Question

In which of the following options the order of arrangement does not goes with the
variation of property indicated against it?
A. $A{l^{3 + }} < M{g^{2 + }} < N{a^ + } < {F^ - }$ (increasing ionic size)
B. $B < C < N < O$ (increasing first ionisation enthalpy)
C. $I < Br < Cl < F$ (increasing electron gain enthalpy)
D. $Li < Na < K < Rb$ (increasing metallic radius)

Answer 1

Hint: Here in the options which will have wrong order will be our answer. Although properties are the same for a row or column of periodic table, there are always some exceptions. This is a multiple choice question.

Step by step answer: According to the first option we need to check the ionic size of all the elements. The elements used are aluminium, magnesium, sodium and fluorine. 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.
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:
$B - 2{s^2}2{p^1}$
$C - 2{s^2}2{p^2}$
$N - 2{s^2}2{p^3}$
$O - 2{s^2}2{p^4}$
We know that 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 $B < C < O < N$ . Therefore, this arrangement given in option is incorrect.
According to the third option the order is $I < Br < Cl < F$ in increasing electron gain enthalpy. We know that electron gain enthalpy increases down the group as size of the atom increases. But there is an exception in case of fluorine. Fluorine has 2 shells and therefore there is repulsion between the valence electron and the added electron and because of this its electron affinity value is less than chlorine. Therefore the correct order is $I < Br < F < Cl$ .
According to the fourth 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.
As option A and D are correct and B and C are incorrect.

Therefore the answer is option B and C.

Note: Most properties in the periodic table are the same for a row or down the group but there are always some exceptions because elements behave differently for stability. So one must be careful while finding properties along a period or group.