Q.(a) What could be the IUPAC name and symbol for the element with atomic number 120? [1]

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The correct order of radii is

   (a) Na+ > F > Al+3 > Mg+2

   (b) N-3 < Na+ < Mg+2 < O-2

   (c) Al+3 < Mg+2 < O-2 < S-2

   (d) Mg+2 < N-3 < O-2 < F-

   ›Reveal solutionSolution

   For a set of isoelectronic cations, higher positive charge means smaller radius (more protons pulling in the same electron cloud); for anions of increasing size (e.g. O2- vs S2-, which has an extra electron shell), radius increases going down a group.

   Approximate ionic radii (pm): Al3+ ~ 53, Mg2+ ~ 72, O2- ~ 140, S2- ~ 184.

   Check option (c): Al+3 < Mg+2 < O-2 < S-2

   53 < 72 < 140 < 184 -- this is correctly increasing. Al3+ is smallest (highest positive charge pulling electrons in tightly, and it is a cation from period 3), Mg2+ is next (lower charge than Al3+), O2- is a fairly large anion (extra electron shell relative to a similarly-charged cation), and S2- is the largest here since it is a period-3 anion with one more electron shell than O2-.

   Checking why the others are wrong: …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following is the correct order of size of the given species? I, I⁻, I⁺

   (a) I > I⁻ > I⁺

   (b) I⁺ > I⁻ > I

   (c) I > I⁺ > I⁻

   (d) I⁻ > I > I⁺

   ›Reveal solutionSolution

   Anions are larger than the parent atom and cations are smaller, because the same nuclear charge acts on more or fewer electrons.

   Reasoning: All three species have the same nuclear charge (53 protons for iodine), but different numbers of electrons.

   • $I^-$ has gained an electron, so more electrons share the same nuclear pull, increasing electron-electron repulsion and expanding the electron cloud — largest.
   • $I$ (neutral) has its normal electron count. …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following ions has highest value of ionic radius?

   (a) Li+

   (b) B3+

   (c) O2-

   (d) F-

   ›Reveal solutionSolution

   O2- has the largest ionic radius among the given isoelectronic-type ions.

   Li+, B3+, O2- and F- are all small ions, and O2- and F- are isoelectronic with Ne (10 electrons), while Li+ and B3+ are isoelectronic with He (2 electrons). Among isoelectronic species, radius decreases as nuclear charge (and hence effective pull on the electron cloud) increases: for the 10-electron series, O2- (Z=8) is larger than F- (Z=9); for the 2-electron series, Li+ (Z=3 …

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4. CBSE 2025Set ANNUAL1 mark
   Q.Arrange the following in increasing order of radius: Metallic radius, covalent radius, Vander waal's radius.

   ›Reveal solutionSolution

   Increasing order: covalent radius < metallic radius < van der Waals radius.

   These three radii measure atomic size under different bonding situations:

   • Covalent radius: half the internuclear distance between two atoms joined by a single covalent bond — the tightest, shortest measure since the atoms' electron clouds overlap strongly.
   • Metallic radius: half the internuclear distance between two adjacent atoms in a metallic crystal lattice — larger than covalent radius because metallic bonding is generally weaker/more diffuse than a covalent bond. …

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5. CBSE 2025Set ANNUAL1 mark
   Q.Atomic size increases from left to right in a period.

   ›Reveal solutionSolution

   The statement is False: atomic size decreases (not increases) from left to right across a period.

   Across a period, electrons are added to the same principal shell while the nuclear charge (number of protons) increases with each element. The increasing effective nuclear charge pulls the electron cloud in more strongly, so atomic radius steadily decreases from …

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6. CBSE 2024Set ANNUAL1 markMCQ
   Q.Which of the following cations has the smallest size?

   (a) Na^+

   (b) Mg^2+

   (c) Ca^2+

   (d) Al^3+

   ›Reveal solutionSolution

   Al³⁺ is the smallest cation, being isoelectronic with Na⁺ and Mg²⁺ but with the highest nuclear charge.

