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

Abstract

The problem of unevenness in the laminated film of a tin oxide film and a catalyst layer used in the pretreatment of electroless plating is that it is easy to produce unevenness. According to a method for forming a tin oxide laminated film with a catalyst layer, a uniform catalyst layer can be formed. The method comprises the following steps: preparing a concentrated solution containing a tin compound dissolved in a polar solvent; diluting the concentrated solution with the polar solvent and a fluoride solution to prepare a reaction solution that is a stable divalent tin solution; immersing a treatment object in the reaction solution to form a tin oxide layer on the surface of the treatment object; and immersing the treatment object with the tin oxide layer formed on the surface in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer.

Description

Technical Field

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

[0002] Conventionally, electroless plating of a non-conductive object is performed by loading a catalyst such as palladium (Pd) onto the surface to be treated to form a metal film. However, since the catalyst element is only placed on the surface to be treated, the resulting metal film has low adhesion.

[0003] Therefore, the non-conductive surface to be treated is subjected to a pre-treatment called electroless plating, which includes a sensitization treatment (sensitization treatment) to form a metal oxide film and an activation treatment (activation treatment) to form 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 method of immersing a substrate in a stannous chloride solution, then in an aqueous solution of hydrofluoric acid or a fluoride salt for sensitization, and then in a salt solution of palladium (Pd), silver (Ag), or gold (Au) for activation.

[0006] In addition, regarding the sensitization treatment, it is disclosed that the same effect can be obtained even when hydrofluoric acid or a 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] Problems to be solved by the invention

[0013] These sensitization treatments are performed by immersing the treated material in a divalent tin source for several minutes. Consequently, the tin oxide layer is very thin and sparse. The density of the catalyst activated on this tin oxide layer is also low. In such substrate treatments, the metal film formed by chemical plating is prone to unevenness and has low film adhesion.

[0014] In addition, the following is also considered as a reason for 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 the 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.

[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, becoming 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 present inventors conducted intensive research and discovered a method for forming a tin oxide laminated film comprising a tin oxide layer and a catalyst layer, in which the generation of tetravalent tin ions is suppressed when forming a tin oxide layer and a thicker tin oxide layer is formed, thereby 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 aforementioned tin oxide layer,

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

[0023] Furthermore, 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 its surface 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 be characterized by comprising the following steps:

[0029] A process for 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 its surface 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 at least 2 nm and up to 60 nm, thus allowing for a high catalyst loading. This reduces unevenness and provides high adhesion to the treated surface of the electroless plating film formed on the tin oxide laminated film.

[0035] The tin oxide laminated film with a catalyst layer of the present invention has a thickness such that a cross section can be confirmed by TEM (Transmission Electron Microscope), and the interface expansion area ratio (Sdr) is zero or more and less than 1×10 -4 , therefore, the catalyst layer can be made to exist widely and uniformly.

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

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

[0038] The following drawings and examples illustrate the method for forming a tin oxide laminated film with a catalyst layer of the present invention. It should be noted that the following description illustrates one embodiment of the present invention, and the present invention is not limited to the following description. The following description can be modified 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 the 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 its surface in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer.

[0044] <Objects 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 ceramics, those using aluminum oxide such as alumina, sapphire, aluminum nitride, silicon nitride, silicon carbide, zirconium oxide, yttrium oxide, titanium nitride, barium titanate, and the like as raw materials can be preferably used.

[0048] Glass is an amorphous substrate formed of a silicon dioxide network, and may contain network modifiers (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 can be suitably single crystal silicon or polycrystalline silicon.

