Okay, so I need to figure out the molecular symmetry group of product E based on the given chemical reactions. Let me start by breaking down each reaction step by step and see if I can deduce the structures of all the compounds involved. Then, I can determine the symmetry of E.

First, the question says that solid A reacts with 8 equivalents of gas B to form bright red product C. Then, C reacts with 2 equivalents of gas D to produce extremely hazardous product E. Also, when C reacts with water, it reforms A plus two acids F and G. F is a strong acid and G is a weak acid. D reacts with B in a 1:1 ratio to form H, which is used as a solvent.

Hmm, let's start with the first reaction: A + 8 B → C. Since B is a gas, maybe it's something like ammonia (NH3)? But 8 equivalents of B... that's a lot. Maybe B is a diatomic gas like O2 or N2, but 8 equivalents would imply a stoichiometric ratio. Alternatively, B could be H2. If A is something like Fe(NO3)3 and B is NH3, maybe the reaction forms some compound like Fe(NO3)3·8NH3? But I'm not sure. Alternatively, maybe B is a reducing agent like H2S or H2O. Wait, but the product is a bright red solid. Bright red solids often indicate the presence of iron in a particular oxidation state. For example, Fe(NO3)3 is brownish, but maybe Fe(OH)3 is reddish brown. But that's not bright red. Alternatively, maybe the product is Prussian blue, which is a dark blue crystal. Wait, but the product is bright red. Maybe it's a different iron compound. Alternatively, maybe it's ferric oxide with some ammonia added. Wait, I'm not sure. Let me think again.

Alternatively, maybe B is a gas like H2O vapor. If A is a metal, reacting with water might form a hydroxide. But the problem states that C reacts with water to reform A and two acids. So C must be a compound that, when hydrolyzed, gives back A and two acids. So if A is a metal, then C might be a metal hydroxide or oxide. For example, if A is Fe(NO3)3, then C might be Fe(NO3)3·8H2O? But when C reacts with water, it would decompose back into Fe(NO3)3 and water, but that doesn't explain the acids. Alternatively, maybe C is a complex hydrate. Wait, but the reaction of C with water gives two acids: F and G. So F and G are acids. Let's assume F is a strong acid like HCl or HNO3, and G is a weak acid like CH3COOH or H2SO3. So when C reacts with water, it dissociates into A, F, and G. So maybe C is a compound that can lose two protons to form A and two acids. For example, if A is Fe(NO3)3, then C could be Fe(NO3)3·HCl·H2SO4, but I'm not sure. Alternatively, perhaps C is a salt with two acidic groups. For example, if A is Fe(NO3)3, then C could be Fe(NO3)3·HCl·H2SO4. When C reacts with water, it would release HCl and H2SO4, giving back Fe(NO3)3 and water. But the problem states that E is formed when C reacts with 2 equivalents of gas D. So D is a gas that reacts with C to form E. Also, D reacts with B in a 1:1 ratio to form H, a solvent. If B is a gas and D is another gas, maybe B is H2O vapor and D is HCl gas. Then H would be H2O and HCl, but that's not a solvent. Alternatively, if B is NH3 and D is HCl, then D reacting with B would form NH4Cl gas, but NH4Cl is a solid. Wait, the problem says D reacts with B in a 1:1 ratio to form H, which is used as a solvent. So H is a solvent, maybe water. If B is NH3 and D is H2O, then they react to form NH3·H2O, which is a solution, but that's not a solvent. Alternatively, maybe B is H2 and D is something else. Wait, this is getting confusing. Let me try to list all the possible compounds and reactions step by step.

Starting from the beginning:

1. Reaction 1: A (solid) + 8 B (gas) → C (product). 8 equivalents of gas B. Let's think of B as NH3. Then 8 equivalents would mean 8 moles of NH3. If A is Fe(NO3)3, then maybe the reaction is Fe(NO3)3 + 8 NH3 → Fe(NO3)3·8NH3. But when this reacts with water (Reaction 3), it should decompose back into Fe(NO3)3 and the NH3 would form NH4NO3 or something. But the problem states that when C reacts with water, it produces A and two acids. So if C is Fe(NO3)3·8NH3, then when it reacts with water, it would hydrolyze to Fe(OH)3 and NH4+ and NO3-? But NH4+ is not an acid. Wait, but the problem says two acids. So maybe the hydrolysis releases two acids. Hmm.

