How do I know crystal structures from formula?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Give an explanation why $ceMgF2$ and $ceCaF2$ adopt different structure types.
For example, the structure of calcium fluoride is 'fluorite' and the structure of magnesium fluoride is 'rutile'. I am struggling, just from looking at the formulae, on how I determine this?
inorganic-chemistry crystal-structure solid-state-chemistry
edited Jan 10 at 21:11
andselisk
15k649108
$endgroup$
add a comment |
$begingroup$
Give an explanation why $ceMgF2$ and $ceCaF2$ adopt different structure types.
For example, the structure of calcium fluoride is 'fluorite' and the structure of magnesium fluoride is 'rutile'. I am struggling, just from looking at the formulae, on how I determine this?
inorganic-chemistry crystal-structure solid-state-chemistry
edited Jan 10 at 21:11
andselisk
15k649108
$endgroup$
4
$begingroup$
You don't. That being said, a post factum explanation is somewhat possible.
$endgroup$
– Ivan Neretin
Jan 10 at 18:03
$begingroup$
Just to note that it's a bit misleading to say that calcium fluoride adopts the fluorite structure. Calcium fluoride is fluorite, and the structure was named after the mineral.
$endgroup$
– Gimelist
Jan 12 at 3:42
add a comment |
$begingroup$
Give an explanation why $ceMgF2$ and $ceCaF2$ adopt different structure types.
For example, the structure of calcium fluoride is 'fluorite' and the structure of magnesium fluoride is 'rutile'. I am struggling, just from looking at the formulae, on how I determine this?
inorganic-chemistry crystal-structure solid-state-chemistry
edited Jan 10 at 21:11
andselisk
15k649108
$endgroup$
Give an explanation why $ceMgF2$ and $ceCaF2$ adopt different structure types.
For example, the structure of calcium fluoride is 'fluorite' and the structure of magnesium fluoride is 'rutile'. I am struggling, just from looking at the formulae, on how I determine this?
inorganic-chemistry crystal-structure solid-state-chemistry
inorganic-chemistry crystal-structure solid-state-chemistry
edited Jan 10 at 21:11
andselisk
15k649108
edited Jan 10 at 21:11
andselisk
15k649108
edited Jan 10 at 21:11
andselisk
15k649108
edited Jan 10 at 21:11
andselisk
15k649108
edited Jan 10 at 21:11
andselisk
15k649108
15k649108
asked Jan 10 at 17:23
user69871
4
$begingroup$
You don't. That being said, a post factum explanation is somewhat possible.
$endgroup$
– Ivan Neretin
Jan 10 at 18:03
$begingroup$
Just to note that it's a bit misleading to say that calcium fluoride adopts the fluorite structure. Calcium fluoride is fluorite, and the structure was named after the mineral.
$endgroup$
– Gimelist
Jan 12 at 3:42
add a comment |
4
$begingroup$
You don't. That being said, a post factum explanation is somewhat possible.
$endgroup$
– Ivan Neretin
Jan 10 at 18:03
$begingroup$
Just to note that it's a bit misleading to say that calcium fluoride adopts the fluorite structure. Calcium fluoride is fluorite, and the structure was named after the mineral.
$endgroup$
– Gimelist
Jan 12 at 3:42
4
4
$begingroup$
You don't. That being said, a post factum explanation is somewhat possible.
$endgroup$
– Ivan Neretin
Jan 10 at 18:03
$begingroup$
You don't. That being said, a post factum explanation is somewhat possible.
$endgroup$
– Ivan Neretin
Jan 10 at 18:03
$begingroup$
Just to note that it's a bit misleading to say that calcium fluoride adopts the fluorite structure. Calcium fluoride is fluorite, and the structure was named after the mineral.
$endgroup$
– Gimelist
Jan 12 at 3:42
$begingroup$
Just to note that it's a bit misleading to say that calcium fluoride adopts the fluorite structure. Calcium fluoride is fluorite, and the structure was named after the mineral.
