Amongst the following elements whose electronic configurations are given below, the one having the highest ion

NCERT Exemplar · Q22

Q. Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is

(i) [Ne]3s^2 3p^1

(ii) [Ne]3s^2 3p^3

(iii) [Ne]3s^2 3p^2

(iv) [Ar]3d^10 4s^2 4p^3

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Concept understanding — Isoelectronic Species

Isoelectronic Species – From Intuition to Precision

Imagine you are building atoms with LEGO blocks. Each block is a proton (positive charge) or an electron (negative charge). The number of protons decides which element you have — that's the atomic number $Z$ . The number of electrons decides the charge on the particle.

Now, here is the key idea: two different particles can have the same number of electrons. When that happens, their electron clouds are arranged in the same way. They become isoelectronic.

The Intuition

Think of a neutral neon atom. It has 10 protons and 10 electrons. Now take a sodium atom (11 protons, 11 electrons) and remove one electron. You get $\text{Na}^+$ , which has 11 protons but only 10 electrons. The electron count of $\text{Na}^+$ is exactly the same as that of neutral neon.

Even though $\text{Na}^+$ and $\text{Ne}$ are different elements with different nuclear charges, their electron configurations are identical: $1s^2 2s^2 2p^6$ . They are iso (same) electronic (electron arrangement).

Note

Isoelectronic species share the same number of electrons and therefore the same electronic configuration. They differ in nuclear charge ( $Z$ ).

The Precise Statement

Definition: Two or more atoms, ions, or molecules are said to be isoelectronic if they have the same number of electrons.

That is the entire definition. But the real power comes from what follows: because their electron clouds are identical in structure, their properties — like ionic radii, ionization energy, and chemical behaviour — show clear, predictable trends when you compare them.

How to Identify Isoelectronic Species

Count the electrons in each species.
- For a neutral atom: electrons = atomic number $Z$ .
- For a positive ion: electrons = $Z$ - (charge magnitude).
- For a negative ion: electrons = $Z$ + (charge magnitude).
Compare the counts. If they match, the species are isoelectronic.

Example: Which of these are isoelectronic? $\text{O}^{2-}$ , $\text{F}^-$ , $\text{Na}^+$ , $\text{Mg}^{2+}$ , $\text{Ne}$ .

$\text{O}^{2-}$ : $Z=8$ , electrons = $8+2=10$
$\text{F}^-$ : $Z=9$ , electrons = $9+1=10$
$\text{Na}^+$ : $Z=11$ , electrons = $11-1=10$
$\text{Mg}^{2+}$ : $Z=12$ , electrons = $12-2=10$
$\text{Ne}$ : $Z=10$ , electrons = $10$

All five have 10 electrons. They form an isoelectronic series.

\text{Electrons in an ion} = Z - (\text{charge})

The Critical Consequence: Size Trends

Here is where the concept becomes exam-relevant. In an isoelectronic series, as nuclear charge ( $Z$ ) increases, the ionic radius decreases.

Why? The same number of electrons is pulled more strongly by a larger positive nucleus. The electron cloud shrinks.

For the series above ( $\text{O}^{2-}$ , $\text{F}^-$ , $\text{Na}^+$ , $\text{Mg}^{2+}$ , $\text{Ne}$ ):

Species	$Z$	Electrons	Relative Radius
$\text{O}^{2-}$	8	10	Largest
$\text{F}^-$	9	10	↓
$\text{Ne}$	10	10	↓
$\text{Na}^+$	11	10	↓
$\text{Mg}^{2+}$	12	10	Smallest

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The key idea is that ionisation enthalpy increases across a period and is highest for a half-filled or stable configuration. Among the given options, (B) [Ne]3s²3p³ has a half-filled p-subshell, giving it the highest ionisation enthalpy.

The question asks you to compare ionisation enthalpies of four elements based on their electronic configurations. Ionisation enthalpy is the energy needed to remove the most loosely bound electron from a gaseous atom. The higher the ionisation enthalpy, the more difficult it is to remove an electron.

The trend across a period is that ionisation enthalpy generally increases from left to right. But there are exceptions due to extra stability of half-filled and fully-filled subshells. So you need to look at both the position in the periodic table and the stability of the configuration.

