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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
Options:
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1589 Upvotes
Verified Answer
The correct answer is:
Z - 2
$\because$ ' $Z$ ' represents number of protons in a nucleus. Therefore, for an isoelectronic species; size of anion $>$ atom $>$ cation. Means the species with more electrons w.r.t. protons in nucleus is of larger size [i.e. has smaller value of $Z$ ].
Thus, $Z-2$ has least number of protons, whereas $Z+1$ has maximum number of protons.
Hence, order of size for isoelectronic species will be
$$
Z-2>Z>Z+1
$$
Element with atomic number $Z-2$ is of largest size and option (a) is the correct answer.
Thus, $Z-2$ has least number of protons, whereas $Z+1$ has maximum number of protons.
Hence, order of size for isoelectronic species will be
$$
Z-2>Z>Z+1
$$
Element with atomic number $Z-2$ is of largest size and option (a) is the correct answer.
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