$endgroup$
– Gimelist
Jan 12 at 3:42
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I'll just point out the direction in which you could look for an answer.
In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.
You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.
For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.
For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.
answered Jan 10 at 18:57
voffchvoffch
4955
$endgroup$
add a comment |
$begingroup$
Not really a complete answer, rather an addition to the given answer (which I think you should accept). As of now, it's possible to predict existence of certain compounds, mostly binary and ternary inorganic ionic solids, at the given temperature and pressure, as well as their crystal structure.
There are many computer programs, among which USPEX is arguably the most feature-rich and promoted. It uses Oganov-Glass evolutionary algorithm and works best for network solids (including particles and clusters). Fairly recently there also were some advancements for predicting crystal structures of small organic molecules.
answered Jan 10 at 21:09
andseliskandselisk
15k649108
$endgroup$
2
$begingroup$
Suggestion: The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. ccdc.cam.ac.uk/Community/initiatives/cspblindtests). Which is why I'm going to replace the link set by you by the one pointing to the sixth one -- equally published as open access article on Acta Cryst B.
$endgroup$
– Buttonwood
Jan 10 at 22:01
$begingroup$
@Buttonwood Good catch, thank you for the edit!
$endgroup$
– andselisk
Jan 10 at 22:04
1
$begingroup$
You knew better about USPEX, and me a bit about the contests. It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.
$endgroup$
– Buttonwood
Jan 10 at 22:19
add a comment |
$begingroup$
You can treat it as a "generic problem solving" type of question. @voffch gave some specific ways to do this.
You're asked why they form different structures, but don't have to say anything about their actual structures. This question sounds like something from an introductory undergrad general/inorganic course or high school honors gen chem course. Typical material for such a course discusses ionic structures in terms of sphere packing/ionic radius (keywords: fcc, bcc, hcp) and covers ionic radius in terms of electronegativity and shielding.
You're given:
- Mg(II) and Ca(II) are different cations
- Their fluorides form different structures (e.g. the anion is the same)
- (you should know) Mg and Ca pretty much always are in +2
So the question is implicitly "Of the differences between Mg and Ca cations, which differences make their fluorides have different structures?."
The specific mention of F suggests there's something special about it too. F is pretty "extreme" all the way in its corner so it was probably chosen as a "superlative," e.g. "most electronegative," "smallest, "biggest occupational hazard" or something like that.
At this point, you can just compile a list of all the things you've learned about metal/halide ions over the relevant time period and pick ones that affect geometry. Also consider what you've been taught about crystal structures.
It could play out like this:
- The course covered elemental trends in ionic radius, and reasons why (electronegativity, shielding)
- The course covered simple ionic structures (sphere packing)
- Sphere packing is only affected by radius
- Mg, Ca have different radii. Both are "big"
- F is "small"
- The reason the structures are different is because Mg, Ca have different radii
- (Only if the course covered this) Mg(II) is [bigger/smaller] than Ca(II) because [electronegativity, orbitals, etc.]
By way of explanation, you can explain radius trends + draw a sketch of the different packings, approximately to scale.
If your course covered different material-- Compare Mg, Ca in terms of the stuff your course covered.
answered Jan 11 at 20:06
jnj16180340jnj16180340
212
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "431"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f107762%2fhow-do-i-know-crystal-structures-from-formula%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'll just point out the direction in which you could look for an answer.
In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.
You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.
For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.
For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.
answered Jan 10 at 18:57
voffchvoffch
4955
$endgroup$
add a comment |
$begingroup$
I'll just point out the direction in which you could look for an answer.
In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.
You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.
For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.
For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.
answered Jan 10 at 18:57
voffchvoffch
4955
$endgroup$
add a comment |
$begingroup$
I'll just point out the direction in which you could look for an answer.
In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.
You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.
For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.
For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.
answered Jan 10 at 18:57
voffchvoffch
4955
$endgroup$
I'll just point out the direction in which you could look for an answer.
