Tin oxide laminated film with catalyst layer and method for forming same

Abstract

There has been a problem that unevenness tends to occur in a laminated film of a tin oxide film and a catalyst layer used in a pretreatment of an electroless plating treatment. According to the method for forming a tin oxide laminated film with a catalyst layer, a uniform catalyst layer can be formed, the method comprising: a step for preparing a concentrated solution in which a tin compound is dissolved in a polar solvent; a step for preparing a reaction solution, which is a stable divalent tin solution, by diluting the concentrated solution with the polar solvent and a fluoride solution; a step in which an object to be treated is immersed in the reaction liquid, and a tin oxide layer is formed on the surface of the object to be treated; and a step in which the object to be treated, the surface of which has been provided with the tin oxide layer, is immersed in a catalyst solution containing catalyst metal ions, and a catalyst layer is formed on the tin oxide layer.

Description

Technical Field

[0001] The present invention relates to a method for forming a laminated film in which a tin oxide film is formed on an insulating substrate such as a resin substrate, a ceramic substrate, a glass substrate, or a silicon substrate, or a metal film is formed on a tin oxide film. Background Art

[0002] Conventionally, chemical plating of a non-conductive object to be processed is performed as follows: a catalyst such as palladium (Pd) is loaded on the surface to be processed to form a metal film. However, since the element that becomes the catalyst is only loaded on the surface to be processed, the formed metal film has low adhesion.

[0003] Therefore, the non-conductive surface to be treated is subjected to a pre-treatment called chemical plating, which includes a sensitization treatment (sensitization treatment) for forming a metal oxide film and an activation treatment (activation treatment) for forming a catalyst on the metal oxide film.

[0004] Zinc oxide is easy to handle and is often used as a metal oxide film, but zinc oxide has a problem of very low resistance to chemicals. Therefore, tin has begun to be used as a metal oxide film.

[0005] Patent Document 1 discloses a scheme in which a substrate is immersed in a stannous chloride solution, then immersed in an aqueous solution of hydrofluoric acid or a fluoride salt to perform a sensitization treatment, and then immersed in a salt solution of palladium (Pd), silver (Ag) or gold (Au) to perform an activation treatment.

[0006] In addition, regarding the sensitization treatment, it is disclosed that the same effect can be obtained even if a hydrofluoric acid or fluoride solution is mixed with a stannous chloride solution.

[0007] In addition, Patent Document 2 discloses the use of a mixed solution of SnF 2 and HF as a sensitization treatment.

[0008] Prior art literature

[0009] Patent Literature

[0010] Patent Document 1: Japanese Patent Application Laid-Open No. 59-074270

[0011] Patent Document 2: Japanese Patent Application Laid-Open No. 08-281957 Summary of the invention

[0012] Problem that the invention aims to solve

[0013] In these sensitization treatments, the treated object is immersed in a divalent tin source for several minutes, so the tin oxide layer is very thin and can be said to be sparse. The density of the catalyst in the activation treatment on the tin oxide layer is also low. The metal film formed by chemical plating in the base treatment in this case is prone to unevenness, and the adhesion of the film is also low.

[0014] In addition, the following is also considered as a cause of the unevenness of the catalyst layer. First, the catalyst metal is reduced by SnO, which is a divalent tin compound in the tin oxide film, and is thus loaded on the tin oxide film. However, when the tin compound used to obtain the divalent tin ions is dissolved in an aqueous solution, the divalent tin ions are oxidized to tetravalent tin ions and become SnO2 due to the dissolved oxygen in the water that serves as a medium. The tetravalent SnO2 does not have a reducing power, and the catalyst metal cannot be loaded on the tin oxide film. It is believed that: this SnO2 is incorporated into the tin oxide film, and the subsequent reduction of the catalyst metal becomes impossible to load.

[0015] In addition, the following method can also be cited as a cause of unevenness in the catalyst layer: divalent tin ions are oxidized and precipitated in the aqueous solution in the form of SnO2, and attached to the treated surface in the form of precipitates, thereby hindering the uniformity of catalyst metal loading.

[0016] In addition, the precipitated SnO2 is incorporated into the oxide film formed on the treated surface and becomes the nucleus of film growth, which rapidly increases the film growth rate and significantly makes it difficult to control the thickness of the oxide film.

[0017] Solutions for solving problems

[0018] The inventors of the present invention have conducted intensive research and found that when a tin oxide layer is formed, the generation of tetravalent tin ions is suppressed and a thicker tin oxide layer is formed, thereby providing a tin oxide laminated film based on a tin oxide layer and a catalyst layer, and a method for forming such a tin oxide laminated film having a catalyst layer without unevenness on the tin oxide film.

[0019] More specifically, the tin oxide laminated film with a catalyst layer of the present invention is characterized by having:

[0020] A tin oxide layer formed on the surface of the insulator, and

[0021] a catalyst layer formed on the tin oxide layer,

[0022] The tin oxide layer has a thickness of 2 nm to 60 nm.

[0023] In addition, the method for forming a tin oxide laminated film with a catalyst layer of the present invention comprises the following steps:

[0024] A process for preparing a concentrated solution of a tin compound dissolved in a polar solvent;

[0025] The concentrated solution is diluted with the polar solvent and the fluoride solution to prepare a reaction solution as a stable divalent tin solution;

[0026] The step of immersing the object to be treated in the reaction solution to form a tin oxide layer on the surface of the object to be treated; and

[0027] A step of immersing the object to be treated having a tin oxide layer formed on the surface thereof in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer.

