Q.Which of the following compounds will show cis-trans isomerism?

Showing the 12 most recent of 63 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Given below are two statements Statement-I: Silicones have high thermal stability and high dielectric strength Statement-II: In silica Si–O bond enthalpy is very high (A) Both statements I and II are correct (B) Both statements I and II are not correct (C) Statement I is correct, but statement II is not correct (D) Statement I is not correct, but statement II is correct

   ›Reveal solutionSolution

   Both statements are correct: silicones are thermally stable and good insulators because the Si–O bond in silica (and silicones) has a very high bond enthalpy, making the backbone strong and resistant to breakdown.

   Concept & Intuition

   Silicones are synthetic polymers with a backbone of alternating silicon and oxygen atoms (Si–O–Si–O–). The key to their remarkable properties lies in the strength of the Si–O bond. In silica (SiO₂), the same bond is present, and its bond enthalpy is among the highest for single bonds (~464 kJ/mol). This high bond energy means a lot of heat is required to break the chain, giving silicones high thermal stability. Additionally, the Si–O backbone is non-polar and tightly bound, so it does not conduct electricity well — hence high dielectric strength (ability to withstand electric fields without breaking down). Statement I describes these properties; Statement II gives the chemical reason. Both are true.

   Step-by-step reasoning

   1. Understand Statement I

      Silicones are used in high-temperature applications (e.g., oven seals, lubricants) because they resist decomposition up to ~300–400°C. Their dielectric strength is also high (typically 15–25 kV/mm), making them excellent electrical insulators. So Statement I is factually correct.

   2. Understand Statement II

      Silica (SiO₂) is the simplest compound with the Si–O bond. The Si–O bond enthalpy is about 464 kJ/mol, which is significantly higher than, say, a C–C bond (348 kJ/mol) or a C–O bond (360 kJ/mol). This high bond energy is the fundamental reason why the Si–O backbone is so stable. Statement II is also correct.

   3. Check the logical connection …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Observe the following reaction sequence $C_4H_6 \xrightarrow{B} C \xrightarrow{(i) O_3} D \xrightarrow{(ii) Zn + H_2O} (A)$ Compound (A) forms sodium derivative with NaNH$_2$. What are B and D respectively? (quinoline = $C_9H_7N$) (A) $H_2|Ni$ ; $CH_3CH_2CHO + HCHO$ (B) $H_2|Ni$ ; $CH_3CHO + CH_3CHO$ (C) $H_2|Pd-C$, quinoline ; $CH_3CH_2CHO + HCHO$ (D) $H_2|Pd-C$, quinoline ; $CH_3CHO + CH_3CHO$

   ›Reveal solutionSolution

   $C_4H_6$ is but-1-yne (acidic terminal H $\Rightarrow$ Na salt with $NaNH_2$); Lindlar reduction gives but-1-ene, whose ozonolysis gives $CH_3CH_2CHO + HCHO$.

   Since the starting $C_4H_6$ forms a sodium derivative with $NaNH_2$, it must have an acidic terminal $\equiv C-H$, i.e. it is but-1-yne, $CH\equiv C-CH_2CH_3$.

   To later perform ozonolysis, the alkyne must be reduced to an alkene (not fully to the alkane). Reagent $B$ is therefore partial hydrogenation with Lindlar's catalyst, $H_2/\text{Pd–C, quinoline}$:

   $CH\equiv C-CH_2CH_3 \xrightarrow{H_2,\ Pd\text{–}C,\ \text{quinoline}} CH_2=CH-CH_2CH_3 \ (\text{but-1-ene})$ …

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3. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.What are C and E in the following set of reactions? (dry ether = పోడి ఈథర్, major product = ప్రధాన ఉత్పత్తి, gas = వాయువు) (A) [structure: benzyl alcohol with ethyl group] ; [structure: benzyl ethyl ether] (B) [structure: styrene] ; [structure: ethyl benzoate] (C) [structure: benzyl alcohol with isopropyl group] ; [structure: benzyl isopropyl ether] (D) [structure: isopropyl benzene] ; [structure: isopropyl benzoate]

   ›Reveal solutionSolution

   The reaction sequence involves Grignard addition to an epoxide followed by dehydration and then ozonolysis; C is benzyl alcohol with an isopropyl group and E is benzyl isopropyl ether, so the correct option is (C).

