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+}

 

Answers (1)
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.

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