
What is the electron configuration of \[{{V}^{2+}}\] ?
A. $\left[ Ar \right]4{{s}^{2}}3{{d}^{3}}.$
B. $\left[ Ar \right]4{{s}^{1}}3{{d}^{4}}.$
C. $\left[ Ar \right]3{{d}^{3}}.$
D. $\left[ Ar \right]4{{s}^{2}}3{{d}^{1}}.$
Answer
490.5k+ views
Hint: We know that the electronic configuration is defined as the distribution of electrons into the orbitals of an atom. Every neutral atom consists of a fixed number of electrons which is equal to the number of protons and is called the atomic number.
Complete answer:
As we know that the electrons and protons, an atom also has neutrons which may or may not be equal to the number of the protons. In an atom, the protons and the neutrons lie in the nucleus and have an almost negligible role in governing any chemical reaction. However, the electrons are the ones which lie outside the nucleus of the atom and their precise distribution inside an atom plays a very important role in governing the chemical reactions that the atom is involved in.
Here, we can see that the atomic number of vanadium is \[23\] and it is present in the d-block of elements. This implies that the last electron to be filled in the orbitals of vanadium will be in the d orbital. First, we will see what the configuration of vanadium is, up to argon, or \[18\] electrons. The configuration of argon $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$ which means out of the \[23\] electrons, \[18\] electrons have already been placed in orbitals in a stable manner. Only five electrons are left to be placed now. The electrons are removed from the outermost shell. Configuration of \[V\] is \[\left[ Ar \right]4{{s}^{2}}3{{d}^{3}}.\] Hence electrons are removed from \[4s.\] Thus configuration ${{V}^{2+}}=\left[ Ar \right]3{{d}^{3}}.$
Therefore, the correct answer is option (C).
Note: Remember that the electronic configuration of an atom defines the precise distribution of electrons in an atom. The atomic number of the elements vanadium and add the electrons to the shells according to the Aufbau principle. Take note of the block in which it is present.
Complete answer:
As we know that the electrons and protons, an atom also has neutrons which may or may not be equal to the number of the protons. In an atom, the protons and the neutrons lie in the nucleus and have an almost negligible role in governing any chemical reaction. However, the electrons are the ones which lie outside the nucleus of the atom and their precise distribution inside an atom plays a very important role in governing the chemical reactions that the atom is involved in.
Here, we can see that the atomic number of vanadium is \[23\] and it is present in the d-block of elements. This implies that the last electron to be filled in the orbitals of vanadium will be in the d orbital. First, we will see what the configuration of vanadium is, up to argon, or \[18\] electrons. The configuration of argon $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$ which means out of the \[23\] electrons, \[18\] electrons have already been placed in orbitals in a stable manner. Only five electrons are left to be placed now. The electrons are removed from the outermost shell. Configuration of \[V\] is \[\left[ Ar \right]4{{s}^{2}}3{{d}^{3}}.\] Hence electrons are removed from \[4s.\] Thus configuration ${{V}^{2+}}=\left[ Ar \right]3{{d}^{3}}.$
Therefore, the correct answer is option (C).
Note: Remember that the electronic configuration of an atom defines the precise distribution of electrons in an atom. The atomic number of the elements vanadium and add the electrons to the shells according to the Aufbau principle. Take note of the block in which it is present.
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