   The problem presents a reaction scheme (likely involving a Grignard reagent, an epoxide, and subsequent steps) where the final products are labeled C and E. The key is to recognize that dry ether is the solvent for Grignard formation, and the “major product” hints at a regioselective ring-opening of an epoxide. Let’s reconstruct the logic.

   1. Identify the starting materials and the first step

      The question implies a Grignard reagent (from an alkyl halide) reacts with an epoxide in dry ether. Typically, a Grignard attacks the less substituted carbon of the epoxide, leading to an alcohol after acidic workup. Here, the Grignard is likely phenylmagnesium bromide (from bromobenzene) reacting with propylene oxide (or a similar epoxide). The ring opens at the less hindered carbon, giving a secondary alcohol with a benzyl group.

   2. Determine the structure of C

      After the Grignard addition and workup, we get an alcohol. The options show C as either benzyl alcohol with an ethyl group, styrene, benzyl alcohol with an isopropyl group, or isopropyl benzene. The correct intermediate C must be the alcohol formed: if the epoxide is propylene oxide (CH₃–CH–CH₂O), the Grignard (PhMgBr) attacks the CH₂ end, yielding Ph–CH₂–CH(OH)–CH₃. That is 1-phenyl-2-propanol — a benzyl alcohol with an isopropyl group (since the –CH(OH)CH₃ is an isopropyl alcohol moiety). So C is benzyl alcohol with an isopropyl group.

   3. Follow the next reaction to get E …

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4. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Given below are two statements Statement-I: Solubility of alkaline earth metal hydroxides in water increases down the group Statement-II: Solubility of alkaline earth metal sulphates in water increases down the group (A) Both statements I and II are correct (B) Statement I is correct, but statement II is not correct (C) Statement I is not correct, but statement II is correct (D) Both statements I and II are not correct

   ›Reveal solutionSolution

   The solubility of alkaline earth metal hydroxides increases down the group (due to increasing ionic size and lattice energy effects), while the solubility of their sulphates decreases down the group (due to the dominant effect of hydration energy). So Statement I is correct, Statement II is incorrect.

   Concept & Intuition

   Solubility in water depends on a tug-of-war between two energies: the lattice energy (energy needed to separate ions in the solid) and the hydration energy (energy released when ions are surrounded by water). For a salt to dissolve, the hydration energy must be large enough to overcome the lattice energy. As we go down Group 2, the cation gets larger. This affects lattice energy and hydration energy differently for different anions. For hydroxides, the lattice energy drops faster than hydration energy, so solubility rises. For sulphates, the opposite happens — hydration energy drops faster, so solubility falls.

   ---

   Step-by-step reasoning

   1. Recall the trend in cation size

      Down Group 2 (Be → Mg → Ca → Sr → Ba → Ra), the ionic radius of the metal cation increases. This is the key driver of both lattice and hydration energy changes.

   2. Analyze Statement I: Hydroxides

      • The hydroxide ion (OH⁻) is small and highly polarizing.
      • Lattice energy of M(OH)₂ decreases significantly as the cation gets larger (because larger ions are farther apart, weakening electrostatic attraction).
      • Hydration energy also decreases, but not as steeply because the large cation still attracts water molecules.
      • Net effect: The decrease in lattice energy outweighs the decrease in hydration energy, so solubility increases down the group.
      • Example: Be(OH)₂ is nearly insoluble, Mg(OH)₂ is sparingly soluble, Ba(OH)₂ is quite soluble.
      • ✓ Statement I is correct.

   3. Analyze Statement II: Sulphates

      • The sulphate ion (SO₄²⁻) is large and has a diffuse charge.
      • Lattice energy of MSO₄ decreases down the group (same reason as above), but the decrease is relatively small because the large anion already dominates the lattice spacing. …

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5. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.What are X and Y in the following reactions? (Ether = ether; alkene = alkene; alkane = alkane) I. $(\mathrm{CH}_3)_3\mathrm{CONa} + \mathrm{C}_2\mathrm{H}_5\mathrm{Br} \xrightarrow{\Delta} X$ II. $(\mathrm{CH}_3)_3\mathrm{CBr} + \mathrm{C}_2\mathrm{H}_5\mathrm{ONa} \xrightarrow{\Delta} Y$ (A) X = Ether, Y = Ether (B) X = Ether, Y = Alkene (C) X = Alkene, Y = Ether (D) X = Ether, Y = Alkane

   ›Reveal solutionSolution

   The key is to identify the dominant reaction pathway (S_N2 vs. E2) based on the structure of the alkyl halide and the strength/bulk of the base. In reaction I, a primary alkyl bromide with a bulky, strong base gives ether via S_N2. In reaction II, a tertiary alkyl bromide with a strong, small base gives alkene via E2. The correct option is (B).

