Q.Amongst the following, the most stable complex is

Showing the 12 most recent of 32 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Which of the following exhibit optical isomerism? I. $\left[\mathrm{Fe}(\mathrm{C}_2\mathrm{O}_4)_3\right]^{3-}$ II. $\left[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_4\mathrm{Cl}_2\right]^+$ III. $\left[\mathrm{Co}(\mathrm{NH}_3)_2(\mathrm{en})_2\right]^{3+}$ The correct answer is (only = మాత్రమే) (A) I, II only (B) I, III only (C) II, III only (D) I, II, III

   ›Reveal solutionSolution

   Optical isomerism arises in coordination complexes that lack a plane of symmetry. Complex I and III are chiral and show optical isomerism; complex II is not. The correct answer is (B).

   The key idea is that optical isomerism — the existence of non-superimposable mirror images — requires the complex to be chiral, meaning it has no plane of symmetry. In coordination chemistry, this usually happens when the arrangement of ligands around the metal centre is such that the complex cannot be superimposed on its mirror image.

   Let’s examine each complex one by one.

   1. Complex I: $\left[\mathrm{Fe}(\mathrm{C}_2\mathrm{O}_4)_3\right]^{3-}$

      Here, oxalate ($\mathrm{C}_2\mathrm{O}_4^{2-}$) is a bidentate ligand. Three oxalate ions coordinate to the iron(III) centre, giving an octahedral geometry. Each oxalate occupies two adjacent positions, forming a chelate ring.

      The complex has a tris(bidentate) structure, which is analogous to $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$. This arrangement is well-known to be chiral — it lacks any plane of symmetry because the three chelate rings twist in a propeller-like fashion. The two mirror-image forms (Δ and Λ isomers) are non-superimposable.

      Tip

      A quick test: if a complex has three identical bidentate ligands in an octahedral geometry, it is almost always chiral. The only exception is if the ligand itself is planar and symmetric in a way that creates a plane — but oxalate does not do that here.

      So, I exhibits optical isomerism.

   2. Complex II: $\left[\mathrm{Cr}(\mathrm{H}_2\mathrm{O})_4\mathrm{Cl}_2\right]^+$

      This is an octahedral complex with four water molecules and two chloride ions. The two chlorides can be arranged either cis (adjacent) or trans (opposite).

      • The trans isomer has a plane of symmetry passing through the metal, the two chlorides, and the two waters opposite them. It is achiral.
      • The cis isomer also has a plane of symmetry: the plane that contains the metal, the two chlorides, and the two waters that are opposite each other in the square plane.
      Watch out

      A common mistake is to think that cis isomers are always chiral. In an octahedral complex of the type $\mathrm{MA}_4\mathrm{B}_2$, even the cis form has a plane of symmetry (the plane containing the two B ligands and the two A ligands opposite them). So it is not chiral.

      Therefore, II does not exhibit optical isomerism. …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Among the four complex ions given, identify the pair of ions with same spin only magnetic moment value $\left[\text{Ni}(\text{NH}_3)_6\right]^{2+}$ $\quad$ $\left[\text{MnCl}_6\right]^{4-}$ $\quad$ $\left[\text{FeF}_6\right]^{3-}$ $\quad$ $\left[\text{CoF}_6\right]^{3-}$ I $\quad$ II $\quad$ III $\quad$ IV (A) II, III only (B) II, IV only (C) I, IV only (D) I, III only

   ›Reveal solutionSolution

   Count unpaired electrons in each complex: d⁸ Ni²⁺ → 2, high-spin d⁵ Mn²⁺ → 5, high-spin d⁵ Fe³⁺ → 5, high-spin d⁶ Co³⁺ → 4; the two d⁵ high-spin ions (II and III) have the same spin-only moment √35 BM.

   Concept. Spin-only magnetic moment $\mu = \sqrt{n(n+2)}$ BM, where $n$ = number of unpaired electrons; equal $n$ ⟹ equal moment. Cl⁻ and F⁻ are weak-field ligands (high-spin complexes); NH₃ with Ni²⁺ in an octahedral field still leaves d⁸ with 2 unpaired electrons regardless of pairing, since $t_{2g}^6 e_g^2$ always has 2 unpaired.