Let’s decode each configuration:

Option (A): [Ne]3s²3p¹

This is aluminium (Al). The outermost electron is in the 3p orbital. It’s a single electron in a p-orbital, so it’s relatively easy to remove. Ionisation enthalpy is moderate.
Option (B): [Ne]3s²3p³

This is phosphorus (P). The 3p subshell is exactly half-filled (three electrons, one in each p-orbital). Half-filled subshells have extra stability due to exchange energy and symmetry. Removing an electron disrupts this stable arrangement, so the ionisation enthalpy is higher than expected from the general trend.
Option (C): [Ne]3s²3p²

This is silicon (Si). The 3p subshell has two electrons. It’s less stable than a half-filled configuration, so its ionisation enthalpy is lower than that of phosphorus.
Option (D): [Ar]3d¹⁰4s²4p³

This is arsenic (As). It belongs to the same group as phosphorus (Group 15) but is in the fourth period. Down a group, ionisation enthalpy decreases because the atomic size increases and the outermost electron is farther from the nucleus. So arsenic has a lower ionisation enthalpy than phosphorus. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 29 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$SO_2$
›Reveal solutionSolution
The acidity of an oxide increases with the oxidation state of the central element and its position in the periodic table (higher electronegativity → more acidic). Among the given oxides, $SO_3$ (S in +6 state) is the most acidic, and $K_2O$ (K, a highly electropositive metal) is the most basic. The correct pair is $SO_3$ and $K_2O$ , option (B).

The key idea is that oxides of elements show a trend from basic to amphoteric to acidic as you move from left to right across a period, and also as the oxidation state of the same element increases. This is because the nature of the bond between the element and oxygen changes — more ionic (basic) vs. more covalent (acidic).

Let’s examine each oxide systematically.

Identify the element and its typical oxide nature.

$K_2O$ : Potassium is a Group 1 metal, highly electropositive. Its oxide is strongly ionic and gives $KOH$ in water — a strong base.

$MgO$ : Magnesium is Group 2, also electropositive but less so than K. $MgO$ is basic, but less strongly than $K_2O$ (it gives $Mg(OH)_2$ , a weak base).

$Al_2O_3$ : Aluminium is in Group 13. Its oxide is amphoteric — it reacts with both acids and bases. So it’s neither strongly acidic nor strongly basic.

$P_2O_5$ : Phosphorus is a non-metal (Group 15). $P_2O_5$ is acidic, dissolving in water to give phosphoric acid ( $H_3PO_4$ ).

$SO_2$ and $SO_3$ : Both are oxides of sulphur (Group 16). $SO_2$ gives sulphurous acid ( $H_2SO_3$ , weak acid), while $SO_3$ gives sulphuric acid ( $H_2SO_4$ , strong acid).

Compare acidity among the acidic oxides.

Acidity increases with the oxidation state of the central atom. In $SO_3$ , sulphur is in the +6 oxidation state; in $SO_2$ , it’s +4. Higher oxidation state means greater polarising power on the oxygen, making the O–H bond in the corresponding acid more polar and thus more easily releasing $H^+$ . So $SO_3$ is more acidic than $SO_2$ .

Between $P_2O_5$ and $SO_3$ , sulphur is more electronegative than phosphorus and also in a higher oxidation state (+6 vs. +5). Hence $SO_3$ is the most acidic among all given oxides. …
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TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
$Q. Identify the pairs, in which electron gain enthalpy of the first element is more than that of second element I. F, Br II. Na, Li III. S, O The correct answer is (A) I, II, III (B) I, III only (C) II, III only (D) I, II only$
›Reveal solutionSolution
Electron gain enthalpy (EA) is the energy released when an atom gains an electron; more negative means greater release. The question asks where the first element has a more negative (i.e., larger magnitude) EA than the second. The correct pairs are I (F > Br) and III (S > O), but not II (Na < Li). So the answer is (B).

The Core Concept: What “more than” really means

Electron gain enthalpy is usually negative (energy is released). “More than” here means more negative — that is, a larger release of energy. So we are comparing:

F vs Br: Which releases more energy when gaining an electron?

Na vs Li: Which releases more?

S vs O: Which releases more?

The key is to remember the periodic trends and the exceptions caused by atomic size and electron repulsion.

Step-by-step reasoning

Pair I: F vs Br

Fluorine is the most electronegative element, but its atomic radius is very small. Adding an electron to the 2p subshell causes significant electron–electron repulsion because the electron cloud is compact.

Bromine has a larger 4p orbital; the added electron is farther from the nucleus and experiences less repulsion.

