Q.Which of the following arrangements represent increasing oxidation number of the central atom?

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Hydrolysis of chlorine nitrate gives oxoacid of chlorine $X$ and oxoacid of nitrogen $Y$. The oxidation states of $Cl$ in $X$, $N$ in $Y$ are respectively (A) +1, +3 (B) -1, +3 (C) +1, +5 (D) +3, +5

   ›Reveal solutionSolution

   Hydrolysis of $ClONO_2$ cleaves the Cl–O–N linkage at the O–N bond, giving $HOCl$ (Cl = +1) and $HNO_3$ (N = +5).

   Concept and Intuition

   Chlorine nitrate has the connectivity $Cl-O-NO_2$. Hydrolysis (reaction with water) breaks the weaker $O-N$ bond, inserting an $-OH$ onto each fragment: the $Cl-O-$ portion picks up an H to become $Cl-OH$ (hypochlorous acid), and the $-NO_2$ portion picks up an $-OH$ to become $HO-NO_2$ (nitric acid).

   $ClONO_2 + H_2O \rightarrow HOCl + HNO_3$

   Step-by-Step Solution

   1. Write the hydrolysis: $ClONO_2 + H_2O \rightarrow HOCl (X) + HNO_3 (Y)$.
   2. Oxidation state of Cl in $HOCl$: let it be $x$. $(+1) + (-2) + x = 0 \Rightarrow x = +1$.
   3. Oxidation state of N in $HNO_3$: let it be $y$. $(+1) + y + 3(-2) = 0 \Rightarrow y = +5$. …

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2. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The oxides of nitrogen obtained by the reaction of nitric acid with

   (i) $\mathrm{P_4O_{10}}$,

   (ii) $\mathrm{P_4}$ respectively are (A) NO, $\mathrm{N_2O}$ (B) $\mathrm{N_2O_3}$, NO (C) $\mathrm{N_2O_5}$, $\mathrm{NO_2}$ (D) $\mathrm{NO_2}$, $\mathrm{N_2O}$

   ›Reveal solutionSolution

   P₄O₁₀ dehydrates HNO₃ to its anhydride N₂O₅, while concentrated HNO₃ oxidising P₄ is itself reduced to NO₂. Answer: (C).

   Concept and Intuition

   This question tests two distinct roles nitric acid can play: (i) as a substrate being dehydrated by a strong desiccant (P₄O₁₀), which simply removes water to leave nitric acid's anhydride, versus (ii) as a concentrated oxidising agent reacting with a reducing element (P₄), where HNO₃ itself gets reduced, typically to NO₂ when concentrated.

   Step-by-Step Solution

   1. Reaction (i): $\mathrm{HNO_3}$ with $\mathrm{P_4O_{10}}$. $\mathrm{P_4O_{10}}$ is one of the strongest dehydrating agents known. It removes water from nitric acid molecules: $4\mathrm{HNO_3} + \mathrm{P_4O_{10}} \rightarrow 2\mathrm{N_2O_5} + 4\mathrm{HPO_3}$. This is purely a dehydration (not redox) — nitric acid's acid anhydride, $\mathrm{N_2O_5}$, is formed.
   2. Reaction (ii): $\mathrm{HNO_3}$ with $\mathrm{P_4}$. Here HNO₃ acts as an oxidising agent on elemental phosphorus. Concentrated nitric acid oxidises $\mathrm{P_4}$ to phosphoric acid ($\mathrm{H_3PO_4}$), while HNO₃ itself is reduced. Since it is the concentrated acid reacting with a reducing agent, the reduction product is $\mathrm{NO_2}$ (concentrated HNO₃ is reduced to NO₂, whereas dilute HNO₃ typically gives NO). …

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3. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The composition of a sample of wustite is $\mathrm{Fe_{0.93}O_{1.00}}$. Percentage of iron in the form of $\mathrm{Fe^{3+}}$ ion is nearly (A) 85 (B) 15 (C) 93 (D) 7

   ›Reveal solutionSolution

   This is a non-stoichiometric defect (metal-deficiency) calculation for wustite $\mathrm{Fe_{0.93}O_{1.00}}$; charge balance gives about 15% of the iron as $\mathrm{Fe^{3+}}$.

   Concept and Intuition

   Wustite is a classic example of a metal-deficient non-stoichiometric compound: some $\mathrm{Fe^{2+}}$ sites are vacant, and to keep the crystal electrically neutral, an equivalent number of the remaining Fe ions are oxidised to $\mathrm{Fe^{3+}}$ (each $\mathrm{Fe^{2+}}$ vacancy is compensated by 2 $\mathrm{Fe^{3+}}$ ions). We solve this by simple charge balance.

