Question

Between ${\text{T}}{{\text{i}}^{2 + }} $and ${{\text{V}}^{2 + }} $which ion contains more number of unpaired electrons?

Answer 1

Hint: An unpaired electron is an electron that occupies a single orbital of an atom rather than as part of an electron pair in chemistry. Each of an atom's atomic orbitals (defined by the three quantum numbers n, l, and m) may hold two electrons (electron pair) with opposite spins. Because the creation of electron pairs, whether in the form of a chemical bond or as a lone pair, is typically energetically advantageous, unpaired electrons are extremely uncommon in chemistry, because an entity carrying an unpaired electron is generally quite reactive.

Complete answer:
There are four main types of orbitals, each with a distinctive form, indicated by the letters s, p, d, and f. The nucleus lies at the centre of an s-orbital, a p-orbital is dumbbell-shaped, and four of the five d orbitals are cloverleaf-shaped. The fifth d orbital is in the shape of an extended dumbbell with a doughnut in the centre. An atom's orbitals are structured into multiple levels, or electron shells.
Each d subshell can only carry ten electrons. Each f subshell can only carry 14 electrons at a time. Each g subshell can only store 18 electrons at a time.
Titanium is a chemical element with the atomic number 22 and the symbol Ti. It has an atomic weight of 47.867 daltons. It's a silver-colored transition metal with a low density and great strength. In seawater, aqua regia, and chlorine, titanium is corrosion resistant.
Electronic configuration of Ti: $[{\text{Ar}}]3\;{d^2}4{s^2} $
Electronic configuration of $T{i^{2 + }} $: $[{\text{Ar}}]3\;{d^2} $
No unpaired electron is present in $T{i^{2 + }} $
Vanadium is a chemical element with the atomic number 23 and the symbol V. It's a malleable transition metal that's hard and silvery-grey in colour. The elemental metal is uncommon in nature, but when isolated artificially, the creation of an oxide layer (passivation) protects the free metal from further oxidation.
Electronic configuration of V: $[{\text{Ar}}]3\;{d^3}4{s^2} $
Electronic configuration of ${V^{2 + }} $: $[{\text{Ar}}]3\;{d^3} $
Three unpaired electrons are present in ${V^{2 + }} $.

Note:
The wave behaviour of electrons (or electron pairs) in an atom is described by atomic orbitals, which are mathematical functions. They provide a method for calculating the chances of finding an electron in a certain location around the nucleus of an atom.