If four elements with atomic numbers Z-2,Z-1, | vedclass

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(A) $\because$ $Z$ represents the number of protons in a nucleus. For isoelectronic species,the size depends on the effective nuclear charge. As the n

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$Z-2$

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(A) $\because$ $Z$ represents the number of protons in a nucleus. For isoelectronic species,the size depends on the effective nuclear charge. As the number of protons $(Z)$ decreases for the same number of electrons,the attraction between the nucleus and the electrons decreases,leading to an increase in ionic size. Thus,the species with the smallest atomic number will have the largest size. Comparing the given atomic numbers: $Z-2 Therefore,the ion with atomic number $Z-2$ will have the largest size. Hence,option $(A)$ is the correct answer.

If four elements with atomic numbers $Z-2$,$Z-1$,$Z$ and $Z+1$ are forming isoelectronic ions,the atomic number of the ion having largest size is

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