[0028] Furthermore, the method for forming a tin oxide laminated film with a catalyst layer of the present invention may also be characterized by comprising the following steps:

[0029] A step of preparing a concentrated solution of stannous chloride dissolved in a fluoride solution having a fluoride ion concentration of 0.3 M or more;

[0030] A step of diluting the concentrated solution or adding the fluoride solution to prepare a reaction solution as a stable divalent tin solution;

[0031] The step of immersing the object to be treated in the reaction solution to form a tin oxide layer on the surface of the object to be treated; and

[0032] A step of immersing the object to be treated having a tin oxide layer formed on the surface thereof in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer.

[0033] Effects of the Invention

[0034] The tin oxide laminated film with a catalyst layer of the present invention has a thickness of 2 nm or more and 60 nm, so the catalyst loading is large, so unevenness is not easy to occur, and the adhesion to the treated surface of the chemically plated film formed on the tin oxide laminated film is also high.

[0035] The tin oxide laminated film with a catalyst layer of the present invention has a thickness that allows the cross section to be confirmed by TEM (Transmission Electron Microscope), and the interface expansion area ratio (Sdr) is not less than zero and less than 1×10 -4 , therefore, the catalyst layer can be present widely and uniformly.

[0036] Furthermore, in the formation method of the present invention, since a stable divalent tin solution is used, a uniform and thick tin oxide layer can be formed even by a long-term treatment. BRIEF DESCRIPTION OF THE DRAWINGS

[0037] Figure 1 This is a diagram for explaining the interface expansion area ratio. DETAILED DESCRIPTION

[0038] The following is a description of the method for forming a tin oxide laminated film with a catalyst layer of the present invention by showing the accompanying drawings and examples. It should be noted that the following description illustrates an embodiment of the present invention, and the present invention is not limited to the following description. The following description can be changed within the scope of the present invention. In addition, in the following description, "up" refers to the direction away from the treated surface serving as a reference, and "down" refers to the direction close to the treated surface.

[0039] The method for forming a tin oxide laminated film with a catalyst layer of the present invention comprises the following steps:

[0040] A process for preparing a concentrated solution of a tin compound dissolved in a polar solvent;

[0041] The concentrated solution is diluted with the polar solvent and the fluoride solution to prepare a reaction solution as a stable divalent tin solution;

[0042] The step of immersing the object to be treated in the reaction solution to form a tin oxide layer on the surface of the object to be treated; and

[0043] A step of immersing the object to be treated having a tin oxide layer formed on the surface thereof in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer.

[0044] <Object to be processed>

[0045] Examples of the object to be processed include insulators and insulators with a metal layer formed in advance on the surface. Specifically, materials such as resins, ceramics, glass, and silicon can be used.

[0046] As the resin, those made of fluorine-based resins such as polyimide resin, methacrylic resin, epoxy resin, liquid crystal polymer, polycarbonate resin, PFA, PTFE, and ETFE can be suitably used. In addition, the resin can contain glass fiber to improve mechanical strength.

[0047] As the ceramic, those using as raw materials aluminum oxide such as aluminum oxide, sapphire, aluminum nitride, silicon nitride, silicon carbide, zirconium oxide, yttrium oxide, titanium nitride, barium titanate, etc., can be preferably used.

[0048] Glass is an amorphous substrate formed of a silicon dioxide network, and may contain network formers (network-forming oxides) such as aluminum, boron, and phosphorus, and network modifiers (network-modifying oxides) such as alkali metals, alkaline earth metals, and magnesium.

[0049] Silicon may be suitably single crystal silicon or polycrystalline silicon.

[0050] <Surface to be treated>

[0051] The processed surface is the surface of the processed object, and may be the inner wall of a depression, hole, or through hole formed on the surface of the processed object as long as the liquid can reach it. Of course, the processed surface is the part where the metal film is to be formed.

[0052] <Divalent tin compounds>

[0053] As the divalent tin compound not containing fluorine, stannous chloride, tin sulfate, and tin nitrate can be suitably used. It should be noted that as long as these tin compounds are contained, tin compounds containing fluorine and other elements can also be contained. It should be noted that the divalent tin compound can be used in the range of 0.01 Mol to 1.0 Mol, more preferably in the range of 0.025 Mol to 0.5 Mol, and most preferably in the range of 0.05 Mol to 0.25 Mol.

[0054] <Fluoride>

[0055] As the fluoride, hydrofluoric acid (HF), ammonium bifluoride ((NH4)HF2) or ammonium fluoride (NH4F) can be suitably used. If only fluorides other than these are used, a catalyst layer with a uniform surface state cannot be obtained. The concentration of the fluoride can be used in the range of 0.01Mol to 2.5Mol in the state of the final reaction solution, more preferably 0.025Mol to 2.0Mol, and most preferably 0.05Mol to 1.0Mol. It should be noted that basically, the entire reaction solution only needs to have fluorine that completely dissolves more than 1mmol / L of divalent tin ions.

[0056] <Solvent>

[0057] As the solvent, water and / or a polar solvent can be suitably used. In addition, the water can contain a chelating agent. Here, the polar solvent can suitably use alcohols such as ethanol, carboxylic acids such as acetic acid, ketones such as acetone, etc.

[0058] <Stable divalent tin solution>

[0059] In the present invention, a stable divalent tin solution is used. In the present invention, a tin oxide layer much thicker than the metal oxide layer obtained by sensitization treatment is formed on the treated surface. Therefore, it is necessary to immerse the treated object in the treatment solution for a long time. Therefore, during this period, it is necessary to have a reaction solution in which no precipitate of substances containing tin occurs. In addition, even if there is no precipitate, when divalent tin is oxidized to tetravalent tin, it is impossible to carry out loading based on adsorption of the catalyst layer. Therefore, it is necessary to have a reaction solution in which divalent tin exists stably for a long time. It is referred to as a stable divalent tin solution.

