The electronic configuration $$1{s^2},2{s^2}2{p^5},3{s^1}$$ shows lowest ionisation energy because this configuration is unstable due to the presence of one electron in $$s$$-orbital. Hence, less energy is required to remove the electron.
222.
Which of the following properties of isotopes of an element is different?
$$Cu = 3{d^{10}}4{s^1},Ag = 4{d^{10}}5{s^1},Au = 4{f^{14}}5{d^{10}}6{s^1}$$ In all the above given cases, unpaired $$s$$ - electron has to be removed. In the case of $$Cu,$$ a $$4s$$ electron is to be removed which is closer to the nucleus than the $$5s$$ electron of $$Ag .$$
So, $$I{E_1}$$ of $$Cu > I{E_1}$$ of $$Ag.$$
However, in case of $$Au,$$ due to imperfect screening effect of $$14{e^ - }s$$ of $$4f$$ orbitals, the nuclear charge increases and therefore $$5s\,{e^ - }$$ of $$Au$$ is more tightly held.
Thus, the order of $$I{E_1}$$ is $$Cu > Ag < Au.$$
224.
The element with the highest first ionization potential is
Electron gain enthalpy becomes more negative across a period while it becomes less negative in a group. However, electron gain enthalpy of $$O$$ or $$F$$ is less than that of the succeeding element. When electron is added to $$n =2,$$ the repulsion is more than when it is added to $$n=3$$ in case of $$Cl$$ or $$S.$$
228.
The screening effect of inner electrons of the nucleus causes
A
decrease in the ionization energy
B
increase in the ionization energy
C
no effect on the ionization energy
D
increases the attraction of the nucleus for the electrons
No explanation is given for this question. Let's discuss the answer together.
230.
The first $$\left( {{\Delta _i}{H_1}} \right)$$ and second $$\left( {{\Delta _i}{H_2}} \right)$$ ionization enthalpies $$\left( {{\text{in}}\,kJ\,mo{l^{ - 1}}} \right)$$ and the electron gain enthalpy $$\left( {{\Delta _{eg}}H} \right)\left( {{\text{in}}\,kJ\,mo{l^{ - 1}}} \right)$$ of the elements (i), (ii), (iii),(iv) and (v) are given below.
$$\eqalign{
& {\text{Element}}\,\,\,\,{\Delta _i}{H_1}\,\,\,{\Delta _i}{H_2}\,\,\,\,\,\,{\Delta _{eg}}H \cr
& \left( {\text{i}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,520\,\,\,\,\,\,\,\,\,7300\,\,\,\,\,\, - 60 \cr
& \left( {{\text{ii}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,419\,\,\,\,\,\,\,\,\,3051\,\,\,\,\,\, - 48 \cr
& \left( {{\text{iii}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1681\,\,\,\,\,\,\,3374\,\,\,\,\, - 328 \cr
& \left( {{\text{iv}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1008\,\,\,\,\,\,\,1846\,\,\,\,\, - 295 \cr
& \left( {\text{v}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2372\,\,\,\,\,\,\,5251\,\,\,\,\,\, + 48 \cr} $$
The most reactive metal and the least reactive non-metal of these are respectively
(i) represents $$Li,$$ (ii) represents $$K$$
(iii) represents $$Br,$$ (iv) represents $$I$$
(v) represents $$He$$
So, amongst these, (ii) represents most reactive metal and (v) represents least reactive non-metal.