Of the following ion with smallest ionic radius is

K^{+}

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C a^{2 +}

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A l^{3 +}

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T i^{4 +}

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Solution

The correct option is C $A l^{3 +}$
Aluminum belongs to third period while $K$ , $C a$ and $T i$ belongs to fourth period. so $A l^{3 +}$ will be having smallest ionic radii.

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Similar questions

Which of the following sets contain only isoelectronic ions?
(i) $Z n^{2 +}, C a^{2 +}, G a^{3 +}, A l^{3 +}$

(ii) $K^{+}, C a^{2 +}, S c^{3 +}, C l^{-}$

(iii) $P^{3 -}, S^{2 -}, C l^{-}, K^{+}$

(iv) $T i^{4 +}, A r, C r^{3 +}, v^{5 +}$

Which of the following ion has smallest ionic radius

A l^{3 +}, N a^{+}, C a^{2 +}

N a^{+}

1.02 \circ A

i n \circ A

M g^{2 +}

A l^{3 +}

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