Result: The electron gain enthalpy of fluorine is actually less negative than that of chlorine, but compared to bromine, fluorine is still more negative.

Experimental values: F ≈ –328 kJ/mol, Br ≈ –325 kJ/mol.

So F has a more negative EA than Br → True.

Pair II: Na vs Li

Both are alkali metals. Their electron gain enthalpies are negative but small (they don’t really want an extra electron).

Lithium has a very small atomic radius; adding an electron to the 2s orbital is relatively unfavorable due to high electron density.

Sodium has a larger radius; the added electron goes into a 3s orbital farther from the nucleus, with less repulsion.

Result: Na has a more negative EA than Li.

Values: Li ≈ –60 kJ/mol, Na ≈ –53 kJ/mol (less negative). Wait — careful: Actually Li is about –60, Na about –53. So Li is more negative than Na.

Therefore, Na’s EA is less than Li’s → False (the first element Na does NOT have a more negative EA than Li).

Pair III: S vs O

Oxygen is small; adding an electron to the 2p subshell causes strong repulsion. …
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TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
$Q. In which of the following, elements are arranged in the correct order of their first electron gain enthalpy values? (A) S < O < Br < I (B) O < S < I < Br (C) I < S < O < Br (D) Br < I < O < S$
›Reveal solutionSolution
The first electron gain enthalpy (EA) becomes more negative (more energy released) as we move left to right across a period and up a group, but oxygen and fluorine are exceptions due to their small size causing electron-electron repulsion. The correct order is O < S < I < Br, which corresponds to option (B).

The key concept here is first electron gain enthalpy — the energy change when a neutral gaseous atom gains one electron to form a negative ion. A more negative value means the atom releases more energy and thus has a higher tendency to gain an electron. The periodic trends are:

Across a period: EA becomes more negative (left to right) because nuclear charge increases, pulling the added electron more strongly.

Down a group: EA generally becomes less negative (top to bottom) because the added electron goes into a larger orbital, farther from the nucleus, so attraction weakens.

But there’s a famous twist: oxygen and fluorine have less negative EA than their heavier congeners (sulfur and chlorine) because their small size causes strong electron-electron repulsion when adding an extra electron into a compact orbital.

Let’s apply this to the elements given: S, O, Br, I.

Identify the group and period positions

O and S are in Group 16 (chalcogens).

Br and I are in Group 17 (halogens).

Halogens have the most negative EA in their periods because they need just one electron to complete an octet.

Chalcogens have less negative EA than halogens in the same period because adding an electron to a half-filled p-subshell is less favorable.

Compare within each group

For Group 16: EA of O is less negative than EA of S (due to repulsion in the small 2p orbital). So S has a more negative EA than O.

For Group 17: EA of Br is more negative than EA of I (normal trend: smaller atom, stronger attraction). So Br has a more negative EA than I.

Cross-group comparison

Halogens (Group 17) always have more negative EA than chalcogens (Group 16) in the same period.

But here we mix periods: S (period 3) vs Br (period 4) vs I (period 5). …
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TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$\begin{array}{|c|c|c|} \hline \text{Sl. No.} & \text{Molecules} & \text{Number of lone pairs of electrons on central atom} \\ \hline \text{I} & \text{SO}_2, \text{PbCl}_2, \text{SF}_4 & 1 \\ \text{II} & \text{XeF}_4, \text{H}_2\text{O}, \text{BrF}_3 & 2 \\ \text{III} & \text{CCl}_4, \text{XeF}_6, \text{IF}_7 & 0 \\ \hline \end{array}$
›Reveal solutionSolution
The key idea is to count lone pairs on the central atom using the formula (valence electrons – bonding electrons)/2. Only sets I and II are correctly matched; set III is wrong because XeF₆ has one lone pair, not zero. The correct answer is (A).

The question tests your ability to determine the number of lone pairs on the central atom in a molecule. The central atom is the one that forms the most bonds (usually the least electronegative, except hydrogen). Lone pairs are pairs of valence electrons not involved in bonding. The standard method: find the total valence electrons on the central atom, subtract the number of electrons used in bonds (each bond uses 2 electrons, but for counting lone pairs we care about the central atom’s share — each bond contributes 1 electron from the central atom to the bonding pair), then divide the remainder by 2.

