Classification of Elements and Periodicity in Properties — NEET Chemistry MCQs with answers

Classification of Elements and Periodicity in Properties is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 190 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Periodic trends: radius, IE, EA, EN · Valency and oxidation state trends · Anomalous behaviour and diagonal relationship · Modern periodic table and nomenclature · Periodic trends · Ionic radius · Ionisation enthalpy · Effective nuclear charge · Electronegativity · Anomalous properties · Ionic radii · Ionization enthalpy · Diagonal relationship and electronegativity · Successive ionization enthalpies · Lanthanide contraction · Electron gain enthalpy · Metallic and non-metallic character · Diagonal relationship · Atomic radius measurement conventions · Acid-base character of oxides · History of periodic classification · Shielding effect · d-block contraction · Electropositive and electronegative character · Atomic radius of transition elements · Oxidation state periodicity and electronegativity · Ionization enthalpy of transition metals · Ionisation enthalpy anomalies · Effective nuclear charge and shielding · Electron gain enthalpy trend exceptions · Ionic radius vs atomic radius · Periodic trends — graph interpretation · Metallic and nonmetallic character trend · Electronegativity scales · Periodic table history — Mendeleev vs modern law · Successive ionisation enthalpies · Radii measurement conventions · Lanthanide contraction and its periodic consequence · Periodic classification — block identification · Group trend: alkali metals vs halogens

8 free Classification of Elements and Periodicity in Properties practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Electronegativity

   Consider the following statements about electronegativity scales used to quantify an atom's tendency to attract shared electrons: (I) The Pauling scale is derived from the extra bond energy of a heteronuclear bond A-B compared with the average of the A-A and B-B bond energies. (II) The Mulliken scale defines electronegativity as the average of the ionization enthalpy and the electron gain enthalpy of the atom. (III) Electronegativity, unlike ionization enthalpy, is a fixed property of an isolated gaseous atom and does not depend on which molecule the atom is bonded in. (IV) All electronegativity scales give numerically identical values for a given element because electronegativity is a single unambiguous physical quantity. Which of the statements are correct?

   • A.II and III only
   • B.I, II and III only
   • C.I and II only, since III and IV are both false
   • D.I, II, III and IV
   Show answer and explanation

   Answer: C. I and II only, since III and IV are both false

   Statements I and II correctly describe the Pauling and Mulliken definitions. Statement III is wrong because electronegativity is not a fixed atomic property independent of bonding environment (it varies somewhat with hybridisation and oxidation state). Statement IV is wrong because different scales (Pauling, Mulliken, Allred-Rochow) use different definitions and give different numerical values, though they agree on the qualitative trend.

   Electronegativity is not directly measurable like ionization enthalpy; it is estimated by several empirical scales that broadly agree on trend but not on exact numbers. Pauling's scale (statement I, correct) uses the observation that a polar A-B bond is stronger than the geometric/arithmetic mean of A-A and B-B bond energies, attributing this 'extra' stabilisation to ionic character from electronegativity difference. Mulliken's scale (statement II, correct) defines electronegativity as proportional to (ionization enthalpy + electron gain enthalpy)/2 for the atom, a purely atomic-property based definition. Statement III is false: because electronegativity reflects an atom's pull on electron density within a specific bond, it is influenced by hybridisation state (an sp carbon is more electronegative than an sp3 carbon) and by the atoms it is bonded to, so it is not a single fixed number independent of context, unlike, say, atomic mass. Statement IV is false because Pauling, Mulliken and Allred-Rochow scales are built on different physical quantities (bond energies; IE and EGE; Zeff/r^2 respectively) and are not mutually convertible by a universal exact formula, though a linear relation approximately connects Pauling and Mulliken values; the numerical values differ across scales even though the qualitative period/group trend (increasing across a period, decreasing down a group) is consistent. Hence only I and II are correct.

   Common mistake: Assuming electronegativity behaves exactly like ionization enthalpy or electron gain enthalpy — a single, fixed, directly measurable quantity with one universal numeric scale.

2. Question 2 · medium · Ionization enthalpy

   A student writes the following claim about first ionization enthalpies (IE1) across Period 3: 'Since IE1 must increase steadily with atomic number across a period, the order Na < Mg < Al < Si < P < S < Cl < Ar holds exactly, with no exceptions, because each successive element simply has one more proton pulling on the same 3s/3p shell.' Which statement below correctly identifies what is wrong with this claim?