[0060] First, when a divalent tin source is dissolved in water, it is oxidized to tetravalent tin due to the dissolved oxygen in the water and precipitates as tin oxide, which sometimes makes the solution turbid. Therefore, a stable divalent tin solution can be obtained by dissolving the divalent tin in a polar solvent that does not contain dissolved oxygen. However, in this case, a divalent tin salt needs to be dissolved in a polar solvent.

[0061] It should be noted that "containing no dissolved oxygen" here only requires that the dissolved oxygen content is 0.5 mg / L or less. In addition, during film formation, the substrate is immersed in the solution, but in order to avoid contact with air, it is preferred that oxygen repelling means such as nitrogen purge be applied to the liquid surface.

[0062] Stannous chloride is soluble in water, ethanol, methanol, and acetone as long as it is within the range confirmed by the applicant. In addition, tin acetate is soluble in ethanol.

[0063] Then, even if it is water with dissolved oxygen, by adding fluoride in advance and then adding divalent tin, a stable divalent tin solution can be obtained. Even if stannous chloride is dissolved in water, by immediately adding fluoride or borofluoride, a stable divalent tin solution can be obtained. However, according to the embodiment described later, it is expected to add fluoride in advance. This is because the time from the association of divalent tin with water to the reaction with fluoride is preferably short.

[0064] In addition, according to the embodiments described later, the fluorine concentration of the fluoride solution with which stannous chloride is initially associated can be 0.3 M or more, more preferably 0.5 M or more. When initially associated with a high-concentration fluoride solution, even if the concentration is diluted later, the stable time becomes longer, and the surface roughness can be reduced. In other words, the interface expansion area ratio described later can be reduced. On the other hand, if the fluorine concentration of the fluoride solution with which stannous chloride is initially associated is low, the surface of the formed tin oxide film becomes rough. That is, the interface expansion area ratio becomes larger.

[0065] Furthermore, with respect to the fluorine concentration and tin concentration of the fluoride solution initially associated with stannous chloride, it is desirable that the fluorine concentration is 3 times or less and 0.8 times or more relative to the tin concentration. When the tin concentration is excessively higher than the fluorine concentration or the fluorine concentration is excessively higher than the tin concentration, an oxide film is not formed or even if an oxide film is formed, the surface roughness is deteriorated (the interface expansion area ratio becomes larger).

[0066] The interface expansion area ratio Sdr described in detail later can be said to be an evaluation method for indicating the unevenness of the catalyst layer. Considering that the unevenness of the catalyst layer is affected by the surface properties of the tin oxide layer, if it is believed that the main reason for the surface properties of the tin oxide layer is the tetravalent tin oxidized by dissolved oxygen, then the expansion surface area ratio Sdr is considered to be the degree to which the generation of tetravalent tin is suppressed during the film formation of the oxide film. Therefore, the interface expansion area ratio of the catalyst layer on the oxide film obtained in 4 hours of film formation (the film thickness of tin oxide is 40 to 60 nm) is less than 1×10-4 Such a reaction solution can also be called a "stable divalent tin solution".

[0067] <Catalyst Metals>

[0068] As the catalyst metal, noble metals such as palladium, platinum, gold, and silver can be suitably used. The catalyst metal is formed into a solution is called a catalyst solution. Regarding the concentration of the catalyst solution, for example, when the catalyst is palladium, it can be preferably used at 0.005g / L to 1g / L, more preferably 0.05g / L to 1.0g / L, and most preferably 0.1g / L to 1.0g / L. It should be noted that, in more detail, when the catalyst metal is palladium, chemical plating can be performed at 0.005g / L to 1.0g / L, but in the case of silver, chemical plating cannot be performed if it is not 0.05g / L to 1.0g / L.

[0069] <Film formation conditions>

[0070] The formation of the tin oxide layer and the catalyst layer in the present invention is suitable for being carried out between 30°C and 50°C, respectively. In addition, the formation time (also referred to as the processing time) of the tin oxide layer takes 1 to 6 hours, and the thickness of the tin oxide layer can be made 2nm to 60nm. By thickening the layer thickness, the unevenness disappears. In addition, due to the divalent tin ions in the film, a large number of catalyst atoms are adsorbed on the surface of the tin oxide layer and firmly loaded. It should be noted that the formation of the catalyst layer can be about 2 to 4 minutes at room temperature.

[0071] <Tin oxide laminated film>

[0072] The tin oxide laminated film is composed of a tin oxide layer formed on the treated surface and a catalyst layer formed on the tin oxide layer. The thickness of the tin oxide layer is 2nm to 60nm. The cross section of the tin oxide layer can be clearly observed by TEM. That is, the thickness of 2nm refers to the identification limit based on TEM. In the catalyst layer, the presence of catalyst elements can be detected by mass spectrometry, etc., but cannot be confirmed in cross-sectional observation. The formation time is short, so it is considered to be an almost completely isolated island state. However, in this specification, the catalyst part is also referred to as a "layer" as one that is formed sequentially on the treated surface.

[0073] <Interface expansion area ratio>

[0074] The catalyst layer is supported on the surface of the tin oxide laminated film, but the film thickness of the catalyst layer is 0.2 to 0.5 nm, so that it almost becomes an independent island. However, when the catalyst layer is formed on the tin oxide layer with a size of about a conventional slide glass (26 mm × 76 mm), it is colored, so the uniformity can be observed visually. Therefore, by making the index of surface roughness consistent with the visual judgment, it can be known that the interface expansion area ratio (Sdr) is suitable as a judgment of uniformity.

[0075] The interface expansion area ratio (Sdr) is an indicator of how much the actual surface area is increased relative to the area of ​​the defined area, and is also defined in ISO25178. Figure 1 As shown in the figure, the cross section is considered to be a right isosceles triangle with the hypotenuse as the base. Let the depth be 1, the side forming the right angle be a, and the base be c. Let the area of ​​one inclined surface be sa (sa·1=2a), then the area S' of the inclined surface is 2sa, and the area S of the base is c·1=(c). In this way, Sdr is calculated using formula (1).