A quick formula:

Lone pairs on central atom = (Valence electrons of central atom – Number of bonds formed by central atom) / 2

This works because each bond uses one electron from the central atom. The rest stay as lone pairs.

Let’s check each set one by one.

Set I: SO₂, PbCl₂, SF₄ — claimed lone pairs = 1

SO₂: Central atom S has 6 valence electrons. It forms 2 double bonds (each double bond counts as 2 bonds for the central atom, but in the formula we count the number of bonds, not bond order — careful: each bond to the central atom uses 1 electron from it, regardless of single/double/triple). Actually, for SO₂, S forms 2 sigma bonds (one to each O) and has one lone pair. Let’s do it properly: S has 6 valence electrons. It forms 2 sigma bonds (using 2 electrons) and one pi bond (which doesn’t use a central atom electron for the sigma framework — but the pi bond uses 2 electrons from S? No, in SO₂, S uses 2 electrons for sigma bonds, and the remaining 4 electrons become 2 lone pairs? That gives 2 lone pairs, but the actual structure has 1 lone pair. Wait — the correct Lewis structure: S is central, double-bonded to one O, single-bonded to the other O (with a negative charge on that O), and S has one lone pair. The formal charge method: S has 6 valence, forms 3 bonds (one double = 2 bonds, one single = 1 bond, total 3 bonds), so lone pairs = (6 – 3)/2 = 1.5? That’s not integer. The issue: the formula works for neutral molecules where the central atom obeys octet. For SO₂, the best structure has S with 1 lone pair and a double bond to one O and a coordinate bond? Actually, the standard: S in SO₂ has 1 lone pair. Using the formula: valence electrons of S = 6, number of bonds (sigma bonds) = 2 (since there are two O atoms, each forms a sigma bond), plus one pi bond doesn’t count as a separate bond for this formula? The correct approach: count the number of atoms bonded to central atom (2) plus the number of lone pairs? That’s circular. Let’s use the reliable method: total valence electrons in molecule = 6 + 6×2 = 18. Subtract 2 for each bond (there are 2 S–O bonds, but one is double? Actually, the molecule has 2 bonds total? No, in SO₂, there are 2 sigma bonds and 1 pi bond, total 3 bonds. Each bond uses 2 electrons, so 6 electrons used in bonding. Remaining 12 electrons go as lone pairs: on each O, there are 2 lone pairs (4 electrons each), total 8 electrons on O, leaving 4 electrons on S, which is 2 lone pairs? That gives 2 lone pairs on S, which is wrong. The correct Lewis structure: S has 1 lone pair, one O has 2 lone pairs, the other O has 3 lone pairs (due to a formal charge). The discrepancy arises because of resonance. For exam …
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TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$\mathrm{O_2^+}$
›Reveal solutionSolution
$\text{O}_2^+$ (15 electrons) has 10 bonding and 5 antibonding electrons, giving a ratio $2:1$ .

$\text{O}_2^+$ has $16-1 = 15$ electrons. Its molecular-orbital configuration is
$\sigma_{1s}^{2}\,\sigma_{1s}^{*2}\,\sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^{2}\,\pi_{2p_x}^{2}\pi_{2p_y}^{2}\,\pi_{2p_x}^{*1}$
Counting the electrons:

Bonding: $\sigma_{1s}^2 + \sigma_{2s}^2 + \sigma_{2p_z}^2 + (\pi_{2p_x}^2+\pi_{2p_y}^2) = 2+2+2+4 = 10$

Antibonding: $\sigma_{1s}^{*2} + \sigma_{2s}^{*2} + \pi_{2p_x}^{*1} = 2+2+1 = 5$ …
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TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$\begin{array}{|c|c|} \hline \text{List-1 (Atomic number of the element)} & \text{List-2 (Group)} \\ \hline \text{A 53} & \text{I 16} \\ \text{B 34} & \text{II 2} \\ \text{C 38} & \text{III 17} \\ \text{D 49} & \text{IV 13} \\ & \text{V 7} \\ \hline \end{array}$
›Reveal solutionSolution
To determine the group of an element, we find its electronic configuration and identify the number of valence electrons. For atomic numbers 53, 34, 38, and 49, the corresponding groups are 17, 16, 2, and 13 respectively, leading to the match (B).

The group number of an element in the periodic table is primarily determined by its electronic configuration, specifically the number of valence electrons. For main group elements (s-block and p-block), the group number can be found as follows:

For s-block elements (Groups 1 and 2): The group number is equal to the number of valence electrons.

