Q.

(a) Account for the following:
(i) Transition metals show variable oxidation states.
(ii) Zn, Cd and Hg are soft metals.
(iii) E° value for the Mn3+/Mn2+ couple is highly positive (+ 1·57 V) as compared to Cr3+/Cr2+ . 56/1 12
(b) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.

OR

(a) Following are the transition metal ions of 3d series:
Ti4+, V2+, Mn3+, Cr3+ (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) 2MnO4+16H++5S2
(ii) KMnO4 heat 

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Detailed Solution

(a) (i) ns-electrons, (n-1) d electron has similar energy. They both can be removed easily. Therefore, transition elements show variable oxidation state.
(ii) These elements have completely filled d orbital due to which they form weak metal-metal bonds. Therefore, Zn, Cd, and Hg are soft metals.
(iii) When Mn³+ is converted to Mn2+ it attains the stable half-filled d5 configuration. Which results in large E0 value. While Cr3+ (d3) when converted into Cr 2+(d2) it goes to less stable configuration.
(b) The atomic radius of both lanthanoids and actinoids decreases with increase in atomic number.
All the Lanthanoids are non-radioactive except promethium while all the actinoids are radioactive.

OR

(a)
(i) V2+ and Cr3+ are most stable as they both have the [Ar]3d3 or 𝑡32g stable electronic configuration.
(ii) Mn3+ ions are the strongest oxidising agent in the aqueous solution because after converting to Mn2+ they get stable half-filled d5 electronic configuration.
(iii) Ti4+ has no electron in d orbital. No d-d transition is possible in this case. Therefore, it is colourless.
(b)
(i) 2MnO4+16H++5S22Mn2++8H2O+5S
(ii) 2KMnO4 heat K2MnO4+MnO2+O2

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