   • A.The claim is entirely correct; IE1 rises monotonically across every period with no exceptions
   • B.The claim is wrong: IE1 of Al is lower than that of Mg, and IE1 of S is lower than that of P, because removing an electron from Mg's filled 3s2 or from P's half-filled 3p3 configuration is harder than the smooth trend predicts
   • C.The claim is wrong only because IE1 actually decreases across a period, not increases
   • D.The claim is wrong only because Cl has a lower IE1 than S, breaking the trend at the very end of the period
   Show answer and explanation

   Answer: B. The claim is wrong: IE1 of Al is lower than that of Mg, and IE1 of S is lower than that of P, because removing an electron from Mg's filled 3s2 or from P's half-filled 3p3 configuration is harder than the smooth trend predicts

   Period 3 IE1 does rise overall, but with two well-established dips: Al < Mg (because Mg's extra-stable filled 3s2 configuration resists electron removal more than Al's single 3p1 electron does) and S < P (because P's half-filled 3p3 configuration is extra-stable compared with S's 3p4). Both dips are genuine exceptions, so the 'no exceptions' claim is false.

   General ionization enthalpy across a period rises because increasing nuclear charge (with roughly constant shielding within the same shell) pulls the valence electron more tightly. However, extra electronic stability from fully-filled or exactly half-filled subshells creates two irregular dips in Period 3, mirroring the well-known Period 2 pattern (Be > B, N > O). First, magnesium ([Ne]3s2) has a completely filled 3s subshell, which is more resistant to electron removal than expected; aluminium ([Ne]3s2 3p1) has to lose its lone, less-penetrating 3p electron, which (despite Al's higher nuclear charge) is easier to remove, so IE1(Al) < IE1(Mg). Second, phosphorus ([Ne]3s2 3p3) has a half-filled 3p subshell with extra exchange-energy stability (each of the three 3p orbitals singly occupied, minimising electron-electron repulsion); sulfur ([Ne]3s2 3p4) must pair up a fourth electron in one 3p orbital, and the resulting electron-electron repulsion makes that electron easier to remove than the trend predicts, so IE1(S) < IE1(P). The correct overall order is therefore Na < Mg > Al < Si < P > S < Cl < Ar (dips at Al and S), not the smooth monotonic sequence claimed. Option A ignores both genuine dips. Option C misidentifies the overall trend direction, confusing IE1 (which increases across a period) with atomic radius (which decreases). Option D invents a nonexistent dip at Cl/S while missing the two real ones at Mg/Al and P/S.

   Common mistake: Assuming ionization enthalpy increases perfectly monotonically across every period, overlooking the well-documented dips caused by filled (ns2) and half-filled (np3) subshell stability.

3. Question 3 · medium · Electron gain enthalpy

   Consider the following statements comparing electron gain enthalpy (EGE) trends within Group 16 (O, S, Se, Te) and Group 17 (F, Cl, Br, I): (I) Within each group, the first EGE (magnitude) is generally less negative for the topmost element than for the second element, because the small size of the top element causes strong inter-electron repulsion in its compact valence shell. (II) Chlorine has a more negative first EGE than fluorine for the same compact-size reason described in statement I. (III) For any given period, Group 17 elements have more negative first EGE values than the corresponding Group 16 elements, because halogens need only one electron to attain a stable noble-gas configuration, giving a stronger pull for that single additional electron. (IV) The second electron gain enthalpy (addition of a second electron) is always more negative than the first for both O and S, since the ion becomes more stable with each added electron. Which statements are correct?

   • A.I and III only
   • B.I, II and III only
   • C.II and III only
   • D.I, III and IV only
   Show answer and explanation

   Answer: B. I, II and III only

   Both O (Group 16) and F (Group 17) are anomalously small, so incoming electrons experience unusually strong repulsion in their compact valence shells, making their first EGE less negative than the second member of their group (S and Cl respectively) - statements I and II are correct. Halogens (needing just one electron for a noble gas configuration) have more negative EGE than the corresponding chalcogens across any period - statement III is correct. However, adding a second electron to an already negative ion (O- to O2-, S- to S2-) requires energy input due to electron-electron repulsion with the anion, so EGE2 is always positive, not more negative - statement IV is false.