   Step-by-Step Solution

   1. Let the moles of $\mathrm{Fe^{3+}}$ be $x$ and $\mathrm{Fe^{2+}}$ be $y$, per formula unit. Total iron: $x + y = 0.93$.
   2. Total negative charge from oxide ions: $\mathrm{O^{2-}}$ present = 1.00, contributing charge $= 2 \times 1.00 = 2$.
   3. For electrical neutrality, total positive charge from iron must equal 2: $3x + 2y = 2$. …

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4. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.The amphoteric oxide of Vanadium (V) reacts with alkali and forms an oxoion 'X' and with acid forms an oxoion Y. The oxidation states of 'V' in X and Y are respectively (A) $+2, +5$ (B) $+3, +3$ (C) $+5, +5$ (D) $+5, +2$

   ›Reveal solutionSolution

   This tests the amphoteric behaviour of $V_2O_5$: it dissolves in both alkali and acid without any change in the oxidation state of vanadium, which stays $+5$ in both product ions.

   Concept and Intuition

   $V_2O_5$ sits at the acidic-to-amphoteric borderline among the transition-metal oxides because vanadium is in its highest ($+5$) oxidation state. An oxide being amphoteric means it can react as a base (with acid) or as an acid (with alkali) — but in both cases here, vanadium's oxidation number doesn't change, because these are acid–base (ligand/oxo-exchange) reactions, not redox reactions.

   Step-by-Step Solution

   1. With alkali (e.g. NaOH), $V_2O_5$ behaves as an acidic oxide and dissolves to form the orthovanadate ion:

   $V_2O_5 + 6\,OH^- \rightarrow 2\,VO_4^{3-} + 3\,H_2O$

   Oxidation state of V in $VO_4^{3-}$: $x + 4(-2) = -3 \Rightarrow x = +5$.

   2. With acid (e.g. dilute $H_2SO_4$ or HCl), $V_2O_5$ behaves as a basic oxide and dissolves to form the dioxovanadium(V) cation:

   $V_2O_5 + 2H^+ \rightarrow 2\,VO_2^+ + H_2O$

   Oxidation state of V in $VO_2^+$: $x + 2(-2) = +1 \Rightarrow x = +5$. …

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5. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.The number of mixed oxides from the following is $\mathrm{Al_2O_3}$, $\mathrm{Pb_3O_4}$, $\mathrm{Mn_2O_7}$, $\mathrm{Fe_3O_4}$, $\mathrm{V_2O_5}$, $\mathrm{Sb_2O_3}$ (A) 2 (B) 3 (C) 4 (D) 1

   ›Reveal solutionSolution

   Tests recognising mixed oxides among a list of common oxides; only $\mathrm{Pb_3O_4}$ and

   $\mathrm{Fe_3O_4}$ qualify, giving a count of 2.

   Concept and Intuition

   A mixed oxide is one that can be regarded as a combination (in a fixed ratio) of two simple

   oxides of the same element in different oxidation states. The two classic textbook examples are

   red lead, $\mathrm{Pb_3O_4} = 2\mathrm{PbO}\cdot\mathrm{PbO_2}$ (lead in +2 and +4 states), and

   magnetite, $\mathrm{Fe_3O_4} = \mathrm{FeO}\cdot\mathrm{Fe_2O_3}$ (iron in +2 and +3 states).

   Step-by-Step Solution

   1. $\mathrm{Al_2O_3}$: aluminium is only in the +3 state here — a simple, amphoteric oxide. Not mixed.
   2. $\mathrm{Pb_3O_4}$: lead present as both Pb(II) and Pb(IV); a genuine mixed oxide.
   3. $\mathrm{Mn_2O_7}$: manganese only in the +7 state — a simple, strongly acidic covalent oxide. Not mixed.
   4. $\mathrm{Fe_3O_4}$: iron present as both Fe(II) and Fe(III); a genuine mixed oxide.
   5. $\mathrm{V_2O_5}$: vanadium only in the +5 state — a simple acidic oxide. Not mixed. …

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6. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.Acidification of chromate gives 'Z'. The oxidation state of chromium in 'Z' is (A) +3 (B) +6 (C) +7 (D) +2

   ›Reveal solutionSolution

   This tests the chromate–dichromate equilibrium; acidifying chromate gives dichromate (Z), and chromium's oxidation state remains +6 throughout (it's a structural/pH-driven change, not a redox change).

