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Explain the following: a. Which of the...

Explain the following:
a. Which of the following `Na, Mg, Si` and `P` would have the greatest difference between the `IE_(1)` and `IE_(2)`? Explain?
b. The `EN's` of `B, Al` and `Ga` are `2.0, 1.5` and `1.6` respectively. The trends is not regular.Explain?
c. `Li_(2)CO_(3)` decomposes on heating but other alkali metal barbonate ( e.g. `Na_(2)CO_(3)`) does not. Explain?
d. Expalin why `Cu^(o+)` is found only is solid state and not solutions.
e. `Be` of `N` have extremely low value of `EA` (i.e. less negative value)against the trend. Explain?
f. Arrange the following in decreasing order of their properties indicated:
`Na^(o+)Mg^(2+)` and `Al^(3+)`
i. Ionic mobillity in `H_(2)O`
ii. Size of ions
iii. Standard reduction protential `(E_(M^(x+(aq.)))^(ɵ)//M_(s))`
iv. Extent of hydration
v. Hydration energy
vi. Size of hydrated ions
g. In iodomentry. why `KI` is not added dropwise to an acidified solution of `KMnO_(4)` but reverse is done?
h. Why the decrease in size between `Li` and `Be` is much greater than that between `Na` and `Mg` or `K` and `Ca`?
i. `K_(2)CO_(3)` is less soluble than `Cs_(2)CO_(3)` or `Rb_(2)CO_(3)` while among group `2` elements `MgCO_(3)` is more soluble than `BaCO_(3)`.
j. Expalin the decreasing order of solubility of sulphate of group `2` elements.
k. `'EA'` of `Cl` is the highest among the halogens, yet `F` is the strongest oxidising agent'. Why?

