Q.Match the properties given in Column I with the metals given in Column II.
Column I (Property):

Showing the 12 most recent of 38 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Identify the element, which does not belong to fifth period of long form of periodic table (A) Pd (B) Pt (C) Cd (D) Ag

   ›Reveal solutionSolution

   The period of an element in the periodic table is determined by the highest principal quantum number ($n$) of its electrons. Palladium (Pd), Cadmium (Cd), and Silver (Ag) are in the 5th period, while Platinum (Pt) is in the 6th period. The element that does not belong to the fifth period is (B) Pt.

   The long form of the periodic table arranges elements based on their atomic number and recurring chemical properties. Elements with similar properties are placed in the same group, while elements with the same number of electron shells are placed in the same period.

   The period number directly corresponds to the principal quantum number ($n$) of the outermost electron shell. For example, elements in the 1st period have their valence electrons in the 1st shell ($n=1$), elements in the 2nd period have their valence electrons in the 2nd shell ($n=2$), and so on. Therefore, to identify which element does not belong to the fifth period, we need to determine the period for each given element.

   1. Understand the concept of a period:

      The period number of an element in the periodic table is equal to the highest principal quantum number ($n$) occupied by its electrons. This means that for an element in the 5th period, its outermost electrons will be in the 5th shell (i.e., $n=5$).

   2. Determine the period for each option:

      We can determine the period by looking at the atomic number and the electron configuration, or by simply recalling their positions in the periodic table.

      • (A) Palladium (Pd):

        • Atomic number ($Z$) = 46.
        • Its electron configuration is $[Kr] 4d^{10}$. The highest principal quantum number is 4 for the $d$-subshell, but the valence shell is $5s$ (even if empty in this specific exception). The element is located in the 5th period.

      • (B) Platinum (Pt):

        • Atomic number ($Z$) = 78.
        • Its electron configuration is $[Xe] 4f^{14} 5d^9 6s^1$. The highest principal quantum number occupied by electrons is $n=6$ (for the $6s$ orbital). Therefore, Platinum belongs to the 6th period.

      • (C) Cadmium (Cd):

        • Atomic number ($Z$) = 48.
        • Its electron configuration is $[Kr] 4d^{10} 5s^2$. The highest principal quantum number occupied by electrons is $n=5$ (for the $5s$ orbital). Therefore, Cadmium belongs to the 5th period.

      • (D) Silver (Ag):

        • Atomic number ($Z$) = 47. …

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2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.In which of the following compounds, the size of manganese is maximum? (A) KMnO$_4$ (B) K$_2$MnO$_4$ (C) Mn$_2$O$_7$ (D) MnO$_2$

   ›Reveal solutionSolution

   The size of manganese is largest when it has the lowest oxidation state, because higher oxidation states pull electrons in more tightly. Among the given compounds, Mn in MnO₂ has the lowest oxidation state (+4), so (D) MnO₂ is the answer.

   The key concept here is oxidation state and its effect on ionic radius. When an atom loses more electrons (higher oxidation state), the remaining electrons are held more tightly by the increased nuclear charge, shrinking the atom or ion. So, to find where manganese is largest, we need the compound where Mn has the lowest positive oxidation state.

   Let’s determine the oxidation state of Mn in each compound:

   1. KMnO₄: Potassium is +1, oxygen is –2 (four oxygens = –8). Let Mn be $x$.

      $(+1) + x + (-8) = 0 \Rightarrow x = +7$.

      Mn is in the +7 oxidation state — very high, so the ion is small.

   2. K₂MnO₄: Two potassiums = +2, four oxygens = –8.

      $(+2) + x + (-8) = 0 \Rightarrow x = +6$.

      Mn is +6 — still high, but slightly lower than +7.

   3. Mn₂O₇: Oxygen is –2, seven oxygens = –14. Let each Mn be $x$.

      $2x + (-14) = 0 \Rightarrow x = +7$.

      Again, Mn is +7 — same as in KMnO₄.

   4. MnO₂: Oxygen is –2, two oxygens = –4.

      $x + (-4) = 0 \Rightarrow x = +4$.

