Q.Transition metals have incomplete d-subshell either in neutral atom or in their ions. The presence of partly filled d-orbitals in their atoms makes transition elements different from that of the non-transition elements. With partly filled d-orbitals, these elements exhibit certain characteristic properties such as display of a variety of oxidation states, formation of coloured ions and entering into complex formation with a variety of ligands. The transition metals and their compounds also exhibit catalytic properties and paramagnetic behaviour. The transition metals are very hard and have low volatility. An examination of the $E^\circ_{(M^{2+}/M)}$ values shows the varying trends : $E^\circ_{(M^{2+}/M)}$/V : V $-1·18$, Cr $-0·91$, Mn $-1·18$, Fe $-0·44$, Co $-0·28$, Ni $-0·25$, Cu $+0·34$, Zn $-0·76$ Answer the following questions :

Showing the 12 most recent of 35 on this concept.

1. CBSE 2026Set 56/3/11 markMCQ
   Q.In aqueous solution, $Cr_2O_7^{2-}$ ion converts to which of the following in alkaline medium ? (A) $Cr^{3+}$ (B) $CrO_4^{2-}$ (C) $CrO$ (D) $CrO_3$

   ›Reveal solutionSolution

   In alkaline medium, dichromate ($Cr_2O_7^{2-}$) converts to chromate ($CrO_4^{2-}$) without any change in oxidation state — it’s a simple acid-base equilibrium, not a redox reaction. The correct option is (B).

   The key to this question lies in understanding that the conversion of dichromate to chromate is not a redox reaction — the oxidation state of chromium remains +6 throughout. Many students instinctively think of reduction to $Cr^{3+}$ because they associate dichromate with strong oxidizing behaviour, but that only happens in acidic medium. In alkaline conditions, the chemistry is entirely different.

   Let’s walk through the reasoning step by step.

   1. Recall the oxidation state of chromium in dichromate.

      In $Cr_2O_7^{2-}$, each oxygen is -2, so total from seven oxygens is -14. The ion has a -2 charge, so the sum of oxidation states of the two chromium atoms must be +12. Hence each Cr is in the +6 state.

   2. Now consider the alkaline medium.

      When you add a base (like NaOH) to a solution of $K_2Cr_2O_7$, the dichromate ion reacts with hydroxide ions. The reaction is:

   $Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O$

   Notice that the oxidation state of Cr in $CrO_4^{2-}$ is also +6 (four oxygens at -2 give -8, charge -2, so Cr = +6). No electrons are transferred — this is an acid-base equilibrium, not a redox change.

   3. Why does this happen? Dichromate exists in equilibrium with chromate, and the position depends on pH. In acidic solution, the equilibrium shifts toward dichromate; in alkaline solution, it shifts toward chromate. The reaction is:

   $2CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$

   Adding $OH^-$ removes $H^+$, pulling the equilibrium to the left — producing chromate. …

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2. CBSE 2026Set 56/3/11 markMCQ
   Q.On electrolysis of very dilute aqueous solution of NaCl using platinum electrodes : (A) $H_2$ gas is evolved at anode. (B) Na is produced at cathode. (C) $O_2$ gas is evolved at anode. (D) $H_2$ gas is evolved at cathode.

   ›Reveal solutionSolution

   In very dilute aqueous NaCl with inert Pt electrodes, water’s reduction to $H_2$ at the cathode and water’s oxidation to $O_2$ at the anode outcompete the NaCl reactions. So $H_2$ is produced at the cathode and $O_2$ at the anode — making option (C) and (D) correct.

   Why standard electrode potentials decide the outcome

   Electrolysis is a battle of competing half-reactions. At each electrode, the species that is easier to oxidise (at the anode) or easier to reduce (at the cathode) will react first. “Easier” means having a more positive reduction potential for reduction, or a more negative reduction potential for oxidation (equivalently, a more positive oxidation potential).

   For a very dilute aqueous solution of NaCl, the possible species are:

   • Cathode (reduction): $Na^+$ ions and $H_2O$ molecules.
   • Anode (oxidation): $Cl^-$ ions and $H_2O$ molecules.

   We compare their standard reduction potentials (at 298 K, 1 M concentration, 1 atm pressure). But remember: concentration matters. In very dilute NaCl, $[Cl^-]$ is tiny, which shifts the actual potential of the chlorine half-reaction significantly.

   ---

   Step-by-step reasoning

   1. What happens at the cathode?

   Two reduction half-reactions compete:

   $\begin{aligned} Na^+ + e^- &\rightarrow Na(s) \quad E^\circ = -2.71\ \text{V} \\[4pt] 2H_2O + 2e^- &\rightarrow H_2(g) + 2OH^- \quad E^\circ = -0.83\ \text{V} \end{aligned}$

   The reduction of water to hydrogen gas has a much less negative (i.e., more positive) standard potential. Even though the actual potential for water reduction depends slightly on pH (here neutral to slightly basic), it remains far above $-2.71\ \text{V}$. So water is reduced preferentially.

