Z=5\) vs \(\displaystyle Z=4\) ), so it "should" hold its electrons more tightly. But nuclear charge is not the whole story — orbital energy and shielding matter too. B's \(\displaystyle 2p^1 \) electron is both higher in energy and better shielded, so it comes off more easily than one of Be's snugly-held \(\displaystyle 2s^2 \) electrons. That is why the actual values run \(\displaystyle \Delta_iH(\text{Be}) = 899 \text{ kJ/mol} > \Delta_iH(\text{B}) = 801 \text{ kJ/mol} \), even though B sits to the right of Be in the period.(ii) Why O has a lower \(\displaystyle \Delta_iH \) than both N and F\[\text{N: } 1s^2\,2s^2\,2p^3 \; (2p_x^1\,2p_y^1\,2p_z^1) \qquad \text{O: } 1s^2\,2s^2\,2p^4 \; (2p_x^2\,2p_y^1\,2p_z^1) \]N's \(\displaystyle 2p^3 \) 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 \(\displaystyle 2p_x^2 \).The step that's easy to miss: this isn't about O having a weaker pull from the nucleus — O's nuclear charge ( \(\displaystyle Z=8\) ) is larger than N's ( \(\displaystyle Z=7\) ), so on nuclear charge alone O's \(\displaystyle \Delta_iH \) "should" be higher. What actually happens is that the two electrons forced into the same \(\displaystyle 2p_x \) 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 \(\displaystyle \Delta_iH(\text{O}) = 1314 \text{ kJ/mol} < \Delta_iH(\text{N}) = 1402 \text{ kJ/mol} \).F, \(\displaystyle 1s^2\,2s^2\,2p^5 \; (2p_x^2\,2p_y^2\,2p_z^1) \), also has paired 2p electrons and so carries the same kind of internal repulsion that O does. But F's nuclear charge ( \(\displaystyle Z=9\) ) 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: \(\displaystyle \Delta_iH(\text{F}) = 1681 \text{ kJ/mol} > \Delta_iH(\text{O}) \). 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 \(\displaystyle 2s^2\) subshell, while B's last electron is in the higher-energy, more shielded \(\displaystyle 2p^1\) orbital, so Be holds its electron more tightly than B despite having one less proton, giving \(\displaystyle \Delta_iH(\text{Be}) > \Delta_iH(\text{B}) \). (ii) N's \(\displaystyle 2p^3\) is a stable, half-filled subshell with no paired electrons, whereas O's \(\displaystyle 2p^4\) forces two electrons into one orbital; the repulsion between that pair makes O's electron easier to remove than N's, so \(\displaystyle \Delta_iH(\text{O}) < \Delta_iH(\text{N}) \). F also has a paired 2p orbital, but its higher nuclear charge outweighs that repulsion, so \(\displaystyle \Delta_iH(\text{O}) < \Delta_iH(\text{F}) \) as well.

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.\[\text{Na (Z = 11): } 1s^2\,2s^2\,2p^6\,3s^1 \] \[\text{Mg (Z = 12): } 1s^2\,2s^2\,2p^6\,3s^2 \]Step $\displaystyle 1$ — why the first ionization enthalpy, \(\displaystyle \Delta_i H_1 \), of Na is lower than that of MgIn sodium, the electron being removed is the lone \(\displaystyle 3s^1 \) electron. Taking it off leaves \(\displaystyle \text{Na}^+ : 1s^2\,2s^2\,2p^6 \), 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 \(\displaystyle \Delta_i H_1(\text{Na}) \) is low ($\displaystyle 496$ kJ mol⁻¹).In magnesium, the electron being removed comes out of a filled subshell, \(\displaystyle 3s^2 \). A filled \(\displaystyle s \)-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 = $\displaystyle 12$ vs Z = $\displaystyle 11$) while the shielding from the inner core is essentially the same. That means the effective nuclear charge \(\displaystyle Z_{\text{eff}} \) felt by the \(\displaystyle 3s \) electrons is higher in Mg. Both effects — greater \(\displaystyle Z_{\text{eff}} \) and the stability of a filled subshell — pull the electron in more tightly, so \(\displaystyle \Delta_i H_1(\text{Mg}) \) is higher ($\displaystyle 737$ 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 $\displaystyle 2$ — why the second ionization enthalpy, \(\displaystyle \Delta_i H_2 \), of Na is higher than that of MgNow look at what is being removed in the second step, i.e., from the +$\displaystyle 1$ ions.\[\text{Na}^+ : 1s^2\,2s^2\,2p^6 \quad (\text{already a stable noble-gas core, like Ne}) \] \[\text{Mg}^+ : 1s^2\,2s^2\,2p^6\,3s^1 \quad (\text{still has one electron left in } 3s) \]To form \(\displaystyle \text{Na}^{2+} \), an electron must be pulled out of the complete, tightly-held \(\displaystyle 2p^6 \) inner shell of \(\displaystyle \text{Na}^+ \) — a shell that is already a stable closed octet held by a nucleus of $\displaystyle 11$ protons acting on far fewer, more contracted electrons. Breaking into that shell costs a huge amount of energy, so \(\displaystyle \Delta_i H_2(\text{Na}) \) is very high ($\displaystyle 4562$ kJ mol⁻¹) — in fact almost ten times \(\displaystyle \Delta_i H_1(\text{Na}) \).To form \(\displaystyle \text{Mg}^{2+} \), the electron removed is the single, loosely-held \(\displaystyle 3s^1 \) electron of \(\displaystyle \text{Mg}^+ \) — exactly the same kind of easy removal Na underwent in its first step, since it leaves behind the same stable Ne-like core, \(\displaystyle 1s^2\,2s^2\,2p^6 \). So \(\displaystyle \Delta_i H_2(\text{Mg}) \) is comparatively modest ($\displaystyle 1450$ kJ mol⁻¹).Since \(\displaystyle 4562 \text{ kJ mol}^{-1} > 1450 \text{ kJ mol}^{-1} \), 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 \(\displaystyle 3s^1\) electron (attaining a stable Ne core), while Mg must remove one electron from a filled, more tightly-held \(\displaystyle 3s^2\) subshell (higher \(\displaystyle Z_{\text{eff}}\), 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 \(\displaystyle 1s^2 2s^2 2p^6\), so removing a further electron means breaking into a complete inner shell, whereas Mg⁺ still has one loosely-held \(\displaystyle 3s^1\) electron left to lose easily to reach the same stable configuration.

