Arrange the following in the given order
(a) Decreasing ionic size, `Mg^(2+),O^(2-),Na^(oplus),F^(ɵ)`
(b )Increasing first ionisaiton energy :`Mg,Al,Si,Na`
(c ) Increasing bond length `F_(2),N_(2),Cl_(2),O_(2)`
(d) The order of their increasing size:
`Cl^(ɵ),S^(2-),Ca^(2+),Al^(3+)`

Explanation

(i)` O^(2-) gt F^(-) gt Na^(+) gt Mg^(2+) (ii) Na < Al < Mg < Si , (iii) N_2 lt O_2 lt F_2 lt Cl_2`
(i) `Mg^(2+) , O^(2-) , Na^(+) " and " F^(-)` are all isoelectronic, has 10 electrons each. Among isoelectronic species, the order of size is cation < neutral < anion. Also, between cations, higher the charge, smaller the size and between anions, greater the negative charge, larger the size. Therefore, the decreasing order of ionic radii : `O^(2-) gt F^(-) gt Na^(+) gt Mg^(2+)`
(ii) First ionisation energy increases from left to right in a period. However, exception occur between group 2 and 13 and group 15 and 16 where trend is reversed on the grounds of stability of completely filled and completely half-filled orbitals. Therefore, Ionisation energy (1st): Na < Al < Mg < Si
(iii) If the atom are from same period, bond length is inversely proportional to bond order. In a group, bond length is related directly to atomic radius. Therefore, bond length `N_2 lt O_2 lt F_2 lt Cl_2`