Question

Find the oxidation number of $$V$$ in$$R{b}_{4}Na\left[H{V}_{10}{O}_{28}\right]$$ .
A. $$+5$$
B. $$+2$$
C. $$-5$$
D. none of these

Answer 1

Hint:Oxidation number is the charge on an atom which appears due to formation of bonds with the neighboring atoms or ligands. It is denoted by $$0$$, positive or negative sign.

Complete step by step answer:
Oxidation number is also termed as oxidation state is equal to the total number of electrons which an atom either gains or loses during the formation of a chemical bond with another atom or molecule.
In order to determine the oxidation number of any atom the number of bonds both ionic and covalent is to be determined first. The number of bonds decides the number of electrons shared by the considered atom in its ionic form.
The steps involved in assigning the oxidation number or oxidation state of an atom or ion in a compound:
i) Add up the constant oxidation state of other atoms/molecules/ions that are bonded with the central metal ion
ii) The total oxidation state of a molecule or ion is equal to the total charge on the molecule or ion.
In compounds the more electronegative atoms gain electrons from a less electronegative atom and possess negative oxidation states. The numerical value of the oxidation state is equal to change in the number of electrons.
The elements in the periodic table vary in their oxidation number or state according to the number of valence shell electrons present. The octet rule must be satisfied during the formation of bonds.
The given compound is $$R{b}_{4}Na\left[H{V}_{10}{O}_{28}\right]$$ in which the central metal atom is $$V$$. This is a coordination complex compound and is a neutral compound. Let $$x$$ is the oxidation state of the metal atom/ion. The ten $$V$$ atoms are attached to four Rubidium, one Sodium, one Hydrogen and twenty eight Oxygen atoms/ions. Thus the equation is
$4(+1)+1+1+10x+28(-2)=0$
$10x=50$
$x=+5$.
Hence the oxidation state or number of $$V$$ in $$R{b}_{4}Na\left[H{V}_{10}{O}_{28}\right]$$ is $$+5$$.

Note: Oxidation number or state may be zero, positive or negative. Oxidation number of alkali metal is fixed as $$+1$$ due to the presence of a single electron in the outermost shell. It has to be an integer and not a fraction number as the number of electrons is always integer.