Question

How do you write the formula of Vanadium (IV) carbonate?

Answer 1

Hint Vanadium has the atomic number 23. It has 2 electrons in K-shell, 8 electrons in L-shell, 11 electrons in M-shell and 2 electrons in the outermost shell which is N-shell.

Complete step by step answer:
- Vanadium is a chemical element which is placed in the fifth group and fourth period. The chemical symbol of vanadium is V.
- It is a transition element with the atomic number 23. It does not show any similarity in physical and chemical properties with Nb although they are present in the same group.
- Vanadium is one of the most abundant metallic elements. It holds the twentieth position as the most abundant element on Earth’s crust.
- Pure vanadium is rare in nature but almost 65 of its various compounds occur naturally.
- Vanadium is obtained by extraction of patronite ores. It is also found in crude oil, coal, oil shale, tar sands deposits and bauxite ore.
- Only two naturally occurring isotopes are available of vanadium which are stable in nature. Apart from these, it has only two synthetic isotopes which are stable.
- The electronic configuration of vanadium is
$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}$
- As we already know, vanadium (IV) consists of +4 cationic charge. And carbonate (CO^{2-}_{3}) is a polyatomic ion.
- Therefore, the molecular formula would be
$V(CO_{3})_{2}$

Note: Vanadium is very useful in various fields mainly due to its unique property. It is a good conductor of electricity and insulator of heat. Vanadium and its compounds show catalytic characteristics and are commercially useful.