Write the symbol and electron configuration for each ion and name the noble gas with the same configuration.
a. nitride _________________________
b. oxide ___________________________
c. sulfide _________________________
d. bromide _________________________
This is just the reverse of your other problem.
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To find the symbol and electron configuration for each ion, we need to determine the number of electrons gained or lost by the element to form the ion.
a. Nitride ion (N3-):
Nitrogen (N) has an atomic number of 7, which means it has 7 electrons in its neutral state.
To become a nitride ion with a charge of 3-, nitrogen gains three electrons to fill its valence shell and attain stability. Thus, the nitride ion has 10 electrons.
The electron configuration for a nitride ion is [He] 2s^2 2p^6, where [He] represents the noble gas configuration of helium (He).
Therefore, the symbol for the nitride ion is N3-, and the noble gas with the same electron configuration is helium (He).
b. Oxide ion (O2-):
Oxygen (O) has an atomic number of 8, meaning it has 8 electrons in its neutral state.
To form an oxide ion with a charge of 2-, oxygen gains two electrons to complete its valence shell and achieve stability. Therefore, the oxide ion has 10 electrons.
The electron configuration for an oxide ion is [He] 2s^2 2p^6, where [He] represents the noble gas configuration of helium (He).
Hence, the symbol for the oxide ion is O2-, and the noble gas with the same electron configuration is helium (He).
c. Sulfide ion (S2-):
Sulfur (S) has an atomic number of 16, indicating that it has 16 electrons in its neutral state.
To become a sulfide ion with a charge of 2-, sulfur gains two electrons to complete its valence shell and achieve stability. Consequently, the sulfide ion has 18 electrons.
The electron configuration for a sulfide ion is [Ne] 3s^2 3p^6, where [Ne] represents the noble gas configuration of neon (Ne).
Thus, the symbol for the sulfide ion is S2-, and the noble gas with the same electron configuration is neon (Ne).
d. Bromide ion (Br-):
Bromine (Br) has an atomic number of 35, which implies it has 35 electrons in its neutral state.
To form a bromide ion with a charge of 1-, bromine gains one electron to fill its valence shell and achieve stability. So, the bromide ion has 36 electrons.
The electron configuration for a bromide ion is [Kr] 4d^10 5s^2 5p^6, where [Kr] represents the noble gas configuration of krypton (Kr).
In conclusion, the symbol for the bromide ion is Br-, and the noble gas with the same electron configuration is krypton (Kr).