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Q.

Correct order of ionic radius of following isoelectronic species is:

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a

$F^{-} > N e > N a^{+} > A l^{+ 3} > M g^{+ 2}$

b

$S e^{- 2} > B r^{-} > K r > R b^{+} > S r^{+ 2}$

c

$S^{- 2} > C l^{-} > K^{+} > A r > C a^{+ 2}$

d

$N^{- 3} > O^{- 2} > N e > F > C a^{+ 2}$

answer is A.

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Detailed Solution

For a given series of isoelectronic species, as atomic number increases, radius decreases, $S e^{+ 2} > B r^{+} > K r > R b^{+} > S r^{+ 2}$

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