[0076] [Number 1]

[0077]

[0078] Thus, in the case of a 45-degree inclined surface, when the square root of 2 is substituted for c and 1 is substituted for a, Sdr becomes 0.414. In addition, when there is no increase in the area of ​​the inclined surface relative to the reference surface (area S in this case) (when S'=S), Sdr becomes zero. In more detail, Sdr is expressed by the following formula (2).

[0079] [Number 2]

[0080]

[0081] Here, x and y represent the vertical and horizontal directions of the sample, and z represents the thickness direction.

[0082] When expressed using this index, a state suitable for visual judgment can be judged as an Sdr of less than 1×10- 4 It should be noted that the above index is a prerequisite for film formation, and whether film formation is possible is determined by visual inspection. If film formation is not possible, the treated surface is measured, and therefore Sdr becomes zero.

[0083] In addition, when the processed surface is porous such as ceramics, the Sdr of the processed surface itself is not zero, and the evaluation based on Sdr is not applicable. When the processed surface is formed into a mirror surface, the evaluation based on Sdr is used.

[0084] Example

[0085] Hereinafter, examples of the tin oxide laminated film with a catalyst layer of the present invention will be described.

[0086] A reaction solution using stannous chloride and hydrofluoric acid was prepared, and an oxide film was formed on a glass substrate. Pd was formed as a catalyst layer on the oxide film, and the surface spread area (Sdr) was measured.

[0087] Stannous chloride is dissolved in about 80 ml of a high-concentration fluorine solution to prepare 100 ml of a concentrated solution. Then, water and / or 55% HF are added to the concentrated solution to prepare 40 ml of a reaction solution of a specified concentration. The fluorine element is combined with the divalent tin element in a high-concentration state, thereby obtaining a reaction solution that is a stable divalent tin solution.

[0088] After adjusting the temperature of the reaction solution to the reaction temperature (50°C to 60°C), the glass substrate is immersed and the layer formation time is maintained (4 hours). The thickness of the tin oxide layer after film formation is about 40nm. The glass substrate is then pulled up and immersed in the catalyst liquid to form a catalyst layer. The catalyst liquid used to form the catalyst layer is a 100ppm solution of palladium (Pd). The formation time of the catalyst layer is set to 2 minutes.

[0089] After the catalyst layer is formed, it is dried and the Sdr is measured using a laser microscope. The laser microscope used is the VK-X1100 series manufactured by KEYENCE CORPORATION, and the surface after the catalyst layer is formed is measured. The measurement conditions are as follows: the magnification is 1200 times, the measurement area is divided into 4 parts, and the Sdr of each is calculated to obtain the average value. For the same sample, the Sdr of 2 places is measured, and the average value is taken as the average Sdr of the sample. By comparing with the visual evaluation of the unevenness of the catalyst layer, the expanded surface area ratio Sdr is 1×10 -4 If it is less than 1×10 -4 , then it can be judged that there is no unevenness.

[0090] The following is an explanation of the preparation methods of the examples and comparative examples. In addition, Tables 1 and 2 show their preparation methods and the Sdr values ​​measured after film formation. In addition, the following substances were used for each material. Hydrofluoric acid (HF) was used at a concentration of 55% (density 1.2 g / ml). Stannous chloride (SnCl2) was used as a dihydrate (molecular weight 226.65) with a purity of 97%. Tin (II) fluoride (SnF2: molecular weight 156.71) was used at a purity of 90%. Tin borofluoride (Sn(BF4)2: molecular weight 292.3) was used at a purity of 50%. Ammonium bifluoride (NH4HF2: molecular weight 57.04) was used at a purity of 97%.

[0091] Example 1: 1.53 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 0.62 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 11.6 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 1. The fluorine concentration of the concentrated solution 1 was 0.50 M and the tin concentration was also 0.50 M.

[0092] Next, 35 ml of pure water was added to 5 ml of the concentrated solution 1 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 63.1 mM, and the tin concentration was 62.3 mM.

[0093] Example 2: 0.77 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 0.31 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 5.8 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 2. The fluorine concentration of the concentrated solution 2 was 0.25 M and the tin concentration was 0.25 M.

[0094] Next, 36 ml of pure water was added to 4 ml of the concentrated solution 2 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 25.4 mM, and the tin concentration was 24.9 mM.

[0095] Example 3: 6.05 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 2.32 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 46.5 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 3. The fluorine concentration of the concentrated solution 3 was 2.00 M and the tin concentration was 2.00 M.

[0096] Next, 36 ml of pure water was added to 4 ml of the concentrated solution 3 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 199.6 mM, and the tin concentration was 199.9 mM.

[0097] Example 4: 12.1 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 4.33 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 45 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 4. The fluorine concentration of the concentrated solution 4 was 3.99 M and the tin concentration was 1.93 M.

[0098] Next, 34 ml of pure water was added to 6 ml of the concentrated solution 4 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 598.7 mM, and the tin concentration was 2902.0 mM.

[0099] Example 5: 12.1 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 4.33 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 45 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 4. The fluorine concentration of the concentrated solution 4 was 3.99 M and the tin concentration was 1.93 M.

[0100] Next, 15 ml of pure water was added to 25 ml of the concentrated solution 4 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 2494.4 mM (about 2.5 M), and the tin concentration was 1209.0 mM (about 1.2 M).

[0101] Example 6: 1.53 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 0.62 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 11.6 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 5. The fluorine concentration of the concentrated solution 5 was 0.50 M and the tin concentration was 0.50 M.

[0102] Next, 39.2 ml of pure water was added to 0.8 ml of the concentrated solution 5 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 10.1 mM, and the tin concentration was 10.0 mM.