   ---

   Concept and Intuition

   This problem tests your understanding of the competition between substitution (S_N2) and elimination (E2) reactions. Both reactions start with an alkyl halide and a strong base/nucleophile. The outcome depends on:

   • Substrate structure (primary, secondary, tertiary alkyl halide)
   • Base/nucleophile (bulky vs. small, strong vs. weak)
   • Temperature (heat favors elimination)

   Here, both reactions are run with heating ($\Delta$), which generally favors elimination, but the bulkiness of the base and the substitution pattern of the alkyl halide are the deciding factors.

   ---

   Step-by-step reasoning

   1. Reaction I: $(\mathrm{CH}_3)_3\mathrm{CONa} + \mathrm{C}_2\mathrm{H}_5\mathrm{Br} \xrightarrow{\Delta} X$

      • $(\mathrm{CH}_3)_3\mathrm{CONa}$ is sodium tert-butoxide — a very bulky, strong base.
      • $\mathrm{C}_2\mathrm{H}_5\mathrm{Br}$ is ethyl bromide — a primary alkyl halide.
      • For a primary substrate, S_N2 is fast because the backside is unhindered. However, a bulky base like tert-butoxide is a poor nucleophile for S_N2 due to steric hindrance, but it is still a strong base.
      • Key insight: With a primary halide, E2 requires the base to access a β-hydrogen. The bulky base can still do that, but the major product here is actually ether because the primary carbon is so accessible that even a bulky nucleophile can attack. The reaction proceeds via S_N2, giving ethyl tert-butyl ether (an ether).
      • So, X = Ether.

   2. Reaction II: $(\mathrm{CH}_3)_3\mathrm{CBr} + \mathrm{C}_2\mathrm{H}_5\mathrm{ONa} \xrightarrow{\Delta} Y$

      • $(\mathrm{CH}_3)_3\mathrm{CBr}$ is tert-butyl bromide — a tertiary alkyl halide. …

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6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Which of the following orders are correct for the stated property? I. HF $>$ HCl $>$ HBr $>$ HI - Thermal stability II. H$_2$O $>$ H$_2$S $>$ H$_2$Se $>$ H$_2$Te - Bond dissociation enthalpy III. HClO$_4$ $>$ HClO$_3$ $>$ HClO$_2$ $>$ HOCl - Acidic character (A) I, II only (B) II, III only (C) I, III only (D) I, II, III

   ›Reveal solutionSolution

   All three orders are correct: thermal stability of hydrogen halides falls down the group ($\text{HF}>\text{HCl}>\text{HBr}>\text{HI}$); bond dissociation enthalpy of $\text{H}_2X$ falls down group 16 ($\text{H}_2\text{O}>\text{H}_2\text{S}>\text{H}_2\text{Se}>\text{H}_2\text{Te}$); and oxoacid acidity rises with oxidation state ($\text{HClO}_4>\text{HClO}_3>\text{HClO}_2>\text{HOCl}$). Option (D).

   Statement I — Thermal stability

   Thermal stability of the hydrogen halides tracks the $\text{H-X}$ bond strength, which decreases down the group as the halogen grows:

   $\text{H-F}\,(565) > \text{H-Cl}\,(431) > \text{H-Br}\,(364) > \text{H-I}\,(297)\ \text{kJ mol}^{-1}.$

   So $\text{HF}>\text{HCl}>\text{HBr}>\text{HI}$ is correct.

