Ceramic Bonding

• Recall ceramic bonding: - Mixed ionic and covalent. - % ionic character ( f ) increases with difference in electronegativity → “ionic ceramics” “covalent ceramics” • Large vs small ionic bond character:

CaF2: large SiC: small

•Since ceramics are composed of two or more elements, their crystal structures tend to be more complex than those of metals. 1 Site Selection in Ceramics

Which sites will the ions occupy in the crystalline lattice? 1. Size of sites – Because the electrostatic attractions in ionic bond are isotropic, we should expect ionic (ceramic) solids to form close-packed structures. – However the number of nearest oppositely charged neighbors (CN) in ionically bonded structures are influenced by the relative size of the cation and anion, (ratio of the two) e.g. does the cation fit in the site?

• Analogous to interstitials (OH + TD) fitting between closed-packed sites. – To obtain a stable structure, need to maximize the number of nearest oppositely charged neighbors that form an ionic bond.

• Just like we saw with increasing U' and Vo by increasing the # of N + a 2. Stoichiometry – If all of one type of site is full, the remainder have to go into other types of sites.

• If cations prefer OH sites and they are all full then remaining will go to TD sites 3. Bond Hybridization – The hybrid orbitals can have impact if significant covalent bond character present. 2 • Examples: group IV, group III-V and II-VI compounds. Site Selection Rule 1

• Ceramic structures are composed of electrically charged ions instead of atoms (as in metals). • The metallic ions, or cations, are smaller and positively charged since they give up their valence electrons to the non-metallic, negatively charged ions, or anions. • Two characteristics of the component ions in ceramic materials influence crystal structure for site selection rule #1: A m X p 1- Magnitude of the electrical charge on the each of the component ions. m, p values to •Crystal must be electrically neutral or balanced charges: achieve charge neutrality F- •Chemical formula of a compound indicates the ratio of cations to anions, 2+ Ca anions or composition that achieves this charge balance, e.g. in CaF2: cation+ 2 F- ions and 1 Ca2+ ion. F-

2- Relative sizes of the cations (rc) and anions (ra), radius ratio (r) = rc /ra (typically <1). •Each one desires as many nearest neighbors as possible for stability (larger lattice energy, Vo). •Stable ceramic structures form when all anions are in contact with that cation: (a) (b) (c) (d) ------+ + + - + ------stable stable unstable r=0.41 r=0.26 stable? r=0.70 3 •The coordination number (# of anion nearest neighbors for a cation) is related to rc/ra ratio. Site Selection Rule 1 (continued)

Cation Site Size:

• Determine minimum rcation/ranion for OH site (C.N. = 6)

2 ranion + 2rcation = 2 2ranion 2 ranion + 2rcation = 2a r anion + rcation = 2ranion r cation = ( 2 −1)ranion  a = 2r r  anion r = cation = 0.414 Cation in red; anion in blue  ranion  What this physically means:

•If rc/ra < 0.414 for CN=6, then the structure is unstable (anion-anion repulsions, cation too small for this CN), which means it would favor a lower CN, e.g. CN=4.

•If rc/ra > 0.414 for CN=6, then the structure is distorted but still stable up until CN=8 ( rc/ra = 0.732). 4 Site Selection Rule 1 (continued) Cation to Anion radius ratio (r) •How many anions can you arrange around a cation to form a stable structure? •As we just determined for a specific coordination number there is a critical or minimum rc/ra ratio for which linear cation-anion contact is made (based on geometry):

• Show that minimum rc/ra ratio for CN=3 is 0.155 triangular ZnS (zinc blende) Tetrahedral (TD)

•CN=12 for rc/ra ratios>1 (cuboctahedron): 8 triangular/6 square faces; Most ceramics CN = 4, 6, or 8. NaCl •Numbers are based on geometrical considerations and assuming hard (sodium chloride) Octahedral sphere ions, thus ranges are only (O ) approximate, ions can be non-spherical H in anisotropic crystals. CsCl •Also, covalent bonding can be (cesium overriding, e.g. with rc/ra ratio>0.414 chloride) Cubic 5 in which bonding is highly covalent (and Adapted from Table 3.3, 5 directional) have CN=4, instead of CN=6. Callister & Rethwisch 3e. Example Problem for AB compound: Predicting the Crystal Structure of FeO

• On the basis of ionic radii, what crystal structure would you predict for FeO?