In general case, it should be almost impossible to determine the crystal structure by looking at the chemical composition of a substance.
You are, however, asked to explain why different fluorides adopt different structure types, which is a different task. Perhaps, you should take into account the difference (sorry) in the ionic radii of Mg and Ca. Some structures prefer "fat" cations, some - "lean" ones. Most of the time, it's just plain geometry combined with experimental observations. I know the explanation is primitive, and ions are not just some rigid spheres, but the "ionic radii" approach works in many cases.
For many crystal structures there are ranges of ionic radii ratios in which the particular structure is more or less stable. Some of these ratios are so important that they have proper names, such as Goldschmidt tolerance factor for perovskites. Returning to your example, this page lists the cation/anion radii ratios for fluorite (0.73 or above) and rutile (between 0.41 and 0.73). You can check these ratios against the dimensions of ions in magnesium and calcium fluorites.
For the reference - this Shannon table hosted by Imperial College London is an excellent source of the values of ionic radii in oxides and halides. Just one more thing: the reference ratios are likely to be designed to work with so-called Shannon Ionic Radii, not the less-common Shannon Crystal Radii.
answered Jan 10 at 18:57
voffchvoffch
4955
answered Jan 10 at 18:57
voffchvoffch
4955
answered Jan 10 at 18:57
voffchvoffch
4955
answered Jan 10 at 18:57
voffchvoffch
4955
4955
add a comment |
add a comment |
$begingroup$
Not really a complete answer, rather an addition to the given answer (which I think you should accept). As of now, it's possible to predict existence of certain compounds, mostly binary and ternary inorganic ionic solids, at the given temperature and pressure, as well as their crystal structure.
There are many computer programs, among which USPEX is arguably the most feature-rich and promoted. It uses Oganov-Glass evolutionary algorithm and works best for network solids (including particles and clusters). Fairly recently there also were some advancements for predicting crystal structures of small organic molecules.
answered Jan 10 at 21:09
andseliskandselisk
15k649108
$endgroup$
2
$begingroup$
Suggestion: The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. ccdc.cam.ac.uk/Community/initiatives/cspblindtests). Which is why I'm going to replace the link set by you by the one pointing to the sixth one -- equally published as open access article on Acta Cryst B.
$endgroup$
– Buttonwood
Jan 10 at 22:01
$begingroup$
@Buttonwood Good catch, thank you for the edit!
$endgroup$
– andselisk
Jan 10 at 22:04
1
$begingroup$
You knew better about USPEX, and me a bit about the contests. It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.
$endgroup$
– Buttonwood
Jan 10 at 22:19
add a comment |
$begingroup$
Not really a complete answer, rather an addition to the given answer (which I think you should accept). As of now, it's possible to predict existence of certain compounds, mostly binary and ternary inorganic ionic solids, at the given temperature and pressure, as well as their crystal structure.
There are many computer programs, among which USPEX is arguably the most feature-rich and promoted. It uses Oganov-Glass evolutionary algorithm and works best for network solids (including particles and clusters). Fairly recently there also were some advancements for predicting crystal structures of small organic molecules.
answered Jan 10 at 21:09
andseliskandselisk
15k649108
$endgroup$
2
$begingroup$
Suggestion: The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. ccdc.cam.ac.uk/Community/initiatives/cspblindtests). Which is why I'm going to replace the link set by you by the one pointing to the sixth one -- equally published as open access article on Acta Cryst B.
$endgroup$
– Buttonwood
Jan 10 at 22:01
$begingroup$
@Buttonwood Good catch, thank you for the edit!
$endgroup$
– andselisk
Jan 10 at 22:04
1
$begingroup$
You knew better about USPEX, and me a bit about the contests. It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.
$endgroup$
– Buttonwood
Jan 10 at 22:19
add a comment |
$begingroup$
Not really a complete answer, rather an addition to the given answer (which I think you should accept). As of now, it's possible to predict existence of certain compounds, mostly binary and ternary inorganic ionic solids, at the given temperature and pressure, as well as their crystal structure.