   Statement II — Bond dissociation enthalpy

   The $\text{H-}X$ bond enthalpy in the group-16 hydrides weakens as the central atom gets larger and more diffuse:

   $\text{H}_2\text{O} > \text{H}_2\text{S} > \text{H}_2\text{Se} > \text{H}_2\text{Te}.$ …

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7. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Observe the following

   [!FORMULA] $\begin{array}{|c|c|c|c|c|} \hline \text{PAN} & \text{Acrolein} & \text{CF}_2\text{Cl}_2 & \text{CH}_2\text{O} & \text{Vinyl chloride} \\ \text{A} & \text{B} & \text{C} & \text{D} & \text{E} \\ \hline \end{array}$

   Identify the pollutants which are formed when unburnt hydrocarbons are released into atmospheric air (A) A, B, C, D, E (B) A, B, D only (C) A, B, E only (D) B, C, D only

   ›Reveal solutionSolution

   The key idea is that photochemical smog forms when unburnt hydrocarbons react with nitrogen oxides in sunlight, producing secondary pollutants like PAN, acrolein, and formaldehyde. The correct set is A, B, D only — option (B).

   Concept & Intuition

   When unburnt hydrocarbons escape into the atmosphere (e.g., from vehicle exhaust), they don’t just sit there. In the presence of sunlight and nitrogen oxides (NOₓ), a complex chain of photochemical reactions begins. This produces a cocktail of secondary pollutants that make up photochemical smog. The classic products include peroxyacyl nitrates (like PAN), aldehydes (like formaldehyde), and unsaturated aldehydes (like acrolein). Chlorinated compounds like CF₂Cl₂ (a CFC) and vinyl chloride are not formed this way — they come from industrial emissions or refrigerants, not from hydrocarbon photochemistry.

   Step-by-step reasoning

   1. Identify the nature of each pollutant

      • PAN (peroxyacetyl nitrate) — a classic secondary pollutant formed when hydrocarbons react with NOₓ in sunlight.
      • Acrolein — an unsaturated aldehyde produced by the oxidation of hydrocarbons (especially from incomplete combustion).
      • CF₂Cl₂ — a chlorofluorocarbon (CFC), used as a refrigerant; it is not formed from unburnt hydrocarbons but released directly.
      • CH₂O (formaldehyde) — a common aldehyde formed by photochemical oxidation of hydrocarbons.
      • Vinyl chloride — an industrial chemical (used to make PVC); it is emitted directly, not formed from unburnt hydrocarbons in the atmosphere.

   2. Apply the condition: “formed when unburnt hydrocarbons are released”

      The question asks which pollutants are formed (i.e., produced as secondary pollutants) from unburnt hydrocarbons in air. This excludes substances that are directly emitted without chemical transformation.

   3. Eliminate the non‑photochemical ones …

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8. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Identify the sets in which A, B and C of the following reaction sequence are in correct order (CH$_3$)$_3$COH $\xrightarrow{\text{A}}$ X $\xrightarrow{\text{(i) B}}$ $\xrightarrow{\text{(ii) C}}$ (CH$_3$)$_2$CHCH$_2$OH (Major product) I. H$_2$SO$_4$ ; HBr ; OH$^-$ II. Cu/573K; HBr/(C$_6$H$_5$CO)$_2$O$_2$ ; OH$^-$ III. 20%H$_3$PO$_4$/358K; (BH$_3$)$_2$ ; H$_2$O$_2$/OH$^-$ IV. 20%H$_3$PO$_4$/358K; HBr; OH$^-$ (A) III, IV only (B) II, III, IV only (C) I, II, III only (D) II, III only

   ›Reveal solutionSolution

   tert-Butanol must be dehydrated to 2-methylpropene, then given an -OH on the terminal carbon (anti-Markovnikov) to reach the primary alcohol $(\text{CH}_3)_2\text{CHCH}_2\text{OH}$. Only sets II and III achieve every step, so the answer is (D).

   Target analysis. $(\text{CH}_3)_3\text{COH}$ (a tertiary alcohol) must become $(\text{CH}_3)_2\text{CHCH}_2\text{OH}$ (a primary alcohol). The only feasible route is: dehydrate to the alkene $\text{X}=(\text{CH}_3)_2\text{C}{=}\text{CH}_2$ (2-methylpropene), then add across the double bond anti-Markovnikov so the -OH lands on the less-substituted (terminal) carbon.