Cation Ionic radius (nm) Al 3+ 0.053 • Answer: Fe 2 + 0.077 r 0.077 cation = Fe 3+ 0.069 ranion 0.140 2+ Ca 0.100 = 0.550 Data from Table 3.4, Callister & Rethwisch 3e. Anion based on this ratio, -- CN= 6 because O 2- 0.140 0.414 < 0.550 < 0.732 Cl - 0.181 -- thus, a good choice for the F - 0.133 crystal structure is NaCl 6 Site Selection Rule 1 (continued)

Review of CN increasing with r: DeGraef p.568/661 from Rohrer Pauling hard sphere radii

r = 0.155-0.225 r = 0.225-0.414 r = 0.414-0.732

r = 0.732-1 r > 1 •Carefully chosen examples, Table 1.6, can make radius ratio concept look like an accurate predictive tool. •However, it can be in error (noted above) particularly in complex structures and when the bonding becomes increasingly covalent (where hard sphere model breaks down). •Important reasons for inadequacy of this rule are the assumption of spherically symmetric forces & symmetric coordination, the assumption that atoms have the same size in all chemical environments (when in fact ionic radii change with CN). •The primary problem with the radius ratio rules is that the ions are not rigid and thus fixed ionic 7 radii are not realistic using Pauling’s radii…..→ Site Selection Rule 1 (continued)

•Using our model discussed previously for ionic bonding, it is possible to examine the change in ionic radius with CN. •We begin by assuming that an AB compound can exist in both Rocksalt, RS (CN=6) and CsCl

(CN=8) structures and that the anion size is constant. 12 − ke 2a( n + n )Z Z  s   •We need only calculate r , interionic separation. Recall V ( r ) = 1 2 1 2 + 4N   o 2r r that if    − ke 2ah V ( h ) = + 4Nh12 then let s/r=h and n1=n2=Z1=Z2=1, thus: s •Recall from Class 6/slide 1 (NaCl) that at ro, the derivative of V(h) with respect to r is equal to 0, 2 1 / 11 ke a 11 or  48 N s  where h=s/r = 48Nh ro = s  s  ke 2a  •We can write the ratio of ro in the 8 coordinate structure to ro in the 6 coordinate structure as:

1 / 11 1 / 11 This numerical result has the physical implication r8  N a   8 1.75  o =  CsCl • RS  =  •  = 1.026 that the interionic separation `expands´ by 3% r6  N a  6 1.76 o  RS CsCl    during the switch from a 6 to 8 coordinate configuration. We can use the same method to find that the ion `contracts´ by 3% when it goes from a 6 to 4 coordinate configuration. •These observations justify Shannon’s development of ionic radii that depend on CN. Data can be found in handout and pp.563/564 (656/657) that are based on experimental XRD measurements. •Using his values for the binary compounds in Table 1.6, the predicted and observed CN 8 are in much better agreement than using Pauling’s hard sphere radii, e.g. NiO…. Cation to Anion radius ratio (r) (continued) •The results based on this Table can be generalized to determine restrictions on radius from DeGraef p.664 (1st ed) [or Table 21.6 p.569 (2nd ed)] ratios for non-equiatomic compounds, as summarized in this Table:

•For a compound, AnBm, the CN’s must satisfy the following relation in order to preserve the CN m stoichiometry: A = ex.: CaF2……. CN B n •In addition, we must consider charge balance, e.g. an A2B3 compound would have +3 A atoms and -2 B atoms to maintain charge balance. •Similar to our ro calculations based on N and a: •Ionic size depends on 1) CN: ionic radius increases as number of nearest-neighbor ions of opposite charge increases (Figure on right): 2) Charge on ion influences its radius, e.g. Fe3+ (0.069nm for CN=6) < Fe2+ (0.077nm for CN=6) < Fe atom (0.124nm). →Since when an electron is removed from an atom or ion, the remaining valence electrons become more tightly bound to the nucleus which results in a decrease in ionic radius and vice versa 9 (increase radius when adding electrons). Site Selection Rule 1 (continued)

•Example: CsCl structure. Cs+ cation sits in an 8-fold cubic coordination. The Cl- anion polyhedron is also a cube and neighboring cubes share faces. The CN of Cl- is also 8. The anion and cation radii are almost the same: r + 0.174 Cs = = 0.961 r − 0.181 Cl which is not typical for ionic structures. •Notice the Cl- anions are not close packed. •The Cs+ cation cannot fit in either the octahedral or tetrahedral site, instead it prefers the more open structure Note: CsCl is not a BCC lattice associated with CN=8. as we saw previously with BCC •Cs+ cation and Cl- anions touch along the body diagonal. metals such as W: The lattice constant, a, equals ? which predicts a=0.410 nm which is in excellent agreement with experimentally measured a=0.412 nm. •To determine CsCl size limitations we refer to previous Table for AB compound, where both ions have CN=8. R •CsCl structure is predicted to form with r ranging from 0.732  A  1.37 R •The above ratio is close to one (0.961) which falls near B the middle of this range. 10