In a regular octahedral molecule, $\mathrm{MX}_{6}$ the number of $X-M-X$ bonds at $180^{\circ}$ is
- three
- two
- $\mathrm{six}$
- four
Solution

Thus here bond angles between
$\mathrm{X}_{4}-\mathrm{M}-\mathrm{X}_{2}=180^{\circ}$
$\mathrm{X}_{1}-\mathrm{M}-\mathrm{X}_{3}=180^{\circ}$
$\mathrm{X}_{5}-\mathrm{M}-\mathrm{X}_{6}=180^{\circ}$
Asked in: JEE-TOPICTESTS-CHEMISTRY
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