   Concept and Intuition

   Chromate ($\mathrm{CrO_4^{2-}}$) and dichromate ($\mathrm{Cr_2O_7^{2-}}$) are interconvertible depending on the pH of the solution: chromate (yellow) is the stable form in basic/neutral solution, while dichromate (orange) is the stable form in acidic solution. This is a classic condensation equilibrium (two tetrahedral $\mathrm{CrO_4^{2-}}$ units joining via a shared oxygen to form $\mathrm{Cr_2O_7^{2-}}$) — not an oxidation-reduction reaction, so the oxidation state of Cr doesn't change.

   Step-by-Step Solution

   1. Write the pH-dependent equilibrium: $2\mathrm{CrO_4^{2-}} + 2\mathrm{H^+} \rightleftharpoons \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O}$.
   2. "Acidification of chromate gives Z" means Z = dichromate ion, $\mathrm{Cr_2O_7^{2-}}$.
   3. Calculate the oxidation state of Cr in $\mathrm{CrO_4^{2-}}$: let Cr = $x$. Then $x + 4(-2) = -2 \Rightarrow x = +6$.
   4. Calculate the oxidation state of Cr in $\mathrm{Cr_2O_7^{2-}}$: let Cr = $x$. Then $2x + 7(-2) = -2 \Rightarrow 2x - 14 = -2 \Rightarrow x = +6$. …

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7. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
   Q.One mole of aqueous $\mathrm{S_2O_3^{2-}}$ is completely reacted with 2 moles of $\mathrm{Br_2(l)}$. What is the oxidation state (s) of S in the product formed? (A) 0, +5 (B) +6 (C) -2, +4 (D) +4

   ›Reveal solutionSolution

   $\mathrm{S_2O_3^{2-}}+2\mathrm{Br_2}+\mathrm{H_2O}\to 2\mathrm{SO_2}+4\mathrm{Br^-}+2\mathrm{H^+}$; sulphur is oxidised to $+4$.

   Concept and Intuition

   In thiosulphate $\mathrm{S_2O_3^{2-}}$ the two sulphur atoms have an average oxidation state of $+2$. Each mole of $\mathrm{Br_2}$ is reduced, $\mathrm{Br_2}+2e^-\to 2\mathrm{Br^-}$, accepting $2$ electrons. The amount of oxidant therefore fixes how many electrons leave the sulphur and hence its final oxidation state.

   Step-by-Step Solution

   1. Electrons removed $=2\ \text{mol Br}_2\times 2 = 4$ per mole of $\mathrm{S_2O_3^{2-}}$.
   2. The two S atoms start at a total oxidation number of $2\times(+2)=+4$.
   3. Removing $4$ electrons raises the total to $+8$, i.e. an average of $+4$ per sulphur.
   4. Balancing confirms one product: $\mathrm{S_2O_3^{2-}}+2\mathrm{Br_2}+\mathrm{H_2O}\to 2\mathrm{SO_2}+4\mathrm{Br^-}+2\mathrm{H^+}$ (charge $-2=-2$, O $=4=4$).
   5. In $\mathrm{SO_2}$ sulphur is $+4$.

   Common Mistakes …

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8. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.$\mathrm{X + Y \to}$ oleum The sum of oxidation states of central atom in X and Y is (A) 12 (B) 10 (C) 06 (D) 08

   ›Reveal solutionSolution

   Oleum is made by dissolving SO₃ in H₂SO₄; sulfur is in the +6 oxidation state in both, so the sum is 12.

   Concept and Intuition

   Oleum (fuming sulfuric acid, $H_2S_2O_7$, also called pyrosulfuric acid) is industrially made by absorbing sulfur trioxide gas into concentrated sulfuric acid: $SO_3 + H_2SO_4 \to H_2S_2O_7$. Both starting materials have sulfur as the central atom.

   Step-by-Step Solution

   1. Identify X and Y from the reaction that forms oleum: $X=SO_3$, $Y=H_2SO_4$.
   2. Oxidation state of S in $SO_3$: let $x$ be S's oxidation state; $x + 3(-2)=0 \Rightarrow x=+6$. …

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9. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.The formula of nickel oxide is $\mathrm{Ni_{0.98}O_{1.00}}$. What is the approximate percentage of $\mathrm{Ni^{+2}}$ in it? (A) 92 (B) 94 (C) 96 (D) 98

   ›Reveal solutionSolution

   This tests using charge balance in a non-stoichiometric metal oxide to find the fraction of metal in each oxidation state. Answer: ≈96% Ni2+.

   Concept and Intuition

   In $\mathrm{Ni_{0.98}O_{1.00}}$, nickel is metal-deficient relative to oxygen, so some nickel must be present as $\mathrm{Ni^{3+}}$ to balance the extra negative charge from oxide ions (charge neutrality of the overall compound).

