What are the inverse spinel structures?

  INVERSE SPINEL STRUCTURE (Fe₃O₄-Magnetite):

In Fe₃O₄, Fe⁺² and Fe⁺³ ions are present in the ratio 2:1. it may be considered as having composition FeO.Fe₂O₃. In Fe₃O₄ Oxide arranged in ccp. Fe⁺² ions occupy octahedral voids while Fe⁺³ ions are equally distributed between octahedral and tetrahedral voids

MgFe₂O₄ also has structure similar to magnetite. In this Mg⁺² ions are present in place of Fe⁺² ion in Fe₃O₄. Magnetite has inverse spinet structure.

Related questions;

What are the normal spinel structures?

What are different stacking pattern of atoms in simple cubic cell (SCC) , body cubic cell (BCC) and face centred cell (FCC) unit lattice cell ?

What is the effect of temperature and pressure on crystal structure of Caesium Chloride?

What is the number of atoms on one unit cell of HCP?

If the total number of atoms per unit cell in an hcp structure and a bcc structure gets halved, then ratio of percentage voids in hcp and bcc structures is

If silver iodide crystallizes in a zinc blende structure with I- ions forming the lattice then calculate fraction of the tetrahedral voids occupied by Ag+ ions.

Sodium metal crystallizes In body centered cubic lattice with the cell edge, a= 4.29 A.What is the radius of sodium atom? (III-.JEE 1994)

A metal crystallizes into two cubic phases, face centered cubic (fcc) and body centered cubic (bcc), whose unit cell lengths are 3.5 A° and 3.0 A°, rcspcctively. Calculate the ratio of densities of fcc and bcc. (IIT-JEE 1999)

A metallic element crystallizes into a lattice containing a sequence of layers ABABAB....Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space? ( IIT-.IEF 2006)

The coordination number of Al in the crystalline state of AlCl3 is .... (IIT-JEE 2009)

The number of hexagonal faces that are present in truncated octahedron is .... (IIT-JEE 2011)

What are structural information of Dimomd ?

What are coordination number (nearest neighbour) in a BCC at distance of (a√3/2) and next nearest neighbour at distance (a) and next to next nearest neighbour at distance (a√2) ?.

What are coordination number (nearest neighbour) in a SCC at distance of it lattice parameter (a) and next nearest neighbour at distance (a√2) and next to next nearest neighbour at distance (a√3) ?.

How to calculate radius ratio of square Voids?

How to calculate radius ratio triangular voids?

How to calculate packing fraction or packing efficiency of two dimensional (2D) hexagonal packing solid atoms?

What are the normal spinel structures?

NORMAL SPINEL (AB₂O₄ ) STRUCTURE:

Example of Spinel is a MgAl₂O₄.( mineral) In it oxide ions (O^-2) are arranged in ccp with Mg⁺² ions occupying tetrahedral voids and Al⁺³ ions in a set of octahedral voids.

Many ferrites (such as ZnFe₂O₄) also possess spinel structure. These are very important magnetic materials and are used in telephone and memory loops in computers.

What are different stacking pattern of atoms in simple cubic cell (SCC) , body cubic cell (BCC) and face centred cell (FCC) unit lattice cell ?

"AA AA AA...", the letters refer to the repeating order of the layers, starting with the bottom layer.

"AB AB AB ..." Type of stacking pattern present  a in body centred cubic (BCC).The considerable space shown between the spheres is misleading: spheres are closely packed in bcc solids and touch along the body diagonal.

"ABC ABC ABC..." type of stacking pattern is of face centered cubic (fcc) or cubic close packing (ccp) type.

What is the effect of temperature and pressure on crystal structure of Caesium Chloride?

Effect of temperature on crystal structure:

Increase of temperature decreases the coordination of number, e.g. upon heating to
760 K, the CsCl type crystal structure having coordination 8:8 changed to NaCl type crystal structures having coordination 6:6.

