Identify three elements and three compounds. How are they similar? How are they different?

1 answer:

6 0

They are similar because they are both pure substances but they are different because compounds have two or more different types of atoms while elements have the same exact type of atoms.

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Draw the lewis structure of sf2 showing all lone pairs.

abruzzese [7]

Answer :

The steps involved in the lewis-dot structure of $SF_2$ are :

First we have to determine the total number of valence electron in $SF_2$ .

Number of valence electrons in S = 6

Number of valence electrons in two F atom = 2(7) = 14

Total number of valence electrons = 6 + 14 = 20 electrons

Placing a bonding pair of electrons between the atoms to form the chemical bonds.

Electrons used in bonding = 4

The remaining electrons are used as a lone pair for central or terminal atoms to achieve an octet to finish the Lewis dot structure.

Electrons used as a lone pair = 20 - 4 = 16

The image is shown below.

6 0

3 years ago

Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropria

Aneli [31]

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

$(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}$