The Ti2 + ion is isoelectronic with the Ca atom. (b) Calculate the number of unpaired electrons for Ca and for Ti²⁺.

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1. First, determine the electron configuration of a neutral calcium (Ca) atom. Calcium has an atomic number of 20, which means it has 20 electrons. The electron configuration is: \[ \text{Ca: } 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 \].
2. Next, identify the electron configuration of the \( \text{Ti}^{2+} \) ion. Titanium has an atomic number of 22, so a neutral titanium atom has 22 electrons. The electron configuration for neutral titanium is: \[ \text{Ti: } 1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2 \].
3. Since \( \text{Ti}^{2+} \) has lost 2 electrons, remove these electrons from the outermost shell. The electron configuration for \( \text{Ti}^{2+} \) becomes: \[ \text{Ti}^{2+}: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 \].
4. Now, determine the number of unpaired electrons in the electron configuration of \( \text{Ca} \). The 4s orbital in calcium is fully paired, so there are no unpaired electrons in calcium.
5. Finally, determine the number of unpaired electrons in \( \text{Ti}^{2+} \). The 3d orbital in \( \text{Ti}^{2+} \) has 2 electrons. Since the 3d orbital can hold up to 10 electrons, these 2 electrons are unpaired.