      Mn is +4 — the lowest oxidation state among the options. …

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3. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.In acidic medium one mole each of $\mathrm{MnO_4^-}$ and $\mathrm{Cr_2O_7^{2-}}$ is reduced by $x$ and $y$ moles of ferrous ions. The sum of $x$ and $y$ is (A) 14 (B) 12 (C) 10 (D) 11

   ›Reveal solutionSolution

   The key is to balance the half‑reactions in acidic medium: each mole of $\mathrm{MnO_4^-}$ accepts 5 electrons, each mole of $\mathrm{Cr_2O_7^{2-}}$ accepts 6 electrons, and each ferrous ion donates 1 electron. Thus $x = 5$, $y = 6$, and $x + y = 11$.

   Concept & Intuition

   The problem is about redox stoichiometry. In acidic solution, permanganate and dichromate are strong oxidizers; ferrous ion ($\mathrm{Fe^{2+}}$) is a common reducing agent that gets oxidized to ferric ion ($\mathrm{Fe^{3+}}$). The number of moles of $\mathrm{Fe^{2+}}$ needed to reduce one mole of oxidizer equals the number of electrons that oxidizer gains. So we just need the electron‑change per mole for each oxidizer, then add them.

   Step‑by‑step reasoning

   1. Determine the electron change for $\mathrm{MnO_4^-}$ in acidic medium The half‑reaction is:

   $\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$

   Manganese goes from oxidation state +7 to +2, a gain of 5 electrons.

   Therefore, 1 mole of $\mathrm{MnO_4^-}$ requires 5 moles of electrons.

   2. Relate electrons to ferrous ions Each $\mathrm{Fe^{2+}}$ loses one electron when oxidized to $\mathrm{Fe^{3+}}$:

   $\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}$

   So 1 mole of $\mathrm{Fe^{2+}}$ supplies exactly 1 mole of electrons.

   Hence, to supply 5 moles of electrons, we need 5 moles of $\mathrm{Fe^{2+}}$.

   Thus $x = 5$.

   3. Determine the electron change for $\mathrm{Cr_2O_7^{2-}}$ in acidic medium The half‑reaction is:

   $\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O}$

   Each chromium goes from +6 to +3, so two chromium atoms gain a total of 6 electrons. …

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4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Work functions of four metals $M_1$, $M_2$, $M_3$ and $M_4$ are $4.8$, $4.3$, $4.75$ and $3.75$ eV respectively. The metals which do not show photoelectric effect when light of wavelength $310$ nm falls on the metals are (A) $M_1$, $M_2$ only (B) $M_1$, $M_3$ only (C) $M_1$, $M_2$, $M_3$ only (D) $M_1$, $M_2$, $M_4$ only

   ›Reveal solutionSolution

   The photoelectric effect occurs only if the incident photon energy exceeds the metal’s work function. For λ = 310 nm, the photon energy is 4.0 eV; metals with work function > 4.0 eV (M₁, M₂, M₃) do not emit electrons. The correct option is (C).

   Concept & Intuition

   The photoelectric effect requires that the energy of a single photon be at least as large as the work function (the minimum energy needed to eject an electron). If the photon energy is too small, no electrons are emitted, no matter how intense the light. So we compute the photon energy for the given wavelength and compare it to each metal’s work function. Metals with work function greater than the photon energy will not show the effect.

   ---

   1. Find the photon energy The energy of a photon is

   $E = \frac{hc}{\lambda}$

   where $h = 4.135667 \times 10^{-15} \ \text{eV·s}$ (Planck’s constant in eV·s), $c = 3.00 \times 10^8 \ \text{m/s}$, and $\lambda = 310 \ \text{nm} = 310 \times 10^{-9} \ \text{m}$.

   Compute:

   $E = \frac{(4.1357 \times 10^{-15})(3.00 \times 10^8)}{310 \times 10^{-9}} \ \text{eV}$

   First, $hc = 1240 \ \text{eV·nm}$ (a handy constant to remember).