There are many computer programs, among which USPEX is arguably the most feature-rich and promoted. It uses Oganov-Glass evolutionary algorithm and works best for network solids (including particles and clusters). Fairly recently there also were some advancements for predicting crystal structures of small organic molecules.
answered Jan 10 at 21:09
andseliskandselisk
15k649108
$endgroup$
Not really a complete answer, rather an addition to the given answer (which I think you should accept). As of now, it's possible to predict existence of certain compounds, mostly binary and ternary inorganic ionic solids, at the given temperature and pressure, as well as their crystal structure.
There are many computer programs, among which USPEX is arguably the most feature-rich and promoted. It uses Oganov-Glass evolutionary algorithm and works best for network solids (including particles and clusters). Fairly recently there also were some advancements for predicting crystal structures of small organic molecules.
answered Jan 10 at 21:09
andseliskandselisk
15k649108
edited Jan 10 at 22:08
answered Jan 10 at 21:09
andseliskandselisk
15k649108
answered Jan 10 at 21:09
andseliskandselisk
15k649108
answered Jan 10 at 21:09
andseliskandselisk
15k649108
15k649108
2
$begingroup$
Suggestion: The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. ccdc.cam.ac.uk/Community/initiatives/cspblindtests). Which is why I'm going to replace the link set by you by the one pointing to the sixth one -- equally published as open access article on Acta Cryst B.
$endgroup$
– Buttonwood
Jan 10 at 22:01
$begingroup$
@Buttonwood Good catch, thank you for the edit!
$endgroup$
– andselisk
Jan 10 at 22:04
1
$begingroup$
You knew better about USPEX, and me a bit about the contests. It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.
$endgroup$
– Buttonwood
Jan 10 at 22:19
add a comment |
2
$begingroup$
Suggestion: The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. ccdc.cam.ac.uk/Community/initiatives/cspblindtests). Which is why I'm going to replace the link set by you by the one pointing to the sixth one -- equally published as open access article on Acta Cryst B.
$endgroup$
– Buttonwood
Jan 10 at 22:01
$begingroup$
@Buttonwood Good catch, thank you for the edit!
$endgroup$
– andselisk
Jan 10 at 22:04
1
$begingroup$
You knew better about USPEX, and me a bit about the contests. It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.
$endgroup$
– Buttonwood
Jan 10 at 22:19
2
2
$begingroup$
Suggestion: The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. ccdc.cam.ac.uk/Community/initiatives/cspblindtests). Which is why I'm going to replace the link set by you by the one pointing to the sixth one -- equally published as open access article on Acta Cryst B.
$endgroup$
– Buttonwood
Jan 10 at 22:01
$begingroup$
Suggestion: The blind tests moderated by the CCDC you mention advanced further than to the fourth round your link points to (cf. ccdc.cam.ac.uk/Community/initiatives/cspblindtests). Which is why I'm going to replace the link set by you by the one pointing to the sixth one -- equally published as open access article on Acta Cryst B.
$endgroup$
– Buttonwood
Jan 10 at 22:01
$begingroup$
@Buttonwood Good catch, thank you for the edit!
$endgroup$
– andselisk
Jan 10 at 22:04
$begingroup$
@Buttonwood Good catch, thank you for the edit!
$endgroup$
– andselisk
Jan 10 at 22:04
1
1
$begingroup$
You knew better about USPEX, and me a bit about the contests. It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.
$endgroup$
– Buttonwood
Jan 10 at 22:19
$begingroup$
You knew better about USPEX, and me a bit about the contests. It's one of the enjoyable ideas of SE, bringing the knowledge of multiple people into one forum, an exchange.
$endgroup$
– Buttonwood
Jan 10 at 22:19
add a comment |
$begingroup$
You can treat it as a "generic problem solving" type of question. @voffch gave some specific ways to do this.
