Chemical Bonding And Molecular Structure

Redox Equation and Calculation

We commence with the equilibrium redox equation:

We aim to determine the quantity of water molecules ( ) generated and the quantity of electrons ( ) engaged in this reaction.

Upon examining the balanced equation, it is evident that 8 moles of water are produced during the reaction. Hence, we can conclude:

In this reaction, hydrogen sulfide ( ) undergoes oxidation to form sulfur ( ). Each molecule of loses 2 electrons during this transformation (as sulfur transitions from an oxidation state of -2 in to 0 in ).

Consequently, for every mole of , 2 moles of electrons are implicated. Given that 5 moles of are participating in the reaction, the overall number of moles of electrons involved amounts to:

Thus, we can affirm:

We are tasked with determining the sum of and , so by adding these two values together, we obtain:

Therefore, .

Total atom with zero oxidation number state are 6

Oxidation States of Atoms

The image provides several chemical structures, and the question asks us to determine the number of atoms with zero oxidation states. Let's analyze each structure:

Structure 1: Bromine Compound

The first structure is a compound containing bromine (Br) and fluorine (F). The oxidation states of the elements are:

• Br is in the +6 oxidation state (as indicated by )
• F is in the -1 oxidation state (as indicated by )
• The oxygen atoms (O) attached to Br are in the -2 oxidation state, as usual for oxygen atoms in oxides

In this structure, there are no atoms with zero oxidation state, as all elements are in their respective oxidation states.

Structure 2: Sulfonate Group (HO-SO3H)

This structure represents a sulfonate group, and we need to calculate the oxidation states of the atoms:

• Hydrogen (H) is in the +1 oxidation state
• Oxygen (O) is in the -2 oxidation state
• Sulfur (S) in this group has an oxidation state of +5

Again, there are no atoms with a zero oxidation state in this structure.

Structure 3: Another Sulfonate Group

This structure is similar to the previous one, so the analysis remains the same. No atoms have a zero oxidation state here either.

Structure 4: Carbon-Carbon Bond (C=C)

The final structure features a carbon-carbon double bond, with two oxygen atoms attached to each carbon atom in a carboxyl group. Let's assign oxidation states:

• Each carbon (C) in this structure is in the 0 oxidation state because it is in a simple double bond with another carbon
• Oxygen (O) attached to the carbon is in the -2 oxidation state, as usual for oxygen in double bonds

Therefore, in this structure, both carbon atoms have a zero oxidation state.

Conclusion

We have identified that in the given structures, there are 6 atoms with a zero oxidation state:

• 2 carbon atoms in the final structure (C=C)
• 4 atoms of sulfur in the sulfonate groups (S=0 in both structures)

Total number of atoms with zero oxidation state: 6

To solve the problem, we analyze the fluorination reaction of phosphorus pentachloride (PCl₅) in a polar organic solvent.

Fluorination of PCl₅:
When PCl₅ reacts with fluorine or a fluorinating agent in polar organic solvents, it often forms ionic species due to ligand exchange and formation of complex ions.

Known species formed:
- The cation [PCl₄]⁺ (tetrachlorophosphonium ion)
- The anion [PF₆]^- (hexafluorophosphate ion)
These ions result from partial substitution of chlorine by fluorine in PCl₅ and subsequent complex formation.

Therefore, the species formed are:

Final Answer:
which corresponds to the second option:

[PCl₄]⁺[PCl₄F₂]^- and [PCl₄]⁺[PF₆]^-

First-Order Reaction Time Calculation

For a first-order reaction, the time is given by:

Given:

Half-life of U-238:

The half-life of U-238 is:

Substituting values into the equation:

Simplifying:

Final Answer:

The value of P is 143.

Analysis of Tetrahedral Geometry in Compounds

Step 1: Analyze Each Compound to Identify Tetrahedral Geometry

1. [Co(CN)₄]⁴⁻: CN is a strong ligand. Due to pairing, it undergoes sp³ hybridization, forming a tetrahedral structure.
2. [Co(CO)₃(NO)]: This forms a trigonal planar geometry due to the coordination environment.
3. XeF₄: Square planar geometry, not tetrahedral.
4. [PCl₄]⁺: Tetrahedral geometry due to sp³ hybridization.
5. [PdCl₄]²⁻: Square planar geometry.
6. [ICl₄]⁻: Square planar geometry.
7. [Cu(CN)₄]³⁻: CN being a strong ligand leads to sp³ hybridization, hence tetrahedral.
8. P₄: Tetrahedral geometry due to its molecular structure.

Step 2: Count the Species with Tetrahedral Geometry

Tetrahedral species: [Co(CN)₄]⁴⁻, [PCl₄]⁺, [Cu(CN)₄]³⁻, and P₄.

Step 3: Final Answer

The total number of tetrahedral species is 3.

To determine which option has at least three molecules that follow the octet rule, we need to evaluate each molecule in the given options. The octet rule states that atoms tend to gain, lose, or share electrons to achieve a stable configuration with 8 electrons in their valence shell (except for hydrogen, which follows the duet rule with 2 electrons).

