Openai/69227557-0850-8003-9c19-8f26825e337d

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gpt-5-1

User: Why is it called electronegativity if it's a measure of the ability to attract not repel electrons?

Why is it called electronegativity if it's a measure of the ability to attract (not repel) electrons?

Assistant: Condensed restatement of your question:

Condensed restatement of your question: Why is the term "electronegativity" used if it refers to attracting electrons rather than repelling them?

Answer: The name makes sense historically and conceptually if you break the word apart:

• "Electro-" → refers to electrons
• "-negativity" → refers to being negative in charge

An atom that strongly attracts electrons tends to gain negative charge (or pull electron density toward itself in a bond). So electronegativity literally describes an element’s tendency to become more negatively charged by drawing electrons toward itself.

It’s not about being negative in personality or repelling electrons—it's about how strongly an atom wants to increase its negative charge.

Why it isn’t called “electro-attractiveness” or something similar: When the concept was developed (Pauling, Mulliken, etc.), the naming followed early chemical intuition: atoms that pulled electrons harder behaved like they “wanted” to be negative. For example:

• Fluorine → highest electronegativity
• It strongly pulls bonding electrons → ends up partially negative (δ–) in bonds.

So the term emphasizes the resulting charge state, not the physical action.

Analogy: Think of it like "greediness": the word describes the state of wanting more, not the action of taking. Electronegativity describes the tendency toward acquiring negative charge.

Bottom line: It’s called electronegativity because the property reflects how strongly an atom tends to draw in electrons and become negatively charged, not because it repels electrons.

If you want, I can give the etymology timeline (Pauling → Mulliken → Allred-Rochow) or a one-sentence version.

User: Can we think of "energy released" as being caused by electrons that would otherwise make the molecule polar and therefo…

Can we think of "energy released" as being caused by electrons that would otherwise make the molecule polar and therefore less stable / more energetic than the stable molecule whose valence orbits are fulfilled?

Assistant: Condensed version of your question:

Condensed version of your question: Is the energy released during bond formation essentially due to electrons avoiding an unstable, polarized, high-energy state and settling into a stable, filled-valence configuration?

You're close in spirit, but not quite in the correct physical framing.

The more accurate view is:

So it isn’t that electrons “would otherwise make the molecule polar,” but rather that bond formation leads to a quantum-mechanically lower energy state where electron density is shared or transferred to minimize electron–electron and nucleus–nucleus repulsion while maximizing electron–nucleus attraction.

When the system drops into that lower-energy configuration, the excess energy must go somewhere → heat, light, or kinetic motion → which we call energy released.

####

• Electrons are in individualized orbitals.
• Each atom has incomplete valence shells and therefore higher potential energy because: - Electrons aren’t in optimal low-energy molecular orbitals. - Attractive forces between nuclei and neighboring electrons don’t yet exist. - Repulsive forces aren’t yet minimized through spatial reorganization.

This state is high-energy, not because it's polar, but because it's poorly optimized electrostatically and quantum-mechanically.

Electrons reorganize into molecular orbitals, where:

• Bonding orbitals increase electron density between nuclei, enhancing attractive forces (lower energy).
• Electrons occupy lower-energy quantum states than they could in isolated atomic orbitals.
• Repulsions (electron–electron and nucleus–nucleus) are reduced by spatial separation and Pauli-allowed orbital alignment.

This reconfiguration lowers potential energy.

The system now has:

• Lower total potential energy.
• Electrons in lower-energy orbital states (σ or π bonds).
• Reduced electrostatic repulsion and increased attractive stabilization.

This lower energy state corresponds to:

• “Filled valence shells”
• Lower chemical potential
• Higher thermodynamic stability

But those concepts are heuristics—not the underlying physical mechanism.

Polarity can matter, but not in the way you phrased.

• If a bond is polar (e.g., C–Cl), the electron density is asymmetric.
• But the stability doesn't come from "avoiding polarity"; rather: >

A polar bond can be more stable than a nonpolar one if:

• The electron cloud shifts into a lower-energy region.
• The resulting bond minimizes system potential energy.

