
How many half-filled orbitals are in a bromine atom?
Answer
554.1k+ views
Hint: To find the number of half-filled orbitals in an atom, we need to know its electronic configuration and how we will fill electrons according to the three rules, namely, Aufbau principle, Pauli exclusion principle and Hund’s multiplicity rule.
Complete step by step answer:
We know that bromine is the member of the halogen family with atomic number 35. It is present as a red-brown fuming liquid at room temperature.
Now, to answer the question, we need to know the electron filling or the electronic configuration of the bromine atom obeying the Aufbau principle, Pauli exclusion principle and Hund’s multiplicity rule. Thus, the electron configuration of bromine is:
\[\left[ {Ar} \right]4{s^2}3{d^{10}}4{p^5}\]
Here, Ar is argon i.e. the nearest noble gas.
By looking at the electronic configuration of bromine, it is very much clear that 4s orbital has two electrons and thus is completely filled. Similarly, 3d orbitals have 10 electrons and thus, all the five d orbitals are completely filled each with two electrons. But in the case of p orbitals, we only have five electrons, and to completely fill all the three p orbitals we need six electrons. Thus, there should be some orbital which is left half-filled with only one electron.
Now, we know that there are three types of p orbitals i.e. \[{p_x},{{ }}{p_y},{{ }}{p_z}\] . Let us consider filling these orbitals according to the Hund’s multiplicity rule then, the \[{p_z}\] orbital will get only one electron and thus it is half-filled.
Hence, there are only one half-filled p orbitals in a bromine atom.
Note: when the value of $n$ is greater than one, the value of l lies up to \[\left( {n - 1} \right)\] and when the value of \[l = 1\], it is called the p orbital. It is dumbbell in shape. There are three types of p orbital i.e. \[{p_x},{{ }}{p_{y,}}{p_z}\], with each pointing in a different direction.
Complete step by step answer:
We know that bromine is the member of the halogen family with atomic number 35. It is present as a red-brown fuming liquid at room temperature.
Now, to answer the question, we need to know the electron filling or the electronic configuration of the bromine atom obeying the Aufbau principle, Pauli exclusion principle and Hund’s multiplicity rule. Thus, the electron configuration of bromine is:
\[\left[ {Ar} \right]4{s^2}3{d^{10}}4{p^5}\]
Here, Ar is argon i.e. the nearest noble gas.
By looking at the electronic configuration of bromine, it is very much clear that 4s orbital has two electrons and thus is completely filled. Similarly, 3d orbitals have 10 electrons and thus, all the five d orbitals are completely filled each with two electrons. But in the case of p orbitals, we only have five electrons, and to completely fill all the three p orbitals we need six electrons. Thus, there should be some orbital which is left half-filled with only one electron.
Now, we know that there are three types of p orbitals i.e. \[{p_x},{{ }}{p_y},{{ }}{p_z}\] . Let us consider filling these orbitals according to the Hund’s multiplicity rule then, the \[{p_z}\] orbital will get only one electron and thus it is half-filled.
Hence, there are only one half-filled p orbitals in a bromine atom.
Note: when the value of $n$ is greater than one, the value of l lies up to \[\left( {n - 1} \right)\] and when the value of \[l = 1\], it is called the p orbital. It is dumbbell in shape. There are three types of p orbital i.e. \[{p_x},{{ }}{p_{y,}}{p_z}\], with each pointing in a different direction.
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