Question

When the following five anions are arranged in order of decreasing ionic radius, the correct sequence is?

• A. $S{e}^{2-},I^{-},B{r}^{-},O^{2-},F^{-}$
• B. $I^{-},S{e}^{2-},O^{2-},B{r}^{-},F^{-}$
• C. $S{e}^{2-},I^{-},B{r}^{-},F^{-},O^{2-}$
• D. $I^{-},S{e}^{2-},B{r}^{-},O^{2-},F^{-}$

Answer 1

The correct option is D

$I^{-},S{e}^{2-},B{r}^{-},O^{2-},F^{-}$

Lets calculate number of electrons for each one of them.

$S{e}^{2-}\to 36electrons$ (Valence $n$ = $4$)

$I^{-}\to \to 54electrons$ (Valence $n$ = $5$)

$B{r}^{-}\to 36electrons$ (Valence $n$ = $4$)

$O^{2-}\to 10electrons$ (Valence $n$ = $2$)

$F^{-}\to 10electrons$ (Valence $n$ = $2$)

$I^{-}$ is biggest in size

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

$S{e}^{2-}$ and $B{r}^{-}$ are isoelectronic, but $n$ = $4$ which is greater than $n$ = $2$, so they both will have greater radii than $O^{2-}$ and $F^{-}$

From these two

$S{e}^{2-}>B{r}^{-}$

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

Here, $N{a}^{+}$ has more charge than others, therefore, $N{a}^{+}$ is the smallest in size.Now, for $O^{2-}$ and $F^{‑}$

$O^{2-}>F^{-}$

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 $H^{-}$,

$L{i}^{+}<H^{-}$, Here, $N{a}^{+}$ has more charge than others, therefore, $N{a}^{+}$ is the smallest in size.

Therefore,

$I^{-}>S{e}^{2-}>B{r}^{-}>O^{2-}>F^{-}$

Chemistry is so simple !!!