So polarity is not inherently destabilizing—sometimes it’s stabilizing.

Example: HF is polar and has a very strong bond (≈ 565 kJ/mol) because the electron density sits closer to fluorine, forming a deep potential well.

Instead of your phrasing:

A more accurate version is:

:

If you want, next step could be: "Explain this using a Lennard-Jones potential diagram or a molecular orbital energy diagram."

User: I want to know about the underlying physical mechanism, not heuristics.

I want to know about the underlying physical mechanism, not heuristics.

Assistant: Restated question:

Restated question: What is the true physical mechanism behind energy release during bond formation—not the heuristic shell rules, but the fundamental physics?

Bond formation releases energy because when two atoms approach, the quantum-mechanical electron wavefunctions reorganize into molecular orbitals whose total system energy is lower than the sum of the isolated atomic states. The reduction in total energy is not metaphorical—it is a change in the Hamiltonian of the system.

Energy is released because the bound state is a lower-energy eigenstate of the full electron-nuclear interaction Hamiltonian.

Nothing more mystical than that.

For two atoms, the relevant Hamiltonian includes:

H^=Te+Tn+Vnn+Vee+Ven\hat{H} = T_e + T_n + V_{nn} + V_{ee} + V_{en}H^=Te​+Tn​+Vnn​+Vee​+Ven​ Where:

• Te,TnT_e, T_nTe​,Tn​: electron and nuclear kinetic energy terms
• VnnV_{nn}Vnn​: nucleus–nucleus repulsion
• VeeV_{ee}Vee​: electron–electron repulsion
• VenV_{en}Ven​: electron–nucleus attraction

When atoms are infinitely separated, the terms are mostly localized. As they approach, the electron wavefunctions overlap and the expectation value of the Hamiltonian changes.

A bond forms only if:

⟨Ψbonded∣H^∣Ψbonded⟩<⟨Ψseparate∣H^∣Ψseparate⟩\langle \Psi_{\text{bonded}} | \hat{H} | \Psi_{\text{bonded}} \rangle < \langle \Psi_{\text{separate}} | \hat{H} | \Psi_{\text{separate}} \rangle⟨Ψbonded​∣H^∣Ψbonded​⟩<⟨Ψseparate​∣H^∣Ψseparate​⟩ i.e., the bonded state is a lower energy eigenstate of the Hamiltonian.

The difference in energy must be emitted as photons or transferred to kinetic motion → that is the “released energy.”

###

Solving the Schrödinger equation for two atoms yields:

• a bonding orbital (lower energy)
• an antibonding orbital (higher energy)

Electrons occupy the bonding orbital, whose energy is lower because the wavefunction places electron probability between the nuclei, increasing negative–positive attraction.

This modifies VenV_{en}Ven​ favorably:

ΔVen<0\Delta V_{en} < 0ΔVen​<0 That's the dominant stabilizing term.

For a Coulombic system in a stable bound state:

E=12Vwhere V<0E = \frac{1}{2} V \quad\text{where } V<0E=21​Vwhere V<0 So if the potential energy becomes more negative during orbital formation, the total energy drops. The system must dispose of that energy → release.

Electrons in a molecular orbital are spatially more delocalized than in isolated atoms.

Heisenberg tells us:

Δx↑⇒Δp↓⇒Te↓\Delta x \uparrow \Rightarrow \Delta p \downarrow \Rightarrow T_e \downarrowΔx↑⇒Δp↓⇒Te​↓ So electron delocalization reduces kinetic energy, contributing to stabilization.

This is not a chemical heuristic—it is straight quantum mechanics.

