In: Chemistry
Which ion in each pair, forms more labile complexes?
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The lability of an ionic complex would indicate the ease with which the complex breaks down when dissolved in a suitable solvent or treated with other reactants.
1. Li and Be occurs in the same period, Be occurring to the right of Li. As we move from left to right along a period, the atomic size decreases. Li+ contains a single positive charge while Be2+ contains dipositive charge. The higher charger and smaller atomic radius of Be means that Be2+ has a higher charge density than Li+. Consequently, Be2+ can bind to anion more strongly and hence, form stable complexes. Li+ will therefore, form the more labile complex.
2. Be and Sr occurs in the same group, Sr occurring much below Be. As we move from top to bottom down a group, the atomic size increases. Both the ions are dipositive; however the larger atomic radius of Sr means that Sr2+ has a lower charge density than Be2+. Consequently, Be2+ can bind to anion more strongly and hence, form stable complexes. Sr2+ will therefore, form the more labile complex.
3. Here, the ionic radius of Co2+ and Co3+ is more or less similar; however, Co3+ has a higher charge density than Co2+ (due to loss of more electrons) and can therefore bind to anions strongly to form a stable complex. Co2+ will form a labile complex.
4. Ga occurs to the right of Ga in the periodic table and as such will have a smaller atomic radius. Moreover, Ga3+ has a higher charge density (due to more charge and smaller size) and will form the more stable complex. Zn2+ will form the labile complex.
5. Ir occurs below Co in the same group; therefore, Ir3+ will have a lower charge density as compared to Co3+ and will form a labile complex.
6. Fe has a larger atomic radius than Co (Fe occurs to the left of Co); therefore, Fe3+ will have a lower charge density and form a labile complex.