Question

Which of the following order of radii is correct?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Lets write the electronic configurations of each species.

$Li\to T{s}^{2}2s^{1}Be\to 1s^{2}2s^2$

$Mg\to 1s^{2}2s^{2}2p^{6}3s^2$

$H^{+}\to 1s^{0}L{i}^{+}\to 1s^2{H}^{-}\to 1s^2$

$O\to 1s^{2}2s^{2}2p^{4}F\to 1s^{2}2s^{2}2p^{5}Ne\to 1s^{2}2s^{2}2p^6$

$N{a}^{+}\to 1s^{2}2s^{2}2p^6{F}^{-}\to 1s^{2}2s^{2}2p^6{O}^{-2}\to 1s^{2}2s^{2}2p^6$

Now, we know that

Along a periodfrom Left to Right, Atomic Radius decreases because effective nuclear charge increases, on the valance electrons resulting in the increased attraction of electrons to the nucleus.

Down the group fromTop to Bottom, Atomic Radius increases because the principal quantum number (n) increases and the valance electrons are farther from the nucleus.

Now, try to remember from the videos, for isoelectronic species, the more cationic charge on the atom, lesser will be then ionic radius.

More the ionic charge on the atom, more will be the ionic radius.

Now, In option (a), the correct order will be $Mg>Li>Be$

Because Mg has principal quantum number ($n$) = $3$ and $Be$ gas $n$ = $2$

Among $Li$ and $Be$, $Li>Be$ because along a period from Left to Right, atomic radius increased

Now, In option (b)

$H^{+}<L{i}^{+}<H^{-}$

For $H^{+}$, there are no electrons, so no atomic radius. In case of $L{i}^{+}$ and $H^{-}$, $L{i}^{+}$ $<$ $H^{-}$,

In option (c), the correct order will be

$Ne>O>F$

Because for noble gases ($Ne$), we consider Van der Waal's radius which is larger than covalent radius.

Between $O$ and $F$, $O>F$ because Along a periodfrom Left to Right, Atomic Radius decreases because effective nuclear charge increases, on the valance electrons resulting in the increased attraction of electrons to the nucleus.

In option (d), the correct order will be

$O^{-2}>F^{-}>N{a}^{+}$

Because, Now, try to remember from the videos, for isoelectronic species, the more cationic charge on the atom, lesser will be then ionic radius.

More the ionic charge on the atom, more will be the ionic radius

. Now, In option (a), the correct order will be $Mg$ $>$ $Li$ $>$ $Be$

Because Mg has principal quantum number ($n$) = $3$ and Be gas $n$ = $2$

Among $Li$ and $Be$, $Li$ $>$ $Be$ because along a period from Left to Right, atomic radius increased

Now, In option (b)

$H^{+}<L{i}^{+}<H^{-}$