Video Transcript
Vanadium(V) solution has an unusual yellow color that differentiates it from other
transition metal ion solutions. What is the electronic configuration of the V5+ ion? (A) Brackets Ar 4s1, (B) brackets Ar 3d1 4s2, (C) brackets Ar 4s2, (D) brackets Ar
3d1 4s1, (E) brackets Ar.
Transition metals can be found in groups three through 11 on the periodic table. These metals have many fascinating properties, including the formation of multiple
ions, some of which form colored solutions. Vanadium is a period four transition metal that forms an ion with an unusual yellow
color in solution. This ion is the V5+ ion.
A transition element is defined as having atoms with incomplete d subshells or can
form cations with incomplete d subshells. It is because of this incomplete d subshell that these elements have unique
properties. The electronic configuration of a vanadium atom is 1s2 2s2 2p6 3s2 3p6 4s2 3d3. Writing out the electronic configuration for elements in period four and beyond is a
lengthy process, so we can use a form of shorthand called condensed notation.
We can use the noble gas in the prior period on the periodic table, which in this
case is argon, to represent the core electrons of the atom. After writing brackets Ar, then we write the remaining subshells, which contain the
valence electrons. Now, when period four transition metals form cations, they lose electrons from the 4s
subshell first, before losing electrons from the 3d subshell.
In this question, we must determine the electronic configuration of the V5+ ion. A charge of five plus indicates that the vanadium atom lost five electrons. First, we will remove two electrons from the 4s subshell, leaving us with the
electronic configuration brackets Ar 3d3. The 4s subshell is now empty, so we’ll have to remove the next three electrons from
the 3d subshell. Since there are only three electrons in the 3d subshell, it is now empty as well,
which leaves us with an electronic configuration of brackets Ar.
If we look at the answer choices, the only one that matches the electronic
configuration we wrote is answer choice (E). Therefore, the electronic configuration of the V5+ ion is (E), brackets Ar.