   Step 1 — oxidation states and d-counts.

   • I. $[\text{Ni(NH}_3)_6]^{2+}$: Ni²⁺ = $3d^8$
   • II. $[\text{MnCl}_6]^{4-}$: Mn²⁺ = $3d^5$
   • III. $[\text{FeF}_6]^{3-}$: Fe³⁺ = $3d^5$
   • IV. $[\text{CoF}_6]^{3-}$: Co³⁺ = $3d^6$

   Step 2 — unpaired electrons (octahedral).

   • I. d⁸: $t_{2g}^6e_g^2$ → $n=2$, $\mu=\sqrt{8}\approx 2.83$ BM …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Match the following List-1 (Process) A Ostwald's process B Haber's process C Deacon's process D Contact process List-2 (Catalyst) I CuCl$_2$ II V$_2$O$_5$ III Iron oxide, K$_2$O, Al$_2$O$_3$ IV Rh gauge The correct answer is (A) A – II, B – III, C – IV, D – I (B) A – IV, B – I, C – II, D – III (C) A – IV, B – III, C – I, D – II (D) A – II, B – IV, C – I, D – III

   ›Reveal solutionSolution

   This is a matching question on industrial processes and their catalysts. The correct pairing is: Ostwald’s process uses a Rh gauge, Haber’s process uses iron oxide with promoters, Deacon’s process uses CuCl₂, and the Contact process uses V₂O₅. The answer is option (C).

   The question tests your recall of the specific catalysts used in four major industrial chemical processes. Each process is defined by the reaction it drives, and the catalyst is chosen to optimize yield under given conditions. Let’s go through each one.

   1. Ostwald’s process – This is used to produce nitric acid ($\text{HNO}_3$) by oxidizing ammonia ($\text{NH}_3$) to nitric oxide ($\text{NO}$). The catalyst is a platinum–rhodium gauze (often called a Rh gauge). So A matches with IV.

   2. Haber’s process – This synthesizes ammonia ($\text{NH}_3$) from nitrogen and hydrogen. The catalyst is finely divided iron (Fe) with promoters like $\text{K}_2\text{O}$ and $\text{Al}_2\text{O}_3$ to enhance activity. So B matches with III. …

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4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.In which of the following sets, the central atom of both the molecules does not obey the octet rule? I. SF$_4$, XeF$_4$ II. SCl$_2$, H$_2$S III. SnCl$_2$, XeO$_3$ The correct answer is (A) I, II, III (B) I, III only (C) II, III only (D) I, II only

   ›Reveal solutionSolution

   Both central atoms disobey the octet only in set I ($\text{SF}_4$, $\text{XeF}_4$ — expanded octets) and set III ($\text{SnCl}_2$ incomplete octet, $\text{XeO}_3$ expanded octet); in set II both S atoms have a normal octet.

   Examine the central atom of each molecule:

   Set I — $\text{SF}_4$: S has 4 bonds + 1 lone pair = 10 electrons → expanded octet, does not obey. $\text{XeF}_4$: Xe has 4 bonds + 2 lone pairs = 12 electrons → expanded octet, does not obey. → both violate.

   Set II — $\text{SCl}_2$: S has 2 bonds + 2 lone pairs = 8 electrons → obeys octet. $\text{H}_2\text{S}$: S has 2 bonds + 2 lone pairs = 8 electrons → obeys octet. → both obey, so this set does NOT qualify. …

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5. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A compound ‘X’ reacts with LiAlH$_4$ in diethylether and gives a colourless, highly toxic gas. This gas when heated with NH$_3$ gives inorganic benzene. X is I. Electron deficient molecule II. Trigonal planar molecule III. On hydrolysis forms octahedral ion Correct answer is (A) I, II only (B) I, II, III (C) II, III only (D) I, III only

   ›Reveal solutionSolution

   The compound 'X' is boron trifluoride ($BF_3$), which is an electron-deficient, trigonal planar molecule. It does not form an octahedral ion on hydrolysis. The correct option is (A).