[0050] <Processed surface>

[0051] The processed surface is the surface of the object to be processed, and can be the inner wall of a depression, hole, or through hole formed on the surface of the object to be processed 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 fluorine-free divalent tin compounds, stannous chloride, tin sulfate, and tin nitrate can be preferably used. It should be noted that as long as these tin compounds are included, tin compounds containing fluorine or other elements may also be included. It should be noted that the divalent tin compound can be used in a range of 0.01 mol to 1.0 mol, more preferably 0.025 mol to 0.5 mol, and most preferably 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 utilized. 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 utilized 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] Solvents

[0057] As solvent, water and / or polar solvent can be suitably utilized. In addition, water can contain chelating agent. Herein, polar solvent can be suitably utilized alcohol such as ethanol, carboxylic acid 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 is formed on the treated surface that is much thicker than the metal oxide layer obtained by sensitization treatment, etc. Therefore, the treated object needs to be immersed in the treatment solution for a long time. Therefore, during this period, the reaction solution must not precipitate substances containing tin. Furthermore, even if there is no precipitate, divalent tin will oxidize to tetravalent tin, making it impossible to load the catalyst layer based on adsorption. Therefore, a reaction solution in which divalent tin is stably present for a long time is required. This 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 by the dissolved oxygen in the water, precipitating as tin oxide, which sometimes causes the solution to become 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 must be dissolved in the polar solvent.

[0061] It should be noted that "does not contain 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 to also apply oxygen repelling means such as nitrogen purge 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 the water contains dissolved oxygen, a stable divalent tin solution can be obtained by adding fluoride in advance and then adding divalent tin. Even if stannous chloride is dissolved in water, a stable divalent tin solution can be obtained by immediately adding fluoride or borofluoride. However, according to the examples described below, it is desirable 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] Furthermore, according to the examples described below, the fluorine concentration of the fluoride solution with which stannous chloride is initially associated can be 0.3 M or higher, more preferably 0.5 M or higher. When initially associated with a high-concentration fluoride solution, the stability time becomes longer even if the concentration is subsequently diluted, which can reduce surface roughness. In other words, the interface expansion area ratio (described below) 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 rougher. In other words, the interface expansion area ratio increases.

[0065] Furthermore, regarding the fluorine concentration and tin concentration of the fluoride solution with which stannous chloride is initially associated, it is desirable that the fluorine concentration be no more than 3 times and no less than 0.8 times the tin concentration. If the tin concentration is excessively higher than the fluorine concentration, or if the fluorine concentration is excessively higher than the tin concentration, an oxide film will not form, or even if an oxide film is formed, the surface roughness will be deteriorated (the interface expansion area ratio will increase).

[0066] The interface expansion area ratio Sdr, which will be described in detail later, can be said to be an evaluation method for expressing 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 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 the 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. A solution of the catalyst metal is referred to as a catalyst solution. Regarding the concentration of the catalyst solution, for example, when the catalyst is palladium, it can be preferably used at 0.005 g / L to 1 g / L, more preferably 0.05 g / L to 1.0 g / L, and most preferably 0.1 g / L to 1.0 g / L. It should be noted that, in more detail, when the catalyst metal is palladium, chemical plating can be performed at 0.005 g / L to 1.0 g / L, but in the case of silver, chemical plating cannot be performed unless the concentration is 0.05 g / L to 1.0 g / L.

[0069] <Film formation conditions>

[0070] The formation of the tin oxide layer and the catalyst layer in the present invention is suitably carried out between 30°C and 50°C. In addition, the formation time of the tin oxide layer (which may also be referred to as the processing time) takes 1 to 6 hours, and the thickness of the tin oxide layer can be made into a thickness of 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 using TEM. In other words, 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 by 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 a part 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, thus forming an almost independent island. However, when the catalyst layer is formed on the tin oxide layer with a size of about a conventional glass slide (26 mm × 76 mm), it is colored, so the uniformity can be observed visually. Therefore, by making the surface roughness index consistent with the visual judgment, it can be seen that the interface expansion area ratio (Sdr) is suitable for judging uniformity.

[0075] The surface expansion 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, 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 of ​​the inclined surface S' is 2sa, and the area of ​​the base S is c·1 = (c). Thus, Sdr is calculated using formula (1).

