What is the electronic geometry of sbr4? enter the electronic geometry of the molecule?

2 answers:

7 0

<span>SBr4 has 4 bonds and 1 lone pair of electrons with a trigonal bipyramidal arrangement with the shape of a seasaw.</span>

Gwar [14]3 years ago

4 0

The valence electron configuration of sulfur S = 3s²3p⁴

Valence electron configuration of Br = 4s²4p⁵

In SBr4, the central Sulfur (S) atom forms 4 covalent bonds with the 4 bromine (Br) atoms. In addition it has 1 lone pair of electrons.

Since the electron geometry not only considers the bond pairs but the lone pairs as well, in the case of SBr4 this would be trigonal bipyramidal.

(Molecular geometry would only consider the 4 covalent bonds which would lend it a distorted tetrahedral structure)

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Hello,

The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

$2NaOH+H_2SO_4-->Na_2SO_4+2H_2O$

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

$n_{NaOH}=7.8gNaOH*\frac{1molNaOH}{40gNaOH}=0.2molNaOH\\n_{H_2SO_4}=4.9gH_2SO_4*\frac{1molH_2SO_4}{98gH_2SO_4}=0.050molH_2SO_4\\$

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

$0.2molNaOH*\frac{1molH_2SO_4}{2molNaOH}=0.098molH_2SO_4$

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

$0.05molH_2SO_4*\frac{1molNa_2SO_4}{1mol H_2SO_4} *\frac{142.04gNa_2SO_4}{1molNa_2SO_4} =7.1gNa_2SO_4$