Step 1:Recall the 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 way V has23 electrons.

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

Based ~ above the figure, V belong in thed-block in duration 4so the electron configuration will contain the4s and 3d subshells.

Recall that:

s–subshell deserve to hold a preferably of 2 electrons

p–subshell deserve to hold a maximum of 6 electrons

d–subshell can hold a best of 10 electrons

f–subshell have the right to hold a maximum of 14 electrons

Starting from 1s, the electron configuration 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 have to subtract 3 e– starting from thehighest power level (4s).

Theelectron configuration for V3+is:

V3+: 1s22s22p63s23p63d2

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Problem Details

How numerous unpaired electrons would you suppose on Vanadium in V2O3

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