Easy Practice — Set 1

Quick Recap — The Periodic Table

• Modern periodic law: properties are a periodic function of atomic number. 7 periods, 18 groups.
• Blocks: s (groups 1-2), p (13-18), d (3-12, transition), f (lanthanoids/actinoids).
• Trends across a period: atomic radius decreases, ionisation enthalpy and electronegativity increase, metallic character decreases.
• Trends down a group: atomic radius and metallic character increase, ionisation enthalpy and electronegativity decrease. Cations $<$ atom $<$ anions.

Beyond-NCERT JEE Essentials

1. Effective nuclear charge and Slater's rules

A valence electron never feels the full nuclear charge Z; it feels a reduced effective charge $Z_{eff} = Z - \sigma$, where the screening (shielding) constant comes from all the other electrons. Slater's rules estimate it for an s or p electron: electrons in shells outside the one considered contribute 0; each other electron in the same (ns, np) group contributes 0.35 (the partner in 1s contributes 0.30); each electron in the (n-1) shell contributes 0.85; and each electron in the (n-2) shell or deeper contributes 1.00, since a filled inner shell screens almost perfectly. (The bookkeeping shifts slightly for d and f electrons, where all inner electrons count 1.00.)

For a feel, take the 3s electron of Na (Z=11): the screening is $\sigma = (8 \times 0.85) + (2 \times 1.00) = 8.8$, so $Z_{eff} = 11 - 8.8 = 2.2$. Running this across period 3 gives a steadily rising valence charge — about 2.2 for Na, 2.85 for Mg and 3.5 for Al. That climb is the real engine behind the shrinking radius and rising ionisation enthalpy across a period.

[JEE Tip] Penetration, and hence screening power, follows the order s > p > d > f: an s electron dives closest to the nucleus, so it screens outer electrons best and is itself held most tightly. This is exactly why, within one shell, orbital energy runs ns < np < nd < nf, and why a p electron is easier to remove than an s electron of the same shell.

2. Ionisation-enthalpy anomalies (learn the WHY)

First ionisation enthalpy rises across a period but dips at two predictable points. JEE tests the reason, so pair each anomaly with its cause.

• Be > B and Mg > Al. The electron pulled from B (or Al) is a p electron, which penetrates less and lies higher in energy than the filled $ns^2$ pair below it, so it leaves more easily; on top of that, Be and Mg enjoy the extra stability of a completely filled $ns^2$ subshell. Both effects push the same way. (First ionisation enthalpies, kJ/mol: Be 899, B 801; Mg 738, Al 577.)
• N > O and P > S. Nitrogen ($2s^2 2p^3$) and phosphorus ($3s^2 3p^3$) have exactly half-filled p subshells — extra exchange-energy stability, every p orbital singly occupied. In oxygen ($2s^2 2p^4$) and sulphur ($3s^2 3p^4$) the fourth p electron must pair up in an already-occupied orbital, and that added electron-electron repulsion helps eject it. (First ionisation enthalpies, kJ/mol: N 1402, O 1314; P 1012, S 1000.)

[JEE Tip] In periods 2 and 3 these reversals mean the group-2 element sits above group 13, and group 15 sits above group 16. Beware of extending the rule blindly: for heavier periods poorer d-shielding shifts the balance, so it need not hold past period 3. A "smooth increase across the whole period" is the classic wrong answer.

3. Electron gain enthalpy — the fluorine anomaly

Electron gain enthalpy $\Delta_{eg}H$ is the energy released when a gaseous atom accepts an electron; more negative means more favourable. Two points go beyond NCERT.

• Chlorine is MORE exothermic than fluorine ($\Delta_{eg}H$: Cl about -349, F about -328 kJ/mol), even though F sits above Cl. Fluorine's electron enters a tiny, compact 2p subshell already dense with electrons, so the incomer meets fierce electron-electron repulsion and less energy is released; chlorine's larger 3p subshell has room to spare. The halogen order of most-negative electron gain enthalpy is therefore Cl > F > Br > I.
• The same "second member beats the first" effect appears in group 16, where sulphur is more exothermic than oxygen, for the identical small-size, high-repulsion reason.

Down a group $\Delta_{eg}H$ generally becomes less negative (the added electron lands farther out and is better shielded), so the period-2 head is the anomaly while the run from period 3 downward is smooth: Cl > Br > I.

[JEE Tip] Do not confuse this with electronegativity. Fluorine is the most electronegative element, yet chlorine — not fluorine — has the most negative electron gain enthalpy. The two properties measure different things.

