Solid MgO has the same crystal structure as NaCl. How many oxide ions surround each Mg * ion as nearest neighbors in MgO? 4 none of these

The number of oxide ions as the nearest neighbors of   ions are known to be as six

Explanation:

The regularity of a crystal structure leads to the idea of space lattice.In order to explain this concept, let us consider a crystal of NaCl, It consists of a perfectly regular arrangement of sodium ions and chlorine ions.

If we represent the position of each Na+ in the crystal by a point marked x the result will be a regular three dimensional network of points. This will be the space lattice of Na+ in the crystal NaCl. The symmetry of the combined lattice determined the symmetry of the crystal as a whole.

The space lattice of a crystal may be considered as built up of a three dimensional basic pattern  called unit cell. The unit cell is a repeat unit which generates the whole pattern in three dimensions of the unit cell.

In Solid MgO , the crystal structure which is used to predict the properties of the material, have the same structure as that of NaCl.

The obtain the structure of a face centered cubic FCC unit cell where the ions occupy the corner of the cube and the center of each face of the cube.

The number of oxide ions as the nearest neighbors of   ions are known to be as six. As a result of that , the coordination number of ions is six.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

the correct answer would be d. hope this !