The molecular geometry of the BR3 Lewis structure is trigonal planar.
The hybridization of a Br3- molecule is sp3d since bromine has 7 valence electrons in the 4th period and requires 2 electrons to complete its octet. Therefore, it forms three bonds in a trigonal bipyramidal molecular geometry.
No, Br3 does not have polar covalent bonds. It is made up of three bromine atoms, which are all part of the same group in the periodic table and have similar electronegativities. As a result, there is no significant difference in electronegativity to create a polar covalent bond.
The formula is C2Br6. This compound is also known as hexabromoethane.
When iron reacts with bromine, it forms iron (III) bromide, a solid compound that appears as a reddish-brown powder. This reaction is highly exothermic, releasing heat energy. Iron bromide is an important compound used in organic synthesis and as a catalyst in certain reactions.
The trihalide ions I3- and Br3- are well known. They are the smallest members of groups of polyhalide ions In- and Brn- The Iodine groups is the best knwn. In comparison only F3- has been isolated and that has proved unstable. A simple explanation is that bromine and iodine have d orbitals which can take part in the bonding wheres fluorine does not. However these compounds have caused a lot of debate and there is no clear cut concensus on the bonding mechanism. If you are really inetested in this follow links to hypervalency, three center 4 electron bonds, Musher.
Linear
The hybridization of a Br3- molecule is sp3d since bromine has 7 valence electrons in the 4th period and requires 2 electrons to complete its octet. Therefore, it forms three bonds in a trigonal bipyramidal molecular geometry.
The electron pair geometry of Br3 (tribromide ion) is trigonal planar. This is due to the presence of three bromine atoms bonded to a central bromine atom, with no lone pairs on the central atom. The arrangement minimizes electron pair repulsion according to VSEPR (Valence Shell Electron Pair Repulsion) theory.
The symbol for Tribromide is Br3.
bronze!
No, Br3 does not have polar covalent bonds. It is made up of three bromine atoms, which are all part of the same group in the periodic table and have similar electronegativities. As a result, there is no significant difference in electronegativity to create a polar covalent bond.
The formula is C2Br6. This compound is also known as hexabromoethane.
The empirical formula for a compound containing 13% magnesium and 87% bromine is MgBr2. This is because the ratio of magnesium to bromine atoms in the compound is 1:2, which corresponds to the formula MgBr2.
There is no anion which is named Tribromide, but the Bromide ion exists. The Tribromide is formed when a compound has 3 Bromine atom attached to an less electronegative element.
When iron reacts with bromine, it forms iron (III) bromide, a solid compound that appears as a reddish-brown powder. This reaction is highly exothermic, releasing heat energy. Iron bromide is an important compound used in organic synthesis and as a catalyst in certain reactions.
The trihalide ions I3- and Br3- are well known. They are the smallest members of groups of polyhalide ions In- and Brn- The Iodine groups is the best knwn. In comparison only F3- has been isolated and that has proved unstable. A simple explanation is that bromine and iodine have d orbitals which can take part in the bonding wheres fluorine does not. However these compounds have caused a lot of debate and there is no clear cut concensus on the bonding mechanism. If you are really inetested in this follow links to hypervalency, three center 4 electron bonds, Musher.
1) First write the bonding sequence. Usually the first atom, other than H is the central atom. In this example Br is the central atom. Write Br first and draw two bonds to connect the 2 extra Br atoms in the Br3-. Br-Br-Br 2) Count the total number of valence electrons : 3 X 7 + 1 = 22 (since bromine belongs to VII group, each Br has 7 valence electrons and add 1 for each - ve charge) 3) Since there are two bonds in the structure drawn in rule 1, subtract 4 bonding electrons (2 electrons for each bond) from the total valence electrons, 22 (22-4 =18). This gives 18 electrons. 4) Distribute these electrons to the atoms in the Br3-. First satisfy the surrounding atoms, making sure that each atom has an octet around it (H will have doublet). Each surrounding Br atom needs 6 electrons to attain octet. Place three pairs of electrons on each of the surrounding Br atom. .. .. :Br - Br - Br: (Note: lone pairs are not placed correctly, and I am not able to edit it) .. .. 12 electrons (2 x 6=12) are used. Still 6 (18-6= 12) electrons are remaining from rule 3. Place them as three pairs around the central Br atom. .. .. .. :Br - Br - Br: .. .. .. .. In this example, the central atom Br has 10 electrons around it, in other words it exceeded the octet. Elements from the third period and below can exceed the octet since they have the d orbitals to accommodate the excess electrons. Since this is an ionic species, draw a square bracket and show the -ve charge of the ion as superscript. .. .. .. _ [:Br - Br - Br:] .. .. .. .. (Note: This editor doesn't let me put the lone pairs on the atoms correctly. I tried to edit and correct it, but goes back to the same way as given above. You please redraw it with three electron pairs around each Br with the -ve charge superscripted.) Pushpa Padmanabhan Lansing Community College