Q.Explain the bonding in coordination compounds in terms of Werner's postulates.

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Which of the following will give maximum number of isomers? (A) $\left[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2\right]^+$ (B) $\left[\mathrm{Ni}(\mathrm{en})(\mathrm{NH}_3)_4\right]^{2+}$ (C) $\left[\mathrm{Ni}(\mathrm{C}_2\mathrm{O}_4)(\mathrm{en})_2\right]$ (D) $\left[\mathrm{Cr}(\mathrm{SCN})_2(\mathrm{NH}_3)_4\right]^+$

   ›Reveal solutionSolution

   The key is to count all possible stereoisomers (geometrical and optical) plus linkage isomers for each complex. The complex with the most isomers is $\left[\mathrm{Cr}(\mathrm{SCN})_2(\mathrm{NH}_3)_4\right]^+$, which gives 6 isomers (three S/N binding combinations, each with cis and trans forms), so the answer is (D).

   Concept & Intuition

   Isomers in coordination chemistry arise from different spatial arrangements of ligands (geometrical isomers) and, for chiral complexes, non-superimposable mirror images (optical isomers). Additionally, ambidentate ligands like SCN⁻ can bind through different atoms (S or N), creating linkage isomers. To find which complex gives the maximum number, we systematically count all distinct isomers for each option.

   ---

   1. Option (A): $\left[\mathrm{Co}(\mathrm{NH}_3)_4\mathrm{Cl}_2\right]^+$

      • This is an octahedral complex with two identical monodentate ligands (Cl) and four identical NH₃ ligands.
      • Geometrical isomers: Only cis and trans arrangements of the two Cl ligands.
      • Optical isomers: Neither cis nor trans is chiral (cis has a plane of symmetry, trans has a center of symmetry).
      • Total isomers = 2.

   2. Option (B): $\left[\mathrm{Ni}(\mathrm{en})(\mathrm{NH}_3)_4\right]^{2+}$

      • en (ethylenediamine) is a bidentate ligand. The complex is octahedral with one en and four NH₃.
      • The en ligand must occupy two adjacent positions (cis), so there is no trans isomer.
      • The complex is not chiral (it has a plane of symmetry through the en ring and opposite NH₃).
      • Total isomers = 1.

   3. Option (C): $\left[\mathrm{Ni}(\mathrm{C}_2\mathrm{O}_4)(\mathrm{en})_2\right]$

      • Oxalate (C₂O₄²⁻) is a bidentate ligand, and en is also bidentate. The complex is octahedral with three bidentate ligands.
      • Geometrical isomers: For three bidentate ligands, only the cis arrangement is possible (all three bidentates must occupy adjacent positions).
      • Optical isomers: The cis arrangement is chiral (no plane of symmetry), so it exists as a pair of enantiomers.
      • Total isomers = 2 (a pair of optical isomers).

   4. Option (D): $\left[\mathrm{Cr}(\mathrm{SCN})_2(\mathrm{NH}_3)_4\right]^+$

      • SCN⁻ is ambidentate: it can bind through sulfur (thiocyanato) or nitrogen (isothiocyanato).
      • This combines linkage isomerism with cis/trans geometrical isomerism. Counting systematically:
        • Both S-bound: cis and trans → 2 isomers.
        • Both N-bound: cis and trans → 2 isomers.
        • One S-bound, one N-bound: the two ligands are now different, so cis and trans are again distinct → 2 isomers.
      • None of these arrangements is chiral: each cis form has a plane of symmetry passing through the two SCN ligands (or the S- and N-bound ligands) and the two opposite NH₃ groups, so there are no additional optical isomers.
      • Total isomers = 2 + 2 + 2 = 6, the maximum among all four options.

   Watch out

   A common mistake is to forget linkage isomerism for ambidentate ligands like SCN⁻. Also, students often assume all cis isomers are chiral — always check for symmetry planes.

   Tip

   For octahedral complexes with two identical monodentate ligands and four others, the number of geometrical isomers is always 2 (cis/trans). Adding an ambidentate ligand multiplies this by the number of distinct binding combinations.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.$S_2O_3^{2-} (aq) + OH^- (aq) \rightarrow SO_4^{2-} (aq) + H_2O (l) + e^-$ After the above half reaction is balanced, which of the following are the coefficients of OH$^-$ and SO$_4^{2-}$ respectively? (A) 8, 3 (B) 6, 2 (C) 10, 2 (D) 5, 2

   ›Reveal solutionSolution

   Balancing the half‑reaction in basic solution gives coefficients 10 for OH⁻ and 2 for SO₄²⁻, so the correct choice is (C).

   We are balancing a half‑reaction in basic solution. The key idea: first balance atoms other than H and O, then balance O by adding H₂O, balance H by adding H⁺ (as if in acid), and finally neutralize H⁺ by adding the same number of OH⁻ to both sides. This method works because every H⁺ added in the acidic step is converted to H₂O when we add OH⁻.

