The correct order of atomic radii of the elem | vedclass

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Question

(D) The atomic radius decreases on moving from left to right in a period because the effective nuclear charge increases while the number of shells rem

Vedclass · Accepted Answer

$O < N < S < P$

Vedclass · Answer

(D) The atomic radius decreases on moving from left to right in a period because the effective nuclear charge increases while the number of shells remains the same. Thus,the size of $O On moving down a group,the atomic radius increases due to the addition of a new shell. Thus,the size of $O Combining these trends,we compare the elements: $O$ and $N$ are in the $2^{nd}$ period,while $S$ and $P$ are in the $3^{rd}$ period.Since $O < N$ and $S < P$,and elements in the $2^{nd}$ period are smaller than those in the $3^{rd}$ period,the overall order is $O < N < S < P$.

$Column\, A$ (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 correct order of atomic radii of the elements $O$,$N$,$S$ and $P$ is

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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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