For p-block elements (Groups 13 to 18): The group number is $10 + \text{number of valence electrons}$ .

For d-block elements (Groups 3 to 12): The group number is the sum of electrons in the outermost s-subshell and the penultimate d-subshell.

We will determine the electronic configuration for each given atomic number to find its group.

Element A: Atomic number $Z = 53$ (Iodine)

The noble gas preceding $Z=53$ is Krypton (Kr), with $Z=36$ .

The electronic configuration after Kr is $5s^2 4d^{10} 5p^5$ .

The outermost shell is the 5th shell, containing $5s^2 5p^5$ electrons.

Number of valence electrons $= 2 + 5 = 7$ .

Since it's a p-block element, its group number is $10 + 7 = 17$ .

Therefore, A (53) matches with List-2 (III) 17.

Element B: Atomic number $Z = 34$ (Selenium)

The noble gas preceding $Z=34$ is Argon (Ar), with $Z=18$ .

The electronic configuration after Ar is $4s^2 3d^{10} 4p^4$ .

The outermost shell is the 4th shell, containing $4s^2 4p^4$ electrons.

Number of valence electrons $= 2 + 4 = 6$ .

Since it's a p-block element, its group number is $10 + 6 = 16$ .

Therefore, B (34) matches with List-2 (I) 16.

Element C: Atomic number $Z = 38$ (Strontium)

The noble gas preceding $Z=38$ is Krypton (Kr), with $Z=36$ .

The electronic configuration after Kr is $5s^2$ .

The outermost shell is the 5th shell, containing $5s^2$ electrons.

Number of valence electrons $= 2$ .

Since it's an s-block element, its group number is $2$ . …
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TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
$Q. When burnt in excess of oxygen, sodium forms a compound X and potassium forms a compound Y. The magnetic natures of X and Y respectively are (A) Both X and Y are paramagnetic in nature (B) X is diamagnetic and Y is paramagnetic in nature (C) X is paramagnetic and Y is diamagnetic in nature (D) Both X and Y are diamagnetic in nature$
›Reveal solutionSolution
When sodium burns in excess oxygen it forms sodium peroxide ( $\text{Na}_2\text{O}_2$ ), which is diamagnetic; potassium forms potassium superoxide ( $\text{KO}_2$ ), which is paramagnetic. So the correct pair is X diamagnetic, Y paramagnetic → option (B).

Concept & Intuition

The magnetic nature of a compound depends on whether it has unpaired electrons. Paramagnetic substances have unpaired electrons and are attracted to a magnetic field; diamagnetic substances have all electrons paired and are weakly repelled.

The key twist here is that sodium and potassium, though both alkali metals, form different oxides when burned in excess oxygen. Sodium gives the peroxide ( $\text{O}_2^{2-}$ ion), while potassium gives the superoxide ( $\text{O}_2^{-}$ ion). The electronic structures of these ions determine the magnetic behaviour.

Identify the compounds formed

Sodium with excess oxygen: $2\text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O}_2$ (sodium peroxide).

Potassium with excess oxygen: $\text{K} + \text{O}_2 \rightarrow \text{KO}_2$ (potassium superoxide). This difference arises because the larger potassium ion stabilises the larger superoxide anion.

Examine the electronic structure of the peroxide ion ( $\text{O}_2^{2-}$ )

The peroxide ion has an O–O single bond. Its molecular orbital configuration (for the valence electrons) is: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^4$

All electrons are paired. Hence $\text{O}_2^{2-}$ is diamagnetic.

Therefore $\text{Na}_2\text{O}_2$ (compound X) is diamagnetic.

Examine the electronic structure of the superoxide ion ( $\text{O}_2^{-}$ )

The superoxide ion has one extra electron compared to neutral $\text{O}_2$ . Neutral $\text{O}_2$ has two unpaired electrons in the $\pi^*$ orbitals. Adding one electron gives $\text{O}_2^{-}$ with the configuration: $(\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^3$ …
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TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
$Q. In second period of the modern periodic table, two elements X and Y have higher first ionization enthalpy values than the preceding and succeeding elements. X and Y are respectively (A) B, C (B) Al, S (C) Be, N (D) Na, S$
›Reveal solutionSolution
The key idea is that in the second period, the elements with anomalously high first ionization enthalpy are beryllium (Be) and nitrogen (N), due to their stable electronic configurations (full s-subshell and half-filled p-subshell). Thus the correct pair is Be and N, which corresponds to option (C).