   ---

   1. Identify the atoms to balance The skeleton is:

   $\text{S}_2\text{O}_3^{2-} + \text{OH}^- \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{O} + e^-$

   Sulfur (S) is unbalanced: left has 2 S, right has 1 S. So put a coefficient 2 in front of SO₄²⁻:

   $\text{S}_2\text{O}_3^{2-} + \text{OH}^- \rightarrow 2\,\text{SO}_4^{2-} + \text{H}_2\text{O} + e^-$

   2. Balance oxygen atoms

      Left: 3 O (from S₂O₃²⁻) + 1 O (from OH⁻) = 4 O total (coefficient of OH⁻ unknown yet).

      Right: 2 × 4 = 8 O (from 2 SO₄²⁻) + 1 O (from H₂O) = 9 O.

      To balance O, we need more O on the left. Add H₂O? No — we add H₂O to the side that needs oxygen. Actually, we balance O by adding H₂O to the side that is short of oxygen.

      Left has 4 O, right has 9 O → left needs 5 more O. So add 5 H₂O to the left? That would add 5 O but also 10 H — messy. Better to use the standard acidic-then-basic method.

   3. Use the acidic method first

      Ignore OH⁻ for a moment. Write the half‑reaction as if in acid:

   $\text{S}_2\text{O}_3^{2-} \rightarrow 2\,\text{SO}_4^{2-}$

   Balance O: left 3 O, right 8 O → add 5 H₂O to left:

   $\text{S}_2\text{O}_3^{2-} + 5\,\text{H}_2\text{O} \rightarrow 2\,\text{SO}_4^{2-}$

   Balance H: left now has 10 H, right has 0 → add 10 H⁺ to right:

   $\text{S}_2\text{O}_3^{2-} + 5\,\text{H}_2\text{O} \rightarrow 2\,\text{SO}_4^{2-} + 10\,\text{H}^+$

   Balance charge: left –2, right 2×(−2) + 10 = +6. Difference = 8 electrons needed on right:

   $\text{S}_2\text{O}_3^{2-} + 5\,\text{H}_2\text{O} \rightarrow 2\,\text{SO}_4^{2-} + 10\,\text{H}^+ + 8e^-$

   4. Convert to basic solution Add 10 OH⁻ to both sides to neutralize the 10 H⁺:

   $\text{S}_2\text{O}_3^{2-} + 5\,\text{H}_2\text{O} + 10\,\text{OH}^- \rightarrow 2\,\text{SO}_4^{2-} + 10\,\text{H}_2\text{O} + 8e^-$

   Cancel 5 H₂O from both sides (5 on left, 10 on right → net 5 H₂O on right):

   $\text{S}_2\text{O}_3^{2-} + 10\,\text{OH}^- \rightarrow 2\,\text{SO}_4^{2-} + 5\,\text{H}_2\text{O} + 8e^-$

   5. Check the given equation The problem’s skeleton had one OH⁻ on left and one H₂O on right, but after balancing we have 10 OH⁻ and 2 SO₄²⁻. The coefficients asked are for OH⁻ and SO₄²⁻ respectively: 10 and 2.

   Tip

   A common mistake is to forget that the H⁺ from the acidic step must be exactly neutralized by OH⁻, which often changes the number of water molecules. Always cancel water after adding OH⁻.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Which of the following aqueous solution has highest freezing point? (A) $0.1\,\text{m}\ \mathrm{Al_2(SO_4)_3}$ (B) $0.1\,\text{m}\ \mathrm{BaCl_2}$ (C) $0.1\,\text{m}\ \mathrm{NH_4Cl}$ (D) $0.1\,\text{m}\ \mathrm{AlCl_3}$

   ›Reveal solutionSolution

   The freezing point is highest for the solution with the smallest total particle concentration after dissociation. Since all are 0.1 m, the one that dissociates into the fewest ions gives the highest freezing point. That is 0.1 m NH₄Cl, which yields 0.2 m particles.

   Concept & Intuition

   Freezing point depression is a colligative property — it depends only on the number of solute particles in solution, not on their identity. The formula is

   $\Delta T_f = i \cdot K_f \cdot m$

   where $i$ is the van’t Hoff factor (number of particles per formula unit after dissociation), $K_f$ is the cryoscopic constant (same for water here), and $m$ is the molality.

   Since all solutions have the same molality (0.1 m), the one with the smallest $i$ will have the smallest $\Delta T_f$, and therefore the highest freezing point. So we just need to count ions produced per formula unit in water.