   Concept and Intuition

   The problem describes a sequence of reactions that helps us identify an unknown compound 'X'.

   1. Reaction with LiAlH$_4$: Lithium aluminium hydride ($LiAlH_4$) is a powerful reducing agent. When it reacts with certain compounds, especially halides of non-metals, it can produce hydrides. The mention of a "colourless, highly toxic gas" is a strong hint.
   2. Reaction with NH$_3$ to form inorganic benzene: This is the crucial clue. "Inorganic benzene" is the common name for borazine ($B_3N_3H_6$). Borazine is formed by the reaction of diborane ($B_2H_6$) with ammonia ($NH_3$) under specific conditions.
      • $3B_2H_6 + 6NH_3 \xrightarrow{\text{heat}} 2B_3N_3H_6 + 12H_2$ This confirms that the "colourless, highly toxic gas" is diborane ($B_2H_6$).
   3. Identifying 'X': Since 'X' reacts with $LiAlH_4$ to produce $B_2H_6$, 'X' must be a boron compound that can be reduced to diborane. A common and industrially important method for preparing diborane involves the reduction of boron halides (like $BF_3$ or $BCl_3$) with $LiAlH_4$. Boron trifluoride ($BF_3$) is a classic example.

   Once 'X' is identified as $BF_3$, we can evaluate the given statements about its properties.

   Step-by-step Solution

   1. Identify the intermediate gas and compound 'X'.

      The problem states that compound 'X' reacts with $LiAlH_4$ to give a colourless, highly toxic gas. This gas then reacts with $NH_3$ to form inorganic benzene ($B_3N_3H_6$).

      The formation of inorganic benzene from a gas and $NH_3$ is characteristic of diborane ($B_2H_6$).

      $3B_2H_6 + 6NH_3 \xrightarrow{\text{heat}} 2B_3N_3H_6 + 12H_2$

      Therefore, the colourless, highly toxic gas is diborane ($B_2H_6$).

      Now, we need to identify 'X'. 'X' reacts with $LiAlH_4$ to produce $B_2H_6$. A common precursor for diborane synthesis using $LiAlH_4$ is boron trifluoride ($BF_3$).

      [!FORMULA]

      $4BF_3 + 3LiAlH_4 \xrightarrow{\text{diethylether}} 2B_2H_6 + 3LiF + 3AlF_3$

      Thus, compound 'X' is $BF_3$.

   2. Evaluate statement I: Electron deficient molecule.

      In $BF_3$, the central boron atom is bonded to three fluorine atoms. Boron has 3 valence electrons, and each fluorine atom contributes one electron to form a covalent bond. Therefore, the boron atom in $BF_3$ has $3 \times 2 = 6$ electrons in its valence shell. Since it has fewer than 8 electrons, it is an electron-deficient molecule and acts as a Lewis acid.

      Statement I is correct.

   3. Evaluate statement II: Trigonal planar molecule.

      To determine the geometry of $BF_3$, we use VSEPR theory. The central boron atom forms three single bonds with three fluorine atoms. There are no lone pairs on the boron atom. …

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6. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Which of the following statements is not correct? (A) The first ionization enthalpy of molecular oxygen is almost identical with that of xenon (B) Chlorine with excess ammonia gives nitrogen as one of the products (C) XeF$_6$ on complete hydrolysis gives Xe, HF and O$_2$ (D) The oxidizing agent used in Deacon’s process is atmospheric oxygen

   ›Reveal solutionSolution

   The key is to check each statement against known chemical facts. Statement (C) is wrong because XeF$_6$ on complete hydrolysis gives XeO$_3$, not Xe and O$_2$. The correct answer is (C).

   The question tests your grasp of p-block chemistry — specifically, the properties of noble gases, reactions of ammonia with chlorine, and industrial processes. Let’s examine each statement one by one.

