NEET
Practice Test 7
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The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is

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Copper crystallises in fcc with a unit cell length of 361 pm.What is the radius of copper atom?

  • Solution

    For fcc unit cell, 4r =\(\sqrt{2}a\)
    r = \(\frac{\sqrt{2} \times 361}{4}\) = 127pm

In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be

  • Solution

    From the given data, we have
    Number of Y atoms in a unit cell = 4
    Number of X atoms in a unit cell = 8 ∴ \(\frac{2}{3}\) = \(\frac{16}{3}\)
    From the above we get the formula of the compound as X16/3Y4 or X4Y3

A metal crystallizes with a face-centered cubic lattice. The edge length of the unit cell is 408 pm. The diameter of the metal atom is :

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A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y–) will be :

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Among the following which one has the highest cation to anion size ratio?

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    \(\frac{M^{+}}{X^{-}}\) is highest in CsF.
    ∴ correct choice : (c)

AB; crystallizes in a body centred cubic lattice with edge length ‘a’ equal to 387 pm. The distance between two oppositely charged ions in the lattice is :

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With which one of the following elements silicon should be doped so as to give p-type of semiconductor ?

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    The semiconductors formed by the introduction of impurity atoms containing one electron less than the parent atoms of insulators are termed as p-type semiconductors. Therefore silicon containing 14 electrons is to be doped with boron containing 13 electrons to give a p-type semi-conductor.

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