   Step‑by‑step reasoning

   1. Determine the van’t Hoff factor for each salt (assuming complete dissociation in dilute aqueous solution).

      • (A) $\mathrm{Al_2(SO_4)_3} \rightarrow 2\,\mathrm{Al^{3+}} + 3\,\mathrm{SO_4^{2-}}$ → total ions = $2+3 = 5$ → $i = 5$
      • (B) $\mathrm{BaCl_2} \rightarrow \mathrm{Ba^{2+}} + 2\,\mathrm{Cl^-}$ → total ions = $1+2 = 3$ → $i = 3$
      • (C) $\mathrm{NH_4Cl} \rightarrow \mathrm{NH_4^+} + \mathrm{Cl^-}$ → total ions = $1+1 = 2$ → $i = 2$
      • (D) $\mathrm{AlCl_3} \rightarrow \mathrm{Al^{3+}} + 3\,\mathrm{Cl^-}$ → total ions = $1+3 = 4$ → $i = 4$

   2. Compare the effective particle concentrations

      Since $m = 0.1$ for all, the total particle molality is $i \times 0.1$:

      • (A) $5 \times 0.1 = 0.5\,\text{m}$
      • (B) $3 \times 0.1 = 0.3\,\text{m}$
      • (C) $2 \times 0.1 = 0.2\,\text{m}$
      • (D) $4 \times 0.1 = 0.4\,\text{m}$

   3. Relate to freezing point

      The larger the particle concentration, the larger the freezing point depression. So the smallest particle concentration gives the smallest depression, hence the highest freezing point.

      Smallest particle concentration: $0.2\,\text{m}$ from NH₄Cl.

   Tip

   A common mistake is to forget that $\mathrm{Al_2(SO_4)_3}$ gives five ions, not four — the subscript 2 on Al and 3 on SO₄ both contribute. Always write the dissociation explicitly.

   Watch out

   If the problem involved a weak electrolyte or ion pairing, the actual $i$ would be less than the ideal. But here all are strong electrolytes in dilute solution, so ideal dissociation is assumed.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Identify the complex ion which does not exist (A) $\left[\mathrm{SiF}_6\right]^{2-}$ (B) $\left[\mathrm{GeCl}_6\right]^{2-}$ (C) $\left[\mathrm{Sn(OH)}_6\right]^{2-}$ (D) $\left[\mathrm{SiCl}_6\right]^{2-}$

   ›Reveal solutionSolution

   The key idea is that the central atom must have available d-orbitals to accommodate six ligands in an octahedral geometry. Silicon lacks accessible d-orbitals for chlorine ligands, so $\left[\mathrm{SiCl}_6\right]^{2-}$ does not exist. The answer is (D).

   The question tests your understanding of coordination chemistry and the availability of d-orbitals in elements of the carbon family (Group 14). For a complex ion with six ligands (octahedral geometry), the central atom must use $sp^3d^2$ hybridization. This requires empty d-orbitals of suitable energy.

   Silicon, germanium, and tin all belong to Group 14. As you go down the group, the size increases and the d-orbitals become more accessible. But here’s the catch: the nature of the ligand also matters. Small, highly electronegative ligands like fluorine can stabilize the high oxidation state and pull electron density, making d-orbital participation feasible even for silicon. Larger, less electronegative ligands like chlorine cannot do this effectively for silicon.

   Let’s examine each option step by step.

   1. Option (A): $\left[\mathrm{SiF}_6\right]^{2-}$

      Silicon is in the +4 oxidation state here. Fluorine is tiny and extremely electronegative. It strongly withdraws electron density, allowing silicon to use its 3d orbitals for $sp^3d^2$ hybridization. This complex is well-known and stable — it exists as salts like $\mathrm{Na_2[SiF_6]}$.

   2. Option (B): $\left[\mathrm{GeCl}_6\right]^{2-}$

      Germanium is larger than silicon and has accessible 4d orbitals. Chlorine, though larger than fluorine, can still coordinate because germanium’s d-orbitals are energetically available. This complex exists, for example as $\mathrm{(NH_4)_2[GeCl_6]}$.

   3. Option (C): $\left[\mathrm{Sn(OH)}_6\right]^{2-}$

      Tin is even larger, with accessible 5d orbitals. The hydroxide ligand is a reasonable ligand for tin(IV). This complex is known — for instance, $\mathrm{K_2[Sn(OH)_6]}$ is a stable compound.

   4. Option (D): $\left[\mathrm{SiCl}_6\right]^{2-}$

      Here’s the problem. Silicon is small, and chlorine is much larger than fluorine. The chlorine atoms would crowd around silicon, causing steric hindrance. More importantly, chlorine is not electronegative enough to pull sufficient electron density from silicon to make its 3d orbitals contract and become available for bonding. Silicon simply cannot expand its octet with chlorine ligands. This complex has never been isolated under normal conditions.

   Watch out

   A common mistake is to think that because $\left[\mathrm{SiF}_6\right]^{2-}$ exists, $\left[\mathrm{SiCl}_6\right]^{2-}$ should too. But fluorine’s small size and high electronegativity are crucial — chlorine lacks both, so silicon cannot form six bonds to it.

