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What is the molecular geometry for CBr4?

Anonymous

∙ 13y ago

Updated: 10/17/2022

whenever carbon atom forms four covalent bonds the basic structure is tetrahedral because carbon in such cases is sp3 hybridized.

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∙ 13y ago

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What is the molecular geometry shape of CBr4?

Tetrahedral

What is the number of lone pairs around the central atom and the molecular geometry of CBr4?

There are no lone pairs and it's tetrahedral.

What is the molecular geometry of SCl6?

The molecular geometry is octahedral.

What is the molecular geometry of C2H2Br2?

The molecular geometry of C2H2Br2 is trigonal planar.The molecular geometry of C2H2Br2 is trigonal planar.

What is the molecular geometry of secl2?

The molecular geometry of secl2 is BENT.

Related Questions

What is the molecular geometry shape of CBr4?

Tetrahedral

What is the number of lone pairs around the central atom and the molecular geometry of CBr4?

There are no lone pairs and it's tetrahedral.

Is CBr4 a molecular?

Yes it is molecular (or covalent) compound

What is the molecular geometry of SCl6?

The molecular geometry is octahedral.

What is the molecular geometry of C2H2Br2?

The molecular geometry of C2H2Br2 is trigonal planar.The molecular geometry of C2H2Br2 is trigonal planar.

What is the molecular geometry of secl2?

The molecular geometry of secl2 is BENT.

What is the molecular geometry of HClO?

The molecular geometry of HClO is bent.

What is the molecular geometry of N2O2?

The molecular geometry of N2O2 is linear.

Is CBr4 ionic or molecular?

CBr4 is a molecular compound because it consists of nonmetallic elements (carbon and bromine) bonded together through covalent bonds. Ionic compounds typically form between a metal and a nonmetal, with electrons being transferred rather than shared.

What is the molecular geometry of IF4?

The molecular geometry of IF4- is square planar.

What is the molecular geometry of NHF2?

The molecular geometry of NHF2 is trigonal pyramidal.

How many molecules are in 393g of CBr4?

The molecular mass of CBr4 is 12.0 + 4(79.9) = 331.6Amount of CBr4 = mass of substance / molecular mass = 393/331.6 = 1.19mol This means that a 393g pure sample contains 1.19 moles of tetrabromomethane. The Avogadro's number is 6.02 x 10^23 So, number of molecules of CBr4 = 1.19 x 6.02 x 10^23 = 7.13 x 10^23

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