化學問題-求過度金屬

Transition Metal 3d
21 Sc Scandium 44.959 g/mol
22 Ti Titanium 47.9 g/mol
23 V Vanadium 50.9415 g/mol
24 Cr Crromium 51.996 g/mol
25 Mn Manganese 54.938 g/mol
26 Fe Iron 55.847 g/mol
27 Co Cobalt 58.9332 g/mol
28 Ni Nickel 58.71 g/mol
29 Cu Copper 63.546 g/mol
30 Zn Zinc 65.38 g/mol

8 O Oxygen 15.9994 g/mol

Let Mt be the required transition metal, k be the number of oxyen atom per metal atom in the resulted oxide.
Mt + (k) * O-- --> Mt O(k) + 2(k) * e-
resulted weight of oxide is 25 + (15.9994 / ( Mt / 25)) g
First we try to determine the possible oxidation states for the metal
If the original metal is 25g,
then if the metal is Mt(+I), the oxide is Mt2O, the range of weight will be in the range
Sc 29.45g to Zn 28.06g which is not the required range
if the metal is Mt(+II), the oxide is MtO, the range of weight will be in the range
Sc 33.90g to Zn 31.12g which is not the required range
if the metal is Mt(+III), the oxide is Mt2O3, the range of weight will be in the range
Sc 38.34g to Zn 34.18g which is not the required range
if the metal is Mt(+IV), the oxide is MtO2, the range of weight will be in the range
Sc 42.79g to Zn 37.24g which is not the required range
if the metal is Mt(+V), the oxide is Mt2O5, the range of weight will be in the range
Sc 47.24g to Zn 40.29g which is the required range
if the metal is Mt(+VI), the oxide is MtO3, the range of weight will be in the range
Sc 51.69g to Zn 43.35g which is the required range
if the metal is Mt(+VII), the oxide is Mt2O7, the range of weight will be in the range
Sc 56.14g to Zn 46.41g which is not the required range
so the metal will only be Mt(+V) or Mt(+VI) as the resulted oxide weight is 44.63g

From the characteristic of those transition metal, we know that their oxidation state is related to their electronic structures.
For those ten transition metals, common electronic structure 1s2 2s2 2p6 3s2 3p6
21 Sc 3d1 4s2
22 Ti 3d2 4s2
23 V 3d3 4s2
24 Cr 3d5 4s1
25 Mn 3d5 4s2
26 Fe 3d6 4s2
27 Co 3d7 4s2
28 Ni 3d8 4s2
29 Cu 3d10 4s1
30 Zn 3d10 4s2
For oxidation states of +V or +VI to exist, the required metal can only be 23 V / 24 Cr / 25 Mn / 26 Fe / 27 Co.

For oxidation states of +V, weight of metal oxide (Mt2O5) with 25g of metal is
21 Sc 47.24g
22 Ti 45.88g
23 V 44.63g
24 Cr 44.23g
25 Mn 43.20g
26 Fe 42.91g
27 Co 41.97g
28 Ni 42.03g
29 Cu 40.74g
30 Zn 40.29g

For oxidation states of +VI, weight of metal oxide (MtO3) with 25g of metal is
21 Sc 51.69g
22 Ti 50.05g
23 V 48.56g
24 Cr 48.08g
25 Mn 46.84g
26 Fe 46.49g
27 Co 45.36g
28 Ni 45.44g
29 Cu 43.88g
30 Zn 43.35g

After excluding of Sc, Ti, Ni, Cu, and Zn, we get
For oxidation states of +V, oxide weight
23 V 44.63g
24 Cr 44.23g
25 Mn 43.20g
26 Fe 42.91g
27 Co 41.97g
For oxidation states of +VI, oxide weight
23 V 48.56g
24 Cr 48.08g
25 Mn 46.84g
26 Fe 46.49g
27 Co 45.36g
As a result, the metal is 23 V Vanadium, with oxidation states of +V form oxide V2O5.