Diamond: four bonds per carbon in a rigid network
도입됨| English | 의미 |
|---|---|
| diamond/ˈdaɪəmənd/ | 다이아몬드 |
| rigid network/ˈrɪdʒɪd ˈnetwɜːk/ | 강성 네트워크 |
What would explain this observation?
- Diamond 金刚石 and graphite both contain carbon, but their bonding arrangements differ. The element name alone does not predict hardness or conductivity.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- In diamond each carbon atom makes four covalent bonds with other carbon atoms in a giant three-dimensional structure. Strong connections extend through the solid, giving a rigid network 刚性网络 and making diamond very hard. Very much energy is needed to overcome many covalent bonds, giving a very high melting point in the specified GCSE account. Its outer electrons are involved in covalent bonds, so it has no delocalised electrons to carry charge and does not conduct electricity.
- diamond: A carbon form with each atom covalently bonded to four others in a giant three-dimensional structure; rigid network: An extended arrangement whose strong connections resist changes in shape.
Why does diamond not conduct electricity?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Every carbon in diamond uses all four outer electrons in covalent bonds, so there are no delocalised electrons to carry charge. It certainly contains electrons, and its atoms are neutral.
Match each technical term to its precise meaning.
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Explain one property with the appropriate connection: resistance to deformation follows the rigid many-direction network; high melting point follows energy needed to overcome many strong covalent bonds; non-conductivity follows the absence of mobile charged carriers. A drawing with a carbon surrounded by four neighbours shows local coordination, not a complete diamond molecule.
- Compare a tetrahedral local model with a graphite-layer model. Count four bonds around an interior diamond carbon and trace continuation beyond the fragment. Mark a two-dimensional picture as a projection of a three-dimensional arrangement. Prepared models or images suffice; there is no need to heat or scratch expensive specimens or infer properties from a gem’s colour.
Which two habits make the investigation or model in this case more defensible?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: a stated large diamond model contains 120 interior-equivalent carbon sites, each with four bond connections. There are 120×4=480 bond-end counts; each bond has two ends, giving 480/2=240 bonds under this explicitly boundary-free counting convention. Simply multiplying by four double-counts each shared connection.
A boundary-free model has 40 carbon sites with four connections each. Dividing shared bond-end counts by two, how many bonds are represented? Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A boundary-free model has 40 carbon sites with four connections each. Dividing shared bond-end counts by two, how many bonds are represented?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
The result is 80 bonds. Known: 40 carbon sites with 4 connections each. Bond-end counts = 40 x 4 = 160; each bond has two ends, so bonds = 160/2 = 80 bonds. Multiplying by four alone would double-count every shared bond.
Check the conclusion and its limits
- Diamond is not an ionic lattice of carbon ions. The atoms do not become larger or lose their electrons entirely. An actual finite fragment has boundaries, so the bond-count formula needs the stated interior-equivalent assumption. No claim about current experimental high-pressure melting conditions is required for this GCSE structure explanation.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Diamond consists of small C₄ molecules held by weak attractions. This claim is false: Diamond is a GIANT covalent network: every carbon bonds to four neighbours in a structure continuing through the whole crystal. No C4 molecules exist in diamond.
Diamond: four bonds per carbon in a rigid network: Explain one property with the appropriate connection: resistance to deformation follows the rigid many-direction network; high melting point follows energy needed to overcome many strong covalent bonds; non-conductivity follows the absence of mobile charged carriers. A drawing with a carbon surrounded by four neighbours shows local coordination, not a complete diamond molecule.
Diamond consists of small C₄ molecules held by weak attractions.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Diamond is a GIANT covalent network: every carbon bonds to four neighbours in a structure continuing through the whole crystal. No C4 molecules exist in diamond.
A carbon form with each atom covalently bonded to four others in a giant three-dimensional structure: write the technical term.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
diamond means A carbon form with each atom covalently bonded to four others in a giant three-dimensional structure.