   Tip

   A quick rule: For Group 14 elements, the ability to form hexacoordinate complexes with halogens decreases as the halogen gets larger. So $\mathrm{SiF_6^{2-}}$ exists, $\mathrm{SiCl_6^{2-}}$ does not, but $\mathrm{GeCl_6^{2-}}$ and $\mathrm{SnCl_6^{2-}}$ do exist because the central atom is larger.

   ✓Final answer

   The complex ion which does not exist is $\left[\mathrm{SiCl}_6\right]^{2-}$, option (D).

5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Consider the following HF, H$_2$O, BeCl$_2$, CO$_2$, BF$_3$, NF$_3$, CCl$_4$, CHCl$_3$ The number of polar molecules is (A) 2 (B) 3 (C) 4 (D) 5

   ›Reveal solutionSolution

   The key idea is that a molecule is polar if it has polar bonds and an asymmetric shape (net dipole moment). After checking each molecule, exactly 4 are polar, so the answer is (C).

   Concept & Intuition

   Polarity depends on two things: (1) the presence of bonds between atoms with different electronegativities (polar bonds), and (2) the molecular geometry not cancelling those bond dipoles. Symmetric shapes like linear, trigonal planar, or tetrahedral can be nonpolar if all bonds are identical and arranged symmetrically. Asymmetric shapes or different substituents leave a net dipole.

   Step-by-step reasoning

   1. HF – Hydrogen and fluorine have a large electronegativity difference. The molecule is diatomic and linear, so the bond dipole is not cancelled. Polar.

   2. H₂O – Bent shape (V-shaped) due to two lone pairs on oxygen. The two O–H bond dipoles do not cancel; they add to a net dipole pointing toward oxygen. Polar.

   3. BeCl₂ – Linear molecule (Be is central, no lone pairs). The two Be–Cl bond dipoles are equal and opposite, cancelling exactly. Nonpolar.

   4. CO₂ – Linear molecule (O=C=O). The two C=O bond dipoles are equal and opposite, cancelling. Nonpolar.

   5. BF₃ – Trigonal planar (B is central, no lone pairs). The three B–F bond dipoles are symmetrically arranged at 120°, so their vector sum is zero. Nonpolar.

   6. NF₃ – Trigonal pyramidal (N has one lone pair). The three N–F bond dipoles do not cancel because the lone pair pushes them downward, creating a net dipole. Polar.

   7. CCl₄ – Tetrahedral (C central, no lone pairs). All four C–Cl bonds are identical and symmetrically arranged; dipoles cancel. Nonpolar.

   8. CHCl₃ – Tetrahedral but with three Cl and one H. The C–Cl and C–H bonds have different polarities, so the symmetry is broken. The bond dipoles do not cancel, giving a net dipole. Polar.

   Tip

   A common pitfall is thinking that all molecules with polar bonds are polar. For example, CCl₄ has four polar C–Cl bonds, but its perfect tetrahedral symmetry makes it nonpolar. Always check geometry!

   Count of polar molecules: HF, H₂O, NF₃, CHCl₃ → 4 molecules.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.For the alkyne with formula $\mathrm{C_6H_{10}}$, the number of alkynes with acidic hydrogens is $x$ and number of alkynes with no acidic hydrogens is $y$. $x$ and $y$ are respectively (A) $2,\,5$ (B) $3,\,4$ (C) $4,\,3$ (D) $5,\,2$

   ›Reveal solutionSolution

   $\mathrm{C_6H_{10}}$ has 4 alkynes with acidic (terminal) hydrogens and 3 without, so $x, y = 4, 3$ — option (C).

   An alkyne has an acidic hydrogen only when the triple bond is terminal ($\equiv\!\text{C–H}$). Enumerating the acyclic hexyne isomers:

   Terminal alkynes (acidic H), $x$:

   1. Hex-1-yne, $\mathrm{HC\equiv C\text{-}CH_2CH_2CH_2CH_3}$
   2. 3-Methylpent-1-yne, $\mathrm{HC\equiv C\text{-}CH(CH_3)CH_2CH_3}$
   3. 4-Methylpent-1-yne, $\mathrm{HC\equiv C\text{-}CH_2CH(CH_3)_2}$
   4. 3,3-Dimethylbut-1-yne, $\mathrm{HC\equiv C\text{-}C(CH_3)_3}$

   $\Rightarrow x = 4$.

   Internal alkynes (no acidic H), $y$:

   1. Hex-2-yne, $\mathrm{CH_3\text{-}C\equiv C\text{-}CH_2CH_2CH_3}$
   2. Hex-3-yne, $\mathrm{CH_3CH_2\text{-}C\equiv C\text{-}CH_2CH_3}$
   3. 4-Methylpent-2-yne, $\mathrm{CH_3\text{-}C\equiv C\text{-}CH(CH_3)_2}$

   $\Rightarrow y = 3$.