Plotting energy vs. internuclear separation produces a Lennard-Jones-type or Morse potential:

E(r)=Ar12−Br6(simplified)E(r) = \frac{A}{r^{12}} - \frac{B}{r^6} \quad (\text{simplified})E(r)=r12A​−r6B​(simplified) Real quantum mechanical treatment yields a potential well:

r=r0⇒dEdr=0andd2Edr2>0r = r_0 \Rightarrow \frac{dE}{dr}=0 \quad \text{and} \quad \frac{d^2E}{dr^2} > 0r=r0​⇒drdE​=0anddr2d2E​>0 The depth of that well is the bond energy.

The energy released upon formation equals the drop to that minimum.

- It is not because electrons “seek full shells.”

• It is not because polarity is minimized.
• It is not because atoms “want to be stable.”

Those are emergent patterns from the quantum physics—not causes.

Next steps could be one of:

• Derive the H₂ molecular orbitals from the LCAO approximation.
• Use Hartree–Fock to compute stabilization energy numerically.
• Examine the role of exchange interaction and Pauli exclusion.

Would you like:

A) The molecular orbital derivation for H₂ using LCAO? B) The virial theorem quantitative application to a real bond? C) A Hartree–Fock energy breakdown for a specific molecule (e.g., HCl)? D) All of the above?

User: What are psi bonded psi separate?

What are psi bonded psi separate?

Assistant: Condensed question:

Condensed question: What do Ψbonded\Psi_{\text{bonded}}Ψbonded​ and Ψseparate\Psi_{\text{separate}}Ψseparate​ mean physically?

They are two different quantum mechanical wavefunctions that describe the same set of nuclei and electrons under two different physical configurations:

• Ψseparate\Psi_{\text{separate}}Ψseparate​: the wavefunction describing the system when the atoms are far apart and not interacting (just isolated atomic orbitals).
• Ψbonded\Psi_{\text{bonded}}Ψbonded​: the wavefunction describing the system when the atoms are close enough that the electrons are shared/delocalized and a chemical bond exists (a molecular orbital).

These are not arbitrary labels—they are different eigenfunctions of the full Hamiltonian of the electron–nuclear system.

###

H^=Te+Tn+Vee+Vnn+Ven\hat{H} = T_e + T_n + V_{ee} + V_{nn} + V_{en}H^=Te​+Tn​+Vee​+Vnn​+Ven​ When the atoms are so far apart that they do not interact, the system factorizes:

Ψseparate=ψA(1)ψB(2)\Psi_{\text{separate}} = \psi_A(1)\psi_B(2)Ψseparate​=ψA​(1)ψB​(2) (or symmetrized/antisymmetrized depending on spin).

This describes:

• electron 1 around nucleus A
• electron 2 around nucleus B
• no molecular bonding
• just two separate atoms.

Ψbonded=12(1+S)[ψA+ψB]\Psi_{\text{bonded}} = \frac{1}{\sqrt{2(1+S)}}\left[\psi_A + \psi_B\right]Ψbonded​=2(1+S)​1​[ψA​+ψB​]

where:

• ψA,ψB\psi_A, \psi_BψA​,ψB​ are atomic orbitals
• S=⟨ψA∣ψB⟩S = \langle \psi_A | \psi_B \rangleS=⟨ψA​∣ψB​⟩ is the overlap integral

This wavefunction has:

• electron probability density between the nuclei
• constructive interference of the atomic wavefunctions
• a lower total energy eigenvalue

This is the bonding orbital.

You evaluate the expectation value of the Hamiltonian in each state:

Eseparate=⟨Ψseparate∣H^∣Ψseparate⟩E_{\text{separate}} = \langle \Psi_{\text{separate}}|\hat{H}|\Psi_{\text{separate}}\rangleEseparate​=⟨Ψseparate​∣H^∣Ψseparate​⟩ Ebonded=⟨Ψbonded∣H^∣Ψbonded⟩E_{\text{bonded}} = \langle \Psi_{\text{bonded}}|\hat{H}|\Psi_{\text{bonded}}\rangleEbonded​=⟨Ψbonded​∣H^∣Ψbonded​⟩ If:

Ebonded<EseparateE_{\text{bonded}} < E_{\text{separate}}Ebonded​<Eseparate​ then the molecule is more stable than separate atoms, and the difference:

ΔE=Eseparate−Ebonded\Delta E = E_{\text{separate}} - E_{\text{bonded}}ΔE=Eseparate​−Ebonded​ must be released as energy (heat, light, kinetic energy of atoms).

