# The correct sequence which shows decreasing order of the ionic radii of the element is$O^{2-}\; \;$    $F^{-}$  $Na^{+}$  $Mg^{2+}$  $Al^{3+}$  $>$ Option 1) $O^{2-}\; \;$ $>$ $F^{-}$ $>$ $Na^{+}$  $>$ $Mg^{2+}$ $>$ $Al^{3+}$ Option 2) $Al^{3+}$ $>$ $Mg^{2+}$ $>$ $Na^{+}$  $>$ $F^{-}$ $>$  $O^{2-}\; \;$ Option 3) $Na^{+}$ $>$ $Mg^{2+}$ $>$ $Al^{3+}$  $>$ $O^{2-}\; \;$ $>$ $F^{-}$ Option 4) $Na^{+}$ $>$ $F^{-}$$>$ $Mg^{2+}$ $>$  $O^{2-}\; \;$ $>$$Al^{3+}$

V Vakul

As we learnt in

Size of isoelectronic species -

Smaller the value of  z/e, larger the size of that species.Smaller z means effective nuclear charge is small hence size is large.

-

For isoelectronic species, as effective nuclear charge increases, ionic radii decreases. Nuclear charge is maximum of the specie with maximum protons .

Order of nuclear charge:

$A1^{3+}>Mg^{2+}>Na^{+}>F^{-}>O^{2-}$

Protons:     $3\:\:\:\:\:\:\:\:\:\:\:\:12\:\:\:\:\:\:\:\:\:\:\:\:11\:\:\:\:\:\:\:\:\:\:\:\:9\:\:\:\:\:\:\:\:\:\:\:\:8$

Electrons:  $10\:\:\:\:\:\:\:\:\:\:10\:\:\:\:\:\:\:\:\:\:10\:\:\:\:\:\:\:\:\:\:10\:\:\:\:\:\:\:\:\:\:10$

Thus increasing order of ionic radii

$O^{2-}>F^{-1}>Na^{+}>Mg^{2+}>A1^{3+}$

Option 1)

$O^{2-}\; \;$ $>$ $F^{-}$ $>$ $Na^{+}$  $>$ $Mg^{2+}$ $>$ $Al^{3+}$

This option is correct.

Option 2)

$Al^{3+}$ $>$ $Mg^{2+}$ $>$ $Na^{+}$  $>$ $F^{-}$ $>$  $O^{2-}\; \;$

This option is incorrect.

Option 3)

$Na^{+}$ $>$ $Mg^{2+}$ $>$ $Al^{3+}$  $>$ $O^{2-}\; \;$ $>$ $F^{-}$

This option is incorrect.

Option 4)

$Na^{+}$ $>$ $F^{-}$$>$ $Mg^{2+}$ $>$  $O^{2-}\; \;$ $>$$Al^{3+}$

This option is incorrect.

Exams
Articles
Questions