$$\eqalign{
& Z = 12,1{s^2}2{s^2}2{p^6}3{s^2}\left( {n = 3} \right) \cr
& Z = 17,1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}\left( {n = 3} \right) \cr} $$
172.
The first, second and third ionization enthalpies of an element are 737, 1450 and 7731 $$kJ\,mo{l^{ - 1}}$$ respectively. What will be the formulae of its oxide and chloride?
The element belongs to group 2, and so it forms $$MO$$ and $$MC{l_2}.$$
173.
If the ionization enthalpy and electron gain enthalpy of an element are $$275$$ and $$86\,kcal\,mo{l^{ - 1}}$$ respectively, then the electronegativity of the element on the Pauling scale is :
On moving down in a group atomic radii increases due to successive addition of extra shell hence $$O < S < Se$$
Further $$As$$ is in group 15 having one less electron in its $$p$$ orbital hence have higher atomic radii than group 16 elements.
i.e., $$O < S < Se < As$$
Proton have very small size, so have large hydration energy. The degree of hydration depends upon the size of the cation. Smaller the size of a cation greater the hydration energy.
177.
Which property decreases from left to right across the periodic table and increases from top to bottom ?
(i) Atomic radius
(ii) Electronegativity
(iii) Ionisation energy
(iv) Metallic character
For the same shell, screening effect decreases in the order : $$s > p > d > f.$$
180.
Consider the following changes :
$$\eqalign{
& \left( {\text{i}} \right)M\left( s \right) \to M\left( g \right) \cr
& \left( {{\text{ii}}} \right)M\left( s \right) \to {M^{2 + }}\left( g \right) + 2{e^ - } \cr
& \left( {{\text{iii}}} \right)M\left( g \right) \to {M^ + }\left( g \right) + {e^ - } \cr
& \left( {{\text{iv}}} \right){M^ + }\left( g \right) \to {M^{2 + }}\left( g \right) + {e^ - } \cr
& \left( {\text{v}} \right)M\left( g \right) \to {M^{2 + }}\left( g \right) + 2{e^ - } \cr} $$
The second ionization energy of $$M$$ could be calculated from the energy values associated with :