Alternatively, maybe C is a complex ion. For example, if A is Al(NO3)3 and B is NH3, then 8 NH3 would form [Al(NH3)6]^3+ and some NH4NO3. But I'm not sure.

Alternatively, maybe A is a transition metal oxide. For example, Fe2O3 reacts with NH3 to form Fe(NH3)6^3+ and H2O. But how many equivalents? Fe2O3 has two Fe^3+ ions. Each Fe^3+ can form a complex with six NH3 molecules. So Fe2O3 + 6 NH3 → Fe(NH3)6^3+ + 3 H2O. But the problem states 8 equivalents. So maybe it's a different ratio. Maybe A is Fe(NO3)3, which has one Fe^3+ ion. Then Fe(NO3)3 + 6 NH3 → Fe(NH3)6^3+ + 3 H2O. But 6 equivalents, not 8. Hmm. Maybe the reaction is with more NH3. Alternatively, maybe B is a different gas. Let's think of B as H2O vapor. If A is Fe(NO3)3, then Fe(NO3)3 + H2O → Fe(NO3)2·H2O + HNO3. But that's not 8 equivalents. Alternatively, maybe B is NH3 and the reaction is A + 8 NH3 → C. If A is Fe(NO3)3, then 8 NH3 would form [Fe(NH3)6]^3+ and 2 NH3 as counterions. Wait, [Fe(NH3)6]^3+ has a charge of +3, so if there's Fe(NO3)3, which is Fe^3+ and 3 NO3^- per formula unit. Then adding 6 NH3 would give [Fe(NH3)6]^3+ and 6 NH3. But Fe(NO3)3 has 3 NO3^- per mole. So 6 NH3 and 3 NO3^- would make 3 HNO3. So the product would be [Fe(NH3)6](NO3)3. But that's only 6 NH3. The problem says 8 equivalents. So maybe there's an extra 2 NH3 somewhere. Alternatively, perhaps the reaction is with 8 NH3, leading to [Fe(NH3)8]^+3. But that's less common. Fe typically forms complexes with 6 NH3. Maybe the reaction is with more NH3, but that's not typical. Alternatively, maybe B is a different gas. Let's move on to the next part.

2. Reaction 2: C reacts with 2 equivalents of gas D to produce E. So C is reacting with 2 moles of D to form E. Then, reaction 3: C reacts with water to form A and two acids F and G. So F is a strong acid, G is a weak acid. So when C is hydrolyzed, it gives back A and two acids. For example, if C is a salt like AlCl3·3H2O, then hydrolysis gives Al(OH)3 and HCl. But in this case, there are two acids. So maybe C is a salt with two acidic groups. For example, if C is AlCl3·HCl·H2SO4, then hydrolysis would give Al(OH)3, HCl, and H2SO4. But that's three acids. Hmm. Alternatively, maybe C is a complex with two acidic ligands. For example, if C is [Al(OH)2Cl]Cl2, then hydrolysis might release Al(OH)3, HCl, and another acid. Not sure.

Alternatively, maybe C is a hydrate of an acid. For example, if A is FeCl3, then C could be FeCl3·2HCl. When C reacts with water, it would release Fe(OH)3 and HCl. But that's only one acid. The problem states two acids. So maybe C is a compound that can release two different acids upon hydrolysis. For example, if C is a metal nitrate with two acidic anions. Let's say C is Fe(NO3)3·HNO3·H2SO4. Then hydrolysis would give Fe(OH)3, HNO3, and H2SO4. But then F and G would be HNO3 (strong acid) and H2SO4 (strong acid), but the problem says G is a weak acid. So that's not matching. Alternatively, maybe G is a weak acid like CH3COOH. So C might be a compound that when hydrolyzed gives A, HCl (strong acid F), and CH3COOH (weak acid G). So maybe C is a salt containing AlCl3 and acetic acid? But AlCl3 is a Lewis acid. Alternatively, maybe C is AlCl3·CH3COOH·3H2O. But I'm not sure.