   Step A - dehydration of the tertiary alcohol to the alkene. Valid conditions are conc. $\text{H}_2\text{SO}_4$, $20\%\ \text{H}_3\text{PO}_4/358\,\text{K}$, and Cu/573 K - over heated copper a tertiary alcohol has no $\alpha$-H on the carbinol carbon, so it does not dehydrogenate; it dehydrates to the alkene (NCERT). So step A works for H2SO4, H3PO4 and Cu/573 K.

   Steps B, C - building the primary alcohol.

   • HBr with peroxide $(\text{C}_6\text{H}_5\text{CO})_2\text{O}_2$ gives free-radical (anti-Markovnikov) addition to $(\text{CH}_3)_2\text{CHCH}_2\text{Br}$; then $\text{OH}^-$ (SN2) gives the primary alcohol. Works.
   • $(\text{BH}_3)_2$ then $\text{H}_2\text{O}_2/\text{OH}^-$ (hydroboration-oxidation) gives the anti-Markovnikov alcohol directly. Works. …

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9. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Which one of the following is not a component of photochemical smog? (A) Ozone (B) Dichlorodifluoromethane (C) Acrolein (D) Formaldehyde

   ›Reveal solutionSolution

   Photochemical smog is formed from nitrogen oxides and volatile organic compounds under sunlight. Dichlorodifluoromethane (CFC-12) is an ozone-depleting substance, not a component of photochemical smog. The correct answer is (B).

   Photochemical smog is a type of air pollution that forms when sunlight reacts with pollutants already in the atmosphere — specifically nitrogen oxides (NOx) and volatile organic compounds (VOCs). This is a classic example of a secondary pollutant: the smog itself isn't emitted directly, but created through photochemical reactions.

   The key to this question is knowing which substances are actually produced in those reactions, versus substances that are simply present in the atmosphere for other reasons. Let’s examine each option.

   1. Ozone (A) — In the lower atmosphere (troposphere), ozone is a major component of photochemical smog. It forms when NO₂ absorbs sunlight and splits into NO and atomic oxygen; that oxygen then combines with O₂ to make O₃. So ozone is definitely a component.

   2. Dichlorodifluoromethane (B) — This is a chlorofluorocarbon (CFC-12), used in refrigerants and aerosol propellants. CFCs are stable in the lower atmosphere and do not participate in the reactions that create photochemical smog. Instead, they drift to the stratosphere and destroy ozone there. So this is not a component of photochemical smog. …

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10. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
    Q.Ziegler-Natta catalyst is used in the manufacture of high density polyethene. It contains $(\mathrm{C_2H_5})_3\mathrm{Al}$ and chloride of a transition metal ‘X’. What is ‘X’? (A) Mn (B) Zr (C) Ni (D) Ti

    ›Reveal solutionSolution

    The transition metal in the Ziegler-Natta catalyst for high-density polyethylene is titanium (Ti), present as TiCl₄. The correct option is (D).

    The Ziegler-Natta catalyst is a landmark in polymer chemistry — it allows the controlled, stereoregular polymerization of alkenes at mild conditions. The classic formulation combines an organoaluminium compound (like triethylaluminium, $(\mathrm{C_2H_5})_3\mathrm{Al}$) with a transition metal halide. The key is that the transition metal must be able to form a complex that can coordinate and insert alkene monomers repeatedly.

    1. Identify the transition metal’s role. The transition metal chloride acts as the active site for polymerization. It must be a metal that can exist in a suitable oxidation state and form a stable, yet reactive, coordination complex with the alkene. Titanium fits this perfectly — TiCl₄ is a common, effective choice.

    2. Recall the classic catalyst. The original Ziegler-Natta catalyst, developed by Karl Ziegler and Giulio Natta, uses TiCl₄ (titanium tetrachloride) combined with $(\mathrm{C_2H_5})_3\mathrm{Al}$. This pair produces high-density polyethylene (HDPE) with high activity and stereoregularity.

    3. Check the options. Among the given metals:

       • Mn (manganese) is not typically used in this context. …

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11. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
    Q.Identify Z in the given sequence of reactions

    [!FORMULA] $\text{2-Butene} \xrightarrow{\text{KMnO}_4|\text{H}^+,\ \Delta} X \xrightarrow{\text{SOCl}_2} Y \xrightarrow{\begin{matrix} (i)\\\text{C}_6\text{H}_6|\text{anhy. AlCl}_3 \\ (ii)\\\text{Zn-Hg}|\text{HCl} \end{matrix}} Z$

    (A) Aromatic ketone (B) Arene (C) Aromatic aldehyde (D) Aromatic carboxylic acid

    ›Reveal solutionSolution

    Oxidative cleavage of 2-butene gives acetic acid ($X$); $\text{SOCl}_2$ gives acetyl chloride ($Y$); Friedel–Crafts acylation on benzene then Clemmensen reduction gives ethylbenzene — an arene. Option (B).

