JEE Mains · Chemistry · STD 12 - 4. d and f- block elements
Niobium ( Nb ) and ruthenium \((\mathrm{Ru})\) have " \(x\) " and " \(y\) " number of electrons in their respective 4 d orbitals. The value of \(x+y\) is ______.
- A 11
- B 22
- C 33
- D 44
Answer & Solution
Correct Answer
(A) 11
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{Z}=41 \rightarrow \mathrm{Nb} \text { (Niobium) : }[\mathrm{Kr}]_{36} 4 \mathrm{~d}^4 5 \mathrm{~s}^1 \\ & \text { Number of electron in } 4 \mathrm{~d}=4=\mathrm{x} \\ & \mathrm{Z}=44 \rightarrow \mathrm{Ru} \text { (Ruthenium) }[\mathrm{Kr}]_{36} 4…
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