That difference is the bond energy.

====

Energy ====
│
│        Ψ_separate   ← higher energy (atoms apart)
│
│
│_________________     ← energy released
│
│   Ψ_bonded       ← lower energy (bond formed)
│
└─────────────────────── internuclear distance →

- Ψseparate\Psi_{\text{separate}}Ψseparate​ is the quantum state describing noninteracting atoms,

• Ψbonded\Psi_{\text{bonded}}Ψbonded​ is the quantum state describing the same atoms after their electronic wavefunctions reorganize into a lower-energy molecular orbital,

and comparing their energies tells us whether a chemical bond forms and how much energy is released.

If you want, the next logical step is: “Show the explicit energy lowering term by expanding the LCAO normalization and overlap integrals (Coulomb integral JJJ, exchange integral KKK) for H₂.”

User: What attracts two atoms to form a bond?

What attracts two atoms to form a bond? Will they simply not react/bond if they are far enough apart? At what point as they draw nearer to each other will the reaction take place? What fundamentally triggers the reaction?

Assistant: Condensed version of your question:

Condensed version of your question: What physically attracts two atoms into forming a bond, why do they not bond at long distance, and what exact event/threshold triggers the reaction?

At the most fundamental level, the attraction comes from the Coulomb force within the electron–nucleus system:

F=14πϵ0q1q2r2F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}F=4πϵ0​1​r2q1​q2​​ Specifically:

• The electrons of one atom are attracted to the nucleus of the other.
• If the electronic wavefunctions overlap constructively, the system can reorganize into a lower-energy quantum state (a molecular orbital).

This attraction competes with:

• electron–electron repulsion
• nucleus–nucleus repulsion

A bond forms only if the net energy decrease (from electron–nucleus attraction + quantum delocalization) outweighs the repulsive terms.

Because at long distances:

• The electron wavefunctions do not overlap significantly.
• The Coulomb attraction between nuclei and the other atom’s electrons falls off as 1/r21/r^21/r2.
• The system’s quantum mechanical state remains essentially the direct product of isolated atomic orbitals.

Without overlap, there's no possibility of forming a lower-energy molecular eigenstate—so there is no bond, just weak long-range interactions (van der Waals, dispersion).

Mathematically:

S=⟨ψA∣ψB⟩≈0⇒no bonding orbital availableS = \langle \psi_A | \psi_B \rangle \approx 0 \quad \Rightarrow \quad \text{no bonding orbital available}S=⟨ψA​∣ψB​⟩≈0⇒no bonding orbital available

Bonding becomes energetically favorable when electron wavefunction overlap becomes non-zero.

This occurs when interatomic separation approaches approximately one Bohr radius scale (10⁻¹⁰ m)—in practice, when atoms are close enough that the overlap integral SSS is significant.

At this distance:

• The bonding molecular orbital becomes lower in energy than the separated-atom state.
• The system now has a quantum mechanically available state that is lower in total energy.

This distance is where the attractive part of the potential energy curve dominates.

Two things must happen simultaneously:

Atoms must be close enough that:

ΔE=Ebonded−Eseparate<0\Delta E = E_{\text{bonded}} - E_{\text{separate}} < 0ΔE=Ebonded​−Eseparate​<0 That is: a lower-energy quantum state becomes available.

To transition into the lower-energy state, the system must shed energy:

• via photon emission
• via collisions with nearby particles
• via redistribution into vibrational/rotational kinetic energy

If the excess energy cannot be removed, the atoms approach → accelerate → and then rebound (like charged elastic balls).

