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JEE Main 202522 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Evening Shift
Subject: Chemistry
Chapter: d and f Block Elements

JEE Main 2025 · 22 January · Evening Shift · Chemistry · d and f Block Elements

Niobium ( Nb ) and ruthenium (Ru) have " x " and " y " number of...

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 ______ -.

Correct answer

0

Step-by-step solution

\begin{aligned} & \mathrm{Z}=41 \rightarrow \mathrm{Nb} \text{ (Niobium) : }[\mathrm{Kr}]_{36} 4 \mathrm{ d}^4 5 \mathrm{ s}^1 \\ & \text{ Number\text{ of } electron in } 4 \mathrm{ d}=4=\mathrm{x} \\ & \mathrm{Z}=44 \rightarrow \mathrm{Ru} \text{ (Ruthenium) }[\mathrm{Kr}]_{36} 4 \mathrm{ d}^7 5 \mathrm{ s}^1 \\ & \text{ Number of electron in } 4 \mathrm{ d}=7=\mathrm{y} \\ & \mathrm{x}+\mathrm{y}=11\end{aligned}

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