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Why is the vanadium (3+) ion paramagnetic? I know that the electron configuration of vanadium is [Ar]4s23d3 [A r] 4 s 2 3 d 3. None of the electrons in the 3d subshell are paired. Once it loses these three electrons, shouldn't the remainder of the electrons be paired? How can VX3+ V X 3 + be paramagnetic if it loses all its unpaired electrons?
4. Electrons always fill in the lowest energy configuration possible. Cr and Cu, as well as Cu and Ag, are exceptions in the "typical" filling order. In the case of Cr and Cu, they are stabilized by having 2 half filled orbitals, which maximizes exchange energy and minimizes electron repulsion. In their case, the energy to promote an s electron ...
The electron configurations for vanadium is: $\ce{1s^2 2s^2 2p^2 3s^2 3p^6 3d^3 4s^2}$, and the electron configuration for chromium is: $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1}$. In looking at the electron configurations of two consecutive elements, I noticed that an electron in the $\ce{4s}$ orbital "jumped" to the higher energy $\ce{3d}$ orbital.
The overall energy for every possible electronic configuration must be determined separately for every combination of the number of electrons and number of protons in an atom, and whichever has the lowest energy is the ground state, irrelevant what the result is for the element or ion next to it. $\endgroup$
The correct electron configuration for vanadium is [Ar] 3d3 4s2. This configuration shows that vanadium has 23 electrons, with two in the 4s orbital and three in the 3d orbital orbital. 1s2 2s2 ...
In this site when they explain about Vanadium electronic configuration it has been mentioned that Vanadium in its unipositive ion state have 4 electrons in $\mathrm{3d}$ rather than 3 electrons in $\mathrm{3d}$ and 1 electron in $\mathrm{4s}$ which supports the idea that electrons is transferred from $\mathrm{4s}$ to $\mathrm{3d}$.
30. The melting and boiling points of transition elements increase from scandium (1530 ∘ C) to vanadium (1917 ∘ C). They increase because as we go across the group, we have more unpaired (free) electrons. But at chromium (1890 ∘ C) however, the melting point decreases even though it has more unpaired electrons than the previous atoms.
Closed 8 years ago. As far as my knowledge goes, transition metal complex ions show a color due to a transfer of electrons between the levels of split d d -orbitals. In the case of VX5+ V X 5 + and CrX6+ C r X 6 + however, there are no electrons in the d d -orbitals, and the electronic configuration of both is similar to Ar A r, yet VX5+ V X 5 ...
Unfortunately, the Aufbau rule cannot predict all electron configuration as it doesn't take into account electron-electron interactions. In the end the Aufbau is only a rule of thumb. Electronic levels have to be found using quantum calculations taking into account electron-electron interactions (not to mention spin orbit coupling).
When Vanadium is ionised it loses the 4s electron first, meaning that it's 3+ ion has a different electron configuration to Calcium despite it being isoelectronic. Can it be explained in terms of radial distribution functions? Maybe the maxima of the 3d and 4s become further apart as the effective nuclear charge increases?