QJEEM05Ch734f: JEE Main 2005 Chemistry - Class 11 JEEM05Ch734f Solution: In which of the following arrangements the order is NOT…

$\left( a \right)\,\,\,\,\,$ 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.

$\left( b \right)\,\,\,\,\,$ 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 {\rm 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.

$\left( c \right)\,\,\,\,\,$ It is False.

Electronic configuration of

$B\left( 5 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^1}$

$C\left( 6 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^2}$

$N\left( 7 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^3}$

$O\left( 8 \right)\,\,\, = \,\,\,1{s^2}2{s^2}2{p^4}$

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 $N\left( {1{s^2}\,2{s^2}\,2{p^3}} \right)$ has stable half filled $2p$ subshell so it is more stable than $O.$

So, correct order is B < C < O < N

$\left( d \right)\,\,\,\,\,$ It is True.

$A{l^{3 + }},M{g^{2 + }},N{a^ + }\,\,$ and

$\,\,{F^ - }$ are isoelectric.

So all have $10$ electrons. So,

$\,\,{Z \over e}\,\,$ of $A{l^{3 + }} = {{13} \over {10}} = 1.3$

$\,\,{Z \over e}\,\,$ of $M{g^{2 + }} = {{12} \over {10}} = 1.2$

$\,\,{Z \over e}\,\,$ of $N{a^ + } = {{11} \over {10}} = 1.1$

$\,\,{Z \over e}\,\,$ of ${F^ - } = {9 \over {10}} = 0.9$

We know ${z \over e}$ means $z$ no of protons in nucleus is pulling $e$ no. of electrons present in the orbital.

So for $A{l^{3 + }},\,13$ protons is pulling $10$ electron so the size of $A{l^{3 + }}$ will decrease most. That is why more ${z \over e}$ means less size of ion.

So, the correct order is

A{l^{3 + }} < M{g^{2 + }} < N{a^ + } < {F^ - }