The compound $\mathrm{MX}_{4}$ is tetrahedral. The number of $\angle \mathrm{XMX}$ formed in the compound are
- three
- four
- five
- $\mathrm{six}$
Solution

In the tetrahedral model the top \(X\) atom is labeled as A, the bottom three are labeled as B, C and D. The atom \(\mathrm{C}\) is above the plane and the atom \(\mathrm{D}\) is inside the plane.
Thus the bond angles in \(M X_4\) are written as
\(\angle A M B\)
\(\angle A M C\)
\(\angle A M D\)
\(\angle B M C\)
\(\angle B M D\)
\(\angle C M D\)
Asked in: JEE-TOPICTESTS-CHEMISTRY
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