Periodic Table & Periodicity

Those elements are present in periodic table like this - .tg .tg Group 2 Group 16 Group 18 Period 3 Ca ..........

S .........

Ar Period 4 Se Period 5 Ba In periodic table, 1.

From left to right ionization energy increases.

2.

From top to bottom ionization energy decreases.

$\therefore$ Ba < Ca and Se < S and S < Ar $\therefore$ Correct order is - Ba < Ca < Se < S < Ar

Here all of them are isoelectric and for isoelectric species size of anion increases as negative charge increases.

So, correct order is : N3– > O2– > F– $\therefore$ Radius of N3– = 1.71 Å and Radius of O2– = 1.40 Å and Radius of F– = 1.36 Å

Na, H, Ba and Sr produce Peroxide. K, Rb and Cs produces Superoxide. So, SiO2 will be oxide.

From the table you can see ionization enthalpy of A is greater than B in all the cases.

So, B comes below A in the group as in a group from top to bottom ionization enthalpy decreases.

After 1st ionization enthalpy each element become cation by removing a electron.

In A+ and B+, after removing one electron from each element, effective nuclear charge increases as per-electron attraction increase by the neuclers.

So, the removal of next electron will be more difficult, therefore more energy is required for 2nd ionization energy.

From the table you can see 2nd ionization energy is more than first ionization energy for both elements.

But you can see 3rd ionization enthalpy is so much higher than 2nd ionization enthalpy.

It means after 2nd ionization enthalpy outermost shell is empty and in 3rd ionization enthalpy from a new shell electron is removed, as the new shell is closer to the neucleus so the attraction by the nucleus to the electrons of this shell is more and to remove a electron from this shell you have to provide very high energy, that is why 3rd ionization enthalpy is so high.

If A and B are from group 1 then after 1st ionization enthalpy outermost shell will be empty and 2nd electron will be removed from inner shell, so 2nd ionization will be so high.

But from table you can see 2nd ionization energy is around double of 1st ionization energy, it is not so much high than 1st onization energy.

So A and B can't be from group 1.

A and B are from group 2 as in group 2 element, outermost shell has 2 electron and after 2nd ionization enthalpy outermost shell will be empty and in 3rd ionization enethalpy, third electron will removed from innershell so 3rd ionization enthalpy will be very high.

First ionization enthalpies = 496 kJ/mole Second ionization enthalpies = 4560 kJ/mol According to the given information, the difference between first and second ionization enthalpy is very high so Metal belong to 1st group i.e.

Monovalent cation.

Metal hydroxide will be of type, MOH.

MOH + HCl $\to$ MCl + H2O MOH +

H2SO4 $\to$

M2SO4 + H2O So one mole of HCl required to react with one mole MOH. And

mole of H2SO4 required to react with one mole MOH.

Order of electron gain enthalpy (magnitude) is Cl > F > Br > I Note: Electron gain enthalpy increases with electro negativity but chlorine has higher electron gain enthalpy than fluorine (exception).

Electron affinity of second period p-block element is less than third period p-block element due to small size of second period p-block element.

$\therefore$ Electron affinity order : F

Cl Down the group electron affinity decreases due to size increases. $\therefore$ Electron affinity E.A. order : S

Se Li

Na

Electronic configuration of 25Mn = [Ar]3d54s2 25Mn2+ = [Ar]3d54s0 26Fe = [Ar]3d64s2 26Fe2+ = [Ar]3d64s0 27Co = [Ar]3d74s2 27Co2+ = [Ar]3d74s0 28Ni = [Ar]3d84s2 28Ni2+ = [Ar]3d84s0 So third ionisation energy is minimum for Fe.

Non metal oxide $\to$ acidic. Metal oxide $\to$ basic. Al2O3 amphoteric.

On moving left to right in a period of the periodic table, Ionisation energy (I.E.) increases due to increase in effective nuclear charge i.e.

Zeff.

But due to extra stability of fully filled and half-filled electronic configuration of Mg and P required ionisation enthalpy is more from neighbouring elements. i.e.

First ionisation enthalpy order is Al