4. Atomic and ionic radii — the orderings examiners love

• Isoelectronic species (same electron count) are sized purely by nuclear charge: more protons grip the fixed electron cloud harder, so radius falls as Z rises. For the 10-electron set the order is N3- > O2- > F- > Na+ > Mg2+ > Al3+ (roughly 171, 140, 133, 102, 72, 54 pm). The key is the nuclear (proton) charge, not the ionic charge: the ion with the fewest protons (N3-) holds the shared cloud most loosely and is largest.
• For one element, cation < neutral atom < anion. Removing an electron (often stripping a whole outer shell) raises $Z_{eff}$ per electron and contracts the species (Na 186 pm shrinks to Na+ about 102 pm); adding an electron piles on repulsion at unchanged Z and swells it (Cl 99 pm expands to Cl- 181 pm).

[JEE Tip] Lanthanoid contraction (preview). Across the 4f series (La to Lu) the diffuse, poorly shielding 4f electrons let $Z_{eff}$ creep up, so radii shrink steadily. The knock-on effect appears in the d-block: the 5d elements end up almost the same size as the 4d elements above them, so pairs like Zr and Hf, or Nb and Ta, are near-identical in size and notoriously hard to separate.

5. Electronegativity, diagonal relationships, and the inert pair effect

• Scales. Pauling's scale (built from bond energies) is the familiar one, peaking at F = 4.0. Mulliken defined electronegativity from the atom itself as the average of its ionisation energy and electron affinity, $\chi = \frac{IE + EA}{2}$ — sensible, since an atom that both holds its own electrons tightly (high IE) and grabs extra ones eagerly (high EA) should attract a shared pair strongly. Mulliken values (with IE and EA in eV) run roughly 2.8 times the Pauling numbers but track them closely.
• Diagonal relationship. The first element of groups 1, 2 and 13 resembles the element one place down and one place to the right: Li-Mg, Be-Al, B-Si. Moving right raises charge density and electronegativity while moving down lowers them, and along the diagonal the two changes nearly cancel, leaving similar charge-to-size ratio and electronegativity. Worth remembering:
• Li resembles Mg: both combine directly with N2 to give nitrides (Li3N, Mg3N2), both form mainly the normal oxide rather than a peroxide or superoxide, and their carbonates decompose on heating — unlike the other alkali metals.
• Be resembles Al: both are passivated by cold concentrated HNO3, both have amphoteric oxides (BeO, Al2O3) that dissolve in alkali, and both form covalent, Lewis-acidic chlorides.
• B resembles Si: both are metalloids that form weakly acidic oxides and volatile, readily hydrolysed covalent hydrides.
• Inert pair effect. Down groups 13 to 15 the $ns^2$ pair grows reluctant to bond, because the intervening d and f electrons shield it poorly and it is drawn tightly inward. So the (group minus 2) oxidation state becomes the stable one for the heavy members: Tl+ over Tl3+, Pb2+ over Pb4+, Bi3+ over Bi5+. A direct consequence is that the higher oxidation states of these heavy elements are strong oxidisers (PbO2, Bi(V), Tl3+).

[JEE Tip] Successive-IE jump = group finder. A sudden large jump between the n-th and (n+1)-th ionisation enthalpies means the (n+1)-th electron is being torn out of a noble-gas core; the atom therefore had exactly n valence electrons. A jump after $IE_2$ flags a group-2 metal; a jump after $IE_3$ flags group 13. Read the JUMP, never $IE_1$ alone.

Worked Examples — Beyond-NCERT Essentials

Example 1: Arrange Be, B, C and N in increasing order of first ionisation enthalpy.

Solution: Principle: across period 2 the first ionisation enthalpy rises with $Z_{eff}$, but a filled $2s^2$ subshell (Be) and the poorer penetration of the 2p electron (B) create the anomaly Be > B. Reasoning: ordering by $Z_{eff}$ alone would give Be < B < C < N. Correct it at boron: boron's outermost electron is a loosely held, poorly penetrating 2p electron, and losing it leaves boron's filled $2s^2$ core untouched, so B needs LESS energy than the fully filled Be beside it. Carbon and nitrogen continue the normal rise, nitrogen being further stabilised by its half-filled $2p^3$ set. Answer: B < Be < C < N (kJ/mol: 801, 899, 1086, 1402).

Example 2: For the isoelectronic species O2-, F-, Na+ and Mg2+, identify the largest and smallest, and give the full radius order.

Solution: Principle: isoelectronic species share the same electron count (here 10), so size is fixed by nuclear charge alone — radius falls as Z rises. Reasoning: the nuclear charges are O 8, F 9, Na 11, Mg 12. The 8 protons of oxygen grip the shared 10-electron cloud most weakly, giving the largest ion; the 12 protons of magnesium grip it hardest, giving the smallest. Answer: largest O2-, smallest Mg2+; full order O2- > F- > Na+ > Mg2+. Caution: what fixes the size is the nuclear (proton) count, not the ionic charge — the two merely move oppositely within an isoelectronic set, so the "more negative means bigger" shortcut must never be carried over to non-isoelectronic comparisons.