Concept and Intuition

Ionization enthalpy generally increases across a period as nuclear charge increases. However, there are two well-known exceptions in the second period:

Beryllium (Be) has a higher first ionization enthalpy than boron (B) because Be has a filled 2s² subshell (stable), while B has one electron in the 2p subshell (easier to remove).

Nitrogen (N) has a higher first ionization enthalpy than oxygen (O) because N has a half-filled 2p³ subshell (extra stability), while O has one paired electron in a 2p orbital (repulsion makes removal easier).

These two elements (Be and N) are the ones that “break the trend” in the second period. The question asks for the pair that fits this description.

Step-by-step reasoning

Identify the second period elements

The second period runs from Li (Z=3) to Ne (Z=10): Li, Be, B, C, N, O, F, Ne.

Recall the general trend of first ionization enthalpy across a period

It increases from left to right due to increasing nuclear charge and decreasing atomic radius. So we expect: Li < Be < B < C < N < O < F < Ne.

Spot the anomalies

Be (1s²2s²) has a higher ionization enthalpy than B (1s²2s²2p¹). Removing an electron from a filled s-subshell requires more energy than removing a p-electron.

N (1s²2s²2p³) has a higher ionization enthalpy than O (1s²2s²2p⁴). The half-filled p-subshell is stable; in O, the fourth p-electron is paired, causing repulsion that lowers the energy needed to remove it.

Check the other elements

Li and Na are not in the second period (Na is third period).

Al and S are in the third period, not the second. …
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TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
$Q. In second period of the modern periodic table, two elements X and Y have higher first ionization enthalpy values than the preceding and succeeding elements. X and Y are respectively (A) Al, S (B) B, C (C) Be, N (D) Na, S$
›Reveal solutionSolution
In the second period, the elements with anomalously high first ionization enthalpy are Be (group 2) and N (group 15), due to fully filled s-orbital and half-filled p-orbital stability, respectively. The correct option is (C).

The key concept here is ionization enthalpy trends and their exceptions in the periodic table. Generally, first ionization enthalpy increases across a period as nuclear charge increases and atomic radius decreases. However, there are two well-known dips: one at group 13 (where the electron is removed from a p-orbital for the first time, which is easier than from a filled s-orbital) and another at group 16 (where the electron is removed from a doubly occupied p-orbital, which is easier due to electron-electron repulsion). This means the elements just before these dips—group 2 and group 15—have higher ionization enthalpies than their neighbors. In the second period, those elements are beryllium (Be) and nitrogen (N).

Let’s walk through the reasoning step by step:

Recall the general trend: Across period 2 (Li → Ne), first ionization enthalpy generally increases. But the increase is not smooth. The actual order is:

Li < Be > B < C < N > O < F < Ne

The “>” signs mark places where an element has a higher value than the element after it.

Identify the two anomalies:

Be (group 2) has a higher ionization enthalpy than B (group 3). Reason: Be has a filled 2s² subshell; removing an electron from a stable, filled s-orbital requires more energy than removing the single 2p¹ electron from B.

N (group 15) has a higher ionization enthalpy than O (group 16). Reason: N has a half-filled 2p³ subshell (all electrons unpaired, maximum exchange energy); O has one doubly occupied p-orbital (2p⁴), and the electron‑electron repulsion makes it easier to remove one electron.

Check the question’s wording: “Two elements X and Y have higher first ionization enthalpy values than the preceding and succeeding elements.”

For Be: preceding = Li (lower), succeeding = B (lower) → Be fits. …
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$Q. The order of negative standard potential values of Li, Na, K is (A) Li > Na > K (B) Na > K > Li (C) Li > K > Na (D) K > Na > Li$
›Reveal solutionSolution
Standard reduction potentials: Li $^+$ /Li $=-3.04$ V, Na $^+$ /Na $=-2.71$ V, K $^+$ /K $=-2.93$ V. Ordering these from most negative to least negative gives Li > K > Na, so the correct option is (C).

