Question

In which of the following options the order of arrangement does not goes with the variation of property indicated against it?

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> AIPMT 2016

• A. $I<Br<Cl<F$ (increasing electron gain enthalpy)
• B. $Li<Na<K<Rb$ (increasing metallic radius)
• C. $A{l}^{3+}<M{g}^{2+}<N{a}^{+}<F^{-}$ (increasing ionic size)
• D. $B<C<O<N$ (increasing first ionisation enthalpy)

Answer 1

The correct option is A $I<Br<Cl<F$ (increasing electron gain enthalpy)

According to the second option the elements are Boron, carbon, nitrogen and oxygen and they all are in the same 2nd row of periodic table. Electronic configuration for all the atoms are:
$B-2s^{2}2p^1$
$C-2s^{2}2p^2$
$N-2s^{2}2p^3$
$O-2s^{2}2p^4$
Since moving left to right ionization enthalpy increases but here there is an exception as p-orbital is half filled in nitrogen which makes it stable and therefore it is difficult to remove electrons from nitrogen atoms. So, nitrogen will have the highest amount of enthalpy. So, the correct order is $B<C<O<N$ Given (a) option is correct
Option (b) :
Electron gain enthalpy of F is less than Cl because in F small size of 2p orbital result in high electron density so inter electronic repulsion is high.
$I<Br<F<Cl$(electron gain enthalpy)
Option b is incorrect.
Option (c):
According to this option the four elements are in the same group and we know that metallic size increases down the group. Therefore, the order given in option is correct and the option is correct.
Option(d):
The elements aluminium, magnesium and sodium are in the same 3rd row and fluorine is in a different row which is the 2nd row of the periodic table. Ionic size of fluorine ion is highest due to negative charge followed by sodium ion, magnesium ion and aluminium ion. Therefore this order in option is correct.