   Watch out

   A common mistake is to think that because $Na^+$ is present, sodium metal will plate out. But sodium’s reduction potential is so negative that water (even in neutral solution) is reduced first. Sodium metal would instantly react with water anyway — it’s never produced in aqueous electrolysis.

   Result at cathode: $H_2$ gas is evolved. This matches option (D).

   ---

   2. What happens at the anode?

   Two oxidation half-reactions compete (written as reductions for comparison):

   $\begin{aligned} Cl_2(g) + 2e^- &\rightarrow 2Cl^- \quad E^\circ = +1.36\ \text{V} \\[4pt] O_2(g) + 4H^+ + 4e^- &\rightarrow 2H_2O \quad E^\circ = +1.23\ \text{V} \end{aligned}$ …

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3. CBSE 2026Set 56/1/11 markMCQ
   Q.Assertion (A) : Actinoids show wide range of oxidation states. Reason (R) : Actinoids are radioactive in nature. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

   ›Reveal solutionSolution

   The assertion that actinoids show a wide range of oxidation states is true, but the reason given — that they are radioactive — is not the correct explanation. The correct explanation lies in the small energy gap between 5f, 6d, and 7s orbitals, which allows many electrons to participate in bonding. So the answer is option (B).

   The question tests your understanding of why actinoids (elements 90–103, from thorium to lawrencium) exhibit so many different oxidation states. Many students memorise that “actinoids show variable oxidation states” and also know they are radioactive, so they assume the second explains the first. That’s a trap.

   Let’s break it down properly.

   1. Is Assertion (A) true?

      Yes. Actinoids display a remarkably wide range of oxidation states. For example, uranium shows +3, +4, +5, and +6; neptunium and plutonium go from +3 to +7. This is far more varied than most d-block elements. The reason is that the 5f, 6d, and 7s orbitals are very close in energy. Electrons from all three can be lost with relatively little energy cost, so many different oxidation numbers become accessible.

   2. Is Reason (R) true?

      Yes, actinoids are indeed radioactive. All actinoid nuclei are unstable and decay over time. So the reason statement is factually correct.

   3. Does the radioactivity explain the wide range of oxidation states?

      No. Radioactivity is a nuclear property — it depends on the instability of the nucleus (proton/neutron ratio, nuclear binding energy). Oxidation states are an electronic property — they depend on how easily electrons are lost from the outer orbitals. These two phenomena are completely independent.

      Watch out

      A common mistake is to think that because both statements are true, the reason must be the explanation. But correlation is not causation. Radioactivity does not cause variable oxidation states; the orbital energy structure does.

   4. What actually causes the wide range of oxidation states in actinoids? …

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4. CBSE 2026Set 56/2/11 markMCQ
   Q.Consider the following reaction : $Zn(s) + Ag_2O(s) + H_2O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^-(aq)$ Given : $E^o_{Ag^+/Ag} = 0.80$ V, $E^o_{Zn^{2+}/Zn} = -0.76$ V, $1\,F = 96500$ C mol$^{-1}$ $\Delta_r G^o$ for the above reaction is : (A) $-301.080$ kJ mol$^{-1}$ (B) $+310.080$ kJ mol$^{-1}$ (C) $-326.070$ kJ mol$^{-1}$ (D) $-375.060$ kJ mol$^{-1}$

   ›Reveal solutionSolution

   Zinc is oxidised and silver is reduced, giving $E^o_{cell} = 0.80 - (-0.76) = 1.56$ V with $n = 2$. Then $\Delta_r G^o = -nFE^o_{cell} = -301.080$ kJ mol$^{-1}$, which is option (A).

   The standard Gibbs energy of a cell reaction is linked to its standard cell potential by

   $\Delta_r G^o = -nFE^o_{cell}$

   so we first find $E^o_{cell}$, then $n$, and finally $\Delta_r G^o$.

   1. Identify the electrodes. Zinc is oxidised (anode) and silver is reduced (cathode):

   $\text{Anode:}\quad Zn \rightarrow Zn^{2+} + 2e^-$

   $\text{Cathode:}\quad Ag_2O + H_2O + 2e^- \rightarrow 2Ag + 2OH^-$

   2. Standard cell potential. Using the given reduction potentials,

   $E^o_{cell} = E^o_{cathode} - E^o_{anode} = 0.80 - (-0.76) = 1.56\ \text{V}$

   3. Electrons transferred. Each half-reaction involves 2 electrons, so $n = 2$. …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following oxidation states is common for all lanthanoids?