   ✓Final answer

   Option (C): $x, y = 4, 3$.

7. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.IUPAC name of $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2\mathrm{O}\right)\mathrm{Cl}\right]\mathrm{Cl}_2$ is (A) Tetraammineaquachlorocobalt (III) dichloride (B) Tetraamminechloroaquacobalt (III) chloride (C) Tetraammineaquachlorocobalt (III) chloride (D) Aquachlorotetraamminecobalt (III) chloride

   ›Reveal solutionSolution

   The complex is a cationic coordination entity with cobalt in the +3 oxidation state; the correct IUPAC name lists ligands alphabetically (ignoring prefixes) and ends with the counterion. The answer is Tetraammineaquachlorocobalt(III) chloride.

   The first thing to recognise is that this is a coordination compound with a complex cation and two chloride counterions. The square brackets enclose the coordination sphere, and the $\mathrm{Cl}_2$ outside the brackets tells us there are two chloride ions balancing the charge. So the name must end with "chloride" — not "dichloride" — because the counterion is simply chloride, and the number is implied by the formula or stated separately if needed.

   Now, the central metal is cobalt. To name it correctly, we need its oxidation state. The complex inside the brackets is $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{H}_2\mathrm{O}\right)\mathrm{Cl}\right]^{2+}$ (since the two external chlorides contribute $-2$ charge, the cation must be $+2$). Ammonia ($\mathrm{NH}_3$) and water ($\mathrm{H}_2\mathrm{O}$) are neutral ligands; chloride inside the coordination sphere is anionic ($\mathrm{Cl}^-$). Let the oxidation state of Co be $x$. Then:

   $x + 4(0) + 0 + (-1) = +2 \quad\Rightarrow\quad x - 1 = +2 \quad\Rightarrow\quad x = +3$.

   So it's cobalt(III).

   The ligands are: four ammines ($\mathrm{NH}_3$), one aqua ($\mathrm{H}_2\mathrm{O}$), and one chloro ($\mathrm{Cl}^-$). In IUPAC nomenclature, ligands are named in alphabetical order regardless of charge or number. The prefixes (tetra-, etc.) are ignored when alphabetising. So we compare the ligand names: "ammine", "aqua", "chloro". Alphabetically: ammine (a), aqua (a), chloro (c). But "ammine" and "aqua" both start with 'a' — we go to the second letter: 'm' vs 'q'. 'm' comes before 'q', so ammine comes before aqua. Then chloro comes last.

   Thus the order is: tetraammineaquachloro.

   Watch out

   A common mistake is to write "tetraamminechloroaqua" because chlorido ligands are sometimes placed before aqua in older conventions, or because students list by negative charge first. IUPAC rules are strictly alphabetical by ligand name (ignoring prefixes), so aqua comes before chloro.

   Now assemble the name: Tetraammineaquachlorocobalt(III) chloride. Notice that the metal name is attached directly after the ligands (no space), and the oxidation state is in Roman numerals in parentheses. The counterion is named as a separate word: "chloride".

   Let's check the options:

   • (A) Tetraammineaquachlorocobalt (III) dichloride — "dichloride" is wrong; there are two chlorides, but the name of the anion is just "chloride". The number is not indicated in the name unless it's a special case (like "dichloride" is used for certain compounds, but here the standard IUPAC name uses "chloride").
   • (B) Tetraamminechloroaquacobalt (III) chloride — wrong order: chloro before aqua.
   • (C) Tetraammineaquachlorocobalt (III) chloride — correct.
   • (D) Aquachlorotetraamminecobalt (III) chloride — wrong order (aqua before ammine) and also "tetraammine" is placed after aqua and chloro, which is not the standard way; the numerical prefix should come before the ligand name it modifies, but the alphabetical ordering of ligand names is still violated.

   Tip

   A quick way to verify: write the ligand names in alphabetical order ignoring prefixes: ammine, aqua, chloro. Then insert the numerical prefixes: tetraammine, aqua, chloro. Then attach to the metal: tetraammineaquachlorocobalt(III). Then add the counterion: chloride. That matches option (C) exactly.

   ✓Final answer

   The correct option is (C) Tetraammineaquachlorocobalt(III) chloride.

8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.Choose the correct statements from the following I) In vapour phase BeCl$_2$ exists as chlorobridged dimer II) BeSO$_4$ is readily soluble in water III) BeO is completely basic in nature IV) BeCO$_3$, being unstable, is kept in the atmosphere of CO$_2$ V) BeCO$_3$ is less soluble among all the carbonates of group 2 elements (A) II, III, IV (B) I, II, IV (C) I, IV, V (D) II, III, V

   ›Reveal solutionSolution

   Statements I, II and IV are correct; III and V are false. The correct option is (B).

