The compound $M X_{6}$ is octahedral. Find the number of $90^{o}$ X - M - X angles in the compound.(If number is X then enter X/3)

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Solution

The compound $M X_{6}$ is octahedral. The number of $X - M - X$ angles in the compound is $12$ . This is shown below. $4 X$ atoms and $M$ are in one plane. They form $4 X - M - X$ angles. One $X$ is above the plane. It forms $4 X - M - X$ angles with $4 X$ in the plane. One $X$ is below the plane. It forms $4 X - M - X$ angles with $4 X$ in the plane.

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Similar questions

In a regular octahedral molecule, $M X_{6}$ the number of X-M-X bonds at an angle of $180^{\circ}$ is

M X_{6}

180^{0}

{M X}_{4}

X - M - X

lim x \to m \frac{x^{3} - m^{3}}{x - m} = 3

m

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x

60^{o}

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