Chemistry · d and f Block Elements

JEE Main 2024 — 27 January, Shift 2 — Question 75

Total number of ions from the following with noble gas configuration is \qquad .

Sr2+(Z=38),Cs+(Z=55),La2+(Z=57)Pb2+\mathrm{Sr}^{2+}(\mathrm{Z}=38), \mathrm{Cs}^{+}(\mathrm{Z}=55), \mathrm{La}^{2+}(\mathrm{Z}=57) \mathrm{Pb}^{2+} (Z=82),Yb2+(Z=70)(\mathrm{Z}=82), \mathrm{Yb}^{2+}(\mathrm{Z}=70) and Fe2+(Z=26)\mathrm{Fe}^{2+}(\mathrm{Z}=26)

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

Noble gas configuration =ns2np6=\text{n}{{\text{s}}^{2}}\text{n}{{\text{p}}^{6}}

\begin{array}{*{35}{r}}{} & \left[ \text{S}{{\text{r}}^{2+}} \right]=\left[ \text{Kr} \right]\\{} & \left[ \text{C}{{\text{s}}^{+}} \right]=\left[ \text{Xe} \right] \\{} & \left[ \text{Y}{{\text{b}}^{2+}} \right]=\left[ \text{Kr} \right]4\text{ }\!\!~\!\!\text{ }{{\text{d}}^{10}}4{{\text{f}}^{14}}5\text{ }\!\!~\!\!\text{ }{{\text{s}}^{2}}5{{\text{p}}^{6}}\\{} & \left[ \text{L}{{\text{a}}^{2+}} \right]=\left[ \text{Xe} \right]5\text{ }\!\!~\!\!\text{ }{{\text{d}}^{1}} \\{} & \left[ \text{ }\!\!~\!\!\text{ P}{{\text{b}}^{2+}} \right]=\left[ \text{Xe} \right]4{{\text{f}}^{14}}5\text{ }\!\!~\!\!\text{ }{{\text{d}}^{10}}6\text{ }\!\!~\!\!\text{ }{{\text{s}}^{2}} \\{} & \left[ \text{F}{{\text{e}}^{2+}} \right]=\left[ \text{Ar} \right]3\text{ }\!\!~\!\!\text{ }{{\text{d}}^{6}} \\\end{array}

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Exam
JEE Main 2024
Subject
Chemistry
Chapter
d and f Block Elements
Topic
Introduction and Properties of Transition Elements
Total number of ions from the following with noble gas configuration… | JEE Main 2024 PYQ with Solution · DhiX AI