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arindameducationusc
 one year ago
Conditions for For forming Electrovalent or Ionic Bond
Tutorial and Doubt at Ending
arindameducationusc
 one year ago
Conditions for For forming Electrovalent or Ionic Bond Tutorial and Doubt at Ending

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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.31) Number of valency electrons> One atom should possess 1,2 or 3 valency electrons while the other atom should have 5,6,or 7 valency electrongs. 2)Difference in electronegativity> Example, the electronegativity of sodium is 0.9 and that of fluorine is 4.0 so their difference is 9.1, both will readily form an electrovalent bond. 3) Overall decrease in Energy (a) Lower the value of ionisation energy of an atom, greater will be the ease of formation of the cation from it, i.e, one atom should have low value of ionisation energy. \[A+ionisation \space energy = A ^{+} +e ^{}\] (b) Higher the value of electron affinity of the atom, greater the ease of the anion from it, i.e, other atom should have high value of electron affinity. \[B+e ^{}= B ^{}+electron \space affinity \] (c) Lattice Energy> The Energy released when the requisite number of positive and negative ions are condensed into crystal to form one mole of the compound is called lattice energy.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.3The value of lattice energy depends on the charge present on the two ions and distance between them \[F=(1\div4\pi \in _{0}K)*q _{1}q _{2}/(r ^{+}_{A}+r ^{}_{B})^{2}\] Higher the value of F, greater shall be the stability of the ionic compound and hence greater shall be the strength of the ionic bond. For example NaCl is more stable than CsCl.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.3The above was a tutorial, revision. If it was useful, you can medal me. \[\huge\color{green}{My~ doubt ~now{}}\] Lattice energy is given by BornLande equation \[L.E=\frac{ N _{0} MZ ^{+}Z ^{}e ^{2}}{ r }*\left[ 1\frac{ 1 }{ n } \right]\] where N0=Avogadro's number M=Madelung constant (depends on crystal structure) Z+ and Z are charges on cation and anion e=charge on the electron r=radius of cation+radius of anion n=Born Constant (depends on the electronic configuration of ions). Please give the derivation of Born Lande Equation and please explain me... You can discuss about the constants. Will definitely Medal.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.3@taramgrant0543664

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.0I can not unfortunately derive the full equation (I think), but I can derive the attraction contribution, to the lattice energy, and the math is rather straight forward; The way to attack this problem first to consider there being two main contributions to the potential energy, an attraction and a repulsion term (the repulsion comes from the overlap of atomic orbitals) Looking at the attraction term we will assume the total Coulombic potential energy is the sum of all electrostatic contributions normalized with the smallest distance to the neighbor charge: \[\Large E_{attract}=\frac{ 1 }{ 4 \pi \epsilon_0 }\sum_{i}^{}\frac{ z_1 z_2e^2}{ d_{ij}/d_0 }\] We remove the first and lowest term from sum (a negative contribution) and now define the sum to instead yield: \[\Large E_{attract}=\frac{ z_1 z_2e^2 }{ 4\pi \epsilon_0 d_0 }\sum_{i}^{}\frac{ d_0 }{ d_{ij} }\] The sum can be noticed to convergent, depending on the crystal geometry and we therefore call this a constant \(M\) \[\Large E_{attract}=\frac{ z_1 z_2 e^2 }{ 4 \pi \epsilon_0 d_0 }M\] To find the lattice energy per mole of ions all we gotta do is to multiply with Avogadro's constant: \[\Large E_{attract}=\frac{N_{A} z_1 z_2 e^2 }{ 4 \pi \epsilon_0 d_0 }M\] The repulsion term I guess they have written like the LJ contribution that is: \[\Large E_{repul}=\frac{ A }{ r^n }\] When all those considerations has been done we need to the minimized energy, that is : \[\Large \frac{ d(E_{attract}+E_{repul}) }{ d \sf d }=0\]

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.0But I bet there are good websites that can explain it way better than I just did now. What I write is mostly just considerations

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.3From second last step it was difficult to understand.... from E replusion. If you can get an easier explanation, please let me know @Frostbite Thank you.....

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.0Born and Lande made a really simply approximation of the repulsion term which is basically just the repulsion term of the Mie potential so to say. That is the repulsive interaction between the lattice ions would be proportional to \(1/r^{n}\) with \(n\) called the Born constant, and has a value between 512 depending on how far the repulsion should work... so to say.

Frostbite
 one year ago
Best ResponseYou've already chosen the best response.0Depending on what repulsion term you use, you get different results. For example, the BornMayer equation use this repulsion term instead: \[E=N_A C' \exp(\frac{ d }{ d* })\]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.3what is d and d* and C'???? @Frostbite

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0If you posted this at the very end of August I would know the exactly how to derive it cause I would have the notebook I accidentally left at school. Frostbite did a really good job with the attraction contribution part. C' and d* are constants. The value of C' is not needed because it cancels out in the BornMayer Equation and d* is commonly taken as 34.5pm d=rcation+ranion, r is radius so d is distance from centre of cation to centre of anion

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0This is a great tutorial so we can use this equation to compare the strength of ionic compounds relative to each other? Is there a quicker way to make this comparison?

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.3Thank you @Photon335 This is all I know.
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