Solve a difficult chemistry problem about the structure of matter in the second year of high school!

x-h，y-n，z-o，w-na

1 na，n，o，h

2 1s22s22p63s1

3 ch4,nh3,h2o,hf,ne,as (arsenic)4 a1=,a2=naoh,b1=hno2,b2=hno35 I don't know the answer, it's about six times, the conditions are not enough.

6 Na+ = OH - greater than H+

7 Nitrous acid is a weak acid, nitric acid is a strong acid, I wonder if this counts as an explanation?

To be honest, this question is not small, and I am not sure if it is. But there is a big problem with this question, and there are many things involved in it. If it is a middle school exam question, it can only be said that it is not rigorous. Good test questions must stand up to the scrutiny of knowledge.

There's a bit of a thunder upstairs.

If it is the ground state, according to the extranuclear electrons z is he, y is li, and w is na. And because x is the most abundant element, x is h.

2) h 1s1 . he 1s2 . li 1s2 2s1. na 1s2 2s2 2p6 3s1

3) z and y have two or three electrons outside the nucleus of the ground state, respectively, which makes it clear that y is the lithium element z is helium.

y3- contains 6 electrons. The atom is a c molecule. The difference between the two electronic layers is whether the question below k and the above is one, if so, the landlord should have copied the question wrong.

These 4 elements do not produce the inscription matter.

If it has nothing to do with the above question, there are too many answers, and the landlord checks to see if it is wrong.

For example, there are two or three electrons outside the nucleus of the ground state of z and y--- I suspect that I have made a mistake, and I am also a sophomore in high school, and I have never seen a problem that would tell the elements so directly. So guess if there are two or three lone electrons outside the nucleus.

Downstairs is the right solution, the landlord should be careful in copying the question.

The first floor is wrong......Carboxyl C is the sp2 hybrid, while methylene c is the sp3 hybrid......The carbonyl group is planar, so the carboxyl group c is sp2 hybridized; Methylene is a V shape so it is sp3 hybridized.

Glycine is H2NCH2COOH, i.e., -aminoacetic acid, which has 1 bond (C=O) in the bonds (2 N-H, 1 N-C, 2 C-H, 1 C-C, 2 C-O, 1 O-H).

H2NCH2COOH + HCl H3N(+)CH2COOH·CL- (or h2NCH2COOH·HCl).

Carboxyl carbon sp3 hybrid, methylene carbon sp2 hybrid. There are nine bonds in glycine, one bond. Its hydrochloride structure is shortened as: Hoocch2NH3 Cl- where the positive sign is marked in the upper right corner of the nitrogen atom.

Fix it. Apologize, I made a mistake in a hurry. The way of hybridization is reversed. Criticism is. Hehe, it's a pleasure to have a problem with the strongman bluepromiser.

1) According to the chemical formula C6H10, the degree of unsaturation = (6*2+2-10) 2=22) can be oxidized by potassium permanganate, and can fade the carbon tetrachloride solution of bromine - there are double or triple bonds 3) but it does not react with dilute sulfuric acid under the catalysis of mercury salt - there is no triple bond 4) Hydrolysis under the reduction of ZN powder only gives an open-chain compound with a molecular formula of C6H10O2 without branching - indicating that it is a cyclic hydrocarbon.

May be 1-cyclohexene, 1-methylcyclopentene, 1,2-dimethylcyclobutene (unstable).

It can be oxidized by potassium permanganate, and can fade the carbon tetrachloride solution of bromine - there are unsaturated bonds but do not react with dilute sulfuric acid under mercury salt catalysis - it is a double bond, not a triple bond, and hydrolysis under the reduction of ZN powder only gives an open-chain compound with a molecular formula of C6H10O2 without branching - no branched chain.

According to the chemical formula C6H10 – there is only one double bond.

It may be 1-hexene or 2-hexene or 3-hexene.

1.f na electronegativity f is the largest of all the elements. The ionization energy is generally the easier it is to lose electrons, the smaller it is.

2. [ne]3s3

3.ad sodium hexafluoroaluminate itself is a ligand, inside is a coordination bond, and the outside is an ionic bond.

4.CO32- is hybridized in sp2, and sp is hybridized in CO2. This is mutually exclusive with valence electrons, the former (4+2) 2 and the latter 4 2

5.The unit cell of CO2 itself is not a three-dimensional centricity, because the spatial orientation of CO2 is not completely consistent. But the title ignores this, and it is enough to be determined at the time. The cube is 12 matches, so 12.

The nearest distance is 1 2 of the diagonal of the bottom, that is, the length of the side of the cube = root 2 a

Density = mass Volume Mass = 4 44 na Volume = 2 roots2 a 3.

h2.~~

4.I don't seem to teach this here, and the hybrid electrons are arranged in 4a squares.

All the atoms of alkynes are in the same straight line, so there are 3 in the same straight line, alkenes have a planar structure, alkynes have a straight line in a plane so 6, and there are 4 in a tetrahedral structure.

This question can only determine that y is o and does not need to be explained.

w is the element with the highest number of unpaired electrons in the same period, and according to the atomic number, w should be in period 3 or 4.

In the third cycle, if you are a high school student and have not learned about the arrangement of s s and p orbitals, it should be Si, because the outer shell has a maximum of 8 electrons, and 4 electrons are semi-filled. If you are a college student, or have learned about the arrangement of s orbital and p orbital, it should be p, because p electron configuration: 1s2 2s2p6 3s2p3, 3 unpaired electrons, that is, s orbital first full and then p orbital.

In the case of the fourth period, the transition element must be involved, and the result should be cr, with its electronic configuration: 1s2 2s2p6 3s2p6d5 4s1, 6 unpaired electrons. If the transition elements are not taken into account, the same as the 3rd cycle.

Z ground state atom has only one unpaired electron and is smaller than w, so there are many answers to z, such as f, na, al (consider s s and p orbitals), cl, k, sc (transition elements).

The results of x and w are related to the results of z and are no longer listed.

There are C-O covalent bonds in B NaCO3 crystals, S-O covalent bonds in sodium sulfate ion crystals, and so on.

c Molecules are made up of atoms and can have covalent bonds such as CO2

d Metal crystals with only metal cations and free electrons.

There are also van der Waals forces in atomic crystals, such as graphite.

Atoms are mainly connected by covalent bonds, such as diamond, silicon dioxide, and aluminum phosphate crystals.

B: For example, copper sulfate crystals, the s-o of sulfate is a covalent bond C: there is a hydrogen bond in the ice, and the ferrocene molecule contains an iron ion, and there is an ionic bond (ferrocene is two cyclopentadiene with an iron sandwiched between them) D:

C contains one cation in ferrocene and no anion. There is also chlorophyll that contains only magnesium ions. a：

Crystals with a spatial network structure formed by strong covalent bonds between adjacent atoms are called atomic crystals.