Alternatively, perhaps C is a hydrate of a bicarbonate. For example, if A is Al2O3, then C could be Al(OH)3·3H2O. When C reacts with water, it would give Al(OH)3 and H2O. But that doesn't produce acids. Hmm. Alternatively, maybe C is a complex with two different counterions that are acidic. For example, if C is [Al(OH)2Cl]Cl2, then upon hydrolysis, it would give Al(OH)3, HCl, and another HCl. But that's still two HCl molecules. But the problem states two different acids. So maybe F is HCl (strong) and G is CH3COOH (weak). So C might be a salt like AlCl3·CH3COOH. When C reacts with water, AlCl3 hydrolyzes to Al(OH)3 and HCl, and CH3COOH stays as CH3COOH. But then the product would be Al(OH)3, HCl, and CH3COOH. So F is HCl (strong) and G is CH3COOH (weak). That fits. So then C is AlCl3·CH3COOH. But then when C reacts with 2 equivalents of gas D, it forms E. What is D? D reacts with B in a 1:1 ratio to form H, which is a solvent. So if B is, say, H2O vapor, then D could be HCl gas. Then D (HCl) reacts with B (H2O) to form HCl gas and H2O as a solvent. But H2O is not a gas. Alternatively, if B is NH3 and D is HCl, then D reacts with B to form NH4Cl, which is a solid, not a solvent. Hmm. Alternatively, if B is H2S and D is HCl, then they react to form H2S + HCl → HCl + H2S (no reaction?), or maybe H2S and HCl react to form HCl and H2S. Wait, maybe they form HCl gas and H2S gas? But that's not a solvent. Alternatively, if B is O2 and D is H2, they react to form H2O, which is a solvent. But then H is H2O. So D is H2 and B is O2 reacting to form H2O. Then, in Reaction 2, C (which is AlCl3·CH3COOH) reacts with 2 moles of H2 to form E. But I don't see how that would form a product. Alternatively, maybe D is H2O and B is HCl. Then D (H2O) and B (HCl) react to form HCl (gas), but that's not a solvent. I'm getting stuck here. Let's try a different approach.

Alternatively, perhaps D is a gas that when reacted with B (another gas) forms water as a solvent. So B and D are gases that react to form water. For example, H2 and O2 reacting in a 2:1 ratio to form H2O. But H2O is a liquid, not a solvent. Alternatively, maybe B is NH3 and D is CH4, but they don't react to form a solvent. Alternatively, if B is H2O vapor and D is something else. Wait, maybe B is H2O vapor and D is HCl gas. Then D (HCl) reacts with B (H2O) to form HCl gas and H2O vapor. But that's not forming a solvent. Alternatively, maybe D is NH3 and B is HCl. Then they react to form NH4Cl, which is a solid. Not helpful. Alternatively, maybe B is H2S and D is HCl. They react to form HCl and H2S. Still not a solvent. Hmm.

Alternatively, maybe B is O2 and D is H2. They react in a 2:1 ratio to form H2O, which is a liquid. But the problem says H is a solvent. H2O as a solvent? Maybe that's possible. So in that case, D is H2 and B is O2. Then H is H2O. Then Reaction 2: C reacts with 2 equivalents of H2 to form E. But H2 is a reducing agent. If C is a compound that reacts with H2, maybe C is an oxidizing agent. For example, if C is Fe3O4, then reacting with H2 would reduce it to Fe. But then E would be Fe. But the question is about the symmetry group of E. If E is Fe, then it's a transition metal, and its symmetry group would depend on the structure. But I'm not sure. Alternatively, maybe C is a metal oxide with acidic properties.

This is getting complicated. Let me try to approach this from the molecular symmetry group perspective. The options are C2v, D∞h, C2, D4h. These are all point groups. C2v is a dihedral group with a vertical mirror plane and a 2-fold rotation axis. D∞h is a dihedral group with an infinite number of mirror planes and a horizontal mirror plane. C2 is a simple 2-fold rotation axis. D4h has a 4-fold rotation axis and multiple mirror planes.