So the key trigger is:

Without dissipation, bonding cannot complete.

This is why:

• cold atoms bond more easily than hot ones
• chemistry slows at low pressure
• catalysts don't force bonding—they provide energy pathways and intermediate states

If the atoms get too close, then:

• nucleus–nucleus repulsion rises steeply (Coulomb)
• Pauli exclusion forces the electron wavefunctions into antibonding configurations

So the total energy curve looks like:

Energy
│         ↑ Pauli + nuclear repulsion
│        /
│       /      ← too close → repulsive
│      /
│     /      minimum (bond length)
│    /
│   /  ← attractive region where bond forms
│  /
│ /
│/ _______________________________________ distance →
   far apart (no overlap)

Two atoms form a bond when they come close enough for their electron wavefunctions to overlap, allowing a lower-energy molecular orbital to exist, and when the system can dissipate the excess energy; before that critical distance there is no significant attraction, and past the equilibrium distance repulsive forces dominate.

If you'd like, next step could be:

• A full Lennard-Jones/Morse potential derivation
• A simulation with actual overlap integrals
• How chemistry differs for ionic, covalent, and metallic cases when viewed at this fundamental level

Which direction do you want to go?

User: You mentioned "when the integral of S is significant" - what determines the significance?

You mentioned "when the integral of S is significant" - what determines the significance?

Assistant: Restated question:

Restated question: What determines when the overlap integral S=⟨ψA∣ψB⟩S = \langle \psi_A | \psi_B \rangleS=⟨ψA​∣ψB​⟩ is “significant,” and how do we quantify that threshold physically—not heuristically?

1. Overlap is not an on/off switch — it becomes significant when it affects energy.

The overlap integral SSS matters because it appears directly in the energy expression for a molecular orbital formed from atomic orbitals.

For a simple diatomic (LCAO approximation), the energy of the bonding orbital is:

E+=HAA+HAB1+SE_+ = \frac{H_{AA} + H_{AB}}{1+S}E+​=1+SHAA​+HAB​​ and the antibonding orbital:

E−=HAA−HAB1−SE_- = \frac{H_{AA} - H_{AB}}{1-S}E−​=1−SHAA​−HAB​​ Here:

• HAAH_{AA}HAA​ is the energy of an isolated atomic orbital
• HAB=⟨ψA∣H^∣ψB⟩H_{AB} = \langle \psi_A | \hat{H} | \psi_B \rangleHAB​=⟨ψA​∣H^∣ψB​⟩ is the resonance/exchange integral
• S=⟨ψA∣ψB⟩S = \langle \psi_A | \psi_B \rangleS=⟨ψA​∣ψB​⟩ normalizes the wavefunction because the orbitals are no longer orthogonal

A bond forms only if:

E+<EseparateE_+ < E_{\text{separate}}E+​<Eseparate​ So overlap is significant when it materially lowers the energy eigenvalue.

2. So what is the quantitative threshold?

A practical physical threshold emerges:

∣HAB∣HAA≳S\frac{|H_{AB}|}{H_{AA}} \gtrsim SHAA​∣HAB​∣​≳S Meaning:

The overlap is significant when it is large enough that the change in energy due to constructive interference of the wavefunctions exceeds the normalization penalty.

Typical values:

• S<0.05S < 0.05S<0.05: negligible overlap → no bonding interaction
• 0.05≤S≤0.200.05 \le S \le 0.200.05≤S≤0.20: moderate overlap → weak bonding (van der Waals, H-bonding, weak covalent)
• S≈0.20–0.35S \approx 0.20\text{–}0.35S≈0.20–0.35: strong covalent bonding regime
• S>0.40S > 0.40S>0.40: rare; orbitals so close Pauli repulsion dominates

So “significant” does not mean geometric closeness only — it means:

ΔE=Ebonded−Eseparate<0\Delta E = E_{\text{bonded}} - E_{\text{separate}} < 0ΔE=Ebonded​−Eseparate​<0 and that only happens when the overlap term meaningfully perturbs the Hamiltonian eigenvalues.