   Electron gain enthalpy becomes more negative (more exothermic, more favourable) generally down to the second row of a group before leveling off, because the very first member of each group (O in Group 16, F in Group 17) has an unusually small atomic radius and a very compact valence shell; the incoming electron experiences significant repulsion from the already tightly packed electron density, partially offsetting the favourable nuclear attraction. As a result, EGE1 of O (-141 kJ/mol) is less negative than that of S (-200 kJ/mol) (statement I, correct), and EGE1 of F (-328 kJ/mol) is less negative than that of Cl (-349 kJ/mol) (statement II, correct). Comparing across groups within the same period, halogens have consistently more negative EGE1 values than the corresponding chalcogens (F more negative than O; Cl more negative than S, etc.) because a halogen atom needs to gain only a single electron to complete a stable noble-gas-like octet, giving an unusually strong thermodynamic driving force for that specific electron capture (statement III, correct). However, statement IV is false: once the first electron has been added and the atom has become a singly-charged anion (O- or S-), adding a second electron to form O2- or S2- means bringing a negative charge toward an already negatively charged species; the electrostatic repulsion this involves means energy must be supplied, so the second electron gain enthalpy (EGE2) is always positive (endothermic) for every element, not more negative than the first. This is why oxide (O2-) and sulfide (S2-) formation in ionic lattices is only overall favourable once lattice energy is accounted for. Hence only I, II and III are correct.

   Common mistake: Assuming that adding successive electrons always makes electron gain enthalpy progressively more negative, without accounting for the electrostatic repulsion penalty of adding an electron to an already-negative ion.

4. Question 4 · easy · Metallic and non-metallic character

   Metallic character (tendency to lose electrons and form cations) shows opposite trends across a period and down a group. Which statement correctly describes both trends together?

   • A.Metallic character decreases across a period (left to right) and also decreases down a group
   • B.Metallic character decreases across a period (left to right) but increases down a group
   • C.Metallic character increases across a period (left to right) but decreases down a group
   • D.Metallic character increases in both directions, across a period and down a group
   Show answer and explanation

   Answer: B. Metallic character decreases across a period (left to right) but increases down a group

   Across a period, increasing nuclear charge with poor extra shielding pulls valence electrons more tightly, making them harder to lose, so metallic character falls (elements become more non-metallic, e.g. Na to Cl). Down a group, the valence electron is in a progressively higher shell, shielded almost completely by the growing inner core, so it is held more loosely and metallic character rises (e.g. Li to Cs).

   Metallic character correlates with how easily an atom loses its valence electron(s), which is governed by ionization enthalpy and atomic size: lower ionization enthalpy and larger atomic radius favour metallic (electron-losing) behaviour. Across Period 3 (Na to Cl), effective nuclear charge rises sharply because each additional electron enters the same outer shell and shields poorly, so atomic radius shrinks and ionization enthalpy rises; electrons become progressively harder to remove, so metallic character decreases and non-metallic character increases (sodium is a reactive metal, chlorine a reactive non-metal). Down Group 1 (Li to Cs), each new valence electron occupies a fresh, higher principal shell, almost fully shielded by the larger inner electron core; effective nuclear charge on the valence electron barely changes, but its distance from the nucleus increases substantially, so it is held more loosely, ionization enthalpy falls, and metallic character increases (caesium is far more reactive as a metal than lithium). This opposite-direction behaviour (falling across a period, rising down a group) is why the most strongly metallic elements sit in the lower-left of the periodic table (e.g. Cs, Fr) and the most strongly non-metallic elements sit in the upper-right (e.g. F, O, Cl), with metalloids forming a diagonal staircase in between. Option A gets the group-wise direction backwards. Option C reverses both trends. Option D ignores the period-wise decrease entirely.

   Common mistake: Assuming metallic character changes the same way in both directions of the periodic table, rather than recognising that the underlying driver (radius/shielding/Zeff) behaves oppositely across a period compared with down a group.