You're asked why they form different structures, but don't have to say anything about their actual structures. This question sounds like something from an introductory undergrad general/inorganic course or high school honors gen chem course. Typical material for such a course discusses ionic structures in terms of sphere packing/ionic radius (keywords: fcc, bcc, hcp) and covers ionic radius in terms of electronegativity and shielding.
You're given:
- Mg(II) and Ca(II) are different cations
- Their fluorides form different structures (e.g. the anion is the same)
- (you should know) Mg and Ca pretty much always are in +2
So the question is implicitly "Of the differences between Mg and Ca cations, which differences make their fluorides have different structures?."
The specific mention of F suggests there's something special about it too. F is pretty "extreme" all the way in its corner so it was probably chosen as a "superlative," e.g. "most electronegative," "smallest, "biggest occupational hazard" or something like that.
At this point, you can just compile a list of all the things you've learned about metal/halide ions over the relevant time period and pick ones that affect geometry. Also consider what you've been taught about crystal structures.
It could play out like this:
- The course covered elemental trends in ionic radius, and reasons why (electronegativity, shielding)
- The course covered simple ionic structures (sphere packing)
- Sphere packing is only affected by radius
- Mg, Ca have different radii. Both are "big"
- F is "small"
- The reason the structures are different is because Mg, Ca have different radii
- (Only if the course covered this) Mg(II) is [bigger/smaller] than Ca(II) because [electronegativity, orbitals, etc.]
By way of explanation, you can explain radius trends + draw a sketch of the different packings, approximately to scale.
If your course covered different material-- Compare Mg, Ca in terms of the stuff your course covered.
answered Jan 11 at 20:06
jnj16180340jnj16180340
212
$endgroup$
add a comment |
$begingroup$
You can treat it as a "generic problem solving" type of question. @voffch gave some specific ways to do this.
You're asked why they form different structures, but don't have to say anything about their actual structures. This question sounds like something from an introductory undergrad general/inorganic course or high school honors gen chem course. Typical material for such a course discusses ionic structures in terms of sphere packing/ionic radius (keywords: fcc, bcc, hcp) and covers ionic radius in terms of electronegativity and shielding.
You're given:
- Mg(II) and Ca(II) are different cations
- Their fluorides form different structures (e.g. the anion is the same)
- (you should know) Mg and Ca pretty much always are in +2
So the question is implicitly "Of the differences between Mg and Ca cations, which differences make their fluorides have different structures?."
The specific mention of F suggests there's something special about it too. F is pretty "extreme" all the way in its corner so it was probably chosen as a "superlative," e.g. "most electronegative," "smallest, "biggest occupational hazard" or something like that.
At this point, you can just compile a list of all the things you've learned about metal/halide ions over the relevant time period and pick ones that affect geometry. Also consider what you've been taught about crystal structures.
It could play out like this:
- The course covered elemental trends in ionic radius, and reasons why (electronegativity, shielding)
- The course covered simple ionic structures (sphere packing)
- Sphere packing is only affected by radius
- Mg, Ca have different radii. Both are "big"
- F is "small"
- The reason the structures are different is because Mg, Ca have different radii
- (Only if the course covered this) Mg(II) is [bigger/smaller] than Ca(II) because [electronegativity, orbitals, etc.]
By way of explanation, you can explain radius trends + draw a sketch of the different packings, approximately to scale.
If your course covered different material-- Compare Mg, Ca in terms of the stuff your course covered.
answered Jan 11 at 20:06
jnj16180340jnj16180340
212
$endgroup$
add a comment |
$begingroup$
You can treat it as a "generic problem solving" type of question. @voffch gave some specific ways to do this.
You're asked why they form different structures, but don't have to say anything about their actual structures. This question sounds like something from an introductory undergrad general/inorganic course or high school honors gen chem course. Typical material for such a course discusses ionic structures in terms of sphere packing/ionic radius (keywords: fcc, bcc, hcp) and covers ionic radius in terms of electronegativity and shielding.