   So

   $E = \frac{1240 \ \text{eV·nm}}{310 \ \text{nm}} = 4.0 \ \text{eV}.$

   Tip

   Memorize $hc \approx 1240 \ \text{eV·nm}$ — it turns wavelength-in-nm to energy-in-eV instantly.

   2. Compare with each work function
      • $M_1$: work function = 4.8 eV → 4.8 > 4.0 → no photoelectric effect.
      • $M_2$: work function = 4.3 eV → 4.3 > 4.0 → no photoelectric effect. …

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5. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.Identify the correct orders regarding atomic radii i. Cl > F > Li ii. P > C > N iii. Tm > Sm > Eu iv. Sr > Ca > Mg (A) i, ii, iii only (B) ii, iv only (C) ii, iii, iv only (D) iii, iv only

   ›Reveal solutionSolution

   Atomic radii generally increase down a group and decrease across a period, but the lanthanide contraction (with the Eu half-filled anomaly) causes exceptions among the f-block elements. After checking each sequence, only orders ii and iv are correct, so the answer is (B).

   Concept & Intuition

   Atomic radius trends are governed by two main forces:

   • Down a group: Adding electron shells increases size.
   • Across a period: Increasing nuclear charge pulls electrons inward, shrinking the atom. For the f-block (lanthanides), the lanthanide contraction — a gradual decrease in radius across the series due to poor shielding of 4f electrons — makes later elements smaller than earlier ones. We must check each pair/sequence against these principles.

   Step-by-step reasoning

   1. Order i: Cl > F > Li

      • Chlorine (Cl) is below fluorine (F) in Group 17, so Cl is larger than F. That part is correct.
      • But fluorine (F) is in Period 2, lithium (Li) is also in Period 2, and across a period atomic radius decreases from left to right. Li (Group 1) is much larger than F (Group 17). So F > Li is false; actually Li > F.
      • Therefore the whole order Cl > F > Li is incorrect.

   2. Order ii: P > C > N

      • Phosphorus (P) is in Period 3, carbon (C) and nitrogen (N) are in Period 2. P has an extra shell, so P is larger than both C and N. That part is fine.
      • Between C and N: both in Period 2, but N has a higher nuclear charge (7 vs. 6), so N is smaller than C. Hence C > N is correct.
      • So P > C > N is correct.

   3. Order iii: Tm > Sm > Eu

      • These are lanthanides: Tm (thulium, atomic number 69), Sm (samarium, 62), Eu (europium, 63).
      • Across the lanthanide series, atomic radius decreases (lanthanide contraction). So later elements (higher Z) are smaller.
      • Tm (Z=69) comes after Sm (Z=62) and Eu (Z=63), so Tm is smaller than both. The order Tm > Sm would mean Tm is larger — that’s false. …

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6. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Among Li, Na, O, S, the elements having the most negative electron gain enthalpy, the least negative electron gain enthalpy respectively are (A) O, Li (B) O, Na (C) S, Na (D) S, Li

   ›Reveal solutionSolution

   Electron gain enthalpy becomes more negative across a period and less negative down a group. Among Li, Na, O, S, the most negative is S and the least negative is Na, so the correct option is (C).

   The concept here is electron gain enthalpy — the energy change when an isolated gaseous atom gains an electron. A more negative value means the atom releases more energy, so it has a stronger tendency to accept an electron. Two trends govern this: across a period, it becomes more negative (nuclear charge increases, pulling in the electron); down a group, it becomes less negative (atomic size increases, shielding weakens the pull). But there's a twist: oxygen is an exception because of its small size and high electron-electron repulsion in the 2p subshell, making its electron gain enthalpy less negative than sulfur's.

   Let's place the elements and reason step by step.

   1. Identify the groups and periods.

      Li and Na are in Group 1 (alkali metals). Li is in period 2, Na in period 3.

      O and S are in Group 16 (chalcogens). O is in period 2, S in period 3.

