(A) : Na⁺ and Al⁺³ are isoelectronic but the magnitude of ionic radius of Al is less than that of Na.
(R) : Effective nuclear charge of Al is greater than that in Na.

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(A) is true but (R) is false

Both (A) and (R) are true but (R) is not the correct explanation of (A)

Both (A) and (R) are true and (R) is the correct explanation of (A)

Both (A) and (R) are false

answer is A.

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Detailed Solution

In case of isoelectronic species, effective nuclear charge increases with increase in atomic number Z (with increase in protons). This results in decrease in size and hence, ionic radius of Al³⁺< Na⁺.

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