Find the unit cell (simple, bcc, fcc), type of holes where the smaller ions are found (tetrahedral, octrahedral, cubic), and number of cations and anions per unit cell for the following compounds:

CdSe
CsI
Li2O
KBr
NaCl
ZnS(I)
i did this and i got most of my answers wrong!!!
and I got this but they're wrong so any help would be great!:
CdSe - fcc, tetrahedral, 3,3
CsI - fcc, octahedral,1,1
Li2O - sc, tetrahedral, 2,2
KBr - bcc, tetrahedral,4,8
NaCl - bcc, cubic,1,2
ZnS(I) - sc, cubic,1,1

CdSe: fcc, tetra, 4, 4

CsI: sc, cubic, 1,1
Li2O: fcc, tetra, 8, 4
KBr: fcc, octa, 4, 4
NaCl: fcc, octa, 4, 4
ZnS: fcc, tetra, 4, 4

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To determine the unit cell type and the type of holes where smaller ions are found, as well as the number of cations and anions per unit cell for the given compounds, we need to consider the arrangement of ions and their coordination in the crystal lattice.

1. CdSe (Cadmium Selenide):
To find the unit cell type for CdSe, we need to consider its crystal structure. CdSe has a zinc blende crystal structure, which belongs to the cubic crystal system (Cubic Close-Packed - ccp or face-centered cubic - fcc). Therefore, the unit cell for CdSe is fcc.

For the type of holes where the smaller ions are found, in the zinc blende structure, CdSe has a tetrahedral coordination of the smaller ions, which in this case are the Se ions. Hence, the type of holes is tetrahedral.

The number of cations and anions per unit cell can be determined by considering the formula of the compound. CdSe has a 1:1 ratio of Cd cations to Se anions. Therefore, there is 1 cation and 1 anion per unit cell.

So, the correct answers for CdSe are: fcc unit cell, tetrahedral holes, and 1 cation and 1 anion per unit cell.

2. CsI (Cesium Iodide):
CsI crystallizes in a body-centered cubic (bcc) structure.

For the type of holes where smaller ions are found, CsI has octahedral holes, where the smaller ions (Iodide ions) are located.

The formula of CsI indicates a 1:1 ratio of Cs cations to I anions. Hence, there is 1 cation and 1 anion per unit cell.

So, the correct answers for CsI are: bcc unit cell, octahedral holes, and 1 cation and 1 anion per unit cell.

3. Li2O (Lithium Oxide):
Li2O has a simple cubic (sc) structure.

For the type of holes where smaller ions are found, Li2O has tetrahedral holes, where the smaller ions (Li ions) are located.

The formula of Li2O indicates a 2:1 ratio of Li cations to O anions. Therefore, there are 2 cations and 2 anions per unit cell.

So, the correct answers for Li2O are: sc unit cell, tetrahedral holes, and 2 cations and 2 anions per unit cell.

4. KBr (Potassium Bromide):
KBr crystallizes in a face-centered cubic (fcc) structure.

For the type of holes where smaller ions are found, KBr has octahedral holes, where the smaller ions (Br ions) are located.

The formula of KBr indicates a 1:1 ratio of K cations to Br anions. Thus, there is 1 cation and 1 anion per unit cell.

So, the correct answers for KBr are: fcc unit cell, octahedral holes, and 1 cation and 1 anion per unit cell.

5. NaCl (Sodium Chloride):
NaCl has a face-centered cubic (fcc) structure.

For the type of holes where smaller ions are found, NaCl has a cubic hole arrangement, where the smaller ions (chloride ions - Cl) are located.

The formula of NaCl shows a 1:1 ratio of Na cations to Cl anions. Therefore, there is 1 cation and 1 anion per unit cell.

So, the correct answers for NaCl are: fcc unit cell, cubic holes, and 1 cation and 1 anion per unit cell.

6. ZnS(I) (Zinc Sulfide):
ZnS(I) has a simple cubic (sc) structure.

For the type of holes where smaller ions are found, ZnS(I) has a cubic hole arrangement, where the smaller ions (Sulfur ions - S) are located.

The formula of ZnS(I) indicates a 1:1 ratio of Zn cations to S anions. Hence, there is 1 cation and 1 anion per unit cell.

So, the correct answers for ZnS(I) are: sc unit cell, cubic holes, and 1 cation and 1 anion per unit cell.

Make sure to double-check your calculations and consider the crystal structures and ion arrangements to arrive at the correct answers.