AIEEE 2005 Chemistry Solved Paper

Section: Chemistry

Question : 6 of 74

Marks: +1, -0

In which of the following arrangements the order is NOT according to the property indicated against it?

[AIEEE 2005]

Li<Na<K<Rb Increasing metallic radius
I<Br<F<Cl Increasing electron gain enthalpy (with negative sign)
B<C<N<O Increasing first ionization enthalpy
Al3+<Mg2+<Na+<F−Increasing ionic size

Solution:

(a) It is True.
Here Li,Na,K and Rb all are belongs to the same group, so their effective nuclear charge is same but in a group from top to bottom the no. of shells increase in a atom. So the radius of atom increases.
(b) It is True.
In entire periodic table Cl (chlorine) has the highest, and in a group it decreases from top to bottom.

So the correct order is I<Br<F<Cl.
Note : Among F and Cl, when an electron is added to the F atom, electron comes to the 2p orbital and for Cl atom electron is added in 3p orbital. As 2p orbital is closer to the nucleus than 3p orbital as 3p orbital is larger in size, so when a new electron comes to 2p orbital then it will face a strong repulsion force by the nucleus than if electron comes to 3p orbital.
So, F have lesser tendency of gaining electron than Cl.Note : Among F and Cl, when an electron is added to the F atom, electron comes to the 2p orbital and for Cl atom electron is added in 3p orbital. As 2p orbital is closer to the nucleus than 3p orbital as 3p orbital is larger in size, so when a new electron comes to 2p orbital then it will face a strong repulsion force by the nucleus than if electron comes to 3p orbital.
So, F have lesser tendency of gaining electron than Cl.
(c) It is False.
Electronic configuration of
B(5)=1s22s22p1
C(6)=1s22s22p2
N(7)=1s22s22p3
O(8)=1s22s22p4
In general in periodic table from left to right effective nuclear change increase and it becomes more difficult to remove electron from an atom. That is why ionization enthalpy increases.
But for those atoms which has half filled or full filled outer most shell, those atoms will be more stable and more energy is needed to remove electron from those atoms. Here N(1s22s22p3) has stable half filled 2p subshell so it is more stable than O.
So, correct order is B<C<O<N
(d) It is True.
Al3+,Mg2+,Na+and
F−are isoelectric.
So all have 10 electrons. So,
‌

of Al3+=‌

=1.3
‌

of Mg2+=‌

=1.2
‌

of Na+=‌

=1.1
‌

of F−=‌

=0.9
We know ‌

means z no of protons in nucleus is pulling e no. of electrons present in the orbital.
So for Al3+,13 protons is pulling 10 electron so the size of Al3+ will decrease most. That is why more ‌

means less size of ion.
So, the correct order is
Al3+<Mg2+<Na+<F−

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