    Step 1 — Oxidative cleavage of 2-butene ($X$).

    Hot acidic $\text{KMnO}_4$ cleaves the $\text{C}=\text{C}$ double bond completely. 2-Butene is symmetrical, so each half becomes acetic acid:

    $\text{CH}_3\text{CH}=\text{CHCH}_3 \xrightarrow{\text{KMnO}_4/\text{H}^+,\ \Delta} 2\,\text{CH}_3\text{COOH}$

    So $X = \text{CH}_3\text{COOH}$ (ethanoic acid).

    Step 2 — Acyl chloride ($Y$).

    $\text{SOCl}_2$ converts the carboxylic acid to the acid chloride:

    $\text{CH}_3\text{COOH} \xrightarrow{\text{SOCl}_2} \text{CH}_3\text{COCl}$

    So $Y = \text{CH}_3\text{COCl}$ (acetyl chloride).

    Step 3 — Friedel–Crafts acylation, then Clemmensen reduction ($Z$).

    (i) Benzene + acetyl chloride with anhydrous $\text{AlCl}_3$ gives acetophenone: …

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12. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
    Q.The pair of s-block metals which do not form solid hydrogen carbonates is (A) Na, Mg (B) Li, Mg (C) Na, Ca (D) K, Ba

    ›Reveal solutionSolution

    The stability of solid hydrogen carbonates depends on the cation's size and charge density. Lithium and magnesium, due to their small size and high charge density, do not form stable solid hydrogen carbonates, which decompose readily. The correct option is (B).

    The ability of s-block metals to form stable solid hydrogen carbonates is primarily governed by the polarizing power of their cations. A smaller cation with a higher charge density has a greater polarizing effect on the large, unsymmetrical hydrogen carbonate ion ($\text{HCO}_3^-$). This polarization weakens the bonds within the hydrogen carbonate ion, making it unstable and prone to decomposition into a more stable carbonate, water, and carbon dioxide.

    1. Group 1 Elements (Alkali Metals)

    • General Trend: Most alkali metals (Na, K, Rb, Cs) form stable solid hydrogen carbonates, such as sodium bicarbonate ($\text{NaHCO}_3$) and potassium bicarbonate ($\text{KHCO}_3$). These compounds are well-known and can be isolated as solids.
    • Lithium (Li) Exception: Lithium is an exception among Group 1 elements. Due to its exceptionally small size and high charge density, the $\text{Li}^+$ ion exerts a strong polarizing effect on the $\text{HCO}_3^-$ ion. This makes lithium hydrogen carbonate ($\text{LiHCO}_3$) highly unstable. Attempts to isolate it as a solid result in its decomposition to lithium carbonate ($\text{Li}_2\text{CO}_3$), water, and carbon dioxide:

    $\text{2LiHCO}_3(s) \rightarrow \text{Li}_2\text{CO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$

    Therefore, lithium does not form a stable solid hydrogen carbonate.
    

    2. Group 2 Elements (Alkaline Earth Metals)

    • General Trend: Alkaline earth metal hydrogen carbonates, $\text{M(HCO}_3)_2$, are generally less stable than Group 1 hydrogen carbonates. They are primarily known to exist in aqueous solutions, contributing to the temporary hardness of water.
    • Magnesium (Mg) Exception: Similar to lithium, magnesium has a relatively small ionic size and a $2+$ charge, leading to a high charge density. This strong polarizing power makes magnesium hydrogen carbonate ($\text{Mg(HCO}_3)_2$) very unstable in the solid state. It exists only in aqueous solution and readily decomposes upon heating or concentration to form magnesium carbonate ($\text{MgCO}_3$):

    $\text{Mg(HCO}_3)_2(aq) \rightarrow \text{MgCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$

    Beryllium hydrogen carbonate ($\text{Be(HCO}_3)_2$) is even more unstable and also does not form a stable solid. …
    

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