Let’s analyze each option:

Option A: CO₂, C₂H₄, NO, and HCl

• CO₂ (Carbon Dioxide): Carbon is the central atom, double-bonded to two oxygen atoms (O=C=O). Carbon has 4 valence electrons and forms two double bonds (8 electrons shared), satisfying the octet rule. Each oxygen has 6 valence electrons, and with a double bond (4 electrons) plus 4 lone pair electrons, they also have 8 electrons. Follows octet rule.
• C₂H₄ (Ethylene): Each carbon is bonded to two hydrogens and double-bonded to the other carbon (H₂C=CH₂). Each carbon has 4 valence electrons, with 4 bonds (8 electrons shared), satisfying the octet rule. Hydrogens have 2 electrons (duet rule). Follows octet rule.
• NO (Nitric Oxide): Nitrogen has 5 valence electrons, oxygen has 6. In NO, there’s a double bond (sometimes represented with a triple bond in resonance structures), but nitrogen ends up with 7 electrons (5 + 2 from the bond) and oxygen with 8. Nitrogen does not satisfy the octet rule due to the odd number of electrons (11 total). Does not follow octet rule.
• HCl (Hydrogen Chloride): Hydrogen has 1 valence electron, chlorine has 7. They form a single bond, giving hydrogen 2 electrons (duet rule) and chlorine 8 electrons (1 bond + 6 lone pair electrons). Follows octet rule.

Result for Option A: CO₂, C₂H₄, and HCl follow the octet rule (3 molecules).

Option B: NO₂, O₃, HCl, and H₂SO₄

• NO₂ (Nitrogen Dioxide): Nitrogen has 5 valence electrons, each oxygen has 6. NO₂ has an odd number of electrons (17), so nitrogen cannot achieve an octet. Resonance structures show nitrogen with 7 electrons. Does not follow octet rule.
• O₃ (Ozone): Each oxygen has 6 valence electrons. In the resonance structures of O₃ (O–O=O ↔︎ O=O–O), the central oxygen has 8 electrons (double bond), and the terminal oxygens share electrons to achieve 8 electrons each. Follows octet rule.
• HCl: As analyzed in Option A, follows octet rule.
• H₂SO₄ (Sulfuric Acid): Sulfur is the central atom, bonded to four oxygens (two with double bonds, two with single bonds to OH groups). Sulfur has 6 valence electrons but forms 6 bonds (12 electrons in its valence shell), exceeding the octet rule (expanded octet, common for elements beyond the second period). Does not follow octet rule.

Result for Option B: O₃ and HCl follow the octet rule (2 molecules).

Option C: BCl₃, NO, NO₂, and H₂SO₄

• BCl₃ (Boron Trichloride): Boron has 3 valence electrons and forms 3 single bonds with chlorine, resulting in 6 electrons around boron. Does not follow octet rule.
• NO: As analyzed in Option A, does not follow octet rule.
• NO₂: As analyzed in Option B, does not follow octet rule.
• H₂SO₄: As analyzed in Option B, does not follow octet rule.

Result for Option C: No molecules follow the octet rule (0 molecules).

Option D: CO₂, BCl₃, O₃, and C₂H₄

• CO₂: As analyzed in Option A, follows octet rule.
• BCl₃: As analyzed in Option C, does not follow octet rule.
• O₃: As analyzed in Option B, follows octet rule.
• C₂H₄: As analyzed in Option A, follows octet rule.

Result for Option D: CO₂, O₃, and C₂H₄ follow the octet rule (3 molecules).

Conclusion

• Option A: 3 molecules follow the octet rule.
• Option B: 2 molecules follow the octet rule.
• Option C: 0 molecules follow the octet rule.
• Option D: 3 molecules follow the octet rule.

Final Answer: Both Options A and D have at least three molecules that follow the octet rule.

Step 1: Determining the Geometry and Lone Pairs of Each Xenon Compound

• (P) XeF₂
  • Central atom: Xe
  • Valence electrons: 8
  • Bonding pairs: 2 (due to 2 F atoms)
  • Lone pairs: (3 lone pairs)
  • Geometry: Trigonal bipyramidal (linear shape due to lone pairs in equatorial positions)
  • Matches: (5)
• (Q) XeF₄
  • Central atom: Xe
  • Valence electrons: 8
  • Bonding pairs: 4 (due to 4 F atoms)
  • Lone pairs: (2 lone pairs)
  • Geometry: Octahedral (square planar shape due to lone pairs at axial positions)
  • Matches: (3)
• (R) XeO₃
  • Central atom: Xe
  • Valence electrons: 8
  • Bonding pairs: 3 (due to 3 O atoms)
  • Lone pairs: (1 lone pair)
  • Geometry: Tetrahedral (pyramidal shape due to lone pair)
  • Matches: (2)
• (S) XeOF₄
  • Central atom: Xe
  • Valence electrons: 8
  • Bonding pairs: 5 (3 O atoms and 2 F atoms)
  • Lone pairs: (0 lone pairs)
  • Geometry: Trigonal bipyramidal (no lone pairs present)
  • Matches: (4)

Step 2: Conclusion

• (P) → (5)
• (Q) → (3)
• (R) → (2)
• (S) → (4)

Thus, the correct answer is (B) P-5, Q-3, R-2, S-4.