The product E's symmetry group depends on its structure. For example, if E is a linear molecule with a plane of symmetry, it might have D∞h symmetry. If it's a planar molecule with a center of inversion, maybe C2v. If it's a molecule with a C2 axis and two mirror planes, then C2v. If it's a tetrahedral or octahedral molecule with higher symmetry, then D4h.

But without knowing E's structure, it's hard to determine. So maybe I need to figure out E's structure from the reactions.

Let me try again. Let's assume the first reaction is A + 8B → C. Let's suppose A is a transition metal salt, like Fe(NO3)3. B is NH3. Then 8 moles of NH3 would form a complex with Fe^3+. Fe(NO3)3 + 8NH3 → [Fe(NH3)6]^3+ + 3HNO3. But that's 6 NH3. If B is H2O, then Fe(NO3)3 + 6 H2O → Fe(NO3)3·6H2O. But the problem says 8 equivalents. Hmm. Alternatively, maybe B is H2O and a different stoichiometry. Or perhaps B is another gas. Let's think of B as H2. Then 8 moles of H2 reacting with A. For example, if A is FeO, then FeO + 8 H2 → Fe + 8 H2O. But FeO is not very common. Alternatively, if A is FeO(OH), then FeO(OH) + H2 → Fe(OH)2 + H2O. But 8 H2 would be needed. Not sure.

Alternatively, perhaps B is a strong acid like HCl. Then 8 equivalents of HCl reacting with A. If A is Na2SiO3 (silica), then 8 HCl + Na2SiO3 → 2 NaCl + SiHCl6. But that's speculative. Alternatively, if A is a metal oxide, like Al2O3, then 8 HCl + Al2O3 → 2 AlCl3 + 4 H2O. But again, not sure.

Alternatively, maybe the first reaction is the formation of a gas-phase complex. For example, A could be a transition metal ion, and B is NH3. Let's say A is Fe(NO3)3. Then Fe(NO3)3 + 8 NH3 → [Fe(NH3)6]^3+ + 3 HNO3. But that's 6 NH3, not 8. Alternatively, maybe the reaction is A + 8 NH3 → [Fe(NH3)6]^3+ + 2 NH3 + 3 HNO3. Wait, that would mean 6 NH3 are used in the complex, 2 NH3 are free, and 3 HNO3 are formed. But the stoichiometry doesn't add up. Alternatively, maybe the reaction is Fe(NO3)3 + 8 NH3 → [Fe(NH3)6]^3+ + 2 NH3 + 3 HNO3. But that would be 6 NH3 in the complex, 2 NH3 in solution, and 3 HNO3. But the total NH3 is 8. So 6 + 2 = 8. So that works. Then C is [Fe(NH3)6]^3+·2NH3·3HNO3? But that seems a bit off. Alternatively, maybe C is [Fe(NH3)6]^3+ and the NH3 and HNO3 are counterions. But not sure.

Alternatively, maybe the first reaction is with a different gas. Let's say B is H2O vapor. Then A reacts with 8 moles of H2O. For example, Fe(NO3)3 + 8 H2O → Fe(NO3)3·8H2O. Then when C reacts with water, it would decompose back into Fe(NO3)3 and H2O. But when C reacts with water, it produces two acids. So maybe Fe(NO3)3·8H2O when hydrolyzed would release Fe(OH)3, HNO3, and something else. But that's only one acid. Alternatively, maybe C is a different compound.

This is getting too speculative. Let me try to look at the problem from the product E's perspective. The question is about the symmetry group of E. The options are C2v, D∞h, C2, D4h. Let's think about common compounds and their symmetry groups. For example:

- CO2 has D∞h symmetry (linear molecule with a plane of symmetry and inversion).
- H2O has C2v symmetry (bent molecule with two mirror planes and a C2 axis).
- NH3 has C3v symmetry (trigonal pyramidal).
- CH4 has Td symmetry (tetrahedral).
- CH2Cl2 has D2h symmetry (bipyramidal).
- Benzene has D6h symmetry.
- For transition metals, the coordination compounds can have various symmetries. For example, [Ni(CO)4] is tetrahedral with Td symmetry. [Ni(CN)4]^2- is square planar with D4h symmetry.

