Question

The compound $$M{X}_4$$ is tetrahedral. The number of $$\mathrm{\angle }XMX$$ angles in the compound is:
A. 3
B. 4
C. 5
D. 6

Answer 1

Hint: The number of bonds gives the information regarding the bond angles in the compound. For this label all the bonded atoms and check for the number of bond pairs and lone pairs.

Complete step by step answer: The compound $$M{X}_4$$ is tetrahedral. That means it resembles methane $$C{H}_4$$ which also has four hydrogen atoms bonded to the central carbon atom.
In order to determine the number of $$\mathrm{\angle }XMX$$ bond angles the four bonded atoms $$X$$ are labeled as A, B, C and D. As the compound is tetrahedral all the bond angles in the compound are at an angle of$${109}^{\circ }{28}^{\prime }$$.
Let us draw the labeled diagram of the compound $$M{X}_4$$.

In the tetrahedral model the top $$X$$ atom is labeled as A, the bottom three are labeled as B, C and D. The atom C is above the plane and the atom D is inside the plane.
The bond angle indicates that the $$\mathrm{\angle }AMD$$ angle is $${109}^{\circ }{28}^{\prime }$$ which is the same as that for $$\mathrm{\angle }XMX$$ angle. Thus the bond angles in $$M{X}_4$$ are written as
$\mathrm{\angle }AMB$
$\mathrm{\angle }AMC$
$\mathrm{\angle }AMD$
$\mathrm{\angle }BMC$
$\mathrm{\angle }BMD$
$\mathrm{\angle }CMD$
Hence the number of $$\mathrm{\angle }XMX$$ bond angles in $$M{X}_4$$ is six
So, the correct answer is “Option A”.

Note: The bond angle of tetrