4 added 116 characters in body edited Dec 3 '18 at 13:22 Kelly Shepphard 6671212 bronze badges Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have a number larger than two, the bond is ionic. $$\begin{array}{lr} \text{Compound} & \Delta(\text{EN})\\\hline \ce{AgF} & 2.03\\ \ce{AgCl} & 1.23\\ \ce{AgBr} & 1.03\\ \ce{CaF2} & 2.96\\ \ce{CaCl2} & 2.16\\ \ce{CaBr2} & 1.96\\ \ce{KCl} & 2.34\\ \ce{NaCl} & 2.18\\ \end{array}$$ For $$\ce{Ag}$$-stuff and $$\ce{KCl}$$ and $$\ce{NaCl}$$ it's obvious: most ionic is least stable. But why do we have $$\ce{Ca}$$-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For $$\ce{Ag}$$, $$\ce{K}$$, $$\ce{Na}$$ (monovalent atoms) it's true. For bivalent atoms, such as $$\ce{Ca}$$, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak ($$\ce{Ca}$$) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as $$\ce{F}$$) would stretch the rubber harder than not very strong (such as $$\ce{Br}$$), delocalizing electrons and making $$\ce{CaF2}$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. [edited]Picture source: http://www.chemhume.co.uk/ASCHEM/Unit%201/Ch3Chemstr/Images%203/electronegativity_values.jpg Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have a number larger than two, the bond is ionic. $$\begin{array}{lr} \text{Compound} & \Delta(\text{EN})\\\hline \ce{AgF} & 2.03\\ \ce{AgCl} & 1.23\\ \ce{AgBr} & 1.03\\ \ce{CaF2} & 2.96\\ \ce{CaCl2} & 2.16\\ \ce{CaBr2} & 1.96\\ \ce{KCl} & 2.34\\ \ce{NaCl} & 2.18\\ \end{array}$$ For $$\ce{Ag}$$-stuff and $$\ce{KCl}$$ and $$\ce{NaCl}$$ it's obvious: most ionic is least stable. But why do we have $$\ce{Ca}$$-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For $$\ce{Ag}$$, $$\ce{K}$$, $$\ce{Na}$$ (monovalent atoms) it's true. For bivalent atoms, such as $$\ce{Ca}$$, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak ($$\ce{Ca}$$) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as $$\ce{F}$$) would stretch the rubber harder than not very strong (such as $$\ce{Br}$$), delocalizing electrons and making $$\ce{CaF2}$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have a number larger than two, the bond is ionic. $$\begin{array}{lr} \text{Compound} & \Delta(\text{EN})\\\hline \ce{AgF} & 2.03\\ \ce{AgCl} & 1.23\\ \ce{AgBr} & 1.03\\ \ce{CaF2} & 2.96\\ \ce{CaCl2} & 2.16\\ \ce{CaBr2} & 1.96\\ \ce{KCl} & 2.34\\ \ce{NaCl} & 2.18\\ \end{array}$$ For $$\ce{Ag}$$-stuff and $$\ce{KCl}$$ and $$\ce{NaCl}$$ it's obvious: most ionic is least stable. But why do we have $$\ce{Ca}$$-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For $$\ce{Ag}$$, $$\ce{K}$$, $$\ce{Na}$$ (monovalent atoms) it's true. For bivalent atoms, such as $$\ce{Ca}$$, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak ($$\ce{Ca}$$) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as $$\ce{F}$$) would stretch the rubber harder than not very strong (such as $$\ce{Br}$$), delocalizing electrons and making $$\ce{CaF2}$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. [edited]Picture source: http://www.chemhume.co.uk/ASCHEM/Unit%201/Ch3Chemstr/Images%203/electronegativity_values.jpg 3 Improved Formatting edited Dec 3 '18 at 12:56 Martin - マーチン♦ 34.9k99 gold badges118118 silver badges245245 bronze badges Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have a number >2larger than two, the bond is ionic. $$AgF - 2.03, AgCl - 1.23, AgBr - 1.03$$ $$CaF_2 - 2.96, CaCl_2 - 2.16, CaBr_2 - 1.96$$ $$KCl - 2.34, NaCl - 2.18$$$$\begin{array}{lr} \text{Compound} & \Delta(\text{EN})\\\hline \ce{AgF} & 2.03\\ \ce{AgCl} & 1.23\\ \ce{AgBr} & 1.03\\ \ce{CaF2} & 2.96\\ \ce{CaCl2} & 2.16\\ \ce{CaBr2} & 1.96\\ \ce{KCl} & 2.34\\ \ce{NaCl} & 2.18\\ \end{array}$$ For Ag$$\ce{Ag}$$-stuff and KCl/NaCl$$\ce{KCl}$$ and $$\ce{NaCl}$$ it's obvious: most ionic is least stable. But why do we have Ca$$\ce{Ca}$$-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For Ag$$\ce{Ag}$$, K$$\ce{K}$$, Na $$\ce{Na}$$ (monovalent atoms) it's true. For bivalent atoms, such as Ca$$\ce{Ca}$$, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak (Ca$$\ce{Ca}$$) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as