If E is a molecule with a high symmetry, like D4h, then it could be something like a square planar or octahedral complex. Alternatively, if E is a linear molecule like CO2, which has D∞h symmetry. If E is a trigonal planar molecule like BF3, which has D3h symmetry (but that's not an option). Alternatively, if E is a tetrahedral molecule like CH4, which has Td symmetry (not an option). If E is a molecule with a C2 axis and two mirror planes, like H2O, which has C2v symmetry.

But the problem states that D reacts with B in a 1:1 ratio to form H, which is a solvent. If B is H2O vapor and D is HCl gas, then they react to form HCl and H2O. But H2O is a liquid, not a solvent. Alternatively, if B is NH3 and D is HCl, they form NH4Cl, which is a solid. If B is O2 and D is H2, they form H2O, which is a liquid. But H is a solvent, so maybe H is H2O. Then D is H2 and B is O2. Then Reaction 2: C reacts with 2 moles of H2 to form E. If C is a compound that reacts with H2, maybe C is a metal oxide. For example, FeO reacts with H2 to form Fe and H2O. But then E would be Fe, which has no symmetry. Alternatively, if C is a metal, like Fe, but that's a solid. Hmm.

Alternatively, maybe E is a diatomic molecule. For example, O2 is symmetric, but its symmetry group is D∞h. But the options include D∞h as one of the choices. So if E is O2, then the symmetry group is D∞h. But how would that fit into the reactions?

Alternatively, if E is a linear molecule like CO2, then D∞h symmetry. Let's see how that could happen. If C is a compound that, when reacted with H2, forms CO2. For example, if C is a carbonate, like Na2CO3. Then reacting with H2 would not form CO2. Alternatively, if C is a carbonate, maybe it reacts with H2 to form CO2 and water. But that would require more H2. If C is Na2CO3, then 2 H2 would produce 2 H2O and 2 CO2. But the problem says C reacts with 2 equivalents of D to form E. Wait, in Reaction 2, C reacts with 2 equivalents of D to form E. If D is H2, then maybe E is CO2, but how?

Alternatively, maybe the reaction is C + 2 H2 → CO2 + 2 H2O. But that's speculative. If E is CO2, then the symmetry group is D∞h. So if that's the case, the answer would be D∞h. But I need to check if that fits with the other information.

But let's think again. If E is CO2, then in Reaction 2, C reacts with 2 moles of H2 to form CO2 and perhaps water. But then how does C look? If C is a compound that when reacted with 2 H2 forms CO2. For example, if C is a metal oxide. Suppose C is FeO. Then FeO + 2 H2 → Fe + 2 H2O. But that's not CO2. Alternatively, if C is FeCO3 (a carbonate), then FeCO3 + 2 H2 → Fe + CO2 + 2 H2O. But that would produce CO2 and H2O. But the problem states that when C reacts with water, it produces A and two acids. So if C is FeCO3, then FeCO3 + H2O → Fe(OH)2 + CO2. But that's only one acid. The problem says two acids. So maybe C is a different compound.

Alternatively, maybe E is a molecule with two oxygen atoms, like O2. If E is O2, then D∞h symmetry. But how would O2 form in the reactions?

Alternatively, if E is a molecule like O=C=O (CO2), which has D∞h symmetry. So if E is CO2, then the answer would be D∞h. But how would that fit into the reactions.

Alternatively, let's think of E as a molecule with a C2 axis and two mirror planes, like H2O. Then the symmetry group is C2v. But how would that form?

Alternatively, if E is a tetrahedral molecule with a C2 axis, like CH4, but that's not an option. The options are C2v, D∞h, C2, D4h.

Another approach: perhaps E is a molecule with a center of inversion. For example, a trigonal planar molecule like BF3 has D3h symmetry, but that's not an option. If E has a center of inversion, its symmetry group would be Dnh with a horizontal mirror plane and a 2-fold rotation axis. For example, H2O has a C2 axis and a vertical mirror plane, but not a horizontal mirror plane. Its symmetry group is C2v. So if E is H2O, the answer is C2v. But how would H2O form from the reactions?

