What is the electron configuration of V3+? [Ar]4s23d3 [Ar]4s23d0 [Ar]4s03d2 [Ar]4s23d5 What are possible correct quantum numbers for two outer most valence electrons in V3+? 3,3,-2, +1/2 AND 3,3,-1, + 1/2 3,2,-2,+ 1/2 AND 3,2,-1, + 1/2 4,1,0 + 1/2 AND 4,1,0, – 1/2 4,0,0+ 1/2 AND 4,0,0, – 1/2 How many unpaired electrons does V3+ have? Three One None Two Is V3+ paramagnetic or diamagnetic? Why? Diamagnetic because it has paired electrons Paramagnetic because it has unpaired electrons Paramagnetic because it has paired electrons Diamagnetic because it has unpaired electrons
The Correct Answer and Explanation is:
To determine the electron configuration of ( \text{V}^{3+} ), we start with the electron configuration of the neutral vanadium (( \text{V} )) atom, which has an atomic number of 23. The ground state electron configuration for vanadium is:
[
\text{V: } [\text{Ar}] 4s^2 3d^3
]
When vanadium loses three electrons to form ( \text{V}^{3+} ), the electrons are removed first from the 4s subshell and then from the 3d subshell. Therefore, the electron configuration for ( \text{V}^{3+} ) becomes:
[
\text{V}^{3+}: [\text{Ar}] 3d^2
]
So, the correct answer is [ \text{Ar} 3d^2 ].
Quantum Numbers for Two Outermost Valence Electrons in ( \text{V}^{3+} )
The two outermost electrons in ( \text{V}^{3+} ) are found in the 3d subshell, where there are two electrons. The possible correct quantum numbers for these electrons can be derived as follows:
- Principal Quantum Number (n): For the 3d subshell, ( n = 3 ).
- Azimuthal Quantum Number (l): For d orbitals, ( l = 2 ).
- Magnetic Quantum Number (m_l): The values of ( m_l ) can range from (-l) to (+l), giving us possible values of -2, -1, 0, +1, and +2. In this case, we can choose any two of these values.
- Spin Quantum Number (m_s): Each electron can have a spin of either ( +\frac{1}{2} ) or ( -\frac{1}{2} ).
From the options provided, the set 3,2,-2,+\frac{1}{2} and 3,2,-1,+\frac{1}{2} accurately reflects the quantum numbers for the two valence electrons in ( \text{V}^{3+} ).
Number of Unpaired Electrons in ( \text{V}^{3+} )
In ( \text{V}^{3+} ), the 3d subshell contains 2 electrons (( 3d^2 )). When filling the 3d orbitals, the two electrons occupy two different orbitals to minimize repulsion (Hund’s rule), resulting in both electrons being unpaired. Therefore, ( \text{V}^{3+} ) has two unpaired electrons.
Paramagnetism vs. Diamagnetism
Since ( \text{V}^{3+} ) has unpaired electrons, it is paramagnetic. Paramagnetic materials are attracted to magnetic fields due to the presence of unpaired electrons, which contribute to a net magnetic moment. In contrast, diamagnetic materials, which have all electrons paired, are weakly repelled by magnetic fields.
In summary:
- Electron Configuration of ( \text{V}^{3+} ): [ \text{Ar} 3d^2 ]
- Quantum Numbers: 3,2,-2,+\frac{1}{2} and 3,2,-1,+\frac{1}{2}
- Unpaired Electrons: Two
- Magnetic Property: Paramagnetic due to the presence of unpaired electrons.