   1. Statement (A): First ionization enthalpy of O$_2$ vs. Xe

      The first ionization enthalpy of molecular oxygen (O$_2$) is about 1175 kJ/mol, while that of xenon is about 1170 kJ/mol. They are indeed very close — almost identical. This is a known fact from periodic trends and noble gas chemistry. So (A) is correct.

   2. Statement (B): Chlorine with excess ammonia

      When chlorine reacts with excess ammonia, the reaction proceeds as:

   $3Cl_2 + 8NH_3 \rightarrow 6NH_4Cl + N_2$

   Nitrogen gas is indeed one of the products. This is a classic reaction — chlorine oxidizes ammonia to nitrogen. So (B) is correct.

   3. Statement (C): XeF$_6$ on complete hydrolysis Xenon hexafluoride undergoes complete hydrolysis to give xenon trioxide:

   $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$

   It does not give elemental xenon and oxygen. That would require a reduction, which doesn’t happen here. So (C) is false — this is the incorrect statement. …

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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If the number of moles of Fe$^{2+}$ ions oxidized by one mole of acidified MnO$_4^-$ is $x$, the number of moles of Fe$^{2+}$ ions oxidised by one mole of acidified Cr$_2$O$_7^{2-}$ is (A) $\dfrac{5x}{8}$ (B) $\dfrac{6x}{5}$ (C) $\dfrac{8x}{5}$ (D) $\dfrac{5x}{6}$

   ›Reveal solutionSolution

   The key is to compare the number of electrons each oxidant accepts per mole: MnO₄⁻ accepts 5 e⁻, Cr₂O₇²⁻ accepts 6 e⁻. Since each Fe²⁺ loses 1 e⁻, the ratio of Fe²⁺ moles oxidized is 6/5 times x, so the answer is (B).

   Concept & Intuition

   This problem is about stoichiometry of redox reactions. The number of moles of Fe²⁺ oxidized by one mole of an oxidant depends solely on how many electrons that oxidant can accept. Each Fe²⁺ gives up exactly one electron (Fe²⁺ → Fe³⁺ + e⁻). So if an oxidant accepts n electrons per mole, it will oxidize n moles of Fe²⁺. The question gives that one mole of MnO₄⁻ oxidizes x moles of Fe²⁺, which tells us x equals the number of electrons MnO₄⁻ accepts. Then we find the same for Cr₂O₇²⁻ and compare.

   Step-by-step reasoning

   1. Determine electrons accepted by MnO₄⁻ In acidic medium, the half-reaction is:

   $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

   So one mole of MnO₄⁻ accepts 5 moles of electrons.

   Since each Fe²⁺ loses 1 electron, one mole of MnO₄⁻ oxidizes 5 moles of Fe²⁺.

   Therefore, $x = 5$.

   2. Determine electrons accepted by Cr₂O₇²⁻ In acidic medium, the half-reaction is:

   $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

   So one mole of Cr₂O₇²⁻ accepts 6 moles of electrons.

   Hence, one mole of Cr₂O₇²⁻ oxidizes 6 moles of Fe²⁺. …

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8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Which one of the orders is correctly matched with the property mentioned against it? (A) $\mathrm{H_2S < H_2O < H_2Se < H_2Te}$ (Boiling point) (B) $\mathrm{N_2O < NO < N_2O_3 < N_2O_4 < N_2O_5}$ (Acidic nature) (C) $\mathrm{HI < HCl < HBr < HF}$ (Acidic nature) (D) $\mathrm{H_2O < H_2S < H_2Se < H_2Te}$ (Bond angle)

   ›Reveal solutionSolution

   The key idea is to check each trend against the known periodic and molecular property rules. Only option (B) correctly matches the increasing acidic nature of nitrogen oxides with increasing oxidation state of nitrogen.

   Concept & Intuition

   This question tests your understanding of periodic trends in boiling points, acidic strength, and bond angles. For each series, you must recall the underlying physical or chemical reason—like hydrogen bonding, oxidation state effects, or lone-pair repulsion—and see if the given order matches. A single mismatch disqualifies the option.

