3. circle the ones that are isoelectric with $ca^{2+}$. (2 points)\na. $ar$ b. $kr$ c. $k^{+}$ d. $cl^{-}$…
3. circle the ones that are isoelectric with $ca^{2+}$. (2 points)\na. $ar$ b. $kr$ c. $k^{+}$ d. $cl^{-}$ e. $br^{-}$
Answer
Explanation:
Step1: Calculate the number of electrons in (Ca^{2+})
Calcium ((Ca)) has an atomic number of (20). (Ca^{2+}) has lost (2) electrons. So the number of electrons in (Ca^{2+}) is (20 - 2=18).
Step2: Calculate the number of electrons for each option
- Option a: (Ar) Argon ((Ar)) has an atomic number of (18). So the number of electrons in (Ar) is (18).
- Option b: (Kr) Krypton ((Kr)) has an atomic number of (36). So the number of electrons in (Kr) is (36).
- Option c: (K^{+}) Potassium ((K)) has an atomic number of (19). (K^{+}) has lost (1) electron. So the number of electrons in (K^{+}) is (19 - 1 = 18).
- Option d: (Cl^{-}) Chlorine ((Cl)) has an atomic number of (17). (Cl^{-}) has gained (1) electron. So the number of electrons in (Cl^{-}) is (17+ 1=18).
- Option e: (Br^{-}) Bromine ((Br)) has an atomic number of (35). (Br^{-}) has gained (1) electron. So the number of electrons in (Br^{-}) is (35 + 1=36).
Answer:
a. (Ar), c. (K^{+}), d. (Cl^{-})