   Beryllium is anomalous in group 2 because of its very small size and high polarising (charge/size) power, giving covalent character and a diagonal relationship with aluminium.

   I) BeCl$_2$ exists as a chloro-bridged dimer in the vapour phase - TRUE.

   On heating the solid (polymeric) chloride, the vapour first forms a chloro-bridged dimer $\text{Be}_2\text{Cl}_4$, which dissociates to the linear $\text{Cl-Be-Cl}$ monomer only at high temperature ($\sim1200\ \text{K}$). So over the ordinary vapour range the bridged dimer exists.

   II) BeSO$_4$ is readily soluble in water - TRUE.

   The tiny Be$^{2+}$ ion has a very high hydration enthalpy that exceeds the lattice enthalpy, so BeSO$_4$ (and MgSO$_4$) are highly soluble, unlike the heavier, insoluble BaSO$_4$.

   III) BeO is completely basic - FALSE.

   BeO is amphoteric: it dissolves in acids to give Be$^{2+}$ salts and in strong alkali to give beryllate, e.g. $\text{Na}_2[\text{Be(OH)}_4]$. Only the heavier group-2 oxides are purely basic.

   IV) BeCO$_3$ is unstable and is kept under a CO$_2$ atmosphere - TRUE.

   BeCO$_3$ decomposes easily to BeO and CO$_2$; storing it under CO$_2$ suppresses this decomposition.

   V) BeCO$_3$ is the least soluble group-2 carbonate - FALSE.

   Carbonate solubility decreases down the group, so BeCO$_3$ is the most soluble, not the least.

   Correct statements: I, II, IV.

   ✓Final answer

   The correct statements are I, II and IV. The correct option is (B).

   ANSWER: B

9. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
   Q.Solubility of A$_3$X$_4$ in pure water is ‘S’ mol L$^{-1}$. Its solubility product is (A) $S^7$ (B) $108\, S^5$ (C) $5184\, S^7$ (D) $6912\, S^7$

   ›Reveal solutionSolution

   The solubility product ($K_{sp}$) for a sparingly soluble salt A$_3$X$_4$ is derived by considering its dissociation into $3A^{4+}$ and $4X^{3-}$ ions. If its solubility is $S$ mol L$^{-1}$, then $[A^{4+}] = 3S$ and $[X^{3-}] = 4S$. Substituting these into the $K_{sp}$ expression, $K_{sp} = [A^{4+}]^3 [X^{3-}]^4$, yields $K_{sp} = 6912 S^7$.

   When a sparingly soluble ionic compound dissolves in water, it establishes an equilibrium between the undissolved solid and its constituent ions in solution. The solubility product constant, $K_{sp}$, quantifies this equilibrium. It is a measure of how much of the solid dissolves to form a saturated solution.

   The key idea is to understand how the stoichiometry of the ionic compound dictates the relative concentrations of its ions in solution. If 'S' represents the molar solubility of the compound (i.e., the number of moles of the compound that dissolve per liter of solution), then the concentration of each ion in the saturated solution will be a multiple of 'S', determined by the coefficients in the balanced dissociation equation.

   For a general sparingly soluble salt $A_mX_n$, its dissociation equilibrium is:

   $A_mX_n(s) \rightleftharpoons mA^{n+}(aq) + nX^{m-}(aq)$

   If the molar solubility of $A_mX_n$ is $S$ mol L$^{-1}$, then at equilibrium:

   $[A^{n+}] = mS$

   $[X^{m-}] = nS$

   The solubility product expression is then given by:

   $K_{sp} = [A^{n+}]^m [X^{m-}]^n$

   Substituting the concentrations in terms of $S$:

   $K_{sp} = (mS)^m (nS)^n = m^m n^n S^{m+n}$

   This general formula is what we will apply to the specific compound A$_3$X$_4$.

   1. Write the dissociation equilibrium for A$_3$X$_4$:

      The compound A$_3$X$_4$ dissociates into its constituent ions. To maintain charge neutrality, if there are 3 A ions and 4 X ions, and the formula is A$_3$X$_4$, then the charge on A must be $+4$ and the charge on X must be $-3$.

      So, the dissociation equilibrium is:

      $A_3X_4(s) \rightleftharpoons 3A^{4+}(aq) + 4X^{3-}(aq)$

   2. Relate ion concentrations to solubility ($S$):

      If the molar solubility of A$_3$X$_4$ is $S$ mol L$^{-1}$, it means that $S$ moles of A$_3$X$_4$ dissolve in one liter of water. According to the stoichiometry of the dissociation reaction:

      For every 1 mole of A$_3$X$_4$ that dissolves, 3 moles of A$^{4+}$ ions and 4 moles of X$^{3-}$ ions are produced.