[0076] [Number 1]

[0077]

[0078] Therefore, for a 45-degree inclined surface, substituting the square root of 2 for c and 1 for a yields Sdr of 0.414. Furthermore, if there is no increase in the area of ​​the inclined surface relative to the reference surface (in this case, area S) (when S' = S), Sdr becomes zero. More specifically, Sdr is expressed by the following equation (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 one with an Sdr lower than 1×10- 4 It should be noted that the above indicators are the premise of 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 treated surface is porous such as ceramic, the Sdr of the treated surface itself is not zero, and the evaluation based on Sdr is not applicable. When the treated 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. A Pd catalyst layer was formed on the oxide film, and the surface spread area (Sdr) was measured.

[0087] Stannous chloride is dissolved in approximately 80 ml of a high-concentration fluorine solution to prepare 100 ml of a concentrated solution. Water and / or 55% HF are then added to this concentrated solution to prepare 40 ml of a reaction solution of a desired concentration. This allows the fluorine element to associate with the divalent tin element at a high concentration, resulting in 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 maintained for only the layer formation time (4 hours). The thickness of the tin oxide layer after film formation is about 40nm. The glass substrate is then pulled and immersed in the catalyst solution to form a catalyst layer. The catalyst solution used to form the catalyst layer is a 100ppm solution of palladium (Pd). The catalyst layer formation time 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, the Sdr of each part is calculated, and the average value is obtained. For the same sample, the Sdr is measured at 2 locations, 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 describes the preparation methods of the Examples and Comparative Examples. Tables 1 and 2 show their preparation methods and the Sdr values ​​measured after film formation. The following materials were used. 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 point corresponds to the initial fluorine concentration at which tin ions associate. 11.6 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Concentrated Solution 1. This Concentrated Solution 1 had a fluorine concentration of 0.50 M and a tin concentration of 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 having a fluorine concentration of 63.1 mM and a tin concentration of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 5.8 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Concentrated Solution 2. This Concentrated Solution 2 had a fluorine concentration of 0.25 M and a tin concentration of 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 having a fluorine concentration of 25.4 mM and a tin concentration of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 46.5 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Concentrated Solution 3. This Concentrated Solution 3 had a fluorine concentration of 2.00 M and a tin concentration of 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 having a fluorine concentration of 199.6 mM and a tin concentration of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 45 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding concentrated solution 4. This concentrated solution 4 had a fluorine concentration of 3.99 M and a tin concentration of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 45 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding concentrated solution 4. This concentrated solution 4 had a fluorine concentration of 3.99 M and a tin concentration of 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 having a fluorine concentration of 2494.4 mM (approximately 2.5 M) and a tin concentration of 1209.0 mM (approximately 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 point corresponds to the initial fluorine concentration at which tin ions associate. 11.6 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding concentrated solution 5. This concentrated solution 5 had a fluorine concentration of 0.50 M and a tin concentration of 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 having a fluorine concentration of 10.1 mM and a tin concentration of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 3.9 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding concentrated solution 6. This concentrated solution 6 had a fluorine concentration of 0.50 M and a tin concentration of 0.17 M.

[0104] Next, 39.2 ml of pure water was added to 0.8 ml of 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 three 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 point corresponds to the initial fluorine concentration at which tin ions associate. 24 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding concentrated solution 7. This concentrated solution 7 had a fluorine concentration of 2.99 M and a tin concentration of 1.03 M.

[0106] Next, 39.0 ml of pure water was added to 1 ml of 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 three 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 point corresponds to the initial fluorine concentration at which tin ions associate. 28 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding concentrated solution 8. This concentrated solution 8 had a fluorine concentration of 1.01 M and a tin concentration of 1.20 M.

[0108] Next, 39.0 ml of pure water was added to 1 ml of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 29 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding concentrated solution 9. This concentrated solution 9 had a fluorine concentration of 0.83 M and a tin concentration of 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. Further ethanol was then added to prepare a total volume of 100 ml to obtain a concentrated solution 10. The fluorine concentration of this 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. Further ethanol was then added to prepare a total volume of 100 ml to obtain a concentrated solution 10. The fluorine concentration of this 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 added to 1 ml of concentrated solution 10 to obtain a total of 40 ml of 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 three times the tin concentration.

