Niobium (Nb) and ruthenium (Ru) have "x" and "y" number of electrons in their respective 4d orbitals. The value of \( x + y \) is:
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Correct Answer: 11
Approach Solution - 1
To solve for \( x + y \), where \( x \) and \( y \) are the number of electrons in the 4d orbitals of niobium (Nb) and ruthenium (Ru), respectively, we need to refer to the electron configurations of these elements:
- Niobium (Nb): The atomic number of Nb is 41. Its electron configuration is \([Kr] 4d^4 5s^1\). Thus, Nb has 4 electrons in its 4d orbital (\( x = 4 \)).
- Ruthenium (Ru): The atomic number of Ru is 44. Its electron configuration is \([Kr] 4d^7 5s^1\). Therefore, Ru has 7 electrons in its 4d orbital (\( y = 7 \)).
Now, compute \( x + y \):
\( x + y = 4 + 7 = 11 \)
Check that the sum \( 11 \) falls within the specified range \([11, 11]\), confirming that our computed value is correct.
Therefore, the value of \( x + y \) is 11.
Approach Solution -2
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