3. A deeper physical statement

The overlap integral matters because it controls the strength of the exchange interaction, which is a quantum effect with no classical analog. The exchange energy term is approximately:

Eexchange∝HAB1+SE_{\text{exchange}} \propto \frac{H_{AB}}{1 + S}Eexchange​∝1+SHAB​​ and this is what lowers the energy enough to stabilize a covalent bond.

So the real physical meaning of “significant overlap” is:

There is enough wavefunction overlap that the exchange term lowers the total energy more than repulsive Coulomb terms raise it.

4. The point where attraction exceeds repulsion

If we treat the total interaction energy as:

E(r)=Ecoulomb(r)+Eexchange(r)+Ekinetic(r)+EPauli(r)E(r) = E_{\text{coulomb}}(r) + E_{\text{exchange}}(r) + E_{\text{kinetic}}(r) + E_{\text{Pauli}}(r)E(r)=Ecoulomb​(r)+Eexchange​(r)+Ekinetic​(r)+EPauli​(r) then the bonding distance is found by:

dEdr=0\frac{dE}{dr} = 0drdE​=0 This minimum corresponds to precisely the point where overlap has become large enough that:

∣ΔEexchange∣>ΔEPauli+ΔErepulsion|\Delta E_{\text{exchange}}| > \Delta E_{\text{Pauli}} + \Delta E_{\text{repulsion}}∣ΔEexchange​∣>ΔEPauli​+ΔErepulsion​ That is the exact physical threshold.

5. One-sentence definition

Overlap becomes “significant” when the electron wavefunctions are close enough that the exchange integral

𝐻

𝐴

𝐵

H

AB

​

lowers the molecular orbital energy more than the normalization penalty from

𝑆

S and the Coulomb repulsion terms raise it, resulting in a net decrease in the system’s total energy.

If you want to go further:

Pick one:

A. Compute S(r)S(r)S(r) explicitly for two hydrogen 1s orbitals and plot it vs. distance. B. Derive the approximate bonding/antibonding splitting magnitude using HABH_{AB}HAB​ and SSS. C. Connect this to observable reaction rates and activation energy.

Which direction?

gpt-5-1-thinking

User: When considering the energy transfer associated to C, Cl, and H reacting to form methyl chloride, I believe that textbo…

When considering the energy transfer associated to C, Cl, and H reacting to form methyl chloride, I believe that textbooks say that ~1500 kJ of energy is transferred from the individual atoms to the bonds formed between the atoms. Is this accurate? How does this or does this not account for the endothermic or exothermic properties of the rection?

Assistant: Condensed restatement of your question:

Condensed restatement of your question: Textbooks say ~1500 kJ of energy is associated with forming the bonds in CH₃Cl from C, H, and Cl atoms. Is that correct, and how does that relate (or not) to whether the overall reaction is endothermic or exothermic?

For gas-phase atoms:

C(g)+3H(g)+Cl(g)→CH3Cl(g)\text{C(g)} + 3\text{H(g)} + \text{Cl(g)} \rightarrow \text{CH}_3\text{Cl(g)}C(g)+3H(g)+Cl(g)→CH3​Cl(g) The total bond energy in CH₃Cl is roughly:

• 3 × C–H bonds ≈ 3 × 410 kJ/mol ≈ 1230 kJ/mol
• 1 × C–Cl bond ≈ 330–340 kJ/mol

Total ≈ 1.5 × 10³ kJ/mol (≈ 1500 kJ/mol).

By definition, bond enthalpy is the energy required to break a bond. So:

• To break CH₃Cl completely into separate atoms, you must put in ~1500 kJ/mol (endothermic).
• Going the other way (atoms → CH₃Cl), the system releases ~1500 kJ/mol (exothermic).

So yes: “~1500 kJ is transferred from the atoms to the bonds” is basically saying

That is not yet a statement about the full reaction you do in the lab.