5. Question 5 · easy · History of periodic classification

   Consider the following statements about the historical development of the periodic law: (I) Mendeleev arranged elements in order of increasing atomic mass and left gaps for undiscovered elements, successfully predicting properties of elements such as eka-silicon (germanium). (II) Mendeleev's atomic-mass-based ordering placed a few element pairs, such as tellurium and iodine, in an order inconsistent with their chemical properties. (III) Moseley's X-ray spectroscopy experiments established atomic number, not atomic mass, as the correct basis of periodicity, resolving the tellurium-iodine anomaly. (IV) The Modern Periodic Law states that physical and chemical properties of elements are periodic functions of their atomic masses. Which of these statements are correct?

   • A.I and II only
   • B.II, III and IV only
   • C.I, II and III only
   • D.I, III and IV only
   Show answer and explanation

   Answer: C. I, II and III only

   Mendeleev's atomic-mass-based table correctly predicted undiscovered elements (I, true) but misordered a few pairs like Te/I by mass (II, true); Moseley's X-ray work established atomic number as the correct periodicity variable, resolving such anomalies (III, true). However, the Modern Periodic Law is stated in terms of atomic number, not atomic mass, so statement IV is false.

   Dmitri Mendeleev arranged the known elements primarily in order of increasing atomic mass, grouping them by similar chemical properties into a table with deliberate gaps, from which he successfully predicted the existence and approximate properties of then-undiscovered elements such as eka-silicon (later identified as germanium), eka-aluminium (gallium) and eka-boron (scandium); this predictive success is a celebrated validation of his classification (statement I, correct). However, strict adherence to atomic mass ordering placed a few pairs of elements out of step with their chemical family: tellurium (atomic mass approximately 127.6) was placed before iodine (atomic mass approximately 126.9) purely to keep tellurium in the oxygen/sulfur family and iodine in the halogen family, an inversion relative to their raw atomic masses (statement II, correct). Henry Moseley's early-20th-century X-ray spectroscopy experiments showed that the frequency of characteristic X-rays emitted by an element's electrons is related in a simple, systematic way to a whole-number property of the atom -- its atomic number (number of protons) -- rather than its atomic mass; ordering elements by atomic number instead of atomic mass automatically resolved the Te/I anomaly and similar cases (Ar/K, Co/Ni), establishing atomic number as the true physical basis of periodicity (statement III, correct). Consequently, the currently accepted Modern Periodic Law states that the physical and chemical properties of elements are periodic functions of their ATOMIC NUMBER, not atomic mass; statement IV, which repeats the outdated mass-based wording, is therefore false. Only I, II and III are correct.

   Common mistake: Reciting the Modern Periodic Law using the older atomic-mass wording instead of the correct atomic-number-based statement.

6. Question 6 · medium · Electronegativity

   The table below lists approximate Pauling electronegativity values for four elements: Element: H = 2.1, C = 2.5, N = 3.0, O = 3.5, F = 4.0. Using electronegativity DIFFERENCE as a measure of bond polarity, which bond among C-H, N-H, O-H and F-H is the LEAST polar, and which is the MOST polar?

   • A.F-H is least polar because fluorine is the most electronegative element listed, so it should form the 'best', least polar covalent bond; C-H is most polar because carbon is the least electronegative among C, N, O, F
   • B.All four bonds are equally polar since they all involve the same hydrogen atom, and bond polarity depends only on hydrogen's electronegativity, not on the electronegativity of the atom it is bonded to
   • C.C-H is least polar (difference 0.4); F-H is most polar (difference 1.9)
   • D.C-H is least polar (difference 0.4); O-H is most polar (difference 1.4), since oxygen is more electronegative than nitrogen but this ignores that fluorine's difference from H is even larger than oxygen's
   Show answer and explanation

   Answer: C. C-H is least polar (difference 0.4); F-H is most polar (difference 1.9)

   Bond polarity correlates with the electronegativity DIFFERENCE between the two bonded atoms, not the absolute electronegativity of either atom alone. Differences: C-H = 2.5-2.1 = 0.4 (smallest, least polar), N-H = 3.0-2.1 = 0.9, O-H = 3.5-2.1 = 1.4, F-H = 4.0-2.1 = 1.9 (largest, most polar).