Alternatively, if E is a diatomic molecule like O2, symmetry group D∞h. If E is CO2, symmetry group D∞h. If E is BF3, symmetry group D3h (not an option). If E is CH3CH3 (ethane), symmetry group D3h (not an option). If E is NH3, symmetry group C3v (not an option). If E is CH4, symmetry group Td (not an option). If E is CHCl3, symmetry group C3v (not an option). If E is C2H5OH (ethanol), symmetry group C2 (since the molecule has a plane of symmetry cutting through the OH group and the CH2 group). Wait, ethanol has a plane of symmetry: the plane bisecting the molecule between the OH and the other side. So its symmetry group is C2h, but the options don't include that. Alternatively, if E is a planar molecule like H2O, symmetry group C2v.

But how do the reactions lead to E being something like H2O or CO2?

Alternatively, perhaps E is a molecule like [Ni(CO)4], which is tetrahedral with Td symmetry. But that's not an option. Alternatively, if E is a square planar complex like [PtCl4]^2-, which has D4h symmetry. The option D4h is one of the choices. So if E is a square planar molecule, its symmetry group is D4h.

So if E is [PtCl4]^2-, then the answer is D4h. But how would that form from the reactions?

Alternatively, if E is a molecule like [Ni(CN)4]^2-, which is square planar with D4h symmetry. But again, the exact formation of E from the given reactions is unclear.

Alternatively, let's think of E as a diatomic molecule. For example, if E is O2, then symmetry group D∞h. If E is O3 (ozone), symmetry group D∞h. But how would O3 form?

Alternatively, if E is H2O, then symmetry group C2v. If the reactions lead to H2O as E, then the answer is C2v.

But given the options, and the fact that H2O has C2v symmetry, and E is a product of the reactions, perhaps that's the answer. But I need to check if the reactions can lead to H2O as E.

Looking back, in Reaction 2: C reacts with 2 equivalents of gas D to form E. If D is H2 and B is O2, then D (H2) and B (O2) react to form H2O. But H2O is a liquid, not a solvent. Alternatively, if B is H2O vapor and D is Cl2, then they react to form HCl and HOCl, but that's not a solvent. Alternatively, if B is NH3 and D is HCl, they form NH4Cl, which is a solid. Alternatively, if B is O2 and D is H2, they form H2O. Then H is H2O, which is a solvent. Then in Reaction 2, C reacts with 2 moles of H2 to form E. If C is Fe(OH)3, then Fe(OH)3 + 2 H2 → Fe + 2 H2O. But then E would be Fe, which has no symmetry. Alternatively, if C is Fe(OH)3·H2O (from Reaction 3), and reacts with 2 moles of H2, maybe it forms Fe(OH)3 + H2O. Not sure.

Alternatively, maybe E is H2O. Then the symmetry group is C2v. So if the reactions can lead to H2O as E, then the answer is C2v. But how?

Alternatively, if D is H2 and B is O2, then D reacts with B to form H2O. Then in Reaction 2, C reacts with 2 moles of H2 to form E. If C is H2O, then 2 H2 + H2O → 2 H2 + H2O. That doesn't make sense. Alternatively, if C is H2O, then reacting with H2 would not change it. Alternatively, if C is NH3, then NH3 + 2 H2 → NH3. Not helpful.

Alternatively, maybe the first reaction is A + 8B → [Fe(B)6]^3+ and 3B. If B is NH3, then 8 NH3 would form [Fe(NH3)6]^3+ and 3 NH3. That gives a total of 11 NH3, which doesn't fit. Alternatively, maybe B is H2O. Then Fe(NO3)3 + 8 H2O → [Fe(H2O)6]^3+ + 3 HNO3. Then when C ([Fe(H2O)6]^3+·3HNO3) reacts with water, it would release Fe(NO3)3 and H2O. But that's only one acid. The problem states two acids. So perhaps C is a complex with two different ligands that release two acids upon hydrolysis. For example, [Fe(H2O)6]^3+ reacts with water to form Fe(OH)3 and HNO3. But that's only one acid. Alternatively, if C is [Fe(H2O)6]^3+ and the hydrolysis involves another ligand, but I don't see how.

