whenever carbon atom forms four covalent bonds the basic structure is tetrahedral because carbon in such cases is sp3 hybridized.
Tetrahedral
There are no lone pairs and it's tetrahedral.
The molecular geometry is octahedral.
The molecular geometry of secl2 is BENT.
See-saw is the molecular geometry, and trigonal bi-pyramidal is the orbital geometry.
Tetrahedral
There are no lone pairs and it's tetrahedral.
Yes it is molecular (or covalent) compound
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.
The molecular geometry is octahedral.
Two possibilities CHBr = CHBr ( 1,2-dibromoethene) This can have two sub-enantiomers; / 'cis'' or 'trans' Where the two bromine atoms are on the same side of the double bond , and is formula ' (z)1,2-dibromoethene. The other being Where the two bromine atoms are on the opposite side of the double bond , and is formula ' (e)1,2-dibromoethene. CH2 = CBr2 ( 1,1- dibromoethene)
The molecular geometry of secl2 is BENT.
The molecular geometry of HClO is bent.
The molecular geometry of N2O2 is linear.
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
The molecular geometry of IF4- is square planar.
The molecular geometry of NHF2 is trigonal pyramidal.