   2. Compare within Group 16: O vs S.

      Down a group, electron gain enthalpy becomes less negative (i.e., more positive). But O is an exception: its small size causes strong repulsion when adding an electron to the already half-filled 2p⁴ subshell. So O actually has a less negative electron gain enthalpy than S.

      Watch out

      A common mistake is to assume O is more negative because it's smaller. The repulsion in the compact 2p subshell makes O's electron gain enthalpy only about −141 kJ/mol, while S's is −200 kJ/mol. So S is more negative. …

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7. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.The following orbital energies (E) are compared. Identify the correct sets (I) $E_{2s}(\text{H}) = E_{2p}(\text{H})$ (II) $E_{2s}(\text{H}) = E_{2s}(\text{He})$ (III) $E_{2s}(\text{H}) < E_{2s}(\text{He})$ (IV) $E_{3s}(\text{He}) < E_{3s}(\text{H})$ (A) I, II only (B) I, II, III, IV (C) III, IV only (D) I, IV only

   ›Reveal solutionSolution

   The key idea is that in hydrogen, orbitals of the same principal quantum number are degenerate (same energy), but in multi-electron atoms like helium, orbital energies depend on both $n$ and $l$ due to shielding and penetration. The correct sets are I and IV only.

   The question asks you to compare orbital energies across different atoms and subshells. This is a classic test of how well you understand the difference between a hydrogen-like atom (single electron, no electron-electron repulsion) and a multi-electron atom (where electrons shield each other from the nucleus).

   Let’s break it down concept by concept.

   Why does hydrogen have degenerate orbitals? In a hydrogen atom, there is only one electron. The energy of an orbital depends only on the principal quantum number $n$, not on the azimuthal quantum number $l$. So $2s$ and $2p$ have exactly the same energy. That’s statement I.

   Why does helium break that degeneracy? Helium has two electrons. They repel each other. An electron in a $2s$ orbital spends more time closer to the nucleus (it has a higher penetration) than one in a $2p$ orbital. So a $2s$ electron in helium feels a stronger effective nuclear charge and is more tightly bound — lower (more negative) energy — than a $2p$ electron. But here we are comparing across atoms, not within the same atom.

   Now let’s go through each statement one by one.

   1. Statement I: $E_{2s}(\text{H}) = E_{2p}(\text{H})$

      In hydrogen, the energy depends only on $n$. Since both $2s$ and $2p$ have $n=2$, their energies are identical. This is true.

   2. Statement II: $E_{2s}(\text{H}) = E_{2s}(\text{He})$

      Hydrogen has nuclear charge $Z=1$; helium has $Z=2$. For a single electron in a $2s$ orbital, the energy scales as $-Z^2/n^2$. But helium’s $2s$ electron is not alone — the other electron (in $1s$) shields it. The effective nuclear charge felt by the $2s$ electron in helium is much less than 2, but still greater than 1. So the $2s$ orbital in helium is lower in energy (more negative) than in hydrogen. They are not equal. This is false.

   3. Statement III: $E_{2s}(\text{H}) < E_{2s}(\text{He})$

      “Less than” here means more negative (more stable). As argued above, the $2s$ electron in helium feels a larger effective nuclear charge than the $2s$ electron in hydrogen, so its energy is lower (more negative). So $E_{2s}(\text{H})$ is actually higher (less negative) than $E_{2s}(\text{He})$. The inequality is reversed: $E_{2s}(\text{H}) > E_{2s}(\text{He})$. So statement III is false. …

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8. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.Identify the pair of elements in which the difference in atomic radii is maximum (A) C, N (B) O, F (C) P, S (D) Li, Be

   ›Reveal solutionSolution

   The key idea is that atomic radii decrease across a period, and the largest drop occurs between the first two elements of a period. Among the given pairs, Li and Be (period 2) show the maximum difference in atomic radii.

   The question asks for the pair with the maximum difference in atomic radii. Atomic radii generally decrease from left to right across a period due to increasing nuclear charge pulling electrons closer. The decrease is not uniform—it is steepest at the start of a period, where the effective nuclear charge jumps significantly.