In reality, you don’t usually start with isolated atoms. You start from the elements in their standard forms, e.g.:

C(s, graphite)+32H2(g)+12Cl2(g)→CH3Cl(g)\text{C(s, graphite)} + \tfrac{3}{2}\text{H}_2\text{(g)} + \tfrac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{CH}_3\text{Cl(g)}C(s, graphite)+23​H2​(g)+21​Cl2​(g)→CH3​Cl(g) To connect this real reaction to the ~1500 kJ number, conceptually break it into steps:

1. Atomize the reactants (break all their bonds, change phases as needed): - C(s, graphite) → C(g) (sublimation/atomization; costs energy) - ½ Cl₂(g) → Cl(g) (½ of Cl–Cl bond; costs energy) - 3/2 H₂(g) → 3 H(g) (3/2 of H–H bond; costs energy) This is all endothermic. You’re climbing up the energy hill.
2. Form CH₃Cl from atoms: - C(g) + 3H(g) + Cl(g) → CH₃Cl(g) This step is where that ≈1500 kJ/mol is released (exothermic).

The net reaction enthalpy (ΔH_rxn) is:

ΔHrxn=(energy to atomize reactants)−(energy released by forming bonds in CH3Cl)\Delta H_{\text{rxn}} = \text{(energy to atomize reactants)} - \text{(energy released by forming bonds in CH}_3\text{Cl)}ΔHrxn​=(energy to atomize reactants)−(energy released by forming bonds in CH3​Cl) or equivalently (in the usual “bonds broken – bonds formed” form):

ΔHrxn≈∑E(bonds broken)−∑E(bonds formed)\Delta H_{\text{rxn}} \approx \sum E(\text{bonds broken}) - \sum E(\text{bonds formed})ΔHrxn​≈∑E(bonds broken)−∑E(bonds formed) So:

• The 1500 kJ is essentially the “bonds formed” term for the product.
• The reactants also have bonds and cohesive energies (C–C in graphite, H–H, Cl–Cl), which contribute to the “bonds broken” term.

Whether the overall reaction is exothermic or endothermic depends on which side “wins”:

• If bonds formed in products are stronger overall (bigger total bond energy) than those broken in reactants → exothermic (ΔH < 0).
• If bonds broken are stronger → endothermic (ΔH > 0).

Think of it like this:

• Choose a reference where separated atoms = high potential energy.
• When those atoms form CH₃Cl, the system drops by ~1500 kJ/mol in potential energy.
• That energy has to go somewhere → usually to the surroundings as heat and/or light → that’s why bond formation is exothermic.

But:

• The reaction enthalpy you see in tables (ΔH°_f for CH₃Cl) starts from standard states, not from free atoms.
• To go from standard states → atoms → CH₃Cl, you must add the atomization costs and subtract the 1500 kJ bond-formation release.
• Only that final net number tells you how exothermic or endothermic the real reaction is.

1. “Is ~1500 kJ from atoms to bonds accurate?” Roughly, yes: it’s a reasonable approximate total bond energy for CH₃Cl (from isolated gas-phase atoms). It’s the energy you’d need to break the molecule into atoms, and the energy released when you form it from atoms.

1. “How does this relate to endothermic vs exothermic?” That 1500 kJ number is only one side of the accounting (the product side, atoms → CH₃Cl). To know if the real reaction from elemental C, H₂, and Cl₂ is endothermic or exothermic, you must compare: - Energy required to break all bonds / cohesive structures in the reactants versus - Energy released when forming the CH₃Cl bonds. The difference of those two is the reaction enthalpy. The 1500 kJ is baked into the “bonds formed” term but doesn’t by itself tell you whether the overall reaction is exothermic or endothermic.

If you’d like, next step could be: we pick actual approximate bond enthalpies and walk the full numeric estimate for ΔH for C(graphite) + 3/2 H₂ + 1/2 Cl₂ → CH₃Cl to see a concrete number.