   Na⁺, Mg²⁺, and Al³⁺ are all isoelectronic species (10 electrons each, neon configuration). Among isoelectronic ions, size decreases as nuclear charge increases (more protons pull the same number of electrons in more tightly): $Na^+ (Z=11) > Mg^{2+} (Z=12) > Al^{3+} (Z=13)$ in size order, …

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7. CBSE 2024Set ANNUAL1 markMCQ
   Q.Out of the options given below, choose the correct order of atomic/ionic radii of sodium (Na) atom and sodium ion (Na+) in pm

   (a) A) 95, 186

   (b) B) 186, 95

   (c) C) 95, 95

   (d) D) 186, 186

   ›Reveal solutionSolution

   [!TLDR]

   B) 186, 95

   Why

   Na atom radius is about 186 pm; on losing its valence electron to form Na+, the radius shrinks to about 95 pm due to loss of a shell and …

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8. CBSE 2023Set ANNUAL1 markMCQ
   Q.The ionic radii of N^3-, O^2-, F^-, Na^+ follow the order

   (a) N^3- > O^2- > F^- > Na^+

   (b) N^3- > Na^+ > O^2- > F^-

   (c) Na^+ > O^2- > N^3- > F^-

   (d) O^2- > F^- > Na^+ > N^3-

   ›Reveal solutionSolution

   N3-, O2-, F-, and Na+ are all isoelectronic (10 electrons each, like neon), so their radii are governed purely by nuclear charge: more protons pull the same electron cloud in tighter, shrinking the radius as Z increases.

   All four species have exactly 10 electrons (configuration of Ne: 1s²2s²2p⁶):

   • N (Z=7) as N3-
   • O (Z=8) as O2-
   • F (Z=9) as F-
   • Na (Z=11) as Na+ …

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9. CBSE 2023Set ANNUAL1 markMCQ
   Q.The correct order of the atomic size of C, N, P, S follows the order.

   (a) N < C < P < S

   (b) C < N < S < P

   (c) C < N < P < S

   (d) N < C < S < P

   ›Reveal solutionSolution

   Correct increasing size order: N < C < S < P.

   Two periodic trends govern atomic size: it decreases across a period (nuclear charge rises) and increases down a group (a new shell is added). Approximate covalent radii (pm): C = 77, N = 75, P = 110, S = 104.

   • Within period 2: C (77) > N (75), so N < C.
   • Within period 3: P (110) > S (104), so S < P.
   • Period-3 atoms are larger than period-2 atoms. …

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10. CBSE 2022Set ANNUAL1 mark
    Q.Which of the following species will have largest size ? Mg, Mg2+, Al, Al3+

    ›Reveal solutionSolution

    Cations are always smaller than their parent atoms, and Mg is bigger than Al (periodic trend); so the largest species of the four is neutral Mg.

    Two trends combine here:

    1. Removing electrons to form a cation reduces electron-electron repulsion and increases effective nuclear charge per electron, shrinking the species. So Mg > Mg$^{2+}$ and Al > Al$^{3+}$. …

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11. CBSE 2022Set ANNUAL1 mark
    Q.What will be the correct order of hydration energy of the following alkali metal ions: Na+, Rb+, K+, Li+?

    ›Reveal solutionSolution

    Hydration energy decreases as ionic radius increases; among Li+, Na+, K+ and Rb+ (ionic size increasing in that order down Group 1), the correct order of hydration energy is Li+ > Na+ > K+ > Rb+.

    Hydration energy is the energy released when gaseous ions are surrounded by water molecules (ion-dipole attraction). A smaller ion has a higher charge density (same +1 charge concentrated over a smaller volume), so it attracts the partially-negative oxygen ends of water molecules more strongly, releasing more energy on hydration.

    …

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12. CBSE 2021Set ANNUAL1 markMCQ
    Q.Which alkali metal gives hydrated salt?

    (a) Li

    (b) Na

    (c) K

    (d) Cs

    ›Reveal solutionSolution

    Lithium, being the smallest alkali metal ion with the highest charge density, is strongly hydrated in solution and crystallises with water of hydration (e.g. $LiCl \cdot 2H_2O$), unlike Na, K, or Cs salts which are typically anhydrous.

    Step 1 — Why lithium is different: Lithium's ionic radius is the smallest among Group 1 elements. A small ion with a +1 charge has a very high charge density, which lets it polarise and attract surrounding water molecules strongly.

    Step 2 — Consequence: This strong hydration means Li$^+$ salts often crystallise out of aqueous solution carrying water molecules with them (e.g. $LiCl\cdot 2H_2O$), whereas the larger, less charge-dense Na, K, and Cs ions form mostly anhydrous salts.

    …

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