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Chapter 24: Problem 7

Discuss Sidgwick's concept of effective atomic number with the help of suitable example.

Short Answer

Expert verified
In the EAN concept, the central metal atom in a compound tends to hold a number of electrons equaling the nearest noble gas. Applying this to the example of [Cu(NH3)4]²⁺, we found that the calculated EAN was close to the atomic number of the nearest noble gas, thereby verifying Sidgwick's concept.

Step by step solution

01

Understanding EAN Concept

Effective Atomic Number (EAN) posits that the stability of a compound can be due to the central metal atom tending to hold a number of electrons that is similar to the nearest inert gas. EAN = Atomic number of the central atom + 2 x (Number of monovalent atoms) - Charge of the ion or the molecule. This is a central theory in understanding coordination compounds.
02

Choosing Suitable Example

A suitable example to discuss the EAN concept is the complex ion of copper in [Cu(NH3)4]²⁺.
03

Computing EAN

The calculation of EAN for [Cu(NH3)4]²⁺ is as follows: The atomic number of copper (Cu) is 29, there are 4 monovalent atoms (NH3), and the charge of the complex ion is +2. By substituting these values into the EAN formula, we obtain: EAN = 29 + 2*4 -2 = 35. This number is very close to the atomic number of the nearest noble gas, Argon (36), verifying Sidgwick's EAN concept.
04

Interpreting the Results

Since the calculated EAN for copper in the given complex ion is close to the atomic number of the nearest noble gas, it verifies Sidgwick's concept of EAN. This similar number of electrons provides stability to the compound, as it behaves in a similar fashion to a noble gas. Hence, this example appropriately demonstrates the concept of EAN, as it illustrates the tendency of the central atom to gain electronic configuration similar to the nearest noble gas.

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Most popular questions from this chapter

Discuss the salient features of valence bond theory. What are all its applications?

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Explain the following: (a) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) is paramagnetic and coloured. (b) \(\left(\mathrm{Ni}(\mathrm{CN})_{4}\right)^{2-}\) is square planar, but \(\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]^{2-}\) is tetrahedral. (c) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}\) is more stable than \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}\), while \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is more stable than \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\) (d) Octahedral complexes are less stable than the square planar complexes. (e) \(\left[\mathrm{Co}(\mathrm{CN})_{6}\right]^{3-}\) is a low-spin complex, but \(\left[\mathrm{CoF}_{6}\right]^{3-}\) is high-spin complex.

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