Arrange the following in the given order
(a) Decreasing ionic size, `Mg^(2+),O^(2-),Na^(oplus),F^(oplus)`
(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

a. `O^(2-) gt F^(ɵ) gt Na^(o+) gt Mg^(2+)`
b. `Na lt Mg gt Al lt Si`
c. `N_(2) lt O_(2) lt F_(2) lt Cl_(2)`
d. `Ca^(2+) lt Ar lt Cl^(ɵ) lt S^(2-)`

Explanation

a. The decreasing order of ionic size is `O^(2-) gt F^(ɵ) gt Na^(o+) gt Mg^(2+)`
All of these are isoelectronic ions.
In isoelectronic species, as the number of protons (atomic number) goes on increasing, size goes on decreasing due to stronger attraction on the electrons.
b. `Na lt Mg lt Al lt Si`
from left to right, IE increases.
But in magnesium, stable configuration is present therefore its ionisation potential is greater than that of aluminium.
`Mg=3s^(2), Al=3s^(2) 3p^(1)`
`IE_(1)` of `Mg gt IE_(1)` of `Al`
Due to penetration effect. It is easier to remove electron from `3p` orbital than `3s` orbital.
c. The increasing order of bond length is
`N_(2) lt O_(2) lt F_(2) lt Cl_(2)`.
In nitrogen, triple bond is present which has the smallest bond length. In oxygen, double bond is present, which has smaller bond length than single bonds. In chlorine, bond length is larger than in fluorine due to large size of the atom.
d. The increasing order of size is `Ca^(2+) lt Ar lt Cl lt S^(2-)`.
All of these are isoelectronic ions.
In isoelectronic species, as the number of protons (atomic number) goes on increasing, size goes on decreasing due to stronger attraction on the electrons.