Effect of pressure on crystal structure:

Increase of pressure increases the Co – ordination number during crystallization e.g. by applying pressure, the NaCl type crystal structure having 6:6 coordination number changes to CsCl type crystal having coordination number 8:8

Fore more details click ⏬

CAESIUM CHLORIDE (CsCl) STRUCTURE:

SODIUM CHLORIDE (ROCK SALT TYPE) STRUCTURE:

What is the number of atoms on one unit cell of HCP?

ANALYSIS OF HCP UNIT CELL:

Number of effective atoms in HCP unit cell (Z):

Lattice point:  corner- total 12 carbon contribute 1/6 to the unit cell

Lattice point:   face- total face 2 contribute ½

Lattice point: body centre- total atom 3 (100% contribution)

Find the ratio of Fe+3 and Fe+2 in a non Stoichiometric oxide of iron , formulated as Fe_0.90 S_1.0 .

This is a non Stoichiometric compound formed due to variable valency of transition elements.   In the Compound Fe_0.90 S_1.0  both Fe+3 and F+2 ions present,  if number of  Fe+2 ions are  x then number of Fe+3 ions are (0.90-x).

We can find the value of x by oxidation state method as :

If the total number of atoms per unit cell in an hcp structure and a bcc structure gets halved, then ratio of percentage voids in hcp and bcc structures is

In HCP ; Packing efficiency:

Atoms occupy:74%

Empty space(voids): 28% 

In BCC; Packing efficiency:

Atoms occupy: 68% 

Empty space(voids): 32% 

According to the given condition if the total number of atoms per unit cell in a hcp structure and a bcc structure get halved. Then packing efficiency in HCP and BCC respectively given as:

In HCP ; Packing efficiency:

Atoms occupy: 74/2 = 37%

Empty space(voids): (100 - 37) = 63%

In BCC; Packing efficiency:

Atoms occupy: 68/2 = 34%

Empty space(voids): (100 - 34) = 66%

Then new ratio of their voids = 63%/66% = 21/22

If silver iodide crystallizes in a zinc blende structure with I- ions forming the lattice then calculate fraction of the tetrahedral voids occupied by Ag+ ions.

In AgI, if there are nI^- ions, there will be nAg⁺ ions. As I^- ions form the lattice, number of tetrahedral voids = 2n. As there are nAg⁺ ions to occupy these voids, therefore fraction of tetrahedral voids occupied by Ag⁺ ions = n/2n = ½ = 50%.

Sodium metal crystallizes In body centered cubic lattice with the cell edge, a= 4.29 A.What is the radius of sodium atom? (III-.JEE 1994)

A metal crystallizes into two cubic phases, face centered cubic (fcc) and body centered cubic (bcc), whose unit cell lengths are 3.5 A° and 3.0 A°, rcspcctively. Calculate the ratio of densities of fcc and bcc. (IIT-JEE 1999)

The ratio of the densities of fcc to bcc is

A metallic element crystallizes into a lattice containing a sequence of layers ABABAB....Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space? ( IIT-.IEF 2006)

The sequence ABABAB... indicates hcp unit cell. Now, there are 6 atoms per unit cell and volume of the umt cell is 24√2 r3.Thus, the packing fraction is:

The coordination number of Al in the crystalline state of AlCl3 is .... (IIT-JEE 2009)

AlCl3 exist in CCP unit cell with 6 co-ordinate layer structure and coordination number of Al is (6)

The number of hexagonal faces that are present in truncated octahedron is .... (IIT-JEE 2011)

The number of hexagonal faces that are present in truncated octahedron is  (6) this can seen if given fig.

What are structural information of Dimomd ?

Diamond structure is ZnS type structure  in which carbon atoms forms  a face centred cubic (FCC/CCP) lattice as well as  four out of eight (50%) or alternate tetrahedral voids are occupied by carbon atoms. Every atom in this structure is surrounded tetrahedrally by four other. No discrete molecule can be discerned (identified) in diamond .the entire crystal is giant molecule a unit cell of which is shown as below.

Note : Only those atoms which form four covalent bond produce a repeated 3D structure using only covalent bonds.