[0115] Example 13: 23.2 g of stannous chloride was added to 80 ml of ethanol and dissolved. Further ethanol was then added to prepare a total volume of 100 ml to obtain a concentrated solution 10. The fluorine concentration of this 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 added to 1 ml of concentrated solution 10 to obtain a total of 40 ml of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 13.96 g of stannous chloride and 23.4 g of Sn(BF4)2 were added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding a concentrated solution 11. This concentrated solution 11 had a fluorine concentration of 1.01 M and a tin concentration of 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 97% pure NH₄HF₂ was added to 80 ml of pure water and stirred to obtain a fluoride solution. The fluoride concentration of the fluoride solution was 1.25 M. This fluorine concentration corresponds to the initial fluorine concentration for tin ion association (2.50 M). 23.2 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding a concentrated solution 12. This concentrated solution 12 had a fluorine concentration of 1.00 M and a tin concentration of 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 fluoride concentration of the fluoride solution was 0.29 M. This fluorine concentration corresponds to the initial fluorine concentration at which tin ions associate. 5.3 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Then, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 1. This Comparative Example Concentrated Solution 1 had a fluorine concentration of 0.23 M and a tin concentration of 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 having a fluorine concentration of 60.0 mM and a tin concentration of 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 fluoride concentration of the fluoride solution was 0.26 M. This fluoride concentration corresponds to the initial fluoride concentration at which tin ions associate. 5.3 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Then, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 1. This Comparative Example Concentrated Solution 1 had a fluorine concentration of 0.23 M and a tin concentration of 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 having a fluorine concentration of 10.4 mM and a tin concentration of 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 fluoride concentration of the fluoride solution was 0.29 M. This fluorine concentration corresponds to the initial fluorine concentration at which tin ions associate. 5.3 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Then, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 1. This Comparative Example Concentrated Solution 1 had a fluorine concentration of 0.23 M and a tin concentration of 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 having a fluorine concentration of 199.1 mM and a tin concentration of 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 fluoride concentration of the fluoride solution was 0.62 M. This fluoride concentration corresponds to the initial fluoride concentration at which tin ions associate. 11.6 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Then, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 2. This Comparative Example Concentrated Solution 2 had a fluorine concentration of 0.50 M and a tin concentration of 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 having a fluorine concentration of 8.2 mM and a tin concentration of 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 ion content of this tin fluoroborate solution was approximately 1.25 M. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 3. This Comparative Example Concentrated Solution 3 had a boron-fluoride concentration of 1.60 M and a tin concentration of 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 having a fluorine concentration of 200.1 mM and a tin concentration of 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 fluoride concentration of the fluoride solution was 4.33 M. This fluoride concentration corresponds to the initial fluoride concentration at which tin ions associate. 45 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 4. This Comparative Example Concentrated Solution 4 had a fluorine concentration of 3.99 M and a tin concentration of 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. