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+)`

Answer

NA

Explanation

(i) `Mg^(2+), O^(2-), Na^(+)` and `F^(-)` are all isoelectronic, has 10 electrons each. Among isoelectronic species, the order of size is cation lt neutral lt anion.
Also, between cations, higher the charge, smaller the sizer and between anions, greater the negative charge, larger the size. Therefore, the decreasing order of ionic radii: `O^(2-) gt F^(-) Na^(+) gt Mg^(2+)`
(ii) First ionisation energy increases from left to right in a period. However, exception occur betwen group 2 and 13 and group 15 and 16 where trends is reversed on the grounds of stability of completely filled and completely half -filled orbitals. Therefore,
Ionisation energy (Ist ): `Na lt Al lt Mg lt Si`
(iii) If the atoms 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)`