   (a) +2

   (b) +3

   (c) +4

   (d) +5

   ›Reveal solutionSolution

   All lanthanoids show a characteristic +3 oxidation state because it corresponds to a stable, similar electronic configuration achieved after losing the two 6s and one 4f (or 5d) electron.

   Lanthanoids have the general electronic configuration [Xe] 4f^(1-14) 5d^(0-1) 6s2. Removal of the two 6s electrons and one more electron (from 4f or 5d) gives the Ln3+ ion, which is the most stable and commonly observed oxidation state across the entire series, from Ce to Lu.

   …

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6. CBSE 2025Set 56/4/11 markMCQ
   Q.The product of the oxidation of $I^{-}$ with $MnO_4^{-}$ in alkaline medium is : (A) $IO_4^{-}$ (B) $I_2$ (C) $IO^{-}$ (D) $IO_3^{-}$

   ›Reveal solutionSolution

   In alkaline medium, permanganate ($MnO_4^-$) oxidises iodide ($I^-$) to iodate ($IO_3^-$), not to iodine or periodate. The balanced reaction shows $I^-$ loses 6 electrons to form $IO_3^-$, while $MnO_4^-$ gains 3 electrons to form $MnO_2$. The correct product is $IO_3^-$, option (D).

   Why the medium matters

   The oxidation state of iodine in its products depends heavily on the pH of the solution. Permanganate is a powerful oxidising agent, but its reduction product changes with medium:

   • In acidic medium: $MnO_4^- \rightarrow Mn^{2+}$ (gains 5 electrons)
   • In neutral/alkaline medium: $MnO_4^- \rightarrow MnO_2$ (gains 3 electrons)

   This difference in electron gain per mole of permanganate directly affects how far it can oxidise iodide. In alkaline medium, permanganate is a milder oxidising agent (gains only 3 electrons) compared to acidic medium (gains 5 electrons). Yet it still oxidises $I^-$ all the way to $IO_3^-$, not stopping at $I_2$.

   Step-by-step reasoning

   1. Identify the half-reactions

      Iodide ($I^-$) has oxidation state $-1$. The possible products given are:

      • $IO_4^-$: iodine in $+7$ state
      • $I_2$: iodine in $0$ state
      • $IO^-$: iodine in $+1$ state (hypoiodite)
      • $IO_3^-$: iodine in $+5$ state (iodate)

      In alkaline medium, permanganate reduces to $MnO_2$ (manganese in $+4$ state, from $+7$ in $MnO_4^-$).

   2. Balance the oxidation half-reaction

      Iodide going to iodate:

   $I^- \rightarrow IO_3^-$

   Balance oxygen with water (alkaline medium):

   $I^- + 3H_2O \rightarrow IO_3^- + 6H^+$

   Balance charge: left side has $-1$, right side has $-1 + 6 = +5$. Add 6 electrons to right:

   $I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$

   In alkaline medium, add $OH^-$ to neutralise $H^+$:

   $I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$

   So each $I^-$ loses 6 electrons to become $IO_3^-$.

   3. Balance the reduction half-reaction

      Permanganate to manganese dioxide in alkaline medium:

   $MnO_4^- \rightarrow MnO_2$

   Balance oxygen with water:

   $MnO_4^- + 2H_2O \rightarrow MnO_2 + 4OH^-$

   Balance charge: left $-1$, right $-4$. Add 3 electrons to left:

   $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$

   So each $MnO_4^-$ gains 3 electrons.

   4. Combine the half-reactions

      To equalise electrons: multiply reduction half by 2 (gives 6 electrons gained) and oxidation half by 1 (gives 6 electrons lost):

   $2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$

   $I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-$

   Adding:

   $2MnO_4^- + I^- + 4H_2O + 6OH^- \rightarrow 2MnO_2 + IO_3^- + 3H_2O + 8OH^-$

   Cancel $3H_2O$ from both sides and $6OH^-$ from both sides: …

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7. CBSE 2025Set 56/6/11 markMCQ
   Q.For the following question, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Actinoids show wide range of oxidation states. Reason (R) : Actinoids are radioactive in nature.

   ›Reveal solutionSolution

   The assertion that actinoids show a wide range of oxidation states is true, but the reason given — that they are radioactive — does not explain this property. The correct answer is (B).

   The question tests your understanding of why actinoids exhibit variable oxidation states. The key is to separate two distinct facts: actinoids are radioactive, and they do show many oxidation states — but the radioactivity is not the cause of the oxidation state variability.

   Let’s break this down.

