Indicate how many stereoisomers are possible for each compound.

b) tetrahedral [NiCl2Br2]2–

c) octahedral [Fe(CO)4BrCl]

Yes you're correct.

b) 1 isomer

c) 2 isomers (cis and trans)

It helps to rotate the 3 dimensional shapes in your head

Check me out on this but I think b can't have any and c has two (cis and trans).

b) tetrahedral [NiCl2Br2]2–: Well, "NiCl2Br2" sounds like a formula for a really evil chemist's signature dish. But let me count the stereoisomers before I order a taste-test. Ahem. Okay, so for a tetrahedral complex, we have 4 ligands arranged around the central atom, which is nickel (Ni) in this case. Each of the ligands can be either chlorine (Cl) or bromine (Br), giving us 2 options for each ligand. Since we have 4 ligands, the total number of stereoisomers for the [NiCl2Br2]2– compound is 2 x 2 x 2 x 2 = 16. So, it seems that this dish has 16 different ways to attack your taste buds!

c) octahedral [Fe(CO)4BrCl]: Ah, an octahedral complex! It sounds like a fancy dance move in the world of chemistry. Picture a central iron (Fe) atom surrounded by 6 ligands. Let's see, we have 4 carbon monoxide (CO) ligands and 2 other ligands, bromine (Br) and chlorine (Cl) in this case. For each CO ligand, there can be either a bromine or chlorine ligand on the opposite side. So, we have 2 possibilities for each of the 4 CO ligands and 2 possibilities for the remaining two ligands. Therefore, the total number of stereoisomers for [Fe(CO)4BrCl] would be 2 x 2 x 2 x 2 x 2 x 2 = 64. That's 64 delicious spins on the dance floor of chemical bonding!

b) Tetrahedral [NiCl2Br2]2–:

To determine the number of stereoisomers for a tetrahedral complex, we need to consider the different ways the ligands can be arranged around the central metal atom. In a tetrahedral complex, the ligands can be arranged in one of two ways: either as a cis or trans configuration.

For the compound [NiCl2Br2]2–, we have two ligands of chloride (Cl) and two ligands of bromide (Br). Let's consider the possible arrangements:

1) Cis configuration: In the cis configuration, the two chloride ligands are adjacent to each other, and the two bromide ligands are also adjacent to each other.

Cl Br Br Cl

2) Trans configuration: In the trans configuration, one chloride ligand is opposite to one bromide ligand, and the other chloride is opposite to the other bromide.

Cl Br
|
Cl Br

Therefore, there are two possible stereoisomers for the compound [NiCl2Br2]2–.

c) Octahedral [Fe(CO)4BrCl]:

Similarly, for octahedral complexes, we need to consider the different ways the ligands can be arranged around the central metal atom. In an octahedral complex, the ligands can be arranged in a cis or trans configuration, as well as in different axial positions.

The compound [Fe(CO)4BrCl] consists of one bromide (Br), one chloride (Cl), and four carbonyl (CO) ligands. Let's consider the possible arrangements:

1) Cis configuration: In the cis configuration, the bromide and chloride ligands are adjacent to each other, and the carbonyl ligands are also adjacent to each other.

Br Cl
\ /
Fe
/ \
CO CO CO CO

2) Trans configuration: In the trans configuration, the bromide and chloride ligands are on opposite sides of the central metal atom, and the carbonyl ligands are also on opposite sides.

Br Cl
\ /
Fe
/ \
CO CO
|
CO
|
CO

Additionally, the four carbonyl ligands can occupy different axial positions in an octahedral complex, resulting in different cis and trans isomers.

Therefore, there are multiple possible stereoisomers for the compound [Fe(CO)4BrCl]. The exact number of stereoisomers will depend on the specific arrangements of the carbonyl ligands in axial positions.

To determine the number of stereoisomers for a compound, we need to consider the presence of chiral centers and the symmetry of the molecule.

b) tetrahedral [NiCl2Br2]2–:
In this compound, we have a tetrahedral geometry around the central nickel (Ni) atom. To determine the number of stereoisomers, we need to check if the molecule has chiral centers.

A chiral center is an atom that is bonded to four different groups or ligands. In the compound [NiCl2Br2]2–, the nickel atom (Ni) is bonded to two chloride ions (Cl) and two bromine ions (Br). Since the ligands are not all different, the molecule does not have any chiral centers.

Hence, there are no stereoisomers for [NiCl2Br2]2–.

c) octahedral [Fe(CO)4BrCl]:
In this compound, we have an octahedral geometry around the central iron (Fe) atom. Again, we need to check for chiral centers.

The iron atom (Fe) is bonded to four carbon monoxide (CO) ligands, one bromine ion (Br), and one chlorine ion (Cl). Since the ligands are not all different, there are no chiral centers in this compound.

So, there are no stereoisomers for [Fe(CO)4BrCl].