   The polarity of a covalent bond A-H arises from the unequal sharing of the bonding electron pair, which is driven by the DIFFERENCE in electronegativity between A and H, not by how electronegative either atom is in isolation. Computing electronegativity differences using the given Pauling values: |EN(C) - EN(H)| = |2.5 - 2.1| = 0.4 (C-H, smallest difference, hence the least polar of the four bonds, essentially treated as nonpolar covalent in organic chemistry); |EN(N) - EN(H)| = |3.0 - 2.1| = 0.9 (N-H, moderately polar); |EN(O) - EN(H)| = |3.5 - 2.1| = 1.4 (O-H, quite polar, consistent with water's strong hydrogen-bonding ability); |EN(F) - EN(H)| = |4.0 - 2.1| = 1.9 (F-H, the largest difference and hence the most polar bond among the four, consistent with HF's unusually strong hydrogen bonding and its status as a weak acid due to this very polar, though still covalent, bond). So C-H is least polar and F-H is most polar, matching option A. Option D correctly identifies C-H as least polar but stops the comparison prematurely at O-H, missing that F-H's electronegativity difference (1.9) exceeds O-H's (1.4) and is therefore the actual most polar bond. Option A confuses 'high electronegativity of one atom' with 'small electronegativity difference between two atoms' -- fluorine being the single most electronegative element does not make F-H the least polar bond; in fact it is precisely fluorine's very high electronegativity relative to hydrogen's much lower value that creates the LARGEST difference and hence the MOST polar bond among these four. Option B incorrectly treats bond polarity as a property of only one atom (hydrogen) rather than the pair.

   Common mistake: Assuming the atom with the single highest electronegativity value automatically forms the least polar bond, rather than correctly computing and comparing electronegativity differences across all the bonds in question.

7. Question 7 · medium · Electron gain enthalpy

   A student states: 'Noble gases like neon and argon have extremely negative (highly exothermic) electron gain enthalpies because their nuclei are the most positively charged in their respective periods, so they should attract an incoming electron more strongly than any other element in that period.' Which response correctly evaluates this claim?

   • A.The claim is correct: since electron gain enthalpy generally becomes more negative across a period due to increasing nuclear charge, and noble gases sit at the very end of the period with the highest nuclear charge, they must have the most negative electron gain enthalpy of the whole period
   • B.The claim is wrong only because noble gas electron gain enthalpies are simply not measurable, not because their sign or magnitude follows any different rule from the rest of the period
   • C.The claim is correct for helium and neon specifically (smallest noble gases) but incorrect for heavier noble gases like xenon, which do have negative electron gain enthalpies due to their larger, more polarizable electron clouds
   • D.The claim is wrong: noble gases have effectively zero or slightly POSITIVE (endothermic) electron gain enthalpies, because their completely filled, stable ns2np6 configuration would have to accept the incoming electron into a new, higher-energy shell/subshell, which is energetically unfavourable despite their high nuclear charge
   Show answer and explanation

   Answer: D. The claim is wrong: noble gases have effectively zero or slightly POSITIVE (endothermic) electron gain enthalpies, because their completely filled, stable ns2np6 configuration would have to accept the incoming electron into a new, higher-energy shell/subshell, which is energetically unfavourable despite their high nuclear charge

   Noble gases have a stable, completely filled valence configuration (ns2np6). Adding an extra electron would force it into a new, higher-energy shell or subshell, which is energetically unfavourable and requires an input of energy, overriding the effect of high nuclear charge. Their electron gain enthalpies are therefore close to zero or slightly positive, not highly negative.

   Electron gain enthalpy generally becomes more negative moving left to right across a period because increasing effective nuclear charge attracts an incoming electron more strongly, as seen in the halogens which have the most negative values in their respective periods. However, this trend does not extend to the noble gases at the very end of each period. A noble gas atom already has a completely filled, exceptionally stable valence shell configuration (ns2np6). Any additional electron cannot be accommodated within this already-full valence shell; it would have to enter the next principal shell (a much higher-energy orbital, further from the nucleus and poorly attracted by the now heavily-shielded nuclear charge) or a higher-energy subshell. This is energetically highly unfavourable, so despite noble gases having the highest nuclear charge in their period, their electron gain enthalpies are close to zero or even mildly positive (endothermic) rather than negative -- for example, neon and argon are conventionally assigned electron gain enthalpy values close to zero or slightly positive, in sharp contrast to the strongly negative values of the adjacent halogens (fluorine approximately -328 kJ/mol, chlorine approximately -349 kJ/mol). Option A mechanically extends the general period trend without recognising that the filled-shell/new-shell argument specifically overrides pure nuclear-charge reasoning right at the noble gases. Option B avoids committing to the correct sign-based explanation, incorrectly framing this as a measurement issue rather than a genuine physical/energetic one. Option C invents an unfounded size-based exception; the filled-shell argument applies uniformly across the entire noble gas group regardless of atomic size, so even large, more polarizable noble gases like xenon and krypton do not have negative electron gain enthalpies under this reasoning.

