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In order to determine which ions have the ground-state electron configurations mentioned, let's first identify the corresponding elements. The abbreviations 鈥� Ar and Kr 鈥� in the given electron configurations refer to the noble gases Argon (Ar) and Krypton (Kr), respectively. Let's note the respective atomic numbers of these noble gases: Argon: \( Z = 18 \) (18 electrons) Krypton: \( Z = 36 \) (36 electrons) Now, starting from these noble gases, let's determine the elements associated with each configuration.

(a) \([\mathrm{Ar}]\): This configuration is the same as that of Argon, which has 18 electrons. The isoelectronic ion of Argon with the same configuration will have 18 electrons. An ion can have 18 electrons if it gains_3_ electrons (if it's a negative ion) or loses some electrons (if it's a positive ion). Let's determine an ion that can fit this description. Calcium (Ca) has an atomic number of \( Z = 20 \). When Ca loses 2 electrons, it forms the ion Ca\(^{2+}\), which has the same electron configuration as Argon: \( Ca^{2+}: [\mathrm{Ar}] \) (b) \([\mathrm{Ar}] 3 d^{5}\): This configuration has 5 more electrons than Argon which has 18 electrons. Therefore, it is associated with the element having 23 electrons. Vanadium (V) has an atomic number of \( Z = 23 \) (23 electrons). When V loses 2 electrons, it forms the ion V\(^{2+}\) with the following electron configuration: \( V^{2+}: [\mathrm{Ar}] 3d^5\) (c) \([\mathrm{Kr}] 5 s^{2} 4 d^{10}\): This configuration has 12 more electrons than Krypton, which has 36 electrons. Therefore, it is associated with the element having 48 electrons. Cadmium (Cd) has an atomic number of \( Z = 48 \) (48 electrons) and has the electron configuration of \( [\mathrm{Kr}] 5 s^{2} 4 d^{10}\). When Cd loses 2 electrons, it forms the ion Cd\(^{2+}\), which has the given electron configuration: \( Cd^{2+}: [\mathrm{Kr}] 5s^2 4d^10 \) Thus, the ions that have the given ground-state electron configurations are: (a) Ca\(^{2+}\) (b) V\(^{2+}\) (c) Cd\(^{2+}\)