Alternatively, maybe C is a hydrate like [Fe(H2O)6]Cl3. Then when it reacts with water, it would release Fe(OH)3 and HCl. But that's only one acid. The problem says two acids. So maybe C is a complex with two different counterions that release two acids. For example, [Fe(H2O)6]Cl3·HCl. Then hydrolysis would give Fe(OH)3, HCl, and HCl. But that's still only one acid. Alternatively, if C is a hydrate with two different acid groups. For example, [Fe(H2O)6](NO3)3·2HCl. Then hydrolysis would give Fe(OH)3, HNO3, and HCl. That's two acids. So F is HNO3 (strong), G is HCl (weak). Then in Reaction 2, C reacts with 2 equivalents of D to form E. If D is H2 and B is O2, then 2 moles of H2 and O2 would form 2 moles of H2O. But how does that fit? Then E would be H2O. But then the symmetry group would be D∞h, which is an option. So if E is H2O, the answer is D∞h.

But let's check the steps again. If A is Fe(NO3)3, B is H2O, then Reaction 1: Fe(NO3)3 + 8 H2O → [Fe(H2O)6]^3+ + 3 HNO3. Then Reaction 3: [Fe(H2O)6]^3+ + water → Fe(OH)3 + 3 HNO3. But that's only one acid. The problem states two acids. So maybe C is [Fe(H2O)6]^3+·3HNO3·something else. Alternatively, maybe C is a different compound. Let's say C is [Fe(H2O)6](NO3)3·2HCl. Then when it reacts with water, it gives Fe(OH)3, HNO3, and 2 HCl. So F is HNO3 (strong), G is HCl (weak). Then Reaction 2: C reacts with 2 equivalents of D to form E. If D is H2 and B is O2, then they react to form H2O. So C is [Fe(H2O)6](NO3)3·2HCl. Then reacting with 2 H2 (but H2O is already in C), so maybe the reaction is [Fe(H2O)6](NO3)3·2HCl + 2 H2 → E. But how does that form H2O? Because C already has H2O as part of the hydrate. Alternatively, maybe the reaction is [Fe(H2O)6](NO3)3·2HCl + 2 H2O → [Fe(H2O)6](NO3)3 + 2 HCl. But that's just dissociation, not forming a new compound. Alternatively, if D is H2O and B is something else. Wait, the problem says D reacts with B in a 1:1 ratio to form H, the solvent. If B is H2O, then D must be something that reacts with H2O. If D is NH3, then NH3 + H2O → NH4OH (not a solvent). If D is O2, then O2 + H2O → H2O2 (not a solvent). If D is H2, then H2 + H2O → H2O + H2 (no reaction). Alternatively, maybe D is HCl and B is NH3. Then HCl + NH3 → NH4Cl. Not a solvent. Alternatively, if B is O2 and D is H2O, then H2O + O2 → H2O2. Not a solvent. Alternatively, if B is H2O and D is something else. Wait, perhaps B is H2O and D is O2. Then O2 + H2O → H2O2. But H2O2 is a liquid, not a solvent. Alternatively, if B is H2O and D is Cl2. Cl2 + H2O → HCl + HOCl. Not a solvent. Alternatively, if B is H2O and D is NH3. NH3 + H2O → NH4OH. Not a solvent. Alternatively, if B is H2O and D is another gas. Wait, maybe B is H2 and D is Cl2. Then 2 Cl2 + H2 → 2 HCl. But that's a 2:1 ratio, not 1:1. The problem states D reacts with B in a 1:1 ratio. So if B is H2 and D is Cl2, they would react in a 1:1 ratio to form 2 HCl. But the problem says D reacts with B in a 1:1 ratio to form H. So if B is H2 and D is O2, they would react in a 1:1 ratio to form H2O. Then H is H2O, which is a solvent. So in Reaction 2, C (which is [Fe(H2O)6](NO3)3·2HCl) reacts with 2 moles of O2 to form E. But why would O2 react with C? Alternatively, maybe in Reaction 2, C reacts with 2 moles of D (which is O2) to form E. But what would O2 do to C? If C is a complex with water, adding O2 might not change it significantly. Alternatively, maybe the reaction is C + 2 O2 → E. If E is H2O, then O2 is already part of C. Not helpful.

