Q.The common oxidation state shown by the element with atomic number 21 is

Showing the 12 most recent of 22 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.The common oxidation state shown by the element with atomic number 21 is

   (a) +3

   (b) +4

   (c) +5

   (d) Both +3 and +5

   ›Reveal solutionSolution

   The element with Z = 21 is scandium, whose common (and essentially only stable) oxidation state is +3.

   Atomic number 21 corresponds to scandium (Sc) with electronic configuration $[\text{Ar}]\,3d^1\,4s^2$.

   Scandium loses its two $4s$ electrons and its single $3d$ electron to attain the stable, noble-gas [Ar] configuration:

   $\text{Sc} \rightarrow \text{Sc}^{3+} + 3e^-$ …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.What is the maximum oxidation state of Mn in its compounds?

   (a) +4

   (b) +5

   (c) +6

   (d) +7

   ›Reveal solutionSolution

   Manganese shows a maximum oxidation state of +7, equal to the sum of its 4s and 3d valence electrons.

   Manganese has the ground-state electronic configuration [Ar]3d^5 4s^2, giving it 7 electrons in its outermost (4s + 3d) shells. For the early-to-middle members of the 3d transition series, the maximum oxidation state shown is equal to the total number of 4s and 3d electrons, since all of them can, in principle, take part in bonding (this trend peaks around Mn and then declines as d-electrons become increasingly core-like towards the end of the series).

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following transition metals does not show variable oxidation state?

   (a) Ti

   (b) Cr

   (c) Cu

   (d) Sc

   ›Reveal solutionSolution

   Sc has only one electron beyond the noble-gas+d0 core to lose (3d1 4s2 -> Sc3+ is d0), so there is no intermediate oxidation state available; Ti, Cr and Cu all show at least two.

   Scandium's configuration is [Ar]3d¹4s²; losing all three of these electrons gives the very stable, empty-d-subshell Sc³⁺ (3d⁰) ion, which is the only oxidation state scandium is practically found in — it has no partly-filled-d intermediate oxidation state to show v …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which element does not show variable oxidation state ?

   (i) Vanadium

   (ii) Iron

   (iii) Mercury

   (iv) Scandium

   ›Reveal solutionSolution

   Scandium has only one stable, common oxidation state (+3) because losing all three electrons outside its noble-gas-like [Ar] core empties the 3d subshell completely — there is no other accessible, stable configuration.

   Most transition metals show variable oxidation states because both the (n-1)d and ns electrons are close in energy and can be lost in different numbers.

   • Vanadium: shows +2, +3, +4, +5 — clearly variable.
   • Iron: shows +2 and +3 (and rarely +6) — variable.
   • Mercury: shows +1 (as Hg₂²⁺) and +2 — variable. …

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5. CBSE 2024Set 56/3/11 markMCQ
   Q.Which of the following does not show variable oxidation states ? (A) Fe (B) Cu (C) Mn (D) Sc

   ›Reveal solutionSolution

   Transition metals show variable oxidation states when they can lose different numbers of $d$-electrons along with their $s$-electrons. Scandium has only one $d$-electron, giving it essentially one stable oxidation state (+3), while Fe, Cu, and Mn have multiple $d$-electrons that can be removed in different combinations. The answer is (D) Sc.

   Why transition metals show variable oxidation states

   Transition metals are famous for their ability to exist in multiple oxidation states. This happens because their $(n-1)d$ and $ns$ orbitals are close in energy, so electrons from both can participate in bonding. The more $d$-electrons available, the more combinations of electron loss are possible, leading to a richer variety of oxidation states.

   The key is to look at the electronic configuration and see how many electrons can realistically be removed to form stable ions.

   Analyzing each element

   Let's examine the electronic configurations and common oxidation states:

   1. Iron (Fe): [Ar] 3d⁶ 4s²

   Iron can lose its two $4s$ electrons to give Fe²⁺ ([Ar] 3d⁶). It can also lose one more $3d$ electron to give Fe³⁺ ([Ar] 3d⁵), which is particularly stable due to the half-filled $d$-subshell. Higher oxidation states like +4, +5, and +6 exist in certain compounds, though they're less common.

   Common oxidation states: +2, +3 (and higher in special cases)

   2. Copper (Cu): [Ar] 3d¹⁰ 4s¹

   Copper readily loses its single $4s$ electron to form Cu⁺ ([Ar] 3d¹⁰), which has a stable filled $d$-subshell. It can also lose one $3d$ electron to give Cu²⁺ ([Ar] 3d⁹), which is actually more common in aqueous chemistry due to higher hydration energy.

   Common oxidation states: +1, +2

   3. Manganese (Mn): [Ar] 3d⁵ 4s²

   Manganese is the champion of variable oxidation states among first-row transition metals. With five $d$-electrons and two $s$-electrons, it can lose anywhere from two to all seven electrons, giving oxidation states from +2 all the way to +7 (as in permanganate, MnO₄⁻).