[0103] Example 7: 1.53 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 0.62 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 3.9 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 6. The fluorine concentration of the concentrated solution 6 was 0.50 M and the tin concentration was 0.17 M.

[0104] Next, 39.2 ml of pure water was added to 0.8 ml of the concentrated solution 6 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 10.1 mM and the tin concentration was 3.4 mM. The fluorine concentration in the reaction solution was less than 3 times the tin concentration.

[0105] Example 8: 9.06 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 3.36 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 24 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 7. The fluorine concentration of the concentrated solution 7 was 2.99 M and the tin concentration was 1.03 M.

[0106] Next, 39.0 ml of pure water was added to 1 ml of the concentrated solution 7 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 74.7 mM, and the tin concentration was 25.8 mM. The fluorine concentration in the reaction solution was less than 3 times the tin concentration.

[0107] Example 9: 3.06 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 1.22 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 28 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 8. The fluorine concentration of the concentrated solution 8 was 1.01 M and the tin concentration was 1.20 M.

[0108] Next, 39.0 ml of pure water was added to 1 ml of the concentrated solution 8 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 25.2 mM, and the tin concentration was 30.1 mM. The fluorine concentration in the reaction solution was greater than 0.8 times the tin concentration.

[0109] Example 10: 2.51 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 1.00 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 29 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 9. The fluorine concentration of the concentrated solution 9 was 0.83 M and the tin concentration was 1.25 M.

[0110] Next, 0.013 ml of 55% HF and pure water were further added to 1 ml of the concentrated solution 9 to obtain a total of 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 31.4 mM, and the tin concentration was 31.2 mM.

[0111] Example 11: 23.2 g of stannous chloride was added to 80 ml of ethanol and dissolved, and then ethanol was further added to prepare a total volume of 100 ml to obtain a concentrated solution 10. The fluorine concentration of the concentrated solution 10 was 0 M, and the tin concentration was 1.00 M.

[0112] Next, 0.032 ml of 55% HF and pure water were further added to 1 ml of the concentrated solution 10 to obtain a total of 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 26.4 mM, and the tin concentration was 24.9 mM.

[0113] Example 12: 23.2 g of stannous chloride was added to 80 ml of ethanol and dissolved, and then ethanol was further added to prepare a total amount of 100 ml to obtain a concentrated solution 10. The fluorine concentration of the concentrated solution 10 was 0 M, and the tin concentration was 1.00 M.

[0114] Next, 0.09 ml of 55% HF and pure water were further added to 1 ml of the concentrated solution 10 to obtain a total of 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 74.2 mM and the tin concentration was 24.9 mM. The fluorine concentration in the reaction solution was less than 3 times the tin concentration.

[0115] Example 13: 23.2 g of stannous chloride was added to 80 ml of ethanol and dissolved, and ethanol was further added to prepare a total volume of 100 ml to obtain a concentrated solution 10. The fluorine concentration of the concentrated solution 10 was 0 M, and the tin concentration was 1.00 M.

[0116] Next, 0.025 ml of 55% HF and pure water were further added to 1 ml of the concentrated solution 10 to obtain a total of 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 20.6 mM and the tin concentration was 24.9 mM. The fluorine concentration in the reaction solution was greater than 0.8 times the tin concentration.

[0117] Example 14: 3.06 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 1.22 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 13.96 g of stannous chloride and 23.4 g of Sn(BF4)2 were added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a concentrated solution 11. The fluorine concentration of the concentrated solution 11 was 1.01 M and the tin concentration was 1.00 M.

[0118] Next, 35.0 ml of pure water was added to 5 ml of the concentrated solution 11 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 126.2 mM, and the tin concentration was 50.0 mM.

[0119] Example 15: 2.94 g of NH4HF2 with a purity of 97% is added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluorine concentration of the fluoride solution is 1.25 M. The fluorine concentration at this time becomes the fluorine concentration (2.50 M) at which tin ions are initially associated. 23.2 g of stannous chloride is added to the hydrofluoric acid solution and dissolved. Then, pure water is added to prepare a total amount of 100 ml to obtain a concentrated solution 12. The fluorine concentration of the concentrated solution 12 is 1.00 M and the tin concentration is 1.00 M.

[0120] Next, 35.0 ml of pure water was added to 5 ml of the concentrated solution 12 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 125.0 mM, and the tin concentration was 124.7 mM.

[0121] Comparative Example 1: 0.7 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluorine concentration of the fluoride solution was 0.29 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 5.3 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrate 1. The fluorine concentration of the Comparative Example Concentrate 1 was 0.23 M and the tin concentration was 0.23 M.

[0122] Next, 29.6 ml of pure water was added to 10.4 ml of the comparative example concentrated solution 1 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 60.0 mM, and the tin concentration was 59.2 mM.

[0123] Comparative Example 2: 0.7 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluorine concentration of the fluoride solution was 0.26 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 5.3 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a comparative example concentrated solution 1. The fluorine concentration of the comparative example concentrated solution 1 was 0.23 M and the tin concentration was 0.23 M.

[0124] Next, 38.2 ml of pure water was added to 1.8 ml of the comparative example concentrated solution 1 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 10.4 mM, and the tin concentration was 10.3 mM.

[0125] Comparative Example 3: 0.7 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluorine concentration of the fluoride solution was 0.29 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 5.3 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a comparative example concentrated solution 1. The fluorine concentration of the comparative example concentrated solution 1 was 0.23 M and the tin concentration was 0.23 M.

[0126] Next, 5.5 ml of pure water was added to 34.5 ml of the comparative example concentrated solution 1 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 199.1 mM, and the tin concentration was 196.5 mM.

[0127] Comparative Example 4: 1.53 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluorine concentration of the fluoride solution was 0.62 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 11.6 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrate 2. The fluorine concentration of the Comparative Example Concentrate 2 was 0.50 M and the tin concentration was 0.50 M.

