Question

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

• A. Z - 2
• B. Z - 1
• C. Z
• D. Z + 1

Answer 1

Answer: (A)

$\because {}^{\prime }{Z}^{\prime }$ 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 $Z+l$ 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