Lowest Oxidation number of an atom in a compound A 2 B is -2 . The number of an electron in its valence shell is

Chemistry · Periodicity of Elements and Periodic Properties

JEE Main 2024 — 1 February, Shift 1 — Question 79

Lowest Oxidation number of an atom in a compound $\mathrm{A}_{2} \mathrm{~B}$ is -2 . The number of an electron in its valence shell is

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{A}_{2} \mathrm{~B} \rightarrow 2 \mathrm{~A}^{+}+\mathrm{B}^{-2}, \mathrm{~B}^{-2}$ has complete octet in its dianionic form, thus in its atomic state it has 6 electrons

in its valence shell. As it has negative charge, it has acquired two electrons to complete its octet.

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Chemistry
Chapter: Periodicity of Elements and Periodic Properties
Topic: Valency, Diagonal Relationship, Inert Pair Effect & Other Miscellaneous Properties

In case of isoelectronic species the size of F^-, Ne and Na^+ is affected by:

Answer: 6

Step-by-step solution

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