QuestionApril 27, 2025

An atomic anion with a charge of -3 has the following electron configuration: [Ar]4s^23d^104p^6 What is the chemical symbol for the ion? How many electrons does the ion have? How many 3d electrons are in the ion? square square square

An atomic anion with a charge of -3 has the following electron configuration: [Ar]4s^23d^104p^6 What is the chemical symbol for the ion? How many electrons does the ion have? How many 3d electrons are in the ion? square square square
An atomic anion with a charge of -3 has the following electron configuration:
[Ar]4s^23d^104p^6
What is the chemical symbol for the ion?
How many electrons does the ion have?
How many 3d electrons are in the ion?
square 
square 
square

Solution
4.4(292 votes)

Answer

Br ### 36 ### 10 Explanation 1. Determine the neutral atom The electron configuration [Ar]4s^{2}3d^{10}4p^{6} corresponds to a neutral atom with 36 electrons, which is Krypton (Kr). 2. Identify the ion Since the anion has a charge of -3, it means 3 extra electrons are added to the neutral atom. Therefore, the original element before gaining electrons is Bromine (Br), which has 35 electrons. 3. Calculate total electrons in the ion The ion has 36 electrons because it gained 3 electrons from its neutral state. 4. Count 3d electrons The number of 3d electrons in the ion is 10, as indicated by the configuration 3d^{10}.

Explanation

1. Determine the neutral atom<br /> The electron configuration $[Ar]4s^{2}3d^{10}4p^{6}$ corresponds to a neutral atom with 36 electrons, which is Krypton (Kr).<br /><br />2. Identify the ion<br /> Since the anion has a charge of $-3$, it means 3 extra electrons are added to the neutral atom. Therefore, the original element before gaining electrons is Bromine (Br), which has 35 electrons.<br /><br />3. Calculate total electrons in the ion<br /> The ion has 36 electrons because it gained 3 electrons from its neutral state.<br /><br />4. Count 3d electrons<br /> The number of 3d electrons in the ion is 10, as indicated by the configuration $3d^{10}$.
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