      Therefore, in a saturated solution:

      $[A^{4+}] = 3S$ mol L$^{-1}$

      $[X^{3-}] = 4S$ mol L$^{-1}$

   3. Write the expression for the solubility product ($K_{sp}$):

      The solubility product constant is defined as the product of the concentrations of the ions, each raised to the power of its stoichiometric coefficient in the balanced dissociation equation.

      $K_{sp} = [A^{4+}]^3 [X^{3-}]^4$

   4. Substitute the ion concentrations in terms of $S$ into the $K_{sp}$ expression:

      Now, substitute the expressions for $[A^{4+}]$ and $[X^{3-}]$ from Step 2 into the $K_{sp}$ expression from Step 3:

      $K_{sp} = (3S)^3 (4S)^4$

      $K_{sp} = (3^3 \cdot S^3) \cdot (4^4 \cdot S^4)$

      $K_{sp} = (27 S^3) \cdot (256 S^4)$

      $K_{sp} = 27 \times 256 \times S^{3+4}$

      $K_{sp} = 6912 S^7$

      Watch out

      A common mistake is to forget to raise the stoichiometric coefficient to its power along with $S$. For example, writing $3S^3$ instead of $(3S)^3$. Remember that the entire concentration term, including the coefficient, is raised to the power.

   Comparing this result with the given options:

   (A) $S^7$

   (B) $108\, S^5$

   (C) $5184\, S^7$

   (D) $6912\, S^7$

   The calculated solubility product matches option (D).

   ✓Final answer

   The solubility product of A$_3$X$_4$ is $\boxed{6912\, S^7}$.

10. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
    Q.What is the correct order of freezing point of the following solutions? I $0.1\,\text{m}\ \text{Ba}_3(\text{PO}_4)_2$ II $0.1\,\text{m}\ \text{Na}_2\text{SO}_4$ III $0.1\,\text{m}\ \text{NaCl}$ (A) III > II > I (B) I > II > III (C) II > III > I (D) III > I > II

    ›Reveal solutionSolution

    Freezing point depression depends on the number of particles in solution. The more particles, the lower the freezing point. The correct order is III > II > I, so option (A).

    The key concept here is colligative properties — properties that depend only on the number of solute particles, not on their identity. Freezing point depression is one such property: adding a solute lowers the freezing point of a solvent, and the drop is proportional to the total molality of particles in solution.

    For ionic compounds, each formula unit dissociates into a certain number of ions. The van’t Hoff factor $i$ tells us how many particles each mole of solute produces. The freezing point depression is given by $\Delta T_f = i \cdot K_f \cdot m$, where $m$ is the molality. Since $K_f$ and $m$ are the same for all three solutions here (all are $0.1\,\text{m}$ in water), the freezing point is lowest for the solution with the largest $i$, and highest for the one with the smallest $i$.

    Let’s work through each solution.

    1. Solution I: $0.1\,\text{m}\ \text{Ba}_3(\text{PO}_4)_2$ This salt dissociates completely in water:

    $\text{Ba}_3(\text{PO}_4)_2 \rightarrow 3\,\text{Ba}^{2+} + 2\,\text{PO}_4^{3-}$

    That’s $3 + 2 = 5$ ions per formula unit. So $i = 5$.

    The effective particle molality is $0.1 \times 5 = 0.5\,\text{m}$.

    This gives the largest depression, hence the lowest freezing point.

    2. Solution II: $0.1\,\text{m}\ \text{Na}_2\text{SO}_4$ Dissociation:

    $\text{Na}_2\text{SO}_4 \rightarrow 2\,\text{Na}^+ + \text{SO}_4^{2-}$

    That’s $2 + 1 = 3$ ions. So $i = 3$.

    Effective particle molality = $0.1 \times 3 = 0.3\,\text{m}$.

    Freezing point is higher than I, lower than III.

    3. Solution III: $0.1\,\text{m}\ \text{NaCl}$ Dissociation:

    $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$

    That’s $2$ ions. So $i = 2$.

    Effective particle molality = $0.1 \times 2 = 0.2\,\text{m}$.

    This gives the smallest depression, hence the highest freezing point.

    Watch out

    A common mistake is to forget that $\text{Ba}_3(\text{PO}_4)_2$ gives 5 ions, not 2 or 3. Count carefully: the subscript 3 on Ba and 2 on PO$_4$ mean 3 barium ions and 2 phosphate ions.

    So the order of freezing point (highest to lowest) is:

    III (highest) > II > I (lowest).

    ✓Final answer

    The correct order is III > II > I, which corresponds to option (A).

11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.Consider the following about the Tyndall effect I) It is used to distinguish between a true and colloidal solution II) It is possible only when the dispersed medium and dispersed phase differ much in their refractive indices III) It is observed only when the size of colloidal particles is much smaller than the wavelength of the light used The correct statements are (A) I & III only (B) II & III only (C) I, II & III (D) I & II only

    ›Reveal solutionSolution

    The Tyndall effect is the scattering of light by colloidal particles; it distinguishes true from colloidal solutions, requires a refractive index difference, and occurs when particle size is comparable to (not much smaller than) the light’s wavelength. Only statements I and II are correct.

