Complete step by step answer:
Reaction between metallic tin and concentrated nitric acid is given as follows: $Sn + 4HN{O_3} \to {H_2}Sn{O_3} + 4N{O_2} + {H_2}O$
When metallic tin reacts with concentrated nitric acid it forms metastannic acid, nitrogen oxide gas and water.
Reaction between very dilute nitric acid /$6\% $ reacts with tin, the reaction is given as follows:$8Sn + 10HN{O_3} \to N{H_4}N{O_3} + 8SnN{O_3} + 3{H_2}O$
When metallic tin reacts with very dilute nitric acid it forms ammonium nitrate, water and tin nitrate.
Reaction between dilute nitric acid$20\% $ reacts with tin, the reaction is as follows:
\[4Sn + 10HN{O_3} \to N{H_4}N{O_3} + 4Sn{(N{O_3})_2} + 3{H_2}O\]
When metallic tin reacts with dilute nitric acid it forms ammonium nitrate, water and tin $(II)$ nitrate.
Thus when tin reacts with concentrated nitric acid it forms metastannic acid.
So, the correct answer is option A) metastannic acid.
Additional Information:
-Tin is used in coating, plating and polishing of metals.
-It is used in the manufacturers of food containers.
-It is also used as electrodes in the batteries.
-Tin metal is used in the manufacture of alloys such as bronze and copper.
Note: Tin reacts with dilute acids slower than many other metals and always heat is required in order to speed up the reaction. This property means that it can be used as a protective coating on other metals. It reacts faster in the presence of strong acids.
Under acidic conditions and in the presence of chloride ions, Hg(II) is reduced to Hg(I) by Sn(II), forming Hg2Cl2:
Hg2+(aq) + Sn2+(aq) + 2 Cl−(aq)
In excess Sn(IV), Hg(I) is reduced to Hg:
Hg2Cl2(s) + Sn2+(aq)
Sn(IV) is reduced by Fe to Sn(II) in HCl(aq)
[SnCl6]2−(aq) + Fe(s)
Manganese with oxidation steps >2 will be reduced to Mn(II) by Sn(II) under acidic conditions under the formation of Sn(IV), e.g.
MnO2(s) + Sn2+(aq) + 4 H+(aq)
Reaction of tin with peroxide
Sn(II) is easily oxidized to Sn(IV) by hydrogen peroxide under acidic conditions
[SnCl3]−(aq) + H2O2(aq) + 2 H+(aq)
Reaction of tin with sulfide
Sn(II) is precipitated by hydrogensulfide in 0.4M HCl(aq):
Sn2+(aq) + H2S(aq)
In strong acid, the precipitate is dissolved:
SnS(s) + 2 H+(aq) + 3 Cl−(aq)
The precipitate is also dissolved in Na2S2 but not in Na2S:
SnS(s) + S22−(aq)
Sn(IV) is precipitated by hydrogensulfide in moderately acidic solutions [6]:
[SnCl6]2−(aq) + H2S(aq)
The precipitate is dissolved in alkali metal sulfides, forming thiostannate ions, and concentrated hydrochloric acid [6]:
SnS2(s) + S2−(aq)
SnS2(s) + 4 H+(aq) + 6 Cl−(aq)
When an alkali metal hydroxide is added, a white precipitate is formed. The precipitate is not Sn(OH)4, as could be expected. It may be orthostannic acid H2[Sn(OH)6] [6]
Reaction of tin with water
Tin does not react with water under normal conditions. When exposed to steam, tin dioxide, SnO2 and hydrogen are formed.
Sn(s) + 2 H2O(g)
Sn(IV) is precipitated as α-tin acid upon hydrolysis of Sn(IV) solutions:
[SnCl6]2−(aq) + 6 H2O(l)
The precipitate is amphoteric and is dissolved in acids and strong alkali:
H2[Sn(OH)6](s) + 2 OH−(aq)
H2[Sn(OH)6](s) + 4 H+(aq) + 6 Cl−(aq)
Quantitative analysis
Method 3500-Sn B Atomic Absorption Spectrometric Method [2]. A portion of the sample is digested in a combination of acids. The digest is atomized in a graphite tube and resulting absorption of light is measured at 235.5 nm.