The order of magnitude of ionic radii of ions | vedclass

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Question

(D) The ions $Na^{+}$,$Mg^{2+}$,$Al^{3+}$,and $Si^{4+}$ are isoelectronic species,meaning they all have the same number of electrons ($10$ electrons).

Vedclass · Accepted Answer

$Na^{+} > Mg^{2+} > Al^{3+} > Si^{4+}$

Vedclass · Answer

(D) The ions $Na^{+}$,$Mg^{2+}$,$Al^{3+}$,and $Si^{4+}$ are isoelectronic species,meaning they all have the same number of electrons ($10$ electrons).For isoelectronic species,the ionic radius decreases as the nuclear charge (atomic number) increases.The atomic numbers are $Na$ $(11)$,$Mg$ $(12)$,$Al$ $(13)$,and $Si$ $(14)$.As the nuclear charge increases,the electrostatic attraction between the nucleus and the electrons increases,pulling the electron cloud closer to the nucleus.Therefore,the order of ionic radii is $Na^{+} > Mg^{2+} > Al^{3+} > Si^{4+}$.

$Column\, A$ (ion)	$Column\, B$ (radius)
$(i)$ $Li^{+}$	$(p)$ $216\, pm$
$(ii)$ $Na^{+}$	$(q)$ $195\, pm$
$(iii)$ $Br^{-}$	$(r)$ $60\, pm$
$(iv)$ $I^{-}$	$(s)$ $95\, pm$

The order of magnitude of ionic radii of ions $Na^{+}$,$Mg^{2+}$,$Al^{3+}$ and $Si^{4+}$ is

Similar Questions

If four different elements $A, B, C$ and $D$ have electronic configurations as $A = [Ne] 3s^2 3p^4$,$B = [Ne] 3s^2 3p^5$,$C = [Ar] 3d^{10} 4s^2 4p^4$ and $D = [Ar] 3d^{10} 4s^2 4p^5$,identify the element with the largest atomic radius.

In case of isoelectronic species,the size of $F^{-}$,$Ne$,and $Na^{+}$ is affected by:

Which of the following sequences is correct for the decreasing order of ionic radius?

Arrange the following elements in the increasing order of their atomic radii:
$(i)$ $O, N, F, B, Be, Li, C$
$(ii)$ $P, S, Mg, Na, Al, Si, Cl$
$(iii)$ $K, Na, Li, Cs, Rb$
$(iv)$ $Cl, F, I, At, Br$

Match $Column\, A$ with $Column\, B$:
$Column\, A$ (ion) $Column\, B$ (radius)
$(i)$ $Li^{+}$ $(p)$ $216\, pm$
$(ii)$ $Na^{+}$ $(q)$ $195\, pm$
$(iii)$ $Br^{-}$ $(r)$ $60\, pm$
$(iv)$ $I^{-}$ $(s)$ $95\, pm$

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