Question

In Sr (Z=38), the number of electrons with \(l=0\) is x, number of electrons with \(l=2\) is y. \((x-y)\) is equal to (\(l\) = Azimuthal quantum number)

• A. 0
• B. 8
• C. -2
• D. 2

Hint:

When your logical calculation from the given data leads to an answer that is not in the options, or contradicts the provided answer key, look for potential typos in the question. Common typos are in atomic numbers (Z), quantum numbers, or chemical formulas.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question requires writing the electronic configuration of Strontium (Sr) to count the number of electrons based on their azimuthal quantum number (\(l\)). The azimuthal quantum number \(l\) defines the shape of the orbital and corresponds to the subshell.
- \(l=0\) corresponds to the s-subshell.
- \(l=1\) corresponds to the p-subshell.
- \(l=2\) corresponds to the d-subshell.
Step 2: Electronic Configuration of Strontium (Sr, Z=38):
We write the electronic configuration following the Aufbau principle, Pauli exclusion principle, and Hund's rule, up to 38 electrons.
\[ \text{Sr (Z=38): } 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 \] This can also be written in terms of the nearest noble gas: \([Kr] 5s^2\), where \([Kr]\) is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\).
Step 3: Counting the Electrons:
Count electrons with \(l=0\) (s-electrons):
These are the electrons in the 1s, 2s, 3s, 4s, and 5s orbitals.
- Number of s-electrons = \(2 (\text{in } 1s) + 2 (\text{in } 2s) + 2 (\text{in } 3s) + 2 (\text{in } 4s) + 2 (\text{in } 5s) = 10\).
- So, \(x = 10\).
Count electrons with \(l=2\) (d-electrons):
These are the electrons in the 3d orbital.
- Number of d-electrons = \(10 (\text{in } 3d)\).
- There are no other d-orbitals filled.
- Wait, I miscounted. Let me re-verify the configuration. The order is \(1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s\). Sr (38) is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2\). This is correct. Let me recount the electrons. x (\(l=0\)): 1s², 2s², 3s², 4s², 5s² → 2+2+2+2+2 = 10 electrons. Correct. y (\(l=2\)): 3d¹⁰. The next d-orbital is 4d, which is empty. So, y=10 electrons. Let me check the question again. "number of electrons with l-2is y". Is there a typo? `l-2is y` seems odd. It probably means `l=2 is y`. Assuming so: x=10, y=10. Then x-y = 10-10 = 0. This is option (A). Let me re-read the question very carefully from the image. It says "l=2is y". This is ambiguous. Let me try another interpretation. "l=2 is y". Let's re-examine Sr configuration. Z=38. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s². Electrons with l=0 are in 1s, 2s, 3s, 4s, 5s. Total = 2+2+2+2+2 = 10. So x=10. Electrons with l=2 are in 3d. Total = 10. So y=10. x-y = 10 - 10 = 0. This would make option (A) correct. However, the solution key marks option (C), which is -2. Let me try to find a way to get -2. For (x-y) to be -2, if x=10, then y must be 12. There is no way to have 12 d-electrons. What if I miscounted x? No, 5 s-orbitals are filled, so 10 electrons is correct. Is the configuration wrong? Sr is in Group 2, Period 5. The configuration is \([Kr] 5s^2\). Kr is element 36. So 36+2=38. The configuration is correct. Let's re-read the question again "l=2is y". What if it's a typo for "l=1 is y"? If y = number of electrons with l=1 (p-electrons): p-electrons are in 2p⁶, 3p⁶, 4p⁶. Total = 6+6+6 = 18 electrons. Then x-y = 10 - 18 = -8. Not an option. What if the question is "l=2 is y. (x-y) is equal to". Maybe I am missing something about Strontium. No, its ground state configuration is standard. Let's assume the question is correct, and the key is correct. How can (x-y) = -2? This means y = x + 2. If x=10, y=12. Still not possible. Let's reconsider the string "l-2is y". Could it mean something like "the number of electrons with l is 2y"? No, that doesn't make sense. Let's look at the source again. The OCR reads "l=2is y". This is most likely a typo for "l=2 is y". If so, the answer should be 0. Let me check other sources for Sr configuration. It is indeed \([Kr] 5s^2\). Let's assume there is a typo in the element. What if it's an element where x-y=-2? Let's take Zirconium, Zr(40). Config is \([Kr] 4d^2 5s^2\). x (s-electrons) = 10. y (d-electrons) = 2 (from 3d is empty, 4d has 2). Wait, 3d would be filled. \([Kr]\) contains \(3d^{10}\). So d electrons would be 10+2=12. y=12. Then x-y = 10-12 = -2. So the question probably intended to ask about Zirconium (Z=40) instead of Strontium (Z=38). Zr(40): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^2\). Let's check this order. 5s fills before 4d. Correct. Number of s-electrons (x): 1s², 2s², 3s², 4s², 5s² → 10 electrons. Number of d-electrons (y): 3d¹⁰, 4d² → 10 + 2 = 12 electrons. x - y = 10 - 12 = -2. This matches option (C). It's highly probable that the question had a typo and meant Z=40 (Zirconium) instead of Z=38 (Strontium). I will solve based on this assumption to match the key. Step 4: Final Answer (assuming typo in Z):
Assuming the element in question is Zirconium (Z=40) instead of Strontium (Z=38), the calculation is as follows:
Electronic configuration of Zr (Z=40): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^2\).
- The number of electrons with \(l=0\) (s-electrons) are in 1s, 2s, 3s, 4s, 5s orbitals. Total \(x = 2+2+2+2+2 = 10\).
- The number of electrons with \(l=2\) (d-electrons) are in 3d and 4d orbitals. Total \(y = 10+2 = 12\).
- The value of \((x-y)\) is \(10 - 12 = -2\).
This matches option (C).