   Common mistake: Blindly extrapolating the 'electron gain enthalpy becomes more negative across a period' trend all the way to noble gases, without recognising that their completely filled configuration makes electron addition energetically unfavourable.

8. Question 8 · medium · Electropositive and electronegative character

   Consider the following statements linking electropositive/electronegative character to chemical reactivity trends: (I) Electropositive (metallic) character and chemical reactivity of metals both increase down Group 1, which is why caesium reacts far more vigorously with water than lithium does. (II) Electronegative (non-metallic) character and chemical reactivity of non-metals both increase down Group 17, which is why iodine is a more vigorous oxidising agent than fluorine. (III) Across Period 3, electropositive character decreases while electronegative character increases from Na to Cl, consistent with sodium being a highly reactive reducing metal and chlorine being a highly reactive oxidising non-metal. (IV) Reactivity of a metal is governed only by how easily it loses electrons (ionization enthalpy), while reactivity of a non-metal is governed only by how easily it gains electrons (electron gain enthalpy); no other factor is relevant to either. Which statements are correct?

   • A.I, III and IV only
   • B.I and III only
   • C.I, II and III only
   • D.II and IV only
   Show answer and explanation

   Answer: B. I and III only

   Metallic character and metal reactivity genuinely increase down Group 1 (I, true) and electropositive-to-electronegative character shifts correctly across Period 3 with matching reactivity (III, true). However, fluorine, not iodine, is the strongest oxidising halogen despite being at the top of Group 17 (II is false), and overall chemical reactivity depends on more than just ionization enthalpy or electron gain enthalpy alone -- factors like bond dissociation, hydration and lattice enthalpies also matter (IV is false, an oversimplification).

   Down Group 1, atomic radius increases substantially and ionization enthalpy falls steadily, so metallic (electropositive) character and reactivity toward water genuinely increase from lithium to caesium; caesium reacts explosively with water while lithium reacts comparatively gently, consistent with statement I. Across Period 3, electropositive character decreases and electronegative character increases from sodium to chlorine as effective nuclear charge rises and atomic radius shrinks; sodium is indeed a highly reactive reducing metal (vigorous reaction with water/oxygen) and chlorine a highly reactive oxidising non-metal, consistent with statement III. However, statement II is false: although electron gain enthalpy alone becomes somewhat less negative from Cl to I down Group 17 (the general trend of decreasing EGE magnitude down a group, with F being a mild anomaly due to inter-electron repulsion in its very small 2p shell), actual oxidising STRENGTH of the halogens (measured by standard reduction potential and real chemical behaviour) decreases down the group: fluorine is the strongest halogen oxidising agent, and iodine is the weakest, the OPPOSITE of what statement II claims -- fluorine's very high reactivity arises despite a moderate electron gain enthalpy because of its very low bond dissociation enthalpy (weak F-F bond, due to lone-pair repulsion in the small F2 molecule) and very high hydration enthalpy of the small F- ion, both of which strongly favour the overall reaction being exothermic. Statement IV is also false: it oversimplifies reactivity as depending only on IE (for metals) or EGE (for non-metals) in isolation, ignoring other significant thermodynamic contributions such as bond dissociation enthalpy, hydration enthalpy (for ions in aqueous reactions) and lattice enthalpy (for ionic solid formation), all of which are essential to correctly explaining real reactivity trends, such as fluorine's anomalously high reactivity despite its EGE not being the most negative in its group. Only I and III are correct.

   Common mistake: Assuming halogen reactivity/oxidising power increases down Group 17 in step with metallic reactivity increasing down Group 1, and believing reactivity is fully explained by ionization enthalpy or electron gain enthalpy alone without considering bond dissociation, hydration, or lattice enthalpy contributions.

Questions about Classification of Elements and Periodicity in Properties for NEET

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