    The Tyndall effect is a classic optical phenomenon used to tell apart a true solution (like salt water) from a colloidal dispersion (like milk or fog). The key idea is that light scatters when it hits particles of a certain size — but the exact conditions matter. Let’s break down each statement.

    1. Statement I: “It is used to distinguish between a true and colloidal solution.”

       This is correct. In a true solution, solute particles are of molecular size (less than 1 nm) and do not scatter visible light — the beam passes through invisibly. In a colloid, particles are larger (1–1000 nm) and scatter light, making the beam visible (the Tyndall effect). So shining a light through a sample and looking for a visible beam is a simple test.

    2. Statement II: “It is possible only when the dispersed medium and dispersed phase differ much in their refractive indices.”

       This is also correct. Scattering intensity depends on the difference in refractive index between the particles and the surrounding medium. If they have nearly the same refractive index, light passes through with little scattering — the effect is weak or absent. A large difference (e.g., water droplets in air) gives strong scattering.

    3. Statement III: “It is observed only when the size of colloidal particles is much smaller than the wavelength of the light used.”

       This is incorrect. The Tyndall effect is strongest when the particle size is comparable to the wavelength of light (roughly 1/10 to 10 times the wavelength). If particles are much smaller than the wavelength, Rayleigh scattering occurs (like the blue sky), but the Tyndall effect — a visible, milky beam — requires larger particles. If particles are much larger, they simply reflect or refract light (like dust or raindrops), not scatter it in the characteristic Tyndall way.

    Watch out

    A common mistake is to think that smaller particles always scatter more. In fact, for the Tyndall effect, the particles must be large enough (roughly 40–900 nm for visible light) to produce the observable cone of scattered light. Particles much smaller than the wavelength give weak, wavelength-dependent scattering (blue sky), not the strong white beam of the Tyndall effect.

    Thus, only statements I and II are correct.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.According to Werner’s theory, the number of groups bonded to the central metal atom / ion in a coordination complex represent (A) Oxidation state (B) Primary valency (C) Secondary Valency (D) Polyhedron

    ›Reveal solutionSolution

    Werner’s theory distinguishes primary valency (ionic, non-directional) from secondary valency (directional, fixed number of ligands directly bonded to the metal). The number of groups bonded to the central metal atom represents the secondary valency.

    ---

    Werner’s theory was the first successful model of coordination compounds. Before it, chemists struggled to explain why compounds like $\ce{CoCl3 \cdot 6 NH3}$ existed — why would ammonia stick to cobalt chloride in such a fixed ratio? Werner’s key insight was that a metal ion has two kinds of valency:

    • Primary valency (now called oxidation state) — satisfied by negative ions, non-directional, and corresponds to the charge on the metal.
    • Secondary valency (now called coordination number) — satisfied by neutral molecules or negative ions directly bonded to the metal, directional, and fixed in number for a given metal.

    The question asks: “the number of groups bonded to the central metal atom / ion” — that is, the count of ligands directly attached to the metal. That is exactly what Werner called secondary valency.

    ---

    1. Identify what “groups bonded” means

       In a complex like $\ce{[Co(NH3)6]^{3+}}$, six ammonia molecules are directly bonded to cobalt. The number “6” is the count of groups bonded. This is not the charge (oxidation state) nor the shape (polyhedron) itself — it’s the number that determines the shape.

    2. Eliminate the distractors

       • (A) Oxidation state — This is the charge left on the metal after satisfying primary valency. For $\ce{Co^{3+}}$, the oxidation state is +3, not 6.
       • (D) Polyhedron — The arrangement of ligands (e.g., octahedron) is a consequence of the number of groups, not the number itself.
       • (B) Primary valency — This is satisfied by ions that balance charge, not necessarily by groups directly bonded. In $\ce{[Co(NH3)6]Cl3}$, the three Cl⁻ ions satisfy primary valency but are not bonded directly to cobalt — they are outside the coordination sphere.

    3. Confirm with Werner’s own words

       Werner wrote that secondary valency is “the number of groups directly attached to the metal atom.” That is the definition. For $\ce{[Co(NH3)6]Cl3}$, secondary valency = 6, primary valency = 3.

    Tip

    A quick memory trick: Primary = charge (oxidation state), Secondary = number of ligands (coordination number). The question asks for “number of groups bonded” — that’s secondary.

    Watch out

    A common mistake is to confuse “groups bonded” with “ions needed to neutralize charge.” Remember: in $\ce{K4[Fe(CN)6]}$, the four K⁺ ions satisfy primary valency, but the six CN⁻ groups bonded to iron represent secondary valency.

    ---

    ✓Final answer

    The correct option is (C).

    ANSWER: C