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The ions of NaF and MgO have the same nu...

The ions of NaF and MgO have the same number of electrons and inter nuclear distances are about the same (235 pm and 215 pm). Why are then the melting points of NaF and MgO so different `(992^(@)C and 2642^(@)C)`?

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Both the compounds are iso-electronic and inter nuclear distance are also about the same. However, the ions differ in their nature. In MgO, both the ions are divalent in nature `(Mg^(2+)and O^(-))` while in NaF, these are of monovalent nature. `(Na^(+) and F^(-))`. The electrostatic forces among the ions are divalent in nature `(Mg^(2+)and O^(2-))` while in NaF, these are of monovalent nature. `(Na^(+) and F^(-))`. The electrosstatic forces among the ions in the crystal lattice of MgO are stronger than in case of NaF. Therefore, the melting point of MgO is higher `(2642^(@)C)` as compared to NaF `(992^(@)C)`.
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Knowledge Check

  • Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . A proton is 1836 times heavier than an electron and has same charge as that of electron. For what velocity of the proton will its de Broglie wavelength be 4455 Å

    A
    `10^(6) m.s^(-1)`
    B
    `10^(7) m.s^(-1)`
    C
    `3 xx 10^(6) m.s^(-1)`
    D
    `3 xx 10^(7) m.s^(-1)`
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