[0128] Next, 39.36 ml of pure water was added to 0.64 ml of the comparative example concentrated solution 2 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 8.2 mM, and the tin concentration was 8.1 mM.

[0129] Comparative Example 5: 58.5 g of tin fluoroborate (Sn(BF4)2) was added to 80 ml of pure water and stirred to obtain a fluoride solution. The boron fluoride concentration of the fluoride solution was approximately 2.5 M. The tin ions in the tin fluoroborate solution were approximately 1.25 M. Then, pure water was added to prepare a total amount of 100 ml to obtain a comparative example concentrated solution 3. The boron fluoride concentration of the comparative example concentrated solution 3 was 1.60 M, and the tin concentration was 1.00 M.

[0130] Next, 35.0 ml of pure water was added to 5 ml of the comparative example concentrated solution 3 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 200.1 mM, and the tin concentration was 125.1 mM.

[0131] Comparative Example 6: 12.1 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluorine concentration of the fluoride solution was 4.33 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 45 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrate 4. The fluorine concentration of the Comparative Example Concentrate 4 was 3.99 M and the tin concentration was 1.93 M.

[0132] Next, 12 ml of pure water was added to 28 ml of the comparative example concentrated solution 4 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 2793.7 mM, and the tin concentration was 1354.1 mM.

[0133] Comparative Example 7: 23.2 g of stannous chloride was added to 80 ml of pure water and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrated Solution 5. The fluorine concentration of the Comparative Example Concentrated Solution 5 was 0 M and the tin concentration was 1.00 M. That is, Comparative Example 7 was dissolved in pure water and was not associated with the high-concentration fluoride solution.

[0134] Next, 0.032 ml of 55% HF and pure water were added to 1 ml of the comparative example concentrated solution 5 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 26.4 mM, and the tin concentration was 24.9 mM.

[0135] Comparative Example 8: 9.06 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluorine concentration of the fluoride solution was 3.36 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 20 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrated Solution 6. The fluorine concentration of the Comparative Example Concentrated Solution 6 was 3.00 M and the tin concentration was 0.86 M. The fluorine concentration was more than 3 times the tin concentration.

[0136] Next, 39.0 ml of pure water was added to 1 ml of the comparative example concentrated solution 6 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 74.7 mM, and the tin concentration was 21.5 mM.

[0137] Comparative Example 9: 2.9 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluorine concentration of the fluoride solution was 1.15 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 29 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrated Solution 7. The fluorine concentration of the Comparative Example Concentrated Solution 7 was 0.96 M and the tin concentration was 1.25 M.

[0138] Next, 39 ml of pure water was added to 1 ml of Comparative Example concentrated solution 7 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 23.9 mM and the tin concentration was 31.2 mM. The fluorine concentration was 0.8 times or less of the tin concentration.

[0139] Comparative Example 10: 17.4 g (equivalent to 0.10 mol) of tin (II) fluoride (SnF2) was added to 80 ml of pure water and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a comparative example concentrated solution 8. The fluorine source of the comparative example concentrated solution 8 was tin (II) fluoride itself. That is, the fluorine concentration of the initial association of tin ions was approximately 2.50 M. The fluorine concentration of the comparative example concentrated solution 8 was 2.00 M and the tin concentration was 1.00 M.

[0140] Next, 39 ml of pure water was added to 1.0 ml of the comparative example concentrated solution 8 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 50.0 mM, and the tin concentration was 24.9 mM.

[0141] Comparative Example 11: 17.4 g (equivalent to 0.10 mol) of tin (II) fluoride (SnF2) was added to 80 ml of pure water and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrated Solution 8. The fluorine source of this Comparative Example Concentrated Solution 8 is tin (II) fluoride itself. That is, the fluorine concentration of the initial association of tin ions is approximately 2.50 M. The fluorine concentration of Comparative Example Concentrated Solution 8 is 2.00 M and the tin concentration is 1.00 M.

[0142] Next, 32 ml of pure water was added to 8.0 ml of the comparative example concentrated solution 8 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 399.7 mM, and the tin concentration was 199.9 mM.

[0143] Comparative Example 12: 5.0 g (equivalent to 0.03 mol) of tin (II) fluoride (SnF2) was added to 80 ml of pure water and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrated Solution 9. The fluorine source of this Comparative Example Concentrated Solution 9 is tin (II) fluoride itself. That is, the fluorine concentration of the initial association of tin ions is approximately 0.75 M. The fluorine concentration of Comparative Example Concentrated Solution 9 is 0.57 M and the tin concentration is 0.29 M.

[0144] Next, 0.039 ml of 55% HF and pure water were added to 5.0 ml of the comparative example concentrated solution 9 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 103.9 mM, and the tin concentration was 35.9 mM.

[0145] Comparative Example 13: 5.0 g (equivalent to 0.03 mol) of tin (II) fluoride (SnF2) was added to 80 ml of pure water and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain Comparative Example Concentrated Solution 9. The fluorine source of this Comparative Example Concentrated Solution 9 is tin (II) fluoride itself. That is, the fluorine concentration of the initial association of tin ions is approximately 0.75 M. The fluorine concentration of Comparative Example Concentrated Solution 9 is 0.57 M and the tin concentration is 0.29 M.

[0146] Next, 35 ml of pure water was added to 5.0 ml of the comparative example concentrated solution 9 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 71.8 mM, and the tin concentration was 35.9 mM.

[0147] Comparative Example 14: 3.06 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 1.22 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 23.2 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a comparative example concentrated solution 10. The fluorine concentration of the comparative example concentrated solution 10 was 1.01 M and the tin concentration was 1.00 M.

[0148] Next, 0.44 ml of 35% HCl and pure water were added to 1 ml of the comparative example concentrated solution 10 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 25.2 mM, and the tin concentration was 24.9 mM.