   1. Why do actinoids show a wide range of oxidation states?

      The 5f, 6d, and 7s orbitals in actinoids are very close in energy. This means electrons can be removed from any of these orbitals with relatively little energy cost. As you move across the actinoid series, the 5f orbitals gradually become more stable, but early actinoids (like Th, Pa, U, Np, Pu) can lose anywhere from 3 to 7 electrons. For example, uranium shows +3, +4, +5, and +6; plutonium shows +3, +4, +5, +6, and +7. This is the real reason for the wide range — it’s an electronic structure effect, not a nuclear one.

   2. What about radioactivity?

      Yes, all actinoids are radioactive — their nuclei are unstable and decay over time. But radioactivity is a nuclear property, while oxidation states depend on electron configuration. A nucleus decaying does not directly change how many electrons an atom can lose or gain in a chemical reaction. So while both statements are factually true, the reason does not explain the assertion. …

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8. CBSE 2025Set D1 markMCQ
   Q.The electromotive force of the following cell is: Zn | Zn2+ (1M) || Fe2+ (1M) | Fe, given E°Zn2+|Zn = -0.76 V, E°Fe2+|Fe = -0.44 V

   (a) 1.2 V

   (b) 0.32 V

   (c) -1.2 V

   (d) -0.32 V

   ›Reveal solutionSolution

   E(cell) = E(cathode) - E(anode) = -0.44 - (-0.76) = +0.32 V.

   In the cell notation Zn | Zn2+ || Fe2+ | Fe, zinc is the anode (oxidation, written left) and iron is the cathode (reduction, written right). The standard cell EMF is:

   E(cell) = E(cathode) - E(anode)

   E(cell) = E(Fe2+/Fe) - E(Zn2+/Zn)

   E(cell) = (-0.44) - (-0.76) …

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9. CBSE 2025Set A1 mark
   Q.Write True or False: The cell potential is the addition of the electrode potentials (reduction potentials) of the cathode and anode.

   ›Reveal solutionSolution

   Cell potential is obtained by subtracting the anode's reduction potential from the cathode's, not by adding the two reduction potentials.

   The standard cell potential is defined as:

   $E^{\circ}_{cell} = E^{\circ}_{cathode(reduction)} - E^{\circ}_{anode(reduction)}$

   If both electrode potentials are taken as reduction potentials (as the statement specifies), the correct operation is a subtraction (cathode minus anode), not an addition. The 'addition' phrasing is only valid if the anode's contribution is expressed as an oxidation potential (= −reduction potential): then …

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10. CBSE 2025Set ANNUAL1 mark
    Q.What is the common oxidation state of Lanthanoids?

    ›Reveal solutionSolution

    All lanthanoids overwhelmingly favour the +3 oxidation state, since their poorly-bonding 4f electrons are not readily involved, leaving the same outer 5d/6s electrons available across the series.

    Across the entire lanthanide series, the +3 oxidation state is by far the most common and stable one, shown by essentially every lanthanoid. This is because the 4f electrons are deeply buried and well-shielded, taking little part in bonding, while the outer $5d^{0-1}6s^2$ electrons are readily lost to give the stable $Ln^{3+}$ ion. Occasional +2 or +4 states occur only for a few elements whe …

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11. CBSE 2025Set ANNUAL1 mark
    Q.Answer in one word/sentence: Given the standard electrode potentials, arrange these metals in their increasing order of reducting power: K+/K = -2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = -2.37 V.

    ›Reveal solutionSolution

    Reducing power increases as the standard electrode (reduction) potential becomes more negative, so we simply rank the four E° values from most positive to most negative.

    Given standard reduction potentials:

    $K^+/K = -2.93\ V,\quad Mg^{2+}/Mg = -2.37\ V,\quad Hg^{2+}/Hg = +0.79\ V,\quad Ag^+/Ag = +0.80\ V$

    A more negative (or less positive) standard reduction potential means the metal has a greater tendency to lose electrons (be oxidised) — i.e. it is a stronger reducing agent. Conversely, a metal with a highly positive reduction potential prefers to stay reduced (gain electrons), making it a poor reducing agent (like Ag, a "noble" metal).

    …

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12. CBSE 2025Set ANNUAL1 mark
    Q.Write two applications of electrochemical series.

    ›Reveal solutionSolution

    Electrochemical series: predicts reaction feasibility and metal-displacement reactivity.

    The electrochemical series arranges elements/ions in order of their standard reduction potentials ($E°$). Two common applications:

    1. Predicting feasibility of a redox reaction: a reaction is spontaneous if the species with the higher (more positive) reduction potential is reduced while the species with the lower (more negative) reduction potential is oxidized, i.e. $E°_{cell} = E°_{cathode} - E°_{anode} > 0$. …

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