Alternatively, maybe E is H2O, formed from the reaction of C with D (which is H2) and B (which is O2). So D is H2 and B is O2. Then H is H2O. Then Reaction 2: C reacts with 2 moles of H2 to form E. If C is FeO, then FeO + 2 H2 → Fe + 2 H2O. Then E is Fe, which has no symmetry. But the problem states that when C reacts with water, it forms A and two acids. If C is FeO, then FeO + H2O → Fe + H2O. That's only one acid. Alternatively, if C is FeO·H2O, then FeO·H2O + H2O → Fe(OH)2 + H2O. Not helpful.

This is getting really complicated. Let me try to make a different approach. The key is to determine the symmetry group of E. The options are C2v, D∞h, C2, D4h. The symmetry group depends on the structure of E. For example:

- If E is a molecule with a plane of symmetry and a two-fold axis perpendicular to it (C2v), like H2O.
- If E is a linear molecule with a plane of symmetry and an infinite-fold rotation axis (D∞h), like CO2.
- If E is a molecule with a two-fold axis and no other symmetry (C2), like CH3CH3.
- If E is a molecule with multiple planes of symmetry and a high-fold rotation axis (D4h), like a square planar molecule like [Ni(CN)4]^2-.

Given the reactions, especially Reaction 3 where C hydrolyzes to A and two acids (F and G), maybe C is a salt that, when hydrolyzed, produces a metal hydroxide and two acids. For example, if C is a salt like [Fe(NO3)3]·3H2O. Then hydrolysis would give Fe(OH)3 and 3 HNO3. But that's only one acid. The problem says two acids. So maybe C is a salt with two different anions that release two different acids when hydrolyzed. For example, [Fe(NO3)3](NO2)3. Then hydrolysis would release Fe(OH)3, HNO3 (strong), and HNO2 (weak). But that's speculative.

Alternatively, perhaps C is a hydrate with two different counterions. For example, [Fe(NO3)3]·2HCl. Then hydrolysis would give Fe(OH)3, HNO3, and HCl. That's two acids: HNO3 (strong) and HCl (weak). Then in Reaction 2, C (which is [Fe(NO3)3]·2HCl) reacts with 2 moles of D to form E. If D is H2 and B is O2, then 2 moles of H2 and O2 would form 2 moles of H2O. But how does that form E? If C is reacting with O2, maybe O2 is acting as an oxidizing agent. Alternatively, if D is H2O and B is something else. Wait, maybe D is H2O and B is NH3. Then D (H2O) and B (NH3) react to form NH4OH (not a solvent). Alternatively, if D is H2 and B is Cl2, they react to form HCl and Cl2, but that's not relevant.

Alternatively, maybe the reaction is C (which has Fe(NO3)3·2HCl) reacting with 2 moles of H2O to form E. But why would H2O react with C? If C is a hydrate, adding more H2O would just increase the hydrate. But the problem states that D reacts with B to form H, the solvent. If D is H2O and B is NH3, then they form NH4OH, which is a weak base. But NH4OH is not a solvent. Alternatively, if B is O2 and D is H2O, they form H2O2, which is a liquid. Not a solvent.

This is really confusing. Let me try to summarize what I have so far:

- From Reaction 1: A is a metal nitrate, B is a gas that forms a complex with A.
- From Reaction 3: C hydrolyzes to A and two acids (F and G).
- From Reaction 2: C reacts with 2 equivalents of D to form E.
- D reacts with B to form H (a solvent).

Given the time I've spent and the complexity, I think the most plausible answer is that E is a molecule with D4h symmetry, such as a square planar complex. For example, if E is a complex like [Ni(CN)4]^2-, which has D4h symmetry. But how does that fit with the reactions?

Alternatively, if E is H2O, symmetry group C2v. But earlier steps didn't clearly lead to H2O as E. However, if the reactions can be interpreted such that E is H2O, then the answer would be C2v. But I'm not entirely confident.

Alternatively, if E is a molecule like O=C=O (CO2), which has D∞h symmetry. If C is a compound that when reacted with D forms CO2, then D might be O2 and B might be H2, but that doesn't fit the earlier steps.

Given all this confusion, and considering the options, I think the most plausible answer is D4h. Because if E is a square planar complex, like [Ni(CN)4]^2-, which has D4h symmetry. Therefore, the answer is D4h.
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