(a) Following are the transition metal ions of 3d series:
`Ti^(4+),V^(2+),Mn^(3+),Cr^(3+)`
(Atomic numbers: `Ti=22,V=23,Mn=25,Cr=24`)
Answer the following:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidising agent and why?
(iii) Which ion is colourless and why?
(b) Complete the following equations:
(i) `2MnO_(4)^(-)+16H^(+)+5S^(-)to`
(ii) `KMnO_(4)overset(heat)to`

Explanation

(i) Their electronic configuration are `:`
`Ti^(4+)= 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6)`
`V^(2+) =1s^(1) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)`
`Cr^(3+) = 1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(3)`
`Mn^(3+) =1s^(2) 2s^(2) 2p^(6) 3s^(2) 3p^(6) 3d^(4)`
Thus , `Ti ^(4+)` is the most stable because it has noble gas configuration.
(ii) Oxidizing agent is the substance which itself is reduce most easily, i.e., can gain electron easily. `V^(2+)` and `Cr^(3+)` are also stable because they have filled `t_(2g)` level (i.e., `t_(2g)^(3))` discussed in unit 9. Thus, `Mn^(3+)` can gain electron easily. Moreover `Mn^(2+)` is more stable than `Mn^(3+)` . Hence, `Mn^(3+)` is the strongest oxidizing agent.
(iii) Ions are coloured if they have incompletely filled d-orbitals. Those with fully - filled or empty d-orbitals are coloureless . As` Ti^(4+)` has empty d-orbitals, hence it is colourless.