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Chapter 19: Problem 4

Give reasons for (a) Compounds of \(\mathrm{Ti}(+\| \mathrm{II})\) are strong reductants (b) \(\mathrm{CuSO}_{4}\) dissolves in ammonia.

Short Answer

Expert verified
(a) Compounds of Titanium (II) have two d-electrons that can be easily lost to form a more stable state, Titanium (IV), thereby reducing other species and acting as a strong reductant. (b) \(\mathrm{CuSo}_4\) dissolves in ammonia since the \(\mathrm{CuSo}_4\) forms a complex ion with ammonia, making it soluble in the solution.

Step by step solution

01

Reasoning for Titanium(II) as a strong reductant

Titanium shows different oxidation states and Ti(+2) is one of these states. The configuration of Ti(+2) is \([Ar]\, 4s^0 3d^2\) that shows two 3d electrons. The reactivity of Ti(II) compounds can be due to the availability of these two d-electrons. It can easily lose the two d-electrons to form a more stable state Ti(IV) with a configuration of \([Ar]\, 4s^0 3d^0\). In this process, Ti(II) gets oxidized (loses electrons) and hence reduces other species, thus, behaving as a strong reductant.
02

Reasoning for solubility of \(\mathrm{CuSo}_{4}\) in ammonia

\(\mathrm{CuSo}_4\) is weakly acidic, and ammonia is a base. It will therefore react with \(\mathrm{CuSo}_4\) and form complex ions. So, in ammonia solution \(\mathrm{CuSo}_4\) forms a dark blue complex ion [\(Cu(NH_3)_4\) \(H_2O_n\)] \(^{2+}\). This is the primary reason for the solubility of \(\mathrm{CuSo}_4\) in aqueous ammonia.

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