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 5. This Comparative Example Concentrated Solution 5 had a fluorine concentration of 0 M and a tin concentration of 1.00 M. In other words, Comparative Example 7 dissolved in pure water and did not associate 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 having a fluorine concentration of 26.4 mM and a tin concentration of 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 fluoride concentration of the fluoride solution was 3.36 M. This fluorine concentration corresponds to the initial fluorine concentration at which tin ions associate. 20 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Then, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 6. This Comparative Example Concentrated Solution 6 had a fluorine concentration of 3.00 M and a tin concentration of 0.86 M. The fluorine concentration was more than three 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 having a fluorine concentration of 74.7 mM and a tin concentration of 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 fluoride concentration of the fluoride solution was 1.15 M. This fluoride concentration corresponds to the initial fluoride concentration at which tin ions associate. 29 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Then, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 7. This Comparative Example Concentrated Solution 7 had a fluorine concentration of 0.96 M and a tin concentration of 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 less than 0.8 times 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. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 8. The fluorine source in Comparative Example Concentrated Solution 8 was tin(II) fluoride itself. In other words, the initial fluorine concentration associated with the tin ions was approximately 2.50 M. The fluorine concentration in 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 having a fluorine concentration of 50.0 mM and a tin concentration of 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. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 8. The fluorine source in Comparative Example Concentrated Solution 8 was tin(II) fluoride itself. In other words, the initial fluorine concentration associated with the tin ions was approximately 2.50 M. The fluorine concentration in Comparative Example Concentrated Solution 8 was 2.00 M, and the tin concentration was 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 having a fluorine concentration of 399.7 mM and a tin concentration of 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. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 9. The fluorine source in Comparative Example Concentrated Solution 9 was tin(II) fluoride itself. In other words, the initial fluorine concentration of the tin ions was approximately 0.75 M. The fluorine concentration in Comparative Example Concentrated Solution 9 was 0.57 M, and the tin concentration was 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 having a fluorine concentration of 103.9 mM and a tin concentration of 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. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 9. The fluorine source in Comparative Example Concentrated Solution 9 was tin(II) fluoride itself. In other words, the initial fluorine concentration associated with the tin ions was approximately 0.75 M. The fluorine concentration in Comparative Example Concentrated Solution 9 was 0.57 M, and the tin concentration was 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 point corresponds to the initial fluorine concentration at which tin ions associate. 23.2 g of stannous chloride was added and dissolved in this hydrofluoric acid solution. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 10. This Comparative Example Concentrated Solution 10 had a fluorine concentration of 1.01 M and a tin concentration of 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 having a fluorine concentration of 25.2 mM and a tin concentration of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 23.2 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 10. This Comparative Example Concentrated Solution 10 had a fluorine concentration of 1.01 M and a tin concentration of 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 point corresponds to the initial fluorine concentration at which tin ions associate. 23.2 g of stannous chloride was added to this hydrofluoric acid solution and dissolved. Subsequently, additional pure water was added to a total volume of 100 ml, yielding Comparative Example Concentrated Solution 10. This Comparative Example Concentrated Solution 10 had a fluorine concentration of 1.01 M and a tin concentration of 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 having a fluorine concentration of 25.2 mM and a tin concentration of 24.9 mM.