   Common oxidation states: +2, +3, +4, +6, +7

   4. Scandium (Sc): [Ar] 3d¹ 4s²

   Here's the critical case. Scandium has only one $d$-electron. When it forms compounds, it loses both $4s$ electrons and its single $3d$ electron to achieve the stable [Ar] configuration, giving Sc³⁺. …

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6. CBSE 2024Set FZ1 markMCQ
   Q.The transition element in which variable oxidation state is $not$ found, is:

   (a) Sc

   (b) Ti

   (c) V

   (d) Cr

   ›Reveal solutionSolution

   Scandium exhibits only the $+3$ state, so variable oxidation state is not found in Sc → option (a).

   Concept. Transition metals normally show variable oxidation states because their $(n-1)d$ and $ns$ electrons have similar energies, so a variable number can be involved in bonding. The exception is an element that has just one accessible state.

   …

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7. CBSE 2024Set D1 markMCQ
   Q.The maximum oxidation state of chromium is

   (a) +2

   (b) +3

   (c) +4

   (d) +6

   ›Reveal solutionSolution

   Cr has the configuration [Ar]3d5 4s1, i.e. six electrons (5 in 3d + 1 in 4s) available for bonding, so its highest oxidation state is +6.

   Chromium (Z = 24) has electronic configuration [Ar]3d5 4s1. All six electrons in the 3d and 4s subshells can participate in bonding, so chromium can reach the +6 oxidation state, seen in chromate (CrO4^2-) and dichromate (C …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.Element showing the highest number of oxidation states is -

   (a) Mn

   (b) Ni

   (c) Fe

   (d) Cr

   ›Reveal solutionSolution

   Manganese, with the electronic configuration [Ar]3d5 4s2, shows oxidation states ranging from +2 to +7, the widest range of any 3d transition element.

   Among the first transition series, the number of oxidation states shown by an element is generally maximum near the middle of the series, where the largest number of unpaired d and s electrons are available for bonding. …

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9. CBSE 2024Set ANNUAL1 mark
   Q.Which element of the 3d series of the transition metals exhibits the largest number of oxidation states?

   ›Reveal solutionSolution

   Manganese, with its half-filled 3d5 4s2 configuration, can lose varying numbers of electrons to give the widest spread of oxidation states among the first-row (3d) transition metals: +2, +3, +4, +5, +6, and +7.

   Across the 3d transition series, the number of accessible oxidation states generally increases from Sc to Mn (as more d and s electrons become available for bonding/removal) and then decreases again from Fe to Zn (as the increasing nuclear charge makes it harder to remove d electrons and pairing energy effects set in).

   …

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10. CBSE 2023Set A1 mark
    Q.Match the following. Column A item: 'Mn'. Choose its correct match from Column B:

    (a) Ether

    (b) Primary amine

    (c) Lactose

    (d) C12H22O11

    (e) Glucose

    (f) Negative ions

    (g) C6H5SO2Cl

    (h) +7

    ›Reveal solutionSolution

    Manganese exhibits a maximum oxidation state of +7, seen in the permanganate ion.

    Among the given items, 'Mn' pairs with option (h) +7, because manganese's highest possible oxidation state — using all its 3d and 4s electro …

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11. CBSE 2023Set ANNUAL1 mark
    Q.Name a transition element which does not exhibit variable oxidation states.

    ›Reveal solutionSolution

    Most transition elements show variable oxidation states because both $4s$ and partially-filled $3d$ electrons can be lost with comparable ease; scandium is the exception among the first-row transition series, showing only the $+3$ state.

    Scandium has the electronic configuration $[Ar]\,3d^1 4s^2$. On forming compounds, it loses all three of these electrons (the two $4s$ and the single $3d$) to attain the very stable, empty-$d$, noble-gas-like configuration $[Ar]$ (i.e. $Sc^{3+}$). Because this $+3$ ion is exceptionally stable and there is no comparably accessible alternative oxidation state (there is only one $3d$ electron to begin with, unlike elements such as $Mn$ or $Fe$ that have sev …

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12. CBSE 2022Set E1 markMCQ
    Q.In which of the following is the oxidation state of Mn lowest ?

    (a) MnSO4

    (b) MnO2

    (c) Mn3O4

    (d) Mn2O7

    ›Reveal solutionSolution

    Mn is +2 in MnSO4, +4 in MnO2, +8/3 in Mn3O4, and +7 in Mn2O7 — so the lowest is +2 in MnSO4.

    Compute the oxidation state of Mn in each:

    • MnSO4: SO4 is -2, so Mn = +2.
    • MnO2: 2 oxygens give -4, so Mn = +4. …

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