A chemistry contest question, a 08 chemistry contest question

You that"Two of the ligands of c do not have c-n stretch bands"Is there a mistake? I think it's C-NI.

Write the answer first What I did was:

Ni(cn)4](4-) regular tetrahedral structure where ni is 0 valence, and the notation of the four charges is to turn ni and cn out and put a curly brace with 4- written in the upper corner

The oxidation state is 0 and the hybrid form is 3d10

Gas. 4.2K4[Ni(CN)4]+ 6CO=Ni2(CO)6+8KCN Analysis Method 1 The reaction with K added must be a reduction reaction.

The products of the reaction may be negative zero valent and positive monovalent.

Because reactant a is already very stable, and it is a tetrahedral complex, it should be a zero-valent product according to the 16-18 electron rule.

2 cn is a strong field ligand and the rearrangement of electrons takes place.

3 It's not difficult for high school students to do it without learning competitions.

4 I really can't do that if you give it to me, I just came up with this two-bridge ligand

Another should. Solubility magnitude.

That is, the solubility is large, the reaction is generated, and the solubility is small and easy to occur.

The strength of the acidity is judged by experiment or rational analysis.

1.According to the information given.

According to the strong acid to make weak acid, it is good to judge the relative strength.

2.According to the example of structural analysis.

Etoho--H is stronger than ETSHS--H (the absolute value of oxygen electronegativity is larger), so it is more acidic.

Because solvent S--H is easier to break and ionize hydrogen ions.

ET stands for ethyl group.

3.Common to memorize.

4.Look up the table to see that the larger the acid ionization equilibrium constant, the stronger.

Carbonic acidity, hypochlorous acid, phenol, bicarbonate, ......It can be seen that the pH of the concentration of D and E is the smallest, that is, the most hydrogen ions are ionized

pH max: Choice.

Summarizing the law: the metathesis reaction must have the formation of difficult ionization substances....For example....Precipitation, gas, water....

nh4＋

so42－co32－

There must be CO32, SO42 so there is no Ba2 , and from the point of view of the charge balance, there must be another cation, which can only be Na. So C is also right.

The answer is BC

The answer is bc you can go and calculate, the data calculated upstairs is correct, that is:

n（nh4＋）

n(so42－)

n(co32－)=

According to the conservation of charge, it can be known that the positive charge carried by the cation should be equal to the negative charge carried by the anion

If no other ions are present in the solution, then the formula n(NH4) = 2N(SO42)+2N(CO32) holds.

Substituting the above data, we get: <

That is, the cation has less positive charge than the anion, and according to the conservation of charge, it can be seen that the solution contains at least one other cation, and according to the title, the cation is NA

Select D, MXOY, the solid mass is reduced by about 50% after the reaction is complete, and the solids are M after the reaction, so the reduced mass is the mass of the oxygen element in the original MXOY, so the mass of the oxygen element in the original MXOY is about 50%, that is, the relative molecular mass of M in MXOY The relative molecular mass of O.

Bringing D,M2O7 in, the relative molecular mass of M is 110, and the relative molecular mass of O is 112, which is the most similar.

After the D reaction, only M remains, and the mass of the solid decreases by 50%, indicating that the mass fraction of M in the oxide is close to 50 percent, and the D option is the closest. abcd

d option is the closest.

After the reaction, only m remains, and the mass of the solid decreases by 50% 55x=16y x y=2 7