   Let’s examine each pair:

   1. Pair (A): C, N

      Both are in period 2, but they are adjacent elements near the middle of the period. The decrease in radius from C to N is modest because the added proton’s pull is partially offset by electron-electron repulsion in the same shell.

   2. Pair (B): O, F

      Also in period 2, but near the end. The radii are already small, and the difference between O and F is even smaller than between C and N—the contraction tapers off.

   3. Pair (C): P, S

      These are in period 3. While the absolute radii are larger, the difference between adjacent elements in period 3 is similar to that in period 2, but not larger than the biggest drop in period 2.

   4. Pair (D): Li, Be …

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9. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.In group 14 elements, the element with highest melting point is Y and element with lowest melting point is X. X and Y respectively are (A) Si, C (B) Si, Pb (C) C, Sn (D) Sn, C

   ›Reveal solutionSolution

   Melting points fall down Group 14 from carbon to lead, but tin is the one true exception — its melting point (232°C) is actually lower than lead's (327°C) because of tin's weaker metallic bonding. So the lowest-melting element is Sn (X) and the highest-melting is C (Y), matching option (D).

   ---

   Concept & Intuition

   Group 14 (C, Si, Ge, Sn, Pb) spans a non-metal (carbon), two metalloids (Si, Ge), and two metals (Sn, Pb). Carbon exists as a giant covalent solid (diamond) with very strong C–C bonds, giving it an exceptionally high melting point. Moving down the group, the bonding weakens overall, but the fall isn't perfectly smooth — tin's melting point actually dips below lead's because of how their metallic lattices pack.

   ---

   Step-by-step reasoning

   1. Recall the melting points across the group

      Carbon (diamond) ≈ 3550°C, silicon ≈ 1414°C, germanium ≈ 938°C, tin ≈ 232°C, lead ≈ 327°C.

   2. Identify the highest melting point

      Carbon is far ahead of the rest, so Y (highest) = C.

   3. Identify the lowest melting point

      Among the group, tin (232°C) melts at a lower temperature than lead (327°C), so X (lowest) = Sn.

   4. Match with the options

      • (A) Si, C — X = Si is not the lowest.
      • (B) Si, Pb — neither matches. …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.The set containing the elements with positive electron gain enthalpies is (A) S, Se, Te (B) Kr, Xe, Rn (C) Cl, Br, I (D) K, Rb, Cs

    ›Reveal solutionSolution

    Electron gain enthalpy is usually negative (energy released) for most non‑metals, but noble gases and certain heavy elements have positive values because adding an electron requires energy. Among the given sets, only the noble gases Kr, Xe, Rn have positive electron gain enthalpies, so the correct option is (B).

    Concept & Intuition

    Electron gain enthalpy (or electron affinity) is the energy change when an atom in the gas phase gains an electron. A negative value means energy is released (favourable), while a positive value means energy must be supplied (unfavourable). Most elements have negative electron gain enthalpies, but noble gases have completely filled shells, so adding an electron forces it into a higher-energy orbital, requiring energy input. Similarly, some heavy elements with stable half‑filled or fully‑filled subshells can also show positive values, but the most clear‑cut case is the noble gases.

    Step‑by‑Step Reasoning

    1. Recall the trend for electron gain enthalpy

       • Across a period, electron gain enthalpy becomes more negative (more energy released) as we move toward the halogens.
       • Down a group, it generally becomes less negative (or more positive) because the added electron goes into a larger orbital, experiencing less effective nuclear charge.
       • Exception: Noble gases have positive electron gain enthalpies because their octet is complete; adding an electron requires overcoming repulsion and entering a new shell.