You're given:
- Mg(II) and Ca(II) are different cations
- Their fluorides form different structures (e.g. the anion is the same)
- (you should know) Mg and Ca pretty much always are in +2
So the question is implicitly "Of the differences between Mg and Ca cations, which differences make their fluorides have different structures?."
The specific mention of F suggests there's something special about it too. F is pretty "extreme" all the way in its corner so it was probably chosen as a "superlative," e.g. "most electronegative," "smallest, "biggest occupational hazard" or something like that.
At this point, you can just compile a list of all the things you've learned about metal/halide ions over the relevant time period and pick ones that affect geometry. Also consider what you've been taught about crystal structures.
It could play out like this:
- The course covered elemental trends in ionic radius, and reasons why (electronegativity, shielding)
- The course covered simple ionic structures (sphere packing)
- Sphere packing is only affected by radius
- Mg, Ca have different radii. Both are "big"
- F is "small"
- The reason the structures are different is because Mg, Ca have different radii
- (Only if the course covered this) Mg(II) is [bigger/smaller] than Ca(II) because [electronegativity, orbitals, etc.]
By way of explanation, you can explain radius trends + draw a sketch of the different packings, approximately to scale.
If your course covered different material-- Compare Mg, Ca in terms of the stuff your course covered.
answered Jan 11 at 20:06
jnj16180340jnj16180340
212
$endgroup$
You can treat it as a "generic problem solving" type of question. @voffch gave some specific ways to do this.
You're asked why they form different structures, but don't have to say anything about their actual structures. This question sounds like something from an introductory undergrad general/inorganic course or high school honors gen chem course. Typical material for such a course discusses ionic structures in terms of sphere packing/ionic radius (keywords: fcc, bcc, hcp) and covers ionic radius in terms of electronegativity and shielding.
You're given:
- Mg(II) and Ca(II) are different cations
- Their fluorides form different structures (e.g. the anion is the same)
- (you should know) Mg and Ca pretty much always are in +2
So the question is implicitly "Of the differences between Mg and Ca cations, which differences make their fluorides have different structures?."
The specific mention of F suggests there's something special about it too. F is pretty "extreme" all the way in its corner so it was probably chosen as a "superlative," e.g. "most electronegative," "smallest, "biggest occupational hazard" or something like that.
At this point, you can just compile a list of all the things you've learned about metal/halide ions over the relevant time period and pick ones that affect geometry. Also consider what you've been taught about crystal structures.
It could play out like this:
- The course covered elemental trends in ionic radius, and reasons why (electronegativity, shielding)
- The course covered simple ionic structures (sphere packing)
- Sphere packing is only affected by radius
- Mg, Ca have different radii. Both are "big"
- F is "small"
- The reason the structures are different is because Mg, Ca have different radii
- (Only if the course covered this) Mg(II) is [bigger/smaller] than Ca(II) because [electronegativity, orbitals, etc.]
By way of explanation, you can explain radius trends + draw a sketch of the different packings, approximately to scale.
If your course covered different material-- Compare Mg, Ca in terms of the stuff your course covered.
answered Jan 11 at 20:06
jnj16180340jnj16180340
212
edited Jan 11 at 20:11
answered Jan 11 at 20:06
jnj16180340jnj16180340
212
answered Jan 11 at 20:06
jnj16180340jnj16180340
212
answered Jan 11 at 20:06
jnj16180340jnj16180340
212
212
add a comment |
add a comment |
Thanks for contributing an answer to Chemistry Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fchemistry.stackexchange.com%2fquestions%2f107762%2fhow-do-i-know-crystal-structures-from-formula%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
$begingroup$
You don't. That being said, a post factum explanation is somewhat possible.
$endgroup$
– Ivan Neretin
Jan 10 at 18:03
$begingroup$
Just to note that it's a bit misleading to say that calcium fluoride adopts the fluorite structure. Calcium fluoride is fluorite, and the structure was named after the mineral.
$endgroup$
– Gimelist
Jan 12 at 3:42