Explain the following question (based on size of atoms or ions and other periodic properties):
a. Arrange the following species in decreasing order of their sizes//ionic radii.
i. `Ar, K^(o+),Cl^(ɵ),S^(2-)` and `Ca^(2+)`
ii. `Al^(3+), Mg^(2+),Na^(o+),Cl^(ɵ), N^(3-)` and `O^(2-)`
b. What are isoelectronic species? Name the species which are isoelectronic with each of the following atom or ions.
i. `Rb^(o+)` ii. `F^(ɵ)`, iii. `Mg^(2+)` iv. `Ar`
c. Arrange the following species`//`atoms in decreasing order of reducing character.
i. `Na, Mg` and `Al` ii. `Mg, Ca` and `Sr`
iii. `Na, K` and `Rb` iv. `F^(ɵ), Cl^(ɵ),Br^(ɵ)` and `I^(ɵ)`
d. The decreasing order of reactivety of group `1` elements is `Cs gt Rb gt K gt Na gt Li` whereas that of group `17` elements is `(Cl gt F gt Br gt I)`. Explain.
e. Predict the formula of the stable binary compounds that would be formed by the combination of the following pairs of elements:
i. `Mg` and `N`
ii. `Si` and `O`
iii. Elements with `Z=71` and `F`
iv. `P` and `F`
v. `Al` and `I`
vi. `Li` and `O`
f. Ansewer the following by the use of periodic table.
i. Identify the element that would tend to gain two electrons.
ii. Identify the group having metal, non-metal, liquid as well as gas at the room temperature,
iii. Identify the element with five electrons in the outer shell.
iv. Identify the element that would tend to lose two elctrons.

Explanation

All of them are isoelectronic species, with `18` electronic.
`:.` Size of more negative ion `gt` size of inert gas `gt` size of less positive `gt` size of more positive ion.
`:. S^(2-) gt CI^(ɵ) gt Ar gt K^(o+) gt Ca^(2+)`
ii. `N^(3-) gtO^(2-) gt F^(ɵ) gt Na^(o+) gt Mg^(2+) gt Al^(3+)`
(All of them are isoelectronic species with `10` electrons)
b. Isoelectronic species have same number of electrons.
i. `Rb^(o+)(Z=37, "No.of " e^(-)'s=37-1=36)` is isoelectronic with `Sr^(2+)(Z=38, "No.of" e^(-)'s38-2=36)`
ii. `F^(ɵ)(Z=9, "No.of" e^(-)'s=9+1=10)` is isoelectronic with `Ne(Z=10)`
iii. `Mg^(2+)(Z=12, "No. of"e^(-)'s=12-2=10)` is isoelectronic with `Na^(o+)( Z=11, "of" e^(-)'s11-1=10)` and with `Ne(Z=10)`.
iv. `Ar(Z=18)`, is isoelectronic with `Ca^(2+)(Z=20,"No.of" e^(-)'s=20-2=18)`.
c. Elements with low `IE` and large size give have more tendency to undergo oxidation and thus acts as stronger reducing agent.
i. `Na ("group" 1) gt Mg ("group" 2) gt Al ("group" 13 "elements")`. `(IE` increase along the period `(rarr)`. All of them belong to the `3 rd` period].
ii. `Sr gt Ca gt Mg` [All of them belong to group `2` and `IE` decreases down the group `(darr)`].
iii. `Rb gt K gt Na`[Group `1` elements. `IE` decreasaes down group `(darr)`.]
iv. `I^(ɵ) gt Br^(ɵ) gt Cl^(ɵ) gt F^(ɵ)` [Group `17`, size increases and `IE` decreases down the group `(darr)`]
(d) The reactivity of group `1` element is the tendency to lose electrons easily, i.e., to undergo oxidation easily. Low `IE` and large size of group `1` elements, so more reactive is the element. Hence reactivity order of group `1` elements is
`Cs gt Rb gt K gt Na gt Li` [`IE` decreases down the group `(darr)`]
The reactivity of group `17` elements is the tendency to accept electrons easily, i.e, to undergo reduction easily. High is the negative `EA` or `Delta_("eg")H^(ɵ)` of group `17` elements so more reactive is that element (or magnitude of `EA` or `Delta_("eg")H^(ɵ)` decreases down the group except in case of `F` and `Cl)`.
Therefore, reactivity order of group `17` elements is
`Cl gt F gt Br gt I (Delta_("eg")H^(ɵ) = -349 gt -333 gt -325 gt -296 kJ mol^(-1))`
(e) i. `Mg^(2+)` and `N^(3-)` give `Mg_(3)N_(2)` (Magnesium nitride)
ii. `Si^(4+)` and `O^(2-)` give `Si_(2)O_(4) rArr SiO_(2)` (Silicon dioxide)
iii. Elements with `Z = 71` is lutetium `(Lu)` and general oxidation states `(OS)` of lanthanides and actinides is `+3`.
So `Lu^(3+)` and `F^(ɵ)` give `Lu F_(3)` (Lutetium trithoride)
iv. `P^(5+)` and `F^(ɵ)` give `PF_(5)` (Phosphorous pentafluoride)
v. `Al^(3+)` and `I^(ɵ)` give `AII_(3)` (Aluminium triiodide)
vi. `Li^(o+)` and `O^(2-)` give `Li_(2)O` (Lithium monoxide)
(f) i. Only oxygen atom in group `16` has `OS` of `-2` (except in case of peroxides and `OF_(2)`).
The other element of group `16` (e.g., `S, Se, Te, Po`), besides `-2 OS`, they also show `+2, +4` and `+6 OS`.
So, oxygen tends to gain two electrons to attain noble gas stable configuration.
ii. Group `17` elements Metal is `At` (Astatine), non-metal is `Cl` (Chlorine), liquid is `Br_(2)` (Bromine) and gas is `F_(2)` and `Cl_(2)`.
iii. Group `15` elements. The valence electronic configuration of `N` is `2s^(2) 2p^(3)`.
iv. Group `2` elements, e.g., `Mg rarr Mg^(2+)(g) + 2e^(-)`