   Step-by-Step Solution

   1. Let moles of $\mathrm{Ni^{2+}} = a$ and moles of $\mathrm{Ni^{3+}} = b$. Total Ni: $a + b = 0.98$.
   2. Total negative charge from O: $1.00 \times 2 = 2.00$. This must equal total positive charge from Ni: $2a + 3b = 2.00$.
   3. Substitute $a = 0.98 - b$: $2(0.98-b) + 3b = 2.00 \Rightarrow 1.96 + b = 2.00 \Rightarrow b = 0.04$.
   4. So $a = 0.98 - 0.04 = 0.94$. …

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10. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
    Q.The common oxidation states of elements of $15^{th}$ group elements are (A) -3, +3, +5 (B) -3, +2, +5 (C) -1, +3, +5 (D) -2, +3, +5

    ›Reveal solutionSolution

    Group 15 (pnictogens) elements commonly show oxidation states $-3$, $+3$ and $+5$.

    Concept and Intuition

    With 5 valence electrons ($ns^2np^3$), these elements can gain 3 electrons to reach the stable octet ($-3$), or lose/share the 3 unpaired p-electrons ($+3$), or lose/share all 5 valence electrons ($+5$, using both s and p electrons — favoured more for lighter members due to the inert pair effect being weaker there).

    Step-by-Step Solution

    1. Write the valence configuration: $ns^2\,np^3$.
    2. Gaining 3 electrons completes the octet → oxidation state $-3$ (e.g., $NH_3$, $PH_3$). …

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11. AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
    Q.Match the following List-I (Reactions) / List-II (Equivalent weights) A $\mathrm{KMnO_4} \xrightarrow{H^+} \mathrm{Mn^{2+}}$ i) M/8 B $\mathrm{C_2O_4^{2-}} \rightarrow \mathrm{CO_2}$ ii) M/6 C $\mathrm{K_2Cr_2O_7} \rightarrow \mathrm{Cr^{3+}}$ iii) M/5 D $\mathrm{NH_3} \rightarrow \mathrm{NO_3^-}$ iv) M/2 (A) A – (iv); B- (iii); C – (ii); D-

    (i) (B) A – (iii); B- (iv); C – (i); D-

    (ii) (C) A – (iii); B- (iv); C – (ii); D-

    (i) (D) A – (ii); B- (iv); C – (iii); D- (i)

    ›Reveal solutionSolution

    Equivalent weight = molar mass / n-factor, where n-factor is the number of electrons transferred per formula unit in each redox change; working these out gives A–iii, B–iv, C–ii, D–i.

    Concept and Intuition

    For a redox reagent, the equivalent weight is $M/n$ where $n$ is the number of electrons gained or lost per formula unit as it converts from the given species to the given product. This is found from the oxidation-state change of the atom(s) being oxidised/reduced.

    Step-by-Step Solution

    1. A. $KMnO_4 \xrightarrow{H^+} Mn^{2+}$: Mn goes from +7 to +2, a change of 5 electrons. n = 5 → equivalent weight = M/5 = (iii).
    2. B. $C_2O_4^{2-}\to CO_2$: each carbon goes from +3 to +4 (loses 1 electron), and there are 2 carbons, so 2 electrons total. n = 2 → equivalent weight = M/2 = (iv).
    3. C. $K_2Cr_2O_7\to Cr^{3+}$: each Cr goes from +6 to +3 (gains 3 electrons), and there are 2 Cr atoms, so 6 electrons total. n = 6 → equivalent weight = M/6 = (ii). …

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12. AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
    Q.Identify the element that can exhibit negative oxidation states. (A) C (B) Ge (C) Sn (D) Pb

    ›Reveal solutionSolution

    Only carbon, among C, Ge, Sn, Pb, readily shows negative oxidation states, since the group becomes progressively more metallic (and hence more inclined to positive states) going down.

    Concept and Intuition

    Group 14 elements range from the clearly non-metallic carbon at the top to the metallic lead at the bottom. Non-metallic character favours forming compounds where the element gains electrons (negative oxidation state), while metallic character favours losing electrons (positive oxidation state). This trend, combined with the inert-pair effect strengthening the +2 state for the heavier members, means Ge, Sn, and Pb are essentially never seen in negative oxidation states, while carbon commonly is (e.g. $-4$ in methane, $CH_4$, or in ionic carbides like $CaC_2$/$Al_4C_3$).

    Step-by-Step Solution

    1. Carbon (C): highly electronegative for its group, non-metallic, forms $-4$ states (e.g. $CH_4$) and other negative states in hydrides/carbides.
    2. Germanium (Ge): a metalloid, predominantly shows $+2$/$+4$, essentially no negative oxidation states in ordinary chemistry.
    3. Tin (Sn): metallic, shows $+2$/$+4$ only. …

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