Question

an atomic anion with a charge of -1 has the following electron configuration: kr5s²4d¹⁰5p⁶ what is the chemical symbol for the ion? how many electrons does the ion have? how many 5p electrons are in the ion?

Explanation:

Step1: Determine the number of electrons in the ion

• Count electrons from noble - gas core and valence electrons

The noble - gas core \([Kr]\) has \(36\) electrons. The valence - shell configuration \(5s^{2}4d^{10}5p^{6}\) has \(2 + 10+6=18\) electrons. So the total number of electrons in the ion is \(36 + 18=54\) electrons.

Step2: Find the chemical symbol of the neutral atom

• Use the relationship between electrons in ion and neutral atom

Since the ion has a charge of \(- 1\) (\(X^{-}\)), the neutral atom \(X\) has \(N_e(\text{neutral})=N_e(\text{ion})-1\). So the number of electrons in the neutral atom is \(54 - 1 = 53\). The element with \(Z = 53\) (atomic number \(Z\) is equal to the number of electrons in a neutral atom) is iodine (\(I\)). So the ion is \(I^{-}\).

Step3: Determine the number of \(5p\) electrons

• Read from the electron - configuration

From the electron - configuration \([Kr]5s^{2}4d^{10}5p^{6}\), the superscript of \(5p\) is \(6\). So there are \(6\) \(5p\) electrons.

Answer:

• Chemical symbol for the ion: \(I^{-}\)
• Number of electrons in the ion: \(54\)
• Number of \(5p\) electrons in the ion: \(6\)