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, \(\displaystyle \Delta_i H \), is defined for the process\[X(g) \rightarrow X^+(g) + e^- \]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, \(\displaystyle Z_{\text{eff}} \), and by the distance of the valence electron from the nucleus — not by the actual nuclear charge \(\displaystyle Z \) alone. Going down a group, three things change together, and the net effect is that the hold on the valence electron gets weaker.Factor $\displaystyle 1$ — the atomic radius increases. Down a group, each successive element has its valence electron in a shell with a higher principal quantum number \(\displaystyle n \) (period $\displaystyle 2$ uses \(\displaystyle n=2 \), period $\displaystyle 3$ uses \(\displaystyle n=3 \), and so on). A larger \(\displaystyle n \) 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 $\displaystyle 2$ — 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 $\displaystyle 3$ — nuclear charge does increase, but it is outweighed. It is true that \(\displaystyle Z \) (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 $\displaystyle 2$) removes more attraction than the extra protons add, and the increase in size (Factor $\displaystyle 1$) pushes the electron even farther from whatever net charge remains. The result is that the effective nuclear charge,\[Z_{\text{eff}} = Z - \sigma \]where \(\displaystyle \sigma \) 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 \(\displaystyle n \)) 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 \(\displaystyle Z \) 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: ($\displaystyle 1$) atomic size increases as a new principal shell is added at each row, moving the valence electron farther from the nucleus, and ($\displaystyle 2$) 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 \(\displaystyle Z \) 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.

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 \(\displaystyle d\) and \(\displaystyle f\) orbitals, which shield very poorly.First write out where each element's outermost electron sits:\[\text{B: } [He]\,2s^2 2p^1 \qquad \text{Al: } [Ne]\,3s^2 3p^1 \] \[\text{Ga: } [Ar]\,3d^{10} 4s^2 4p^1 \qquad \text{In: } [Kr]\,4d^{10} 5s^2 5p^1 \qquad \text{Tl: } [Xe]\,4f^{14} 5d^{10} 6s^2 6p^1 \]The value that matters for ionisation enthalpy is not the full nuclear charge \(\displaystyle Z\) but the effective nuclear charge \(\displaystyle Z_{eff} = Z - \sigma\), where \(\displaystyle \sigma\) (the shielding constant) is how much of the nuclear pull the inner electrons cancel out for the outermost electron. Down a normal group, \(\displaystyle Z\) rises but so does the number of complete inner shells, so \(\displaystyle \sigma\) rises almost as fast — \(\displaystyle Z_{eff}\) 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, $\displaystyle 801$, $\displaystyle 577$, $\displaystyle 579$, $\displaystyle 558$, $\displaystyle 589$ (all in \(\displaystyle kJ\ mol^{-1}\)), one step at a time: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 \(\displaystyle Z_{eff}\) is the quality of shielding, and \(\displaystyle d\) and \(\displaystyle f\) electrons shield poorly. Whenever a \(\displaystyle d\) or \(\displaystyle f\) 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 \(\displaystyle 3d^{10}\) subshell and Tl follows filled \(\displaystyle 4d^{10}5s^2\) and \(\displaystyle 4f^{14}5d^{10}\) subshells. \(\displaystyle d\) and \(\displaystyle f\) electrons shield the nucleus poorly, so the effective nuclear charge on the outer \(\displaystyle p\) 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.

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 \[X(g) + e^- \rightarrow X^-(g), \qquad \Delta_{eg}H = \text{energy change for this step} \] Here \(\displaystyle X\) is the atom, \(\displaystyle X^-\) the anion formed, and \(\displaystyle \Delta_{eg}H\) (electron gain enthalpy) is negative because energy is usually released when a neutral atom traps an extra electron. A more negative \(\displaystyle \Delta_{eg}H\) 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 FBoth sit in period $\displaystyle 2$, 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 \(\displaystyle Z_{eff}\) felt at the surface rises and the atomic radius shrinks. A smaller, more highly charged atom pulls an incoming electron in more forcefully, so \(\displaystyle \Delta_{eg}H\) 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.\[\Delta_{eg}H(\text{O}) = -141\ \text{kJ mol}^{-1}, \qquad \Delta_{eg}H(\text{F}) = -328\ \text{kJ mol}^{-1} \]F's value is the more negative one, exactly as the period trend predicts.(ii) F or ClHere 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 \(\displaystyle |\Delta_{eg}H|\) — 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.\[\Delta_{eg}H(\text{F}) = -328\ \text{kJ mol}^{-1}, \qquad \Delta_{eg}H(\text{Cl}) = -349\ \text{kJ mol}^{-1} \]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 $\displaystyle 2$, 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.