[0149] Comparative Example 15: 3.06 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 1.22 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 23.2 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a comparative example concentrated solution 10. The fluorine concentration of the comparative example concentrated solution 10 was 1.01 M and the tin concentration was 1.00 M.

[0150] Next, 8.0 ml of KF prepared to 50 g / L and pure water were added to 1 ml of the comparative example concentrated solution 10 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 25.2 mM, and the tin concentration was 24.9 mM.

[0151] Comparative Example 16: 3.06 ml of 55% HF was added to 80 ml of pure water and stirred to obtain a hydrofluoric acid solution. The concentration of the hydrofluoric acid solution was 1.22 M. The fluorine concentration at this time became the fluorine concentration at which the tin ions were initially associated. 23.2 g of stannous chloride was added to the hydrofluoric acid solution and dissolved. Then, pure water was added to prepare a total amount of 100 ml to obtain a comparative example concentrated solution 10. The fluorine concentration of the comparative example concentrated solution 10 was 1.01 M and the tin concentration was 1.00 M.

[0152] Next, 2.0 ml of ascorbic acid adjusted to 50 g / L and pure water were added to 1 ml of the comparative example concentrated solution 10 to obtain 40 ml of a reaction solution. The fluorine concentration in the reaction solution was 25.2 mM, and the tin concentration was 24.9 mM.

[0153] The main data of the concentrated solution of the example is shown in Table 1, the main data of the concentrated solution of the comparative example is shown in Table 2, the main data of the reaction solution of the example is shown in Table 3, and the main data of the reaction solution of the comparative example is shown in Table 4. It should be noted that Tables 3 and 4 also show the fluorine concentration of the concentrated solution when the tin ions are first associated.

[0154] [Table 1]

[0155]

[0156] [Table 2]

[0157]

[0158] [Table 3]

[0159]

[0160] [Table 4]

[0161]

[0162] The following description will be made with reference to Tables 3 and 4. As shown in Examples 1 to 10, if the fluorine concentration of the fluoride solution to which the tin ions are initially associated is 0.3 M or more, it can be said that an interface expansion area ratio Sdr of less than 1×10 -4 In addition, at this time, a suitable catalyst layer can be formed by setting the fluorine concentration of the reaction solution to be the final solution in the range of 10 mM to 2.49 M.

[0163] The fluorine concentration of the reaction solution of Comparative Example 1 in Table 4 is basically the same as that of Example 1 (60 mM), but the interface expansion area ratio Sdr is 1×10 -4 Similarly, the fluorine concentrations of Comparative Example 2 and Example 6, and Comparative Example 3 and Example 3 are basically the same (10 mM and 200 mM, respectively), but the interface expansion area ratio Sdr of the comparative examples is 1×10 -4 In addition, Comparative Example 7 is a case where fluoride ions are not present when preparing the concentrated solution, and fluoride ions are added later when preparing the reaction solution by adding hydrofluoric acid. In this case, the interface expansion area ratio Sdr is also 1×10 -4 above.

[0164] It is believed that if the concentration of the fluoride solution to which the tin ions are initially associated is low, the divalent tin is oxidized to tetravalent tin by the dissolved oxygen in the solution and precipitates as microparticles, which do not contribute to the formation of the catalyst layer and are deposited on the substrate surface, thereby worsening the interface expansion area ratio Sdr (the numerical value becomes higher). In other words, during the film formation time of more than 4 hours, in order to suppress the formation of tetravalent tin to the extent that the unevenness of the catalyst layer formed later has no effect, it must be dissolved in a fluoride solution of 0.3M or more at the beginning.

[0165] Examples 11 to 13 are examples using ethanol as a solvent. In these cases, the only dissolved oxygen mixed in is the water contained in the hydrofluoric acid added later, and no oxidation caused by dissolved oxygen occurs in the concentrated solution. Therefore, the interface expansion area ratio Sdr is stably 1×10 -5 The value of left and right.

[0166] The fluorine concentration of the fluoride solution to which the tin ions of Comparative Example 4 were initially associated was 0.62 M. Based on the above-mentioned embodiment, it is considered that the interface expansion area ratio Sdr is less than 1×10 -4 , but in reality it is 1×10 -4 It is therefore believed that even if the fluorine concentration of the fluoride solution to which the tin ions are initially associated is 0.62 M, when the concentration is too dilute (less than 10 mM) when preparing the reaction solution, the formation of tetravalent tin cannot be suppressed to a level that has no effect on the unevenness of the catalyst layer during the film formation period of more than 4 hours.

[0167] On the other hand, in Comparative Example 6, the fluorine concentration of the reaction solution was set to 2.5 M or more, but the glass substrate was dissolved and film formation was not possible. In Example 5, considering that film formation was possible at about 2.5 M, the fluorine concentration of the reaction solution was considered to be 2.5 M as the limit.

[0168] Example 7 is a case where the concentrated solution is rich in fluorine ions relative to tin ions, and Example 9 is a case where the concentrated solution is lacking in fluorine ions relative to tin ions. On the other hand, Comparative Example 8 is a case where the fluorine ion concentration is 3 times or more relative to tin ions, and Comparative Example 9 is a case where the fluorine ion concentration is 80% or less relative to tin ions. If the fluorine ion concentration is too high relative to tin ions, the tin ions become dilute, and the unevenness of the catalyst layer becomes noticeable. As a result, the interface expansion area ratio Sdr becomes 1×10 -4 In addition, if the concentration of tin ions is dilute (the concentration of tin ions is high) relative to the concentration of fluorine ions, the concentrated solution itself will have a sticky feel, and the catalyst layer formed later will have unevenness (interface expansion area ratio Sdr is 1×10 -4 above).

