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{ "quality": "blurry", "question_count": 5, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor" ], "needs_drawing": false, "concepts_used": [ "Chemical Nomenclature", "Electron Configuration Blocks", "Lewis Structures", "VSEPR Theory" ], "new_concepts": [ "Valence Bond Theory", "Orbital Hybridization" ], "current_concepts": [ "Chemical Nomenclature", "Electron Configuration Blocks", "Lewis Structures", "VSEPR Theory", "Valence Bond Theory", "Orbital Hybridization" ] } ## Identify copper(II) nitrate formula Using the **Chemical Nomenclature** knowledge point: - Copper(II) indicates a copper cation with a $2+$ charge: $\text{Cu}^{2+}$. - Nitrate is a polyatomic anion with a $1-$ charge: $\text{NO}_3^-$. - To balance the charges, two nitrate ions are needed for every copper(II) ion, yielding $\text{Cu(NO}_3)_2$. ## Identify p-block elements forming anions Using the **Electron Configuration Blocks** knowledge point: - We look for elements located in the p-block (Groups 13–18) that typically form anions (nonmetals/metalloids). - Option 1: $\text{Se, P, As}$ are all nonmetals/metalloids in the p-block that form anions ($\text{Se}^{2-}$, $\text{P}^{3-}$, $\text{As}^{3-}$). - Option 2: $\text{Na, Al, Rb}$ contain s-block metals ($\text{Na, Rb}$) and a p-block metal ($\text{Al}$) which form cations. - Option 3: $\text{F, S, Cl}$ are p-block nonmetals forming anions, but the third option listed in the image is $\text{F, S, Cl}$. Let's double check the options: - Option 1: $\text{Se, P, As}$ - Option 2: $\text{Na, Al, Rb}$ - Option 3: $\text{F, S, Cl}$ Both Option 1 and Option 3 contain p-block elements that form anions. Let's re-examine the image text: "Which of the following elements are found in the p-block and can form anions?". $\text{F, S, Cl}$ are classic p-block elements forming monoatomic anions ($\text{F}^-$, $\text{S}^{2-}$, $\text{Cl}^-$). ## Analyze carbonate anion structure Using the **Lewis Structures** and **VSEPR Theory** knowledge points: - Carbonate ($\text{CO}_3^{2-}$) has a central carbon bonded to three oxygen atoms with a steric number of 3 (trigonal planar electron geometry, $sp^2$ hybridization, and $120^\circ$ bond angles). - The double bond consists of a $\sigma$-bond and a $\pi$-bond. The $\pi$-bond is formed by the sideways overlap of unhybridized $p$-orbitals on carbon and oxygen. - Therefore, the correct statement is: "The electron geometry of carbon in a carbonate anion is trigonal planar with a $sp^2$ hybridization and bond angles are $120^\circ$. There is a $\pi$-bond formed between a non-hybridized 'C' p-orbital overlapping with an 'O' p-orbital." ## Determine hybridization of C2H2 Using the **Valence Bond Theory** knowledge point: - Acetylene ($\text{C}_2\text{H}_2$) has a triple bond between the two carbon atoms ($\text{H}-\text{C}\equiv\text{C}-\text{H}$). - Each carbon is bonded to two atoms (one hydrogen, one carbon) and has zero lone pairs, giving a steric number of 2. - A steric number of 2 corresponds to $sp$ hybridization for both carbon atoms. ## Analyze carbon tetrachloride bonding Using the **Valence Bond Theory** and **VSEPR Theory** knowledge points: - Carbon tetrachloride ($\text{CCl}_4$) has a central carbon bonded to four chlorine atoms with a steric number of 4. - This results in a tetrahedral electron geometry and $sp^3$ hybridization. - The single $\text{C}-\text{Cl}$ bonds are $\sigma$-bonds. - An $sp^3$ hybrid orbital has $25\%$ s-character and $75\%$ p-character (1 part s to 3 parts p), which means the $\text{C}-\text{Cl}$ bond is of s-character or described by the overlap of carbon's $sp^3$ hybrid orbital. Let's check the options: - Option 1: "...tetrahedral with a $sp^3$ hybridization. The $\text{C}-\text{Cl}$ bond is of s-character" (incorrect, it is a $\sigma$-bond formed by $sp^3$ hybrid orbital). - Option 2: "...tetrahedral with a $sp^3$ hybridization. The $\text{C}-\text{Cl}$ bond is of $\sigma$-character" (correct, single covalent bonds are $\sigma$-bonds). ## Question 45 (A) $\text{CuNO}_3$

(B) $\text{Cu(NO}_3)_2$

(C) $\text{CuNO}_2$

(D) $\text{Cu}_2\text{NO}_3$ ## Question 46 (A) $\text{Se, P, As}$

(B) $\text{Na, Al, Rb}$

(C) $\text{F, S, Cl}$ ## Question 47 (A) The electron geometry of carbon in a carbonate anion is trigonal planar with a $sp^2$ hybridization and bond angles are <$120^\circ$. There is a $\sigma$-bond formed between a hybridized 'C' p-orbital overlapping with an 'O' p-orbital.

(B) The electron geometry of carbon in a carbonate anion is trigonal planar with a $sp^2$ hybridization and bond angles are $120^\circ$. There is a $\pi$-bond formed between a non-hybridized 'C' p-orbital overlapping with an 'O' p-orbital.

(C) The electron geometry of carbon in a carbonate anion is trigonal planar with a $sp^3$ hybridization and bond angles are <$120^\circ$. There is a $\pi$-bond formed between a non-hybridized 'C' s-orbital overlapping with an 'O' p-orbital. ## Question 48 (A) both carbons are $sp$

(B) one carbon is $sp^2$ and the other carbon is $sp$

(C) both carbons are $sp^2$ ## Question 49 (A) The electron geometry of carbon in carbon tetrachloride is tetrahedral with a $sp^3$ hybridization. The $\text{C}-\text{Cl}$ bond is of s-character

(B) The electron geometry of carbon in carbon tetrachloride is tetrahedral with a $sp^3$ hybridization. The $\text{C}-\text{Cl}$ bond is of $\sigma$-character

(C) The electron geometry of carbon in carbon tetrachloride is tetrahedral with a $sp^2$ hybridization. The $\text{C}-\text{Cl}$ bond is of $\pi$-character

{ "subject": "Natural Science", "question_type": "Multiple Choice", "knowledge_point": [ "Natural Science", "Chemistry", "Orbital Hybridization" ] }

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QUESTION IMAGE

Question

Explanation:

Identify copper(II) nitrate formula

Identify p-block elements forming anions

Analyze carbonate anion structure

Determine hybridization of C2H2

Analyze carbon tetrachloride bonding

Answer:

Question 45

Question 46

Question 47

Question 48

Question 49