Step 1:Recall that for a neutral element,Atomic number= # of protons = # of electrons.

You are watching: How many unpaired electrons are in the vanadium atom

The atomic number of V is 23 which means V has23 electrons.

The location of V in the regular table is presented below:


Based on the figure, V belongs in thed-block in period 4so its electron configuration will contain the4s and also 3d subshells.

Recall that:

s–subshell have the right to hold a best of 2 electrons

p–subshell have the right to hold a preferably of 6 electrons

d–subshell have the right to hold a maximum of 10 electrons

f–subshell can hold a maximum of 14 electrons

Starting indigenous 1s, the electron construction for V is:

V: 1s22s22p63s23p64s23d3

Step 2:In V2O3, the fee of V is +3.

The +3 fee means3 e– to be lost. Therefore, we must subtract 3 e– starting from thehighest power level (4s).

Theelectron configuration for V3+is:

V3+: 1s22s22p63s23p63d2

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Problem Details

How countless unpaired electrons would you intend on Vanadium in V2O3

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