[0169] Example 14 is a case where tin fluoroborate (Sn(BF4)2) is used, and Comparative Example 5 is a case where only tin fluoroborate (Sn(BF4)2) is used. Based on these comparisons, it can be said that tin fluoroborate is not suitable as a fluorine source for the liquid phase film formation method in the present invention.

[0170] Example 15 is the case where ammonium bifluoride (NH4HF2) is used, but the interface expansion area ratio Sdr is less than 1×10 -4 It can be said that ammonium bifluoride (NH4HF2) is suitable as a fluorine source for the liquid phase film forming method in the present invention.

[0171] Comparative Example 14 is a case where the fluorine concentration of the fluoride solution to which the tin ions are initially associated is 0.3 M or more, but hydrochloric acid is added during the preparation of the reaction solution. The interface expansion area ratio Sdr is 1×10 -4 As a result, the unevenness of the catalyst layer formed later becomes noticeable. Hydrochloric acid is a material often used in the liquid phase deposition method (Liquid Phase Deposition; hereinafter also referred to as the "LPD method") containing fluorine, but it becomes the cause of the deterioration of the surface roughness at the level required by the present invention (film formation for 4 hours). As a result, it can be said that the interface expansion area ratio of the present invention achieves a state of low surface roughness (high surface property) compared with the conventional LPD method.

[0172] Comparative Examples 15 and 16 are cases where the fluorine concentration of the fluoride solution to which the tin ions were initially associated was 0.3 M or more, but KF (Comparative Example 15) and ascorbic acid (Comparative Example 16) were added during the preparation of the reaction solution. In either case, no film was formed. It is believed that these additives are not suitable for the film forming method of the present invention.

[0173] As described above, when a tin oxide film is made by LPD method, a reaction solution is prepared by diluting a concentrated solution, thereby obtaining a stable divalent tin solution (reaction solution), and the surface roughness of the catalyst layer (Pd) on the tin oxide film can be made to be less than 1×10 -4 More specifically, when preparing the concentrated solution, the stannous chloride is prepared so that the fluorine concentration at the time of initial addition is 0.3 M or more, and the concentrated solution is diluted to prepare the reaction solution, so that the interface expansion area ratio can be less than 1×10 -4 As a result, even if a long time is applied to film formation, film formation can be achieved without deteriorating the surface roughness.

[0174] It is considered to be a method for inhibiting stannous chloride from being hydrolyzed by dissolved oxygen in the reaction solution. Therefore, even if the fluorine concentration when stannous chloride is initially introduced is lower than 0.3M, if ethanol without dissolved oxygen is used as a solvent, even if it is a long-term film formation, the surface property can also maintain a high (interface expansion area ratio is low) state.

[0175] It should be noted that stannous chloride is suitable as the divalent tin, and hydrogen fluoride can be suitably used as the fluorine source. In addition to hydrofluoric acid, ammonium hydrogen fluoride and ammonium fluoride can also be used as the fluorine source.

[0176] Industrial Applicability

[0177] The method for forming a tin oxide laminated film with a catalyst layer of the present invention can obtain a uniform tin oxide laminated film without unevenness.

Claims

1. A tin oxide laminated film with a catalyst layer, comprising: A tin oxide layer formed on the surface of the insulator, and a catalyst layer formed on the tin oxide layer, The tin oxide layer has a thickness of 2 nm or more and 60 nm or less.

2. The tin oxide laminated film with a catalyst layer according to claim 1, wherein The interface expansion area ratio (Sdr) of the catalyst layer is greater than zero and less than 1×10 -4 .

3. The tin oxide laminated film with a catalyst layer according to any one of claims 1 or 2, wherein The catalyst layer is a metal layer selected from gold, silver, platinum and palladium.

4. A method for forming a tin oxide laminated film with a catalyst layer, comprising the following steps: A process for preparing a concentrated solution of a tin compound dissolved in a polar solvent; The concentrated solution is diluted with the polar solvent and / or the fluoride solution to prepare a reaction solution as a stable divalent tin solution; The step of immersing the object to be treated in the reaction solution to form a tin oxide layer on the surface of the object to be treated; and A step of immersing the object to be treated having a tin oxide layer formed on the surface thereof in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer.

5. The method for forming a tin oxide laminated film with a catalyst layer according to claim 4, wherein: The polar solvent is water and / or ethanol, and the fluoride is any one of hydrofluoric acid, ammonium bifluoride, and ammonium fluoride.

6. The method for forming a tin oxide laminated film with a catalyst layer according to any one of claims 4 or 5, wherein: In the step of diluting the concentrated solution with the polar solvent and / or the fluoride solution to prepare a reaction solution as a stable divalent tin solution, The fluoride ion concentration of the fluoride solution is 0.8 times or more and 3 times or less the divalent tin ion concentration of the tin compound.

7. A method for forming a tin oxide laminated film with a catalyst layer, comprising the following steps: A step of preparing a concentrated solution of stannous chloride dissolved in a fluoride solution having a fluoride ion concentration of 0.3 M or more; The concentrated solution is diluted with the polar solvent and / or the fluoride solution to prepare a reaction solution as a stable divalent tin solution; The step of immersing the object to be treated in the reaction solution to form a tin oxide layer on the surface of the object to be treated; and A step of immersing the object to be treated having a tin oxide layer formed on the surface thereof in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer.

8. The method for forming a tin oxide laminated film with a catalyst layer according to claim 7, wherein: The fluoride is any of hydrofluoric acid, ammonium bifluoride, and ammonium fluoride.

9. The method for forming a tin oxide laminated film with a catalyst layer according to any one of claims 7 or 8, wherein: In the step of preparing a concentrated solution of stannous chloride dissolved in a fluoride solution having a fluoride ion concentration of 0.3 M or more, The fluorine ion concentration is 0.8 times or more and 3 times or less the divalent tin ion concentration based on the stannous chloride.

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