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Consider the following electron configuration. Give four species that in theory, would have this electron configuration (σ3s)2(σ3s*)2(σ3p)2(π3p)4(π3p*)4.

Short Answer

Expert verified

Four species with this electron configuration are Cl2,S22-,SCl-,Ar22+

Step by step solution

01

Step 1- Finding electronic configuration

The given configuration starts with 3s which shows they are valence configurations, So the total number of electrons calculated is 14. So now the species should have 14 valence electrons and should be under the 3rd shell as the configuration is up tothe 3rd shell only.

02

Explanation regarding chlorine

Two chlorine atom makes a chlorine molecule. Each chlorine atom has 7 valence atoms. So, a chlorine molecule will have 14 valence electrons. Also, the valence electronic configuration of chlorine is 3s23p5and they are under the 3rd shell.

03

Explanation regarding S22-

Sulphide ion is formed by two sulphur atoms each sulphur atom has 6 valence electron so, the sulphide ion has 12 electrons while the two negative charges is added so the total is 14 valence electrons. Also, the valence electronic configuration for a sulphur atom is 3s23p4. They are also under 3rd shell.

04

Explanation regarding SCl-

This molecule is formed by sulphur and chlorine atom with a negative charge on it which is added. Sulphur has 6 valence electrons and chlorine has 7 valence electrons and with a negative charge so the total valence electrons are 14. Also, the valence electronic configuration of sulphur atom is 3s23p4and of chlorine atom is 3s23p5. They are under 3rd shell.

05

Explanation regarding Ar22+ 

Each argon atom consists of 8 valence electrons. The ion bears two positive charges which will be subtracted. Hence total valence electrons are 14. The electronic configuration of the argon atom is 3s23p6. They are also under 3rd shell.

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Most popular questions from this chapter

Values of measured bond energies may vary greatly depending on the molecule studied. Consider the following reactions:

NCl3(g)NCl2(g)+Cl(g)ΔH=375kJ/molONCl(g)NO(g)+Cl(g)ΔH=158kJ/mol
Rationalize the difference in the values ofΔHfor these reactions, even though each reaction appears to involve the breaking of only one N-Cl bond. (Hint: Consider the bond order of the NO bond in ONCl and in NO.)

Complete the Lewis structures of the following molecules. Predict the molecular structure, polarity, bond angles, and hybrid orbitals used by the atoms marked by asterisks for each molecule.

Cholesterol C27H46Ohas the following structure

In such shorthand structures, each point where lines meet represents a carbon atom and most H atom are not shown. Draw the complete structure showing all carbon and hydrogen atom (There will be four bonds to each carbon atom). Indicate which carbon atoms use sp2or sp3 hybrid orbital. Are all carbon atoms in the same plane as implied by the structure?

Question: How does molecular orbital theory explain the following observations?

a. H2 is stable, whereas He2 is unstable.

b. B2 and O2 are paramagnetic, whereas C2 , N2and F2 are diamagnetic.

c. N2 has a very large bond energy associated with it.

d. NO+ is more stable than NO- .

The NMR spectra on the right are for the organic compounds C6H12 and C4H10O. Deduce the structures for these compounds. See Exercise 74 for a discussion of the bonding in organic compounds. The structure of C6H12 has one double bond, with the rest being single bonds, and the structure of C4H10O has only single bonds. Note that the TMS reference has been omitted in each spectrum.

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