F$$\ce{F}$$) would stretch the rubber harder than not very strong (such as Br$$\ce{Br}$$), delocalizing electrons and making $$CaF_2$$$$\ce{CaF2}$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have number >2, the bond is ionic. $$AgF - 2.03, AgCl - 1.23, AgBr - 1.03$$ $$CaF_2 - 2.96, CaCl_2 - 2.16, CaBr_2 - 1.96$$ $$KCl - 2.34, NaCl - 2.18$$ For Ag-stuff and KCl/NaCl it's obvious: most ionic is least stable. But why do we have Ca-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For Ag, K, Na (monovalent atoms) it's true. For bivalent atoms, such as Ca, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak (Ca) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as F) would stretch the rubber harder than not very strong (such as Br), delocalizing electrons and making $$CaF_2$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have a number larger than two, the bond is ionic. $$\begin{array}{lr} \text{Compound} & \Delta(\text{EN})\\\hline \ce{AgF} & 2.03\\ \ce{AgCl} & 1.23\\ \ce{AgBr} & 1.03\\ \ce{CaF2} & 2.96\\ \ce{CaCl2} & 2.16\\ \ce{CaBr2} & 1.96\\ \ce{KCl} & 2.34\\ \ce{NaCl} & 2.18\\ \end{array}$$ For $$\ce{Ag}$$-stuff and $$\ce{KCl}$$ and $$\ce{NaCl}$$ it's obvious: most ionic is least stable. But why do we have $$\ce{Ca}$$-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For $$\ce{Ag}$$, $$\ce{K}$$, $$\ce{Na}$$ (monovalent atoms) it's true. For bivalent atoms, such as $$\ce{Ca}$$, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak ($$\ce{Ca}$$) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as $$\ce{F}$$) would stretch the rubber harder than not very strong (such as $$\ce{Br}$$), delocalizing electrons and making $$\ce{CaF2}$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. 2 edited body edited Nov 30 '18 at 11:17 Kelly Shepphard 6671212 bronze badges Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have number >2, the bond is polarionic. $$AgF - 2.03, AgCl - 1.23, AgBr - 1.03$$ $$CaF_2 - 2.96, CaCl_2 - 2.16, CaBr_2 - 1.96$$ $$KCl - 2.34, NaCl - 2.18$$ For Ag-stuff and KCl/NaCl it's obvious: most ionic is least stable. But why do we have Ca-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For Ag, K, Na (monovalent atoms) it's true. For bivalent atoms, such as Ca, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak (Ca) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as F) would stretch the rubber harder than not very strong (such as Br), delocalizing electrons and making $$CaF_2$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have number >2, the bond is polar. $$AgF - 2.03, AgCl - 1.23, AgBr - 1.03$$ $$CaF_2 - 2.96, CaCl_2 - 2.16, CaBr_2 - 1.96$$ $$KCl - 2.34, NaCl - 2.18$$ For Ag-stuff and KCl/NaCl it's obvious: most ionic is least stable. But why do we have Ca-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For Ag, K, Na (monovalent atoms) it's true. For bivalent atoms, such as Ca, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak (Ca) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as F) would stretch the rubber harder than not very strong (such as Br), delocalizing electrons and making $$CaF_2$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have number >2, the bond is ionic. $$AgF - 2.03, AgCl - 1.23, AgBr - 1.03$$ $$CaF_2 - 2.96, CaCl_2 - 2.16, CaBr_2 - 1.96$$ $$KCl - 2.34, NaCl - 2.18$$ For Ag-stuff and KCl/NaCl it's obvious: most ionic is least stable. But why do we have Ca-stuff, ignoring this rule? The problem is ... of rivalry kind. The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For Ag, K, Na (monovalent atoms) it's true. For bivalent atoms, such as Ca, delocalization comes into play: those two little anions, fighting for electron pair. Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak (Ca) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as F) would stretch the rubber harder than not very strong (such as Br), delocalizing electrons and making $$CaF_2$$ more stable. I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made. 1 answered Nov 29 '18 at 16:30 Kelly Shepphard 6671212 bronze badges