    2. Examine each option

       • (A) S, Se, Te – These are chalcogens (Group 16). They all have negative electron gain enthalpies (e.g., S: –200 kJ/mol, Se: –195 kJ/mol, Te: –190 kJ/mol). So not positive.
       • (B) Kr, Xe, Rn – These are noble gases (Group 18). Their electron gain enthalpies are positive: Kr ≈ +60 kJ/mol, Xe ≈ +77 kJ/mol, Rn ≈ +68 kJ/mol. This matches the condition. …

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11. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.The wavenumber of first spectral line of Lyman series of He$^+$ ion is $x$ m$^{-1}$. What is the wavenumber (in m$^{-1}$) of second spectral line of Balmer series of Li$^{2+}$ ion? (A) $\dfrac{9x}{16}$ (B) $\dfrac{16x}{9}$ (C) $\dfrac{8x}{27}$ (D) $\dfrac{27x}{8}$

    ›Reveal solutionSolution

    For a hydrogen-like ion the wavenumber is $\tilde{\nu}=RZ^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$. Fixing $R$ from the He$^+$ Lyman first line and evaluating the Li$^{2+}$ Balmer second line gives $\frac{9x}{16}$, option (A).

    For any one-electron (hydrogen-like) ion the Rydberg formula applies with the nuclear charge $Z$:

    $\tilde{\nu}=RZ^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

    The Rydberg constant $R$ is common to all such ions, so it cancels when we form the ratio of the two lines.

    Step 1 - The given line (He$^+$, $Z=2$).

    The first line of the Lyman series is the transition $n_i=2\to n_f=1$:

    $x=R(2)^2\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=4R\cdot\frac{3}{4}=3R \quad\Rightarrow\quad R=\frac{x}{3}$

    Step 2 - The required line (Li$^{2+}$, $Z=3$).

    The second line of the Balmer series is the transition $n_i=4\to n_f=2$:

    $\tilde{\nu}=R(3)^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=9R\left(\frac{1}{4}-\frac{1}{16}\right)=9R\cdot\frac{3}{16}=\frac{27R}{16}$ …

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12. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.In which of the following ionic pairs, second ion is smaller in size than the first ion? (A) $\mathrm{Al^{3+}},\ \mathrm{Mg^{2+}}$ (B) $\mathrm{F^-},\ \mathrm{Na^+}$ (C) $\mathrm{O^{2-}},\ \mathrm{N^{3-}}$ (D) $\mathrm{Mg^{2+}},\ \mathrm{Na^+}$

    ›Reveal solutionSolution

    The key idea is that ionic size depends on nuclear charge and number of electrons. For isoelectronic species, higher nuclear charge pulls electrons in tighter, making the ion smaller. The pair where the second ion is smaller is (B).

    Concept & Intuition

    When comparing ionic sizes, two main factors matter:

    • Number of electron shells – more shells = larger radius.
    • Effective nuclear charge – for ions with the same number of electrons (isoelectronic), the one with more protons pulls the electron cloud inward more strongly, making it smaller.

    Here, all pairs are either isoelectronic or have different electron counts. We check each pair systematically.

    ---

    1. Pair (A): $\mathrm{Al^{3+}}$ and $\mathrm{Mg^{2+}}$

       • Both have 10 electrons (isoelectronic with neon).
       • $\mathrm{Al^{3+}}$ has 13 protons, $\mathrm{Mg^{2+}}$ has 12 protons.
       • More protons → stronger pull → smaller radius.
       • So $\mathrm{Al^{3+}}$ is smaller than $\mathrm{Mg^{2+}}$.
       • Here the first ion is smaller, not the second. → Not correct.

    2. Pair (B): $\mathrm{F^-}$ and $\mathrm{Na^+}$

       • Both have 10 electrons (isoelectronic with neon).
       • $\mathrm{F^-}$ has 9 protons, $\mathrm{Na^+}$ has 11 protons.
       • More protons → smaller radius.
       • So $\mathrm{Na^+}$ is smaller than $\mathrm{F^-}$.
       • The second ion is smaller. → This fits.

    3. Pair (C): $\mathrm{O^{2-}}$ and $\mathrm{N^{3-}}$

       • Both have 10 electrons (isoelectronic).
       • $\mathrm{O^{2-}}$ has 8 protons, $\mathrm{N^{3-}}$ has 7 protons.
       • More protons → smaller radius.
       • So $\mathrm{O^{2-}}$ is smaller than $\mathrm{N^{3-}}$.
       • Here the first ion is smaller, not the second. → Not correct. …

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