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What is the two dimensional coordination number of a molecule in square close packed layer?
In square close-packed layer, a molecule is in contact with four of its neighbours. Therefore, the two-dimensional coordination number of a molecule in square close-packed layer is 4.
A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Number of close-packed particles = 0.5 × 6.022 × 1023 = 3.011 × 1023
Therefore, number of octahedral voids = 3.011 × 1023
And, number of tetrahedral voids = 2 × 3.011 × 1023 = 6.022 ×1023
Therefore, total number of voids = 3.011 × 1023 + 6.022 × 1023 = 9.033 × 1023
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?
The ccp lattice is formed by the atoms of the element N.
Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N.
According to the question, the atoms of element M occupy _20-11-08_Utpal_12_Chemistry_1_24_html_m87b182d.gif){kind=small}
of the tetrahedral voids.
Therefore, the number of atoms of M is equal to _20-11-08_Utpal_12_Chemistry_1_24_html_m21d22eae.gif){kind=small}
of the number
of atoms of N.
Therefore, ratio of the number of atoms of M to that of N is M: N _20-11-08_Utpal_12_Chemistry_1_24_html_m3affbdf1.gif){kind=small}
_20-11-08_Utpal_12_Chemistry_1_24_html_43161937.gif){kind=small}
Thus, the formula of the compound is M2 N3.
Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and (iii) hexagonal close-packed lattice?
Hexagonal close-packed lattice has the highest packing efficiency of 74%. The packing efficiencies of simple cubic and body-centred cubic lattices are 52.4% and 68% respectively.
An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m−3, what is the nature of the cubic unit cell?
It is given that density of the element, d = 2.7 × 103 kg m−3
Molar mass, M = 2.7 × 10−2 kg mol−1
Edge length, a = 405 pm = 405 × 10−12 m
= 4.05 × 10−10 m
It is known that, Avogadro’s number, NA = 6.022 × 1023 mol−1
Applying the relation,
_20-11-08_Utpal_12_Chemistry_1_24_html_6b706ff9.gif)
This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).
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