[0153] Table 1 below shows the main data of the concentrated solutions of Examples, Table 2 shows the main data of the concentrated solutions of Comparative Examples, Table 3 shows the main data of the reaction solutions of Examples, and Table 4 shows the main data of the reaction solutions of Comparative Examples. Tables 3 and 4 also show the fluorine concentration of the concentrated solutions at the time of initial tin ion association.

[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 initially associate is 0.3 M or higher, it can be said that an interface expansion area ratio Sdr of less than 1×10 -4 In addition, in this case, 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 substantially 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 in which the tin ions initially associate is low, divalent tin is oxidized to tetravalent tin by the dissolved oxygen in the solution, precipitating as fine particles that 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 (increasing the value). In other words, during a film formation time exceeding 4 hours, in order to suppress the formation of tetravalent tin to a level that does not affect the unevenness of the subsequently formed catalyst layer, the initial solution must be dissolved in a fluoride solution of 0.3M or higher.

[0165] Examples 11 to 13 are examples using ethanol as the solvent. In these cases, the only dissolved oxygen introduced was the water contained in the hydrofluoric acid added later, and no oxidation due to dissolved oxygen occurred in the concentrated solution. Therefore, the interface expansion area ratio Sdr was stably 1×10 -5 The value of around.

[0166] The fluorine concentration of the fluoride solution to which the tin ions initially associate in Comparative Example 4 is 0.62 M. Based on the above-mentioned examples, it is considered that the interface expansion area ratio Sdr is less than 1×10 -4 , but in reality it is 1×10 -4 Therefore, it is believed that even if the fluorine concentration of the fluoride solution with which the tin ions initially associate is 0.62 M, if the concentration is too dilute (less than 10 mM) when preparing the reaction solution, the generation of tetravalent tin cannot be suppressed to a level that does not affect 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 higher, but the glass substrate dissolved and film formation was impossible. In Example 5, considering that film formation was possible at approximately 2.5 M, the fluorine concentration of the reaction solution was considered to be 2.5 M as the limit.

[0168] Example 7 shows the case where the concentrated solution was rich in fluorine ions relative to tin ions, while Example 9 shows the case where the concentrated solution was poor in fluorine ions relative to tin ions. On the other hand, Comparative Example 8 shows the case where the fluorine ion concentration was increased to more than 3 times that of the tin ions, while Comparative Example 9 shows the case where the fluorine ion concentration was decreased to 80% or less relative to the tin ions. If the fluorine ion concentration is too high relative to the tin ions, the tin ions become diluted, 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 and fluorine ions is dilute (the concentration of tin ions is high), the concentrated solution itself will feel sticky, and the catalyst layer formed later will be uneven (the interface expansion area ratio Sdr is 1×10 -4 above).

[0169] Example 14 used tin fluoroborate (Sn(BF4)2), while Comparative Example 5 used only tin fluoroborate (Sn(BF4)2). 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 of 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 initially associate is 0.3 M or higher, but hydrochloric acid is added during the preparation of the reaction solution. The interface expansion area ratio Sdr is 1×10 -4 This results in noticeable unevenness in the catalyst layer formed later. Hydrochloric acid is a commonly used material in fluorine-containing liquid phase deposition (LPD) methods, but it contributes to deterioration in surface roughness compared to the level required by the present invention (4 hours of film formation). As a result, the present invention achieves a lower surface roughness (higher surface properties) than the conventional LPD method, due to its interface expansion area ratio.

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

[0173] As described above, when a tin oxide film is formed by the 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 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, thereby making the interface expansion area ratio 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] This is considered to be a method for suppressing the hydrolysis of stannous chloride by the dissolved oxygen in the reaction solution. Therefore, even if the fluorine concentration when stannous chloride is initially added is lower than 0.3 M, if ethanol or the like having no dissolved oxygen is used as a solvent, the surface properties can be maintained high (with a low interface expansion area ratio) even for a long period of film formation.

[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 bifluoride or 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 method for forming a tin oxide laminated film with a catalyst layer, comprising the following steps: A step of dissolving a fluoride in water to a concentration of 0.3 M or higher and then dissolving a tin compound therein to prepare a concentrated solution having a fluoride ion concentration based on the fluoride of 0.8 times or higher and a divalent tin ion concentration based on the tin compound of 3 times or lower; diluting the concentrated solution with water to prepare a reaction solution as a stable divalent tin solution having a fluoride ion concentration of 0.01M to 2.5M; The step of immersing the object to be treated in the reaction solution to form a tin oxide layer with a thickness of 2 nm to 60 nm on the surface of the object to be treated; and The step of immersing the object having a tin oxide layer formed on its surface in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer. The fluoride is any of hydrofluoric acid, ammonium bifluoride, and ammonium fluoride.

2. A method for forming a tin oxide laminated film with a catalyst layer, comprising the following steps: A process for preparing a concentrated solution by dissolving a tin compound in a polar solvent other than water; diluting the concentrated solution with a fluoride solution to prepare a reaction solution as a stable divalent tin solution having a fluoride ion concentration of 0.01M to 2.5M based on the fluoride; The step of immersing the object to be treated in the reaction solution to form a tin oxide layer with a thickness of 2 nm to 60 nm on the surface of the object to be treated; and The step of immersing the object having a tin oxide layer formed on its surface in a catalyst solution containing catalyst metal ions to form a catalyst layer on the tin oxide layer. The fluoride is any of hydrofluoric acid, ammonium bifluoride, and ammonium fluoride.

Images