Procedure tain seven 250 mL beakers and seven stirring rods. Label the beakers 1

Question

Procedure tain seven 250 mL beakers and seven stirring rods. Label the beakers 1-7. Add 75 distilled water to each beaker. 2) Add 1.2 g of NiCl2-6H20 to each beaker and stir until dissolved. Note the color in your laboratory notebook 3) To beakers 1, 4, and 5, add 10 mL of 5 M ammonia. Stir and record any color changes as [Ni(NH3)6]2+ is formed. Cover the beakers with watchglasses. 4) To beakers 2, 4, and 6, add 5 mL of 25% en solution. Note any color changes. [Ni(en)I2+ should form in beaker 2. L + 26112 ram (ligt green 5) Add 5 mL of 25% dien solution to beakers 3, 5, and 7. Note any color changes. [Ni(dien) should form in beaker 3 6) Add 10 mL of 5 M ammonia to beakers 6 and 7. Note any color changes. Questions 1) Why are the reactions of [Ni(en)s]2 and [Ni(dien)2]2 preferred? Based on the equilibrium constants for reactions 2, 4 and 6, what would the expected products be? Conclusion Are the results of the reactions consistent with those predicted by the chelate effect? Explain part by indicating the product and color of each reaction

Explanation / Answer

When we have a complex with ploydentate ligand we will see different values of equllibrium constants even though the donor atom is same.

It is shown that (Ni(en)3)2+ is 10 times more stable than (Ni(NH3)6)3+ although the donor atom is same, this is due to chelate effect.

Chealate effect is defined as the enhanced stability caused by the chelating ligand(poly dentate ligands).

Since ethelene diamine and and diethylenetriamine ligands are bidentate and tridentate ligands respectively, hence the formation of these complexes will be preffered over ammonia and water.

In reaction 2,4,6 the expected product would be (Ni(en)3)2+ complex due to chelate effect.

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