   Step-by-step reasoning

   1. Option (A): $\mathrm{H_2S < H_2O < H_2Se < H_2Te}$ (Boiling point)

      Boiling points of hydrides depend on intermolecular forces. $\mathrm{H_2O}$ has strong hydrogen bonding, so its boiling point is highest in Group 16, not second lowest. The correct order is $\mathrm{H_2S < H_2Se < H_2Te < H_2O}$ (increasing van der Waals forces, then water’s H‑bonding dominates). Hence (A) is wrong.

   2. Option (B): $\mathrm{N_2O < NO < N_2O_3 < N_2O_4 < N_2O_5}$ (Acidic nature)

      Acidity of nitrogen oxides increases with the oxidation state of nitrogen.

      • $\mathrm{N_2O}$: N = +1 (neutral)
      • $\mathrm{NO}$: N = +2 (neutral)
      • $\mathrm{N_2O_3}$: N = +3 (forms nitrous acid, weak acid)
      • $\mathrm{N_2O_4}$: N = +4 (forms a mix, more acidic)
      • $\mathrm{N_2O_5}$: N = +5 (forms nitric acid, strong acid) The given order matches this increasing trend. So (B) is correct.

   3. Option (C): $\mathrm{HI < HCl < HBr < HF}$ (Acidic nature)

      In aqueous solution, acid strength of hydrogen halides increases down the group: $\mathrm{HF < HCl < HBr < HI}$ (due to decreasing bond strength). The given order is reversed. Hence (C) is wrong. …

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9. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.Observe the following complex ions

   [!FORMULA] $\begin{array}{|c|c|c|c|} \hline \left[\text{Mn(CN)}_6\right]^{3-} & \left[\text{Fe(CN)}_6\right]^{3-} & \left[\text{CoF}_6\right]^{3-} & \left[\text{Co(C}_2\text{O}_4)_3\right]^{3-} \\ \text{A} & \text{B} & \text{C} & \text{D} \\ \hline \end{array}$

   Identify the option in which the unpaired electrons in the complex ions are in correct increasing order (A) C, A, B, D (B) B, A, C, D (C) D, A, B, C (D) D, B, A, C

   ›Reveal solutionSolution

   The key is to determine the number of unpaired electrons in each complex by considering the metal’s oxidation state, ligand field strength, and whether the complex is high-spin or low-spin. The correct increasing order is D (0), A (2), B (1), C (4) → option (B).

   We need to compare the number of unpaired electrons in four coordination complexes. The trick is that the same metal can have different spin states depending on the ligand. Strong-field ligands (like CN⁻) cause pairing; weak-field ligands (like F⁻) favor high-spin configurations. Also, oxalate (C₂O₄²⁻) is a strong-field ligand but forms a chelate, which also favors low-spin.

   Let’s work through each complex step by step.

   1. Determine the oxidation state of the metal in each complex.

      • A: $[Mn(CN)_6]^{3-}$ — CN⁻ has charge –1, so total ligand charge = –6. Overall charge = –3, so Mn must be +3.
      • B: $[Fe(CN)_6]^{3-}$ — CN⁻ charge –6, overall –3 → Fe is +3.
      • C: $[CoF_6]^{3-}$ — F⁻ charge –6, overall –3 → Co is +3.
      • D: $[Co(C_2O_4)_3]^{3-}$ — Oxalate (C₂O₄²⁻) has charge –2 each, total –6, overall –3 → Co is +3.

   2. Write the d-electron count for each metal ion.

      • Mn³⁺: Mn atomic number 25, electron configuration [Ar] 4s² 3d⁵. Remove 3 electrons → 3d⁴.
      • Fe³⁺: Fe atomic number 26, [Ar] 4s² 3d⁶. Remove 3 electrons → 3d⁵.
      • Co³⁺: Co atomic number 27, [Ar] 4s² 3d⁷. Remove 3 electrons → 3d⁶.

   3. Consider ligand field strength and geometry.

      All complexes are octahedral (coordination number 6).

      • CN⁻ is a strong-field ligand → large Δ (crystal field splitting) → low-spin configuration.
      • F⁻ is a weak-field ligand → small Δ → high-spin configuration.
      • C₂O₄²⁻ (oxalate) is a moderately strong-field ligand and also chelating, which further stabilizes low-spin for Co³⁺.

