Period 4 transition element that forms 2+ ion with a half‐filled d sub level is
Manganese (Mn)
What is the half-filled d sub-level?
Transition metals are an interesting and challenging group of elements. They have perplexing patterns of electron distribution that don’t always follow the electron-filling rules. Predicting how they will form ions is also not always obvious.
Transition metals belong to the d block, meaning that the d sublevel of electrons is in the process of being filled with up to ten electrons. Many transition metals cannot lose enough electrons to attain a noble-gas electron configuration. In addition, the majority of transition metals are capable of adopting ions with different charges. Iron, which forms either the Fe2+ or Fe3+ ions, loses electrons as shown below.
Some transition metals that have relatively few d electrons may attain a noble-gas electron configuration. Scandium is an example. Others may attain configurations with a full d sublevel, such as zinc and copper.
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Answer:
Li atoms readily give up one electron to form positively charged, Li+ ions. These ions have the same stable electron configuration as the noble gas helium. All Group 1 atoms can lose one electron to form positively charged ions.
Explanation:
Answer: 3
Explanation: I took the k12 quiz
Answer: Yes we agree with the student's claim.
Explanation:
When the molecules are present in smaller size, more reactants can react as decreasing the size increases the surface area of the reactants which will enhance the contact of molecules.Hence, more products will form leading to increased rate of reaction.
On increasing the temperature will make more reactant molecules will have sufficient energies to cross the energy barrier and thus the number of effective collisions increases, thus leading to more products and increased rate of reaction.
When the solution is stirred , the molecule's kinetic energy and thus the rate of reaction increases.
Thus smaller size, stirring and increase of temperature will make the solution quickly.
Answer:
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
Explanation:
Let's consider the unbalanced equation that occurs when phosphoric acid reacts with barium hydroxide to form water and barium phosphate. This is a neutralization reaction.
H₃PO₄(aq) + Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
We will balance it using the trial and error method.
First, we will balance Ba atoms by multiplying Ba(OH)₂ by 3 and P atoms by multiplying H₃PO₄ by 2.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + H₂O(l)
Finally, we will get the balanced equation by multiplying H₂O by 6.
2 H₃PO₄(aq) + 3 Ba(OH)₂(aq) ⇒ Ba₃(PO₄)₂(s) + 6 H₂O(l)