Example 3: Fluorine is the most electronegative element, yet its electron gain enthalpy is less negative than that of chlorine. Explain.

Solution: Principle: electronegativity measures the pull on a shared pair inside a bond; electron gain enthalpy $\Delta_{eg}H$ measures the energy released when an isolated gaseous atom captures a whole extra electron. Different quantities, different trends. Reasoning: fluorine's high electronegativity comes from its small size and high $Z_{eff}$ acting on a bonding pair. But when a free F atom accepts an electron, that electron is forced into the very small, electron-dense 2p subshell, where strong electron-electron repulsion cancels much of the attraction, so less energy is released. Chlorine's 3p subshell is larger and less crowded, so it releases more. Answer: F is most electronegative, but $\Delta_{eg}H$ is more negative for Cl; the halogen order is Cl > F > Br > I.

Example 4: An element has successive ionisation enthalpies (kJ/mol) IE1 = 496, IE2 = 4562, IE3 = 6910. To which group does it belong?

Solution: Principle: a sudden, large jump between $IE_n$ and $IE_{n+1}$ marks the point where a stable noble-gas core is first broken; the number of electrons removed before the jump equals the number of valence electrons, which fixes the group. Reasoning: here $IE_2$ is about nine times $IE_1$ — an enormous jump right after the first electron. So the atom has just one loosely held valence electron, and the second must be torn from a noble-gas core. Answer: one valence electron means group 1 (the data are sodium's). The trap is reading the group from the size of $IE_1$ alone; only the location of the jump is decisive.

Example 5: Which chemical behaviour does lithium share with magnesium but NOT with sodium, and why?

Solution: Principle: the diagonal relationship makes Li resemble Mg (one place down and to the right), because the rise in charge density on going right is offset by the fall on going down, giving the two a similar polarising power. Reasoning: as a consequence, lithium behaves less like its own group and more like magnesium in ways sodium does not share — for instance, Li combines directly with N2 to form the nitride Li3N (as Mg forms Mg3N2), and Li2CO3 decomposes on heating to Li2O and CO2. Sodium forms no nitride with N2, and Na2CO3 is thermally stable. Answer: direct nitride formation and thermal decomposition of the carbonate are shared by Li and Mg but not by Na — classic diagonal-relationship consequences.

Example 6: Using Slater's rules, estimate the effective nuclear charge felt by a 2p electron in fluorine (Z = 9).

Solution: Principle: $Z_{eff} = Z - \sigma$; for a 2p electron, each other electron in the (2s, 2p) group screens 0.35 and each 1s electron (the n-1 shell) screens 0.85. Reasoning: fluorine is 1s2 2s2 2p5, so the chosen 2p electron sees six other (2s, 2p) electrons and two 1s electrons, giving $Z_{eff} = 9 - (6 \times 0.35 + 2 \times 0.85) = 9 - 3.80 = 5.20$. Answer: about 5.2. Note how much larger this is than the value near 2.2 for sodium's outer electron — this is why fluorine grips its electrons so tightly and sits at the top-right of the electronegativity scale.

Example 7: Explain why PbCl2 is more stable than PbCl4, whereas for carbon it is CCl4 that is stable.

Solution: Principle: the inert pair effect — the reluctance of the heavy-element $ns^2$ pair to take part in bonding because the underlying d and f electrons shield it poorly — makes the lower (group minus 2) oxidation state increasingly favoured down group 14. Reasoning: carbon has no inert pair issue, so it readily shows +4 in CCl4. For lead, the 6s2 pair is held tightly and resists involvement in bonding, so Pb prefers +2; Pb(IV) in PbCl4 is unstable and a strong oxidiser, tending to slip back to the +2 state. Answer: PbCl2 (Pb in +2) is the stable chloride while PbCl4 (Pb in +4) is not — a direct inert-pair-effect result; carbon shows the opposite because it has no such stabilised s pair.

Example 8: Element A has a high ionisation energy and a large (favourable) electron affinity; element B has a low ionisation energy and a near-zero electron affinity. On the Mulliken scale, which is more electronegative?

Solution: Principle: Mulliken electronegativity is the atom's own average of ionisation energy and electron affinity, $\chi = \frac{IE + EA}{2}$: an atom is strongly electronegative if it both resists losing its electrons (high IE) and welcomes an extra one (large EA). Reasoning: element A scores high on both terms, so its average is large; element B scores low on both, so its average is small. Answer: A is far more electronegative than B. This is why non-metals near the top-right (high IE, favourable EA) are the most electronegative, while alkali metals (low IE, small EA) are the least.