Question

Which is having higher value for the property against each pair:
$\left(I\right)K,Al-$Ionisation enthalpy
$\left(II\right){NH}_3,{NF}_3-$Bond angle
$\left(III\right)H_{2}O,{NH}_3-$Dipole moment
$\left(IV\right)Ca,Sr-$larger radius
$\left(V\right)F,I-$Electronegativity

Answer 1

$\left(i\right)$ Among $K$ and $Al$, $Al$ is having high inoisation enthalpy due to its small size than $K$ and thus having high effective nuclear charge than $K$ due to which removal of an $e^{-}$ requires more energy.

$\left(ii\right)$ In ${NH}_3$, $N$ is more electronegative due to which the shared $e^{-}$ are towards $N$,

$\therefore$ less $lp-bp$ repulsion hence bond angle more than $N{F}_3$ in which due to the more electronegativity of $F$, the shared pair of $e^{-}$ are towards $F$,

$\therefore$ suffers more $lp-bp$ repulsion thereby decreasing bond angle.

$\left(iii\right)$ Dipole moment of $H_{2}O$ will be higher than $N{H}_3$ due to presence of two lone pair on oxygen atom. Also oxygen being moreelectronegative than nitrogen attract the shared $e^{-}$ more, hence has more dipole moment.

$\left(iv\right)$ Among $Ca$ and $Sr$, $Sr$ is having larger radius as $Sr$ is having more number of shell than $Ca$, so $e^{-}$ pull towards nucleus in case of $Sr$ is less as compared to $Ca$. Hence size or atomic radius of $Sr$ is more than $Ca$.

$\left(v\right)$ In halogen family as we move from $F\to I$, atomic radii increases resulting in decrease of effective numclear charge and thus electronegativity decreases from $F\to I$. $F$ being smallest in size have highest electronegativity value than $I$.