Chemistry · Structure of Atom

JEE Main 2025 — 22 January, Evening Shift — Question 44

Niobium $(\mathrm{Nb})$ and ruthenium ( Ru ) have " $x$ " and " $y$ " number of electrons in their respective $4 d$ orbitals. The value of $x+y$ is____

Answer: 11

Numerical answer — enter this value.

Step-by-step solution

$\quad Z=41 \rightarrow \mathrm{Nb}$ (Niobium) : $[\mathrm{Kr}]_{36} 4 \mathrm{~d}^{4} 5 \mathrm{~s}^{1}$

Number of electron in $4 d=4=x$

$\mathrm{Z}=44 \rightarrow \mathrm{Ru}$ (Ruthenium) $[\mathrm{Kr}]_{36} 4 \mathrm{~d}^{7} 5 \mathrm{~s}^{1}$

Number of electron in $4 \mathrm{~d}=7=\mathrm{y}$

$x+y=11$

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Exam: JEE Main 2025
Paper: 22 January, Evening Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Electronic Configuration: Rules, Aufbau Principle and Exceptions

Given below are two statements, one is labelled as Statement I and the other is labelled as Statement II. Statement (I): A spectral line…

Answer: 11

Step-by-step solution

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