- A${F^ - }$
- B$B^{3+}$
- ✓$O^{2-}$
- D$Li^+$
No. of electrons in $B ^{3+}=8$
No. of electrons in $Li ^{+}=4$
No. of electrons in $F ^{-}=10$
$F ^{-}$and $O ^{2-}$ have highest no. of electrons, thus the effective nuclear charge
$\{\left.\frac{\text { nuclear charge }}{\text { No. of electrons }}\right\}$ for them is less than $B$ and $Li$.
When effective nuclear charge increases, size of an ion decreases.
Amongst $F ^{-}$and $O ^{2-}$, size of $O ^{2-}$ is more than $F ^{-}$as size decreases on moving from left to right in a period.
Therefore, $O ^{2-}$ is largest in size.
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$4 HNO _{3}(l)+3 KCl ( s ) \rightarrow Cl _{2}( g )+ NOCl ( g )+ 2 H _{2} O ( g )+3 KNO _{3}( s )$
The amount of $HNO _{3}$ required to produce $110.0 \;g$ of $KNO _{3}$ is $...... \;g$
(Given : Atomic masses of $H , O , N$ and $K$ are $1 , 16,14$ and $39$ respectively.)
$\begin{array}{*{20}{c}}
{C{H_2} - CH - C{H_2}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,} \\
{OH\,\,\,\,\,\,\,\,OH\,\,\,\,\,\,OH}
\end{array}$
is :