C. Ca and Br because they're metals and nonmetals
Answer:
B) K⁺, Sr²⁺ , O²⁻
Explanation:
Potassium is present in group one. It is alkali metal and have one valance electron.Potassium need to lose its one valance electron and form cation to get complete octet.
That's why it shows K⁺.
Sr is alkaline earth metal. It is present in group two. It has two valance electrons. Strontium needed to lose its two valance electrons and get stable electronic configuration.
When it loses its two valance electrons it shows cation with charge of +2.
Sr²⁺
Oxygen is present in group 16. It has sex valance electrons. It needed two more electrons to complete the octet. That's why oxygen gain two electron and form anion with a charge of -2.
O²⁻
Magnetic moment (spin only) of octahedral complex having CFSE=−0.8Δo and surrounded by weak field ligands can be : Q
To answer this, the Crystal Field Stabilization Energy has to be calculated for a (d3 metal in both configurations. The geometry with the greater stabilization will be the preferred geometry. So for tetrahedral d3, the Crystal Field Stabilization Energy is: CFSE = -0.8 x 4/9 Δo = -0.355 Δo.
[Co(CN)64-] is also an octahedral d7 complex but it contains CN-, a strong field ligand. Its orbital occupancy is (t2g)6(eg)1 and it therefore has one unpaired electron. In this case the CFSE is −(6)(25)ΔO+(1)(35)ΔO+P=−95ΔO+P.
The crystal field stabilization energy (CFSE) (in kJ/mol) for complex, [Ti(H2O)6]3+. According to CFT, the first absorption maximum is obtained at 20,3000cm−1 for the transition.
To learn more about crystal field stabilization energy visit:brainly.com/question/29389010
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Answer:
See explanation
Explanation:
Qualitative analysis in chemistry is a method used to determine the ions present in a solution chiefly by means of chemical reactions.
In this case, I suspect the presence of silver ions and/or barium ions. The first step is to add dilute HCl. This will lead to the precipitation of the silver ion as AgCl. If a white precipitate is formed upon addition of HCl then Ag^+ is present in the solution.
Secondly, I add a carbonate such as NH4CO3(aq). This will cause the barium ions to become precipitated as barium carbonate. Hence, the formation of a white precipitate when NH4CO3(aq) is added to the solution indicates the presence of barium ion in the solution.
