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CBSE Class 11 Chemistry · 10 questions · 24 marks
The modern periodic table is not a list to memorise but a map of electronic configurations. Once you see that the block, period and group of an element follow directly from where its last electron goes, trends in size, ionisation enthalpy and electronegativity stop being arbitrary facts and start being predictions you can make yourself.
An element has the ground-state electronic configuration [Ar] 3d¹⁰ 4s² 4p³. Its position in the periodic table is:
Answer
Period 4, group 15, p-block. The outermost shell has n = 4, so the element lies in period 4. The last electron enters a 4p orbital, so it belongs to the p-block. For a p-block element the group number is 10 + number of valence electrons = 10 + (2 from 4s + 3 from 4p) = 15. The element is arsenic (Z = 33). It is not d-block because the 3d subshell is completely filled and is no longer the differentiating subshell.
Arrange the isoelectronic species Mg²⁺, Na⁺, F⁻ and O²⁻ in order of increasing ionic radius.
Answer
Mg²⁺ < Na⁺ < F⁻ < O²⁻ is the correct order of increasing radius. All four species have 10 electrons, so the shielding is essentially identical and size is decided purely by nuclear charge. The nuclear charges are Mg (12), Na (11), F (9) and O (8). A larger nuclear charge pulls the same 10 electrons in more tightly, so Mg²⁺ with Z = 12 is the smallest and O²⁻ with Z = 8 is the largest. Radius therefore increases as Z decreases.
The first ionisation enthalpy of nitrogen is greater than that of oxygen because:
Answer
Nitrogen has a stable half-filled 2p³ configuration that resists electron removal. Nitrogen is 1s² 2s² 2p³, with one electron in each of the three 2p orbitals — an extra-stable symmetrical arrangement with high exchange energy, so removing an electron is unusually difficult. Oxygen is 1s² 2s² 2p⁴, so one 2p orbital already holds a pair. The inter-electronic repulsion in that pair makes the fourth 2p electron easier to remove, and this outweighs oxygen's slightly higher nuclear charge. The other options are factually wrong or irrelevant to ionisation enthalpy.
Which oxide is amphoteric in nature?
Answer
Al₂O₃ is amphoteric. It reacts with acids to give a salt and water (Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O) and also with strong bases to give an aluminate (Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O), so it behaves both as a base and as an acid. Na₂O and MgO are basic oxides of strongly metallic elements, and SO₃ is an acidic oxide of a non-metal that dissolves in water to give H₂SO₄. Amphoteric behaviour sits at the metal–non-metal borderline of a period.
Assertion (A): The electron gain enthalpy of chlorine is more negative than that of fluorine. Reason (R): The incoming electron experiences less inter-electronic repulsion in the larger 3p subshell of chlorine than in the compact 2p subshell of fluorine.
Answer
Both A and R are true and R is the correct explanation of A. On the general trend down group 17, electron gain enthalpy should become less negative as size increases, so fluorine ought to be the most exothermic. In fact chlorine is (about −349 kJ/mol against fluorine's −328 kJ/mol). The cause is fluorine's very small size: its 2p subshell is already crowded, so an incoming electron suffers strong repulsion from the electrons already present, and this repulsion cancels much of the energy released. Chlorine's 3p subshell is more diffuse, the repulsion is smaller, and more energy is released overall. The reason is therefore the correct explanation.
Define effective nuclear charge. Explain why the atomic radius decreases from sodium to chlorine across period 3.
Answer
Effective nuclear charge (Z_eff) is the net positive charge actually experienced by a valence electron after allowing for the screening or shielding by the inner electrons; approximately Z_eff = Z − S, where S is the shielding constant. From Na to Cl, each successive element adds one proton to the nucleus but the extra electron enters the same third shell. Electrons in the same shell shield one another very poorly, so S rises far more slowly than Z. Z_eff therefore increases steadily across the period, pulling the valence shell inward and shrinking the atomic radius from about 186 pm in Na to about 99 pm in Cl.
The first ionisation enthalpies (in kJ/mol) of four consecutive second-period elements are: Be = 899, B = 801, C = 1086, N = 1402. Account for the drop from Be to B, and comment on the overall trend.
Answer
The overall trend across a period is an increase in ionisation enthalpy, because effective nuclear charge rises while the shell stays the same, holding electrons more tightly. This is why C > B and N > C. The dip from Be (899) to B (801) breaks that trend for a structural reason. Beryllium is 1s² 2s², so its outermost electron is removed from a completely filled, penetrating 2s orbital that lies close to the nucleus and is unusually stable. Boron is 1s² 2s² 2p¹, and its outermost electron sits in a 2p orbital, which is higher in energy, less penetrating, and additionally shielded by the filled 2s pair. So boron's valence electron is both further out and better screened, and it comes away with less energy despite boron having one more proton. This confirms that ionisation enthalpy depends on the type of orbital involved, not on nuclear charge alone.
