Question:
Sphalerite is the ore of metal \( M_1 \) and malachite is the ore of metal \( M_2 \). Atomic numbers of \( M_1 \) and \( M_2 \) are respectively
Sphalerite is the ore of metal \( M_1 \) and malachite is the ore of metal \( M_2 \). Atomic numbers of \( M_1 \) and \( M_2 \) are respectively
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Identify the metals associated with the ores: sphalerite is the ore of zinc (\( \text{Zn} \), atomic number 30), and malachite is the ore of copper (\( \text{Cu} \), atomic number 29). Match the atomic numbers accordingly.
Updated On: May 19, 2025
- 30, 82
- 82, 30
- 30, 29
- 29, 30
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The Correct Option is C
Approach Solution - 1
To solve this, we need to identify the metals associated with the given ores and their atomic numbers:
- Sphalerite: Sphalerite is the primary ore of zinc (\( \text{Zn} \)). The atomic number of zinc is 30. Therefore, \( M_1 = \text{Zn} \), and the atomic number of \( M_1 \) is 30.
- Malachite: Malachite is a common ore of copper (\( \text{Cu} \)). The atomic number of copper is 29. Therefore, \( M_2 = \text{Cu} \), and the atomic number of \( M_2 \) is 29.
Now, let's evaluate the options:
- Option (1): 30, 82 – Incorrect, as 82 corresponds to lead (\( \text{Pb} \)), not copper.
- Option (2): 82, 30 – Incorrect, as 82 is lead (\( \text{Pb} \)), not zinc, and 30 is zinc, not copper.
- Option (3): 30, 29 – Correct, as 30 is zinc (\( M_1 \)) and 29 is copper (\( M_2 \)).
- Option (4): 29, 30 – Incorrect, as 29 is copper (not the metal for sphalerite) and 30 is zinc (not the metal for malachite).
So, the atomic numbers of \( M_1 \) and \( M_2 \) are 30, 29.
- Sphalerite: Sphalerite is the primary ore of zinc (\( \text{Zn} \)). The atomic number of zinc is 30. Therefore, \( M_1 = \text{Zn} \), and the atomic number of \( M_1 \) is 30.
- Malachite: Malachite is a common ore of copper (\( \text{Cu} \)). The atomic number of copper is 29. Therefore, \( M_2 = \text{Cu} \), and the atomic number of \( M_2 \) is 29.
Now, let's evaluate the options:
- Option (1): 30, 82 – Incorrect, as 82 corresponds to lead (\( \text{Pb} \)), not copper.
- Option (2): 82, 30 – Incorrect, as 82 is lead (\( \text{Pb} \)), not zinc, and 30 is zinc, not copper.
- Option (3): 30, 29 – Correct, as 30 is zinc (\( M_1 \)) and 29 is copper (\( M_2 \)).
- Option (4): 29, 30 – Incorrect, as 29 is copper (not the metal for sphalerite) and 30 is zinc (not the metal for malachite).
So, the atomic numbers of \( M_1 \) and \( M_2 \) are 30, 29.
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Approach Solution -2
Sphalerite is the ore of metal \( M_1 \) and malachite is the ore of metal \( M_2 \). Atomic numbers of \( M_1 \) and \( M_2 \) are respectively:
Step 1: Identify the metal from sphalerite:
Sphalerite is a commonly occurring ore of zinc. Its chemical formula is ZnS (zinc sulfide). Therefore, metal \( M_1 \) is zinc.
The atomic number of zinc is 30.
Step 2: Identify the metal from malachite:
Malachite is a green mineral and a carbonate ore of copper. Its formula is CuCO₃·Cu(OH)₂. Thus, metal \( M_2 \) is copper.
The atomic number of copper is 29.
Step 3: Conclusion:
- \( M_1 = \) Zinc → Atomic number = 30
- \( M_2 = \) Copper → Atomic number = 29
Final Answer:
\[ \boxed{30, \ 29} \]
Step 1: Identify the metal from sphalerite:
Sphalerite is a commonly occurring ore of zinc. Its chemical formula is ZnS (zinc sulfide). Therefore, metal \( M_1 \) is zinc.
The atomic number of zinc is 30.
Step 2: Identify the metal from malachite:
Malachite is a green mineral and a carbonate ore of copper. Its formula is CuCO₃·Cu(OH)₂. Thus, metal \( M_2 \) is copper.
The atomic number of copper is 29.
Step 3: Conclusion:
- \( M_1 = \) Zinc → Atomic number = 30
- \( M_2 = \) Copper → Atomic number = 29
Final Answer:
\[ \boxed{30, \ 29} \]
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