Concept and Intuition

Standard electrode potentials measure a metal's tendency to lose electrons (be oxidized) in aqueous solution. Going down Group 1, ionization energy decreases, so a naive guess would expect the potential to become steadily more negative from Li to Cs. The actual values, however, show lithium as an exception:

Li $^+$ /Li: $-3.04$ V

Na $^+$ /Na: $-2.71$ V

K $^+$ /K: $-2.93$ V

Lithium has the most negative potential even though it is the smallest and has the highest ionization energy among these three. This anomaly arises because the potential depends not only on ionization energy but also on hydration enthalpy. Li $^+$ is tiny and highly polarizing, so its hydration enthalpy is exceptionally large — this extra stabilization of the hydrated ion pulls the equilibrium further toward oxidation, making the potential more negative than ionization energy alone would suggest.

Step-by-step reasoning

Recall the standard reduction potentials

Li $^+$ /Li: $-3.04$ V, Na $^+$ /Na: $-2.71$ V, K $^+$ /K: $-2.93$ V.

Order by magnitude of negativity

"Order of negative standard potential values" means ranking from most negative to least negative. Since $-3.04 < -2.93 < -2.71$ , the order from most to least negative is:

$\text{Li } (-3.04) > \text{K } (-2.93) > \text{Na } (-2.71).$

Match with options

This is Li > K > Na, which is option (C). Option (A) — Li > Na > K — would require Na to be more negative than K, which is not the case.

Why the anomaly? …
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$\mathrm{Cl} > \mathrm{F} > \mathrm{Li}$
›Reveal solutionSolution
Atomic radius trends are governed by periodic position (group/period) and, for lanthanoids, by the lanthanoid contraction. The correct orders are ii, iv only, making option (D) the answer.

The key to this question is knowing two separate trends: the general periodic trend for main-group elements, and the special case of the lanthanoid series. Atomic radius increases down a group and decreases across a period. But for the f-block elements, the lanthanoid contraction causes a steady decrease in radius as atomic number increases, overriding the usual group trend.

Let’s examine each statement one by one.

Statement i: $\mathrm{Cl} > \mathrm{F} > \mathrm{Li}$

Fluorine and chlorine are in the same group (17), with chlorine below fluorine, so $\mathrm{Cl} > \mathrm{F}$ is correct. But lithium is in group 1, period 2. Across period 2, atomic radius decreases sharply from Li to F. So $\mathrm{Li} > \mathrm{F}$ , not the reverse. The order given has F > Li, which is false. This statement is incorrect.

Statement ii: $\mathrm{P} > \mathrm{C} > \mathrm{N}$

Phosphorus (period 3, group 15) is below nitrogen (period 2, group 15), so $\mathrm{P} > \mathrm{N}$ is correct. Carbon and nitrogen are both in period 2, with carbon to the left of nitrogen, so $\mathrm{C} > \mathrm{N}$ . But is $\mathrm{P} > \mathrm{C}$ ? Phosphorus is in period 3, carbon in period 2 — the increase down a group (P vs N) is larger than the decrease across a period (C vs N). So yes, P is larger than C. The full order $\mathrm{P} > \mathrm{C} > \mathrm{N}$ is correct.

Statement iii: $\mathrm{Tm} > \mathrm{Sm} > \mathrm{Eu}$ …
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TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
$Q. Identify the number of metalloids from the following Sb, Be, P, Ge, Te, S, Cs, Tc, I (A) 3 (B) 4 (C) 5 (D) 2$
›Reveal solutionSolution
Metalloids are elements with properties intermediate between metals and non-metals. From the given list, the metalloids are Sb, Ge, Te — a total of 3.

The concept here is the classification of elements based on their position in the periodic table and their physical/chemical nature. Metalloids lie along the zigzag line (staircase) that separates metals from non-metals — typically boron, silicon, germanium, arsenic, antimony, tellurium, and sometimes polonium and astatine. You need to check each element against this known set.

Let’s go through the list one by one.

Sb (Antimony) — Located in Group 15, Period 5. It sits right on the metalloid staircase. It is a classic metalloid.

Be (Beryllium) — Group 2, Period 2. An alkaline earth metal. Not a metalloid.

P (Phosphorus) — Group 15, Period 3. A non-metal.

Ge (Germanium) — Group 14, Period 4. Directly on the staircase. A well-known metalloid.

Te (Tellurium) — Group 16, Period 5. Also on the staircase. A metalloid.

S (Sulfur) — Group 16, Period 3. A non-metal.

Cs (Cesium) — Group 1, Period 6. An alkali metal. …
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Q.Amongst the following elements whose electronic configurations are given below, the one having the highest ionisation enthalpy is

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