Copper and calcium both form +2 ions, but copper is far less reactive. Suggest an explanation, taking into account the ground-state electron configurations of these elements and their atomic radii.

Short Answer

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Copper is less reactive than calcium due to its stable and filled 3d orbital, which is resistant to losing electrons. Additionally, copper has a smaller atomic radius, resulting in a stronger attraction between the nucleus and electrons, making it more challenging to remove electrons and thus reducing its reactivity compared to calcium. On the other hand, calcium has a larger atomic radius with weaker attraction between the nucleus and electrons, making it easier to lose electrons and exhibit higher reactivity.

Step by step solution

01

Obtain the Electron Configurations of Copper and Calcium

For a clear comparison between the two elements, we will first write down the electron configurations of copper and calcium. Copper (Cu) has 29 electrons and its electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\]. Calcium (Ca) has 20 electrons and its electron configuration is \[1s^2 2s^2 2p^6 3s^2 3p^6 4s^2\]. Now that we have the electron configurations, we can better understand certain aspects of the reactivity difference between copper and calcium.
02

Discuss the Role of Electron Configuration in Reactivity

Reactivity depends on the ease with which atoms can either lose electrons (to form cations) or gain electrons (to form anions). For copper and calcium, we are focusing on their ability to form +2 ions. Let's examine the difficulties they may face in losing electrons. Copper: - To form a +2 ion, copper has to lose two electrons. One electron can be quite easily removed from the 4s orbital. However, it becomes challenging to remove another electron from the highly stable and filled 3d orbital. This higher stability causes copper to be less inclined to lose electrons and therefore less reactive. Calcium: - Calcium can easily lose its two valence electrons in the 4s orbital to form +2 ions. These electrons are farther from the nucleus and less strongly attracted compared to those in the inner orbitals. So, calcium is more willing to lose electrons and exhibit higher reactivity.
03

Discuss the Influence of Atomic Radii on Reactivity

The atomic radius plays an essential role in the reactivity of elements. Smaller atomic radii indicate stronger attraction between the nucleus and electrons, making it harder to remove electrons, thus lowering reactivity. On the other hand, a larger atomic radius corresponds to weaker attraction and therefore higher reactivity. Copper has an atomic radius of approximately 128 pm, while calcium has an atomic radius of approximately 197 pm. The larger atomic radius of calcium indicates that the electrons in the outermost orbitals are not as strongly attracted to the nucleus as in copper. This weaker attraction increases calcium's tendency to lose electrons and thereby enhances its reactivity. In summary, based on their electron configurations and atomic radii, copper is far less reactive than calcium as copper has a filled and stable 3d orbital, which is resistant to losing electrons. Additionally, copper has a smaller atomic radius, resulting in a stronger attraction between the nucleus and electrons. This stronger attraction makes it more challenging to remove electrons from copper, reducing its reactivity compared to calcium.

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Most popular questions from this chapter

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Na}\) and \(\mathrm{K}\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Na}\) and \(\mathrm{K}\) by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }}\) ? (d) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group? (e) Predict \(Z_{\text {eff }}\) for the outermost electrons in the \(\mathrm{Rb}\) atom based on the calculations for \(\mathrm{Na}\) and \(\mathrm{K}\).

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