   4. Fill the d-orbitals for each complex.

      • A: Mn³⁺ (d⁴) with strong-field CN⁻ → low-spin: t₂g⁴ eg⁰. All four electrons pair in the three t₂g orbitals? Actually, t₂g holds 6 electrons max; with 4 electrons, we get one unpaired (since Hund’s rule applies within t₂g: three orbitals, four electrons → one orbital doubly occupied, two singly occupied → 2 unpaired electrons).
      • B: Fe³⁺ (d⁵) with strong-field CN⁻ → low-spin: t₂g⁵ eg⁰. Five electrons in t₂g: all paired except one → 1 unpaired electron.
      • C: Co³⁺ (d⁶) with weak-field F⁻ → high-spin: t₂g⁴ eg². Four in t₂g (one pair, two singles) and two in eg (both single) → total 4 unpaired electrons.
      • D: Co³⁺ (d⁶) with strong-field oxalate → low-spin: t₂g⁶ eg⁰. All six electrons paired → 0 unpaired electrons. …

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10. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.Observe the following complex ions $\begin{array}{|c|c|c|c|} \hline \left[\mathrm{Mn(CN)_6}\right]^{3-} & \left[\mathrm{Fe(CN)_6}\right]^{3-} & \left[\mathrm{CoF_6}\right]^{3-} & \left[\mathrm{Co(C_2O_4)_3}\right]^{3-} \\ \text{A} & \text{B} & \text{C} & \text{D} \\ \hline \end{array}$ Identify the option in which the unpaired electrons in the complex ions are in correct increasing order (A) D, A, B, C (B) D, B, A, C (C) B, A, C, D (D) C, A, B, D

    ›Reveal solutionSolution

    The number of unpaired electrons depends on the metal’s oxidation state, the ligand field strength (strong vs weak field), and the resulting d-orbital splitting. The correct increasing order is D (0), A (2), B (1), C (4) — which matches option (B).

    The key to this problem is understanding how the crystal field theory determines the electron configuration of the central metal ion in each complex. Two factors matter: the oxidation state of the metal (which tells you the d-electron count) and whether the ligands are strong-field (low-spin) or weak-field (high-spin). For octahedral complexes, strong-field ligands cause pairing of electrons in the lower $t_{2g}$ set, while weak-field ligands leave electrons unpaired as much as possible.

    Let’s work through each complex one by one.

    1. Complex A: $\left[\mathrm{Mn(CN)_6}\right]^{3-}$

       • Manganese is in the +3 oxidation state: $\mathrm{Mn}^{3+}$ has an electron configuration of $[Ar]\,3d^4$.
       • $\mathrm{CN}^-$ is a strong-field ligand, so the complex is low-spin. In an octahedral field, the four d-electrons fill the $t_{2g}$ set first, giving $t_{2g}^4$ — all four electrons are paired in two orbitals, leaving 2 unpaired electrons (since $t_{2g}$ can hold 6 electrons, the fourth electron must pair up).
       • Unpaired electrons: 2.

    2. Complex B: $\left[\mathrm{Fe(CN)_6}\right]^{3-}$

       • Iron is in the +3 oxidation state: $\mathrm{Fe}^{3+}$ has $[Ar]\,3d^5$.
       • Again, $\mathrm{CN}^-$ is strong-field, so low-spin. The five d-electrons fill $t_{2g}$ completely ($t_{2g}^5$), meaning one electron is unpaired in the $t_{2g}$ set (since five electrons in three orbitals leave one unpaired).
       • Unpaired electrons: 1.

