We require to determine the kind of framework of MnO by identify how many atoms are existing on the MnO unit cell.

You are watching: Assuming that the ionic radius of oxygen is 140. pm, estimate the ionic radius of manganese.



Edge size (a) = 4.47×10-8 cm

Density = 5.28 g/cm3

Calculating the volume the the unit cell:

Volume of Unit Cell =a3=(4.47×10-8 cm)3=8.93×10-23cm3

Calculating the number of atoms every unit cell:

The molar fixed of MnO is 70.9374 g/mol.


NaCl → FCC → 4 atoms every unit cell

CsCl → an easy cubic → 1 atom per cell

1 formula unit the MnO = 1 Mn Atom and 1 O atom = 2 atoms

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Problem Details

MnO has actually either the NaCI type structure or the CsCI form structure. The edge length of the MnO unit cabinet is 4.47 x 10-8 cm, and also the thickness of MnO is 5.28 g/cm3.

(a) walk MnO crystalize in the NaCI or the CsCI kind structure?

(b) Assuming the the ionic radius of oxygen is 140. Pm, estimate the ionic radius the manganese.

(c) walk the calculation cation-to-anion radius proportion for MnO substantiate her answer in component (a)? Explain.

The proportion is _____, i beg your pardon _____fall within the limits for pour it until it is full these type of holes. Therefore, us _____ suppose Mn2+ to accounting the holes created by the cubic closet packed range of O2- ions, and also the calculated proportion _____ substantiate the price in part (a).

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