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Vanadium(IV) oxide or vanadium dioxide is an inorganic compound with the formula VO 2.It is a dark blue solid. Vanadium(IV) dioxide is amphoteric, dissolving in non-oxidising acids to give the blue vanadyl ion, [VO] 2+ and in alkali to give the brown [V 4 O 9] 2− ion, or at high pH [VO 4] 4−. [4]
The number indicates the degree of oxidation of each element caused by molecular bonding. In ionic compounds, the oxidation numbers are the same as the element's ionic charge. Thus for KCl, potassium is assigned +1 and chlorine is assigned -1. [4] The complete set of rules for assigning oxidation numbers are discussed in the following sections.
Vanadium(II) oxide is the inorganic compound with the idealized formula VO. It is one of the several binary vanadium oxides. It adopts a distorted NaCl structure and contains weak V−V metal to metal bonds. VO is a semiconductor owing to delocalisation of electrons in the t 2g orbitals.
The Weir formula is a formula used in indirect calorimetry, relating metabolic rate to oxygen consumption and carbon dioxide production. According to original source, it says: [ 1 ] Metabolic rate (kcal per day) = 1.440 (3.9 VO 2 + 1.1 VCO 2 )
From left: [V(H 2 O) 6] 2+ (lilac), [V(H 2 O) 6] 3+ (green), [VO(H 2 O) 5] 2+ (blue) and [VO(H 2 O) 5] 3+ (yellow).. Vanadium compounds are compounds formed by the element vanadium (V). The chemistry of vanadium is noteworthy for the accessibility of the four adjacent oxidation states 2–5, whereas the chemistry of the other group 5 elements, niobium and tantalum, are somewhat more limited to ...
Vanadium(V) oxide (vanadia) is the inorganic compound with the formula V 2 O 5. Commonly known as vanadium pentoxide, it is a dark yellow solid, although when freshly precipitated from aqueous solution, its colour is deep orange. Because of its high oxidation state, it is both an amphoteric oxide and an oxidizing agent.
Vanadyl sulfate is most commonly obtained by reduction of vanadium pentoxide with sulfur dioxide: . V 2 O 5 + 7 H 2 O + SO 2 + H 2 SO 4 → 2 [V(O)(H 2 O) 4]SO 4. From aqueous solution, the salt crystallizes as the pentahydrate, the fifth water is not bound to the metal in the solid.
Oxidation states are unitless and are also scaled in positive and negative integers. Most often, the Frost diagram displays oxidation state in increasing order, but in some cases it is displayed in decreasing order. The neutral species of the pure element with a free energy of zero (nE° = 0) also has an oxidation state equal to zero. [2]