Lattice of Diamond is ZnS type structure.

(1) C- form FCC/CCP (4-atoms)

(2) C- atoms present at the (50%) alternative tetrahedral voids (4-atoms)

(3) Total Number of one lattice unit is eight (8) hence molecular formula of diamond is (C8) (i.e. Z= 8)

(4) Number of C-C bond in lattice cell is = 4×4= 16

(5) Number of C-C bond per carbon atom is 16/8=2

(6) The distance between two Corbin atom is d_C-C = a√3/4 and the radius of carbon atom = dc-c/2 = r_c = a√3/2x4

(8) Packing efficiency (PE = π√3/10= 0.34 or 34%):

(9) Voids = 66 % 

Additional Information: 

Related Questions:

What are the normal spinel structures?

What are the inverse spinel structures?

How to calculate radius ratio of square Voids?

How to calculate radius ratio triangular voids?

How to calculate packing fraction or packing efficiency of two dimensional (2D) hexagonal packing solid atoms?

The coordination number of Al in the crystalline state of AlCl3 is .... (IIT-JEE 2009)

The number of hexagonal faces that are present in truncated octahedron is .... (IIT-JEE 2011)

What are structural information of Dimomd ?

In diamond, carbon atoms occupy FCC/CCP lattice point as well as alternate tetrahedral voids. If edge length of unit cell is 3.56pm, the radius of carbon atom is ?

The edge length of unit cubic cell of diamond is 356.7 pm and this cubic unit cell contains 8 carbon atoms calculate (A) Distance between two carbon atoms, assuming them to spheres in contact (B) Radius of carbon atoms (C) Fraction of the total volume that is occupied by the carbon atoms.

A non Stoichiometric oxide of iron represented as Fe_x O_1.0 , contains one Fe+3 for every three Fe+2 ions , then find the value of x is...

Given that one Fe+3 for every Fe+2 and total ions of Fe+3 and Fe+2 is equal to "x"  then number of Fe+3 is x/4 and Fe+2 is equal to 3x/4.

Finally we can calculate the value of x by balancing Oxidation state of Compound, given as :

What is fractional presence of Fe+2 in Fe0.94 O1.0 ?

Fe0.94 O 1.0 is non stachiometric compound hence Nickel contains both Fe+2 and Fe+3 .

We can calculate ℅  presence of Fe+2  and Fe+3 by oxidation method.

(F+2+Fe+3) O1.O

2x+3(0.94-x)-2=0

Fe+2=x=0.82 or 82% and Fe+3= 8%

What are coordination number (nearest neighbour) in a FCC at distance of (a/√2) and next nearest neighbour at distance (a) and next to next nearest neighbour at distance (a√3/√2) ?.

Coordination Numbers:

What are coordination number (nearest neighbour) in a BCC at distance of (a√3/2) and next nearest neighbour at distance (a) and next to next nearest neighbour at distance (a√2) ?.

Coordination Numbers:

(1) The nearest neighbour distance is half of body diagonal (a) root 3/2 (along body diagonal) thereforecoordination number for a given atom in BCC unit cell is 8.

(2) The next nearest neighbour are 6 at distance (a)Lattice parameter.

(3) 3^rd neighbour (Next to Next nearest neighbour) are(12) at distance (a root 2) ( All corner along face diagonal in x,y and Z plane).

What are coordination number (nearest neighbour) in a SCC at distance of it lattice parameter (a) and next nearest neighbour at distance (a√2) and next to next nearest neighbour at distance (a√3) ?.

Coordination Number:

(1) The nearest neighbour distance is just the lattice parameter (a) therefore coordination number for a given atom in SCC unit cell is 6 (six).

(2) The next nearest neighbour are 12 at distancea/root 2 (each face diagonal in x ,y and Z plane).

(3) 3^rd neighbour (Next to Next nearest neighbour) are(8) at distance a root 3 (each corner along body diagonal.

If edge fraction unoccupied in ideal antiflourite structure is x . Calculate x/ 0.097