    3. Complex C: $\left[\mathrm{CoF_6}\right]^{3-}$

       • Cobalt is in the +3 oxidation state: $\mathrm{Co}^{3+}$ has $[Ar]\,3d^6$.
       • $\mathrm{F}^-$ is a weak-field ligand, so the complex is high-spin. The six d-electrons follow Hund’s rule: they occupy all five d-orbitals singly before pairing. This gives $t_{2g}^4 e_g^2$ — four electrons in $t_{2g}$ (two pairs) and two unpaired in $e_g$, for a total of 4 unpaired electrons.
       • Unpaired electrons: 4. …

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11. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.Identify the correct set from the following (A) Penicillin G – Narrow spectrum – Bacteriostatic (B) Ofloxacin – Broad spectrum – Bactericidal (C) Amoxycillin – Narrow spectrum – Bactericidal (D) Chloramphenicol – Broad spectrum – Bactericidal

    ›Reveal solutionSolution

    The key is to match each antibiotic with its correct spectrum (narrow or broad) and its correct mode of action (bacteriostatic or bactericidal). Only option (B) — Ofloxacin – Broad spectrum – Bactericidal — is entirely accurate.

    The question tests your understanding of two fundamental classifications of antibiotics: spectrum of activity (how many types of bacteria they target) and mode of action (whether they kill bacteria outright or just stop their growth). Getting these right requires knowing specific examples from medicinal chemistry.

    Let’s examine each option one by one.

    1. Option (A): Penicillin G – Narrow spectrum – Bacteriostatic

       Penicillin G is indeed a narrow spectrum antibiotic — it works mainly against Gram-positive bacteria like Streptococcus and Staphylococcus (not Gram-negative ones). But its mode of action is bactericidal, not bacteriostatic. It kills bacteria by inhibiting cell wall synthesis, causing the cell to burst. So the “bacteriostatic” label here is wrong.

    2. Option (B): Ofloxacin – Broad spectrum – Bactericidal

       Ofloxacin is a fluoroquinolone antibiotic. It is broad spectrum, effective against both Gram-positive and Gram-negative bacteria. It works by inhibiting bacterial DNA gyrase and topoisomerase IV, which stops DNA replication and leads to cell death — making it bactericidal. Both descriptors are correct.

    3. Option (C): Amoxycillin – Narrow spectrum – Bactericidal

       Amoxycillin is a penicillin derivative, and like penicillin G, it is bactericidal (inhibits cell wall synthesis). However, it is broad spectrum, not narrow — it works against a wider range of bacteria, including some Gram-negative ones (e.g., E. coli). So “narrow spectrum” is incorrect.

    4. Option (D): Chloramphenicol – Broad spectrum – Bactericidal …

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12. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.The minimum concentration (ppm) of dissolved oxygen in water that is required for the growth of fish is (A) $10$ (B) $8$ (C) $12$ (D) $6$

    ›Reveal solutionSolution

    Fish require a minimum dissolved oxygen concentration of 4–6 ppm for survival, but for healthy growth the standard threshold is 6 ppm, making option (D) the correct answer.

    The question asks about the minimum concentration of dissolved oxygen (DO) in water needed for fish growth. This is a factual environmental science concept, but understanding the why behind the number helps you remember it.

    Fish, like all animals, need oxygen for cellular respiration. They extract dissolved oxygen from water through their gills. The amount of oxygen dissolved in water depends on temperature, salinity, and pressure — colder water holds more oxygen. For fish to grow and thrive, the DO level must be above a critical threshold. Below that, fish become stressed, their growth slows, and they may die.

    The typical benchmark used in water quality standards is that most fish species require at least 4–6 ppm (parts per million) of dissolved oxygen. For "growth" (not just survival), the higher end of this range — 6 ppm — is considered the minimum. This is a standard value taught in Indian exam syllabi (e.g., for NEET, environmental studies, or fisheries science).

    Let’s break it down:

    1. What does "ppm" mean here?

       Parts per million (ppm) for dissolved oxygen is equivalent to milligrams per litre (mg/L). So 6 ppm means 6 mg of oxygen per litre of water.

    2. Why not lower values like 2 or 4 ppm?

       At 2–4 ppm, many fish can survive for short periods but cannot grow or reproduce normally. Growth requires more energy, hence more oxygen. Below 4 ppm, fish are in "oxygen debt" — their metabolic processes slow down.

    3. Why not higher values like 8 or 10 ppm? …

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