Explain the periodic trends in (i) atomic radius, (ii) ionisation enthalpy, and (iii) electronegativity, both across a period and down a group. In each case state the trend, give the reason, and cite one example.
Answer
(i) Atomic radius. Across a period it decreases: protons are added while the added electrons enter the same shell and shield poorly, so Z_eff rises and the shell contracts (Li 152 pm falls to F 64 pm). Down a group it increases: each new element adds a whole new principal shell and the inner shells shield the valence electrons effectively, so the increase in size outweighs the increase in nuclear charge (Li < Na < K < Rb). (ii) Ionisation enthalpy. Across a period it increases, because the rising Z_eff and shrinking radius bind the valence electron more tightly, so more energy is needed to remove it (Na 496 kJ/mol rises to Ar 1521 kJ/mol), with local exceptions at Be > B and N > O caused by filled and half-filled subshell stability. Down a group it decreases, because the valence electron is progressively further from the nucleus and better shielded (Li 520 > Na 496 > K 419 kJ/mol). (iii) Electronegativity. Across a period it increases, since a smaller atom with a higher Z_eff attracts the shared bonding pair more strongly (Na 0.9 rises to Cl 3.0). Down a group it decreases, because the larger distance and greater shielding weaken the pull on a shared pair (F 4.0 > Cl 3.0 > Br 2.8 > I 2.5). All three trends share one root cause: the competition between increasing effective nuclear charge, which pulls electrons in, and increasing distance plus shielding, which lets them go. Across a period the first dominates; down a group the second does.
An element X has atomic number 20 and an element Y has atomic number 17. (i) Write their electronic configurations. (ii) Identify the period, group and block of each. (iii) Predict the formula of the compound they form and the nature of the bonding. (iv) Compare their atomic radii and metallic character with reasons. (v) State the nature of the oxide each forms.
Answer
(i) X (Z = 20): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s², that is [Ar] 4s². Y (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵, that is [Ne] 3s² 3p⁵. (ii) X has its outermost electron in the 4s orbital, so it is in period 4, group 2, s-block (it is calcium). Y has its last electron entering 3p, so it is in period 3, group 17 (10 + 7 valence electrons), p-block (it is chlorine). (iii) X readily loses its two 4s electrons to reach the argon configuration, forming X²⁺; Y readily gains one electron to complete its 3p subshell, forming Y⁻. Charge balance requires two Y⁻ per X²⁺, so the compound is XY₂ (CaCl₂). The bonding is ionic, since a low-ionisation-enthalpy metal transfers electrons outright to a high-electron-affinity non-metal, and the resulting solid is high melting and conducts in the molten state. (iv) X is much larger than Y. X lies one period lower, so it has an extra occupied shell, and it also sits at the far left of its period where Z_eff is low. Y is at the right of period 3 where the high Z_eff has contracted the atom sharply. X is strongly metallic — it loses electrons easily, is a good conductor and is reducing in character — whereas Y is a typical non-metal that gains electrons and acts as an oxidising agent. (v) X forms a basic oxide, XO (CaO), which reacts with water to give the alkali Ca(OH)₂. Y forms acidic oxides such as Cl₂O₇, which dissolve in water to give oxoacids like HClO₄. This contrast illustrates the general change from basic to acidic oxides as one moves left to right across the table.
A materials engineer is choosing a metal to act as a sacrificial coating on steel pipelines buried in damp soil. She narrows the choice to magnesium (group 2, period 3) and zinc (group 12, period 4). She notes that magnesium has the lower ionisation enthalpy and the lower electronegativity of the two. Answer: (a) Which property of the coating metal makes it work as a sacrificial anode? (b) Based on the data given, which metal is the stronger reducing agent, and why? (c) Explain why elements towards the left of a period generally make better sacrificial anodes than those towards the right. (d) Suggest one practical reason zinc might still be chosen over magnesium.
Answer
(a) A sacrificial coating must corrode in place of the steel, so it must give up electrons more readily than iron does — in other words it must be more easily oxidised. The coating is consumed while the iron underneath stays protected. (b) Magnesium is the stronger reducing agent. A lower ionisation enthalpy means less energy is needed to remove its valence electrons, and a lower electronegativity means it holds shared electrons less tightly. Both point to magnesium releasing electrons more readily than zinc, so magnesium is oxidised in preference. (c) Elements on the left of a period have low effective nuclear charge acting on their valence electrons, large atomic radii and few valence electrons to lose in order to reach a noble-gas configuration. Their ionisation enthalpies and electronegativities are therefore low and their metallic character high, so they are oxidised easily. Moving right, Z_eff rises, radii shrink, and elements begin to gain rather than lose electrons, which makes them useless as sacrificial anodes. (d) Magnesium is so reactive that it corrodes very quickly, so the coating would need frequent replacement, and it can waste current in low-resistance soils. Zinc is reactive enough to protect iron but corrodes far more slowly, giving a longer service life at lower cost — which is why galvanising with zinc is the common industrial choice.
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