Text Solution

Verified by Experts

a. Elements having stable noble gas configuration after removal of one electron will have the maximum difference between `IE_(1)` and `IE_(2)` so the elements is `Na`.
Electronic configuration of `Na, Na^(o+)` and `Na^(2+)` ion
`Na=1s^(2),2s^(2)2p^(6),3s^(1)underset("from" 3s)overset(-e^(-))rarrunderset(Na^(2+)(1s^(2),2s^(2)2p^(5)))underset(darr_(-e^(-)"from"2p))(Na^(o+)(1s^(2),2s^(2)2p^(6)))`
`A` jump in `IE` is noticed only when the valence shell changes during the successive removal of electrons.
b. `B, Al` and `Ga` belong to group `13`, and down the group `(darr), EN` decreases. But after `Al`, dur to the imperfect (minimum) shielding effect of `d` electrons, the nuclear charge increase and hence `EN` also increase.
`[["Group" 13,EN],[B,2.0],[Al,1.5],[Ga,1.6],[In,1.7],[TI,1.8]]`
c. Comparable sizes of the ions from the strongest lattice, so crystal lattice of `Li_(2)CO_(3)` is not so strong due to small size of `Li^(o+)` and larger size of `CO_(3)^(2-)`. So `Li_(2)CO_(3)` decomposes on heating. But other alkali metal carbonate (e.g `Na_(2)CO_(3)`), have strong crystal lattice, due to comparable size of cation (e.g `Na^(o+)` ion and `CO_(3)^(2-)` ion). So they are stable to heat.
`{:(Li_(2)CO_(3)overset(Delta)rarrLi_(2)O+CO_(2),,,,),(MgCO_(3)overset(Delta)rarrMgO+CO_(2),,,,):}]{:("Due to diagonal"),("relationship of"),(Li and Mg):}`
`Na_(2)CO_(3),K_(2)CO_(3)etc. overset(Delta)rarr"No action"`
Note: The crystal lattice of `Li_(2)O` is stronger due to comparable sizes of `li^(o+)` and `O^(2-)` ions.
d. In solution `Cu^(o+)` disproportionate to `Cu^(2+)` and `Cu`, but in solid state it does not. So, `Cu^(o+)` ion is found onlu in solid state.
`2Cu_((aq.))^(o+)rarrCu_((aq.))^(2+)+Cu Delta_(r )G^(ɵ)=-ve`
e. Valence electronic configuration of `Be` and `N`:
`Be=2s^(2)2p^(0), Nk=2s^(2)sp^(3)`
In, `Be`, the incoming electron has to be added in `2p` orbitals because `2s` orbital is completely filled an in `N`, it is to be added to half-filled `2p` orbitals, Since half-filled and full-filled orbitals are more stable, therfore incoming electrons will be added with difficulty. Hence `Be` and `N` have low value of `EA` (i.e., less negative value)
f. i. Hydration energy or extent of hydration increases along the period `(rarr)`. Greater the extent of hydrations, lesser is the ionic mobility in aqueous solution:
`:. Na^(o+) gt Mg^(2+) gt Al^(3+)`
ii. The greater the positive charge, the lesser is the size,
`:. Na^(o+) gt Mg ^(2+) gt Al^(3+)`
iii. The greater the positive charge, the higher is the reduction potential value.
`:. E_(Al^(3+(aq.))//A'(S))^(ɵ) gt E_(Mg^(2+)(aq.)//Mg_((S)))^(ɵ) gt E_((Na^(o+)(aq.)//Na_((s))`
iv. Extent of hydration increase along the period due to increase in charge density
`(("Charge")/("Size")"ratio")`
`:. Al^(3+) gt Mg^(2+) gt Na^(o+)`
v. Same explanation as in part (iv) above The greater the extent of hydration the same (negative) energry is released
`:. Al^(3+) gt Mg^(2+) gt Na^(o+)`
vi. The greater the extent of hydration, the larger is the size of that ion (reverse of the size of ions) .
g. Iodometry: The estimation of oxidising substance involving the liberation of `I_(2)` and subsequent volumetric estimation of `I_(2)` is called iodometry, e.g.,
`{:(2Cu^(2+)+4I^(ɵ)rarrCuI_(2)+I_(2),,,,),(2CuI_(2)rarrCu_(2)I_(2)+I_(2),,,,):}`
If `KI` is added dropwise to acidic `KMnO_(4)` solution, then `MnO_(4)^(ɵ)` will oxidise `I^(ɵ)` to `IO_(3)^(ɵ)` instead of `I_(2)`.
Note. Iodometry is the estimation of reducing substance by use of standard `I_(2)`, eg
h. From `Li(2s^(1))` to `Be( 2s^(2))`, the additional electron is added to the `2s` orbital or ( `L` shell) which is quite closer to nucleus whose charge is also increased by `1` unit. So, greater force is experienced by the electron and size decreses.
From `Na(3s^(1))` to `Mg(3s^(-2))` and `K(4s^(1))` to `Ca(4s^(2))`, the electron are added in `3s` and `4s` orbitals ( or `M` and `N` shell) respectively, which are away from nucleus. So, lesser force of attraction is experienced by the electrons and decrease in size is less than that of decrese in size between `Li` and `Be`. Moreover, screening effect also contributes in the small decreases in the size of `Na` and `K` and `K` and `Ca`.
i. `K_(2)CO_(3)` is less soluble than `Cs_(2)CO_(3)` or `Rb_(2)CO_(3)` because `IE` factor out weighs hydration energy factor ( `K` has high `IE` and high hydration energy and `Cs` and `Rb` have low `IE` and how hydration energy).
But it is reverse for group `2` element carbonates.
j.N/A
k. `EA` of `Cl` is more negative `(-349 KJ mol^(-1))` than that of `F(-333 KJmol^(-1))`,. But the reduction potential value or oxidising ability of `F_(2)`
`( i.e., E_(1)/(2)F_(2)(g)F^(ɵ_((aq)).)=2.87V)`
is greater than of `Cl_(2)`
`(i.e. E_((1)/(2)Cl_(2)(g)//Cl^(ɵ)(aq.))=1.36V)`
as shown by Boron-Haber cycle.
(i). `(1)/(2)F_(2_((g))) overset((1)/(2)Delta_(dissH^(ɵ))=(160)/(2) = 80 kJ mol^(-1))rarrF_((g))underset(darrEA = -33kJ mol^(-1))+e^(-)`
`F_((aq))^(ɵ) overset(Delta_(hyd)H^(ɵ)-515 kJ mol^(-1))larr = F_((g))^(ɵ) +aq`
`:. Delta_(r )H^(ɵ)(80-333-515) = -768 kJ mol^(-1)`.
Note: `Delta_(diss)H^(ɵ) of Cl_(2_((g))) gt Br_(2_((g))) gt F_(2_((g))) gt I_(2_((g)))`
(ii) `(1)/(2)Cl_(2_((g))) overset((1)/(2)Delta_(dissH^(ɵ))=(242)/(2) = 121 kJ mol^(-1))rarrCl_((g))underset(darrEA = -349kJ mol^(-1))+e^(-)`
`Cl_((aq))^(ɵ) overset(Delta_(hyd)H^(ɵ) = -422 kJ mol^(-1))larr Cl_((g))^(ɵ) +aq`
`:. Delta_(r )H^(ɵ)(121-349-422)=-650 KJ mol^(-1)`
`F` has low `Delta_("diss")H^(ɵ)`, high (negative) `Delta_("hyd")H^(ɵ)` and less negative `EA` than that of `Cl`. Thus overall position is that `F` has the largest negtive `Delta m H^(ɵ)` value ( i.e.`-768 KJ mol^(-1))` than that of `Cl` (i.e., `-650 KJmol^(-1))`.
So, reaction `((1)/(2)Fe_(2(g)) "to" F_((aq.))^(ɵ))` is more feasible
than that of `((1)/(2) Cl_(2(g)) "to" Cl_((aq.))^(ɵ))`. Therefore ,the more is the negative `Delta_(r )H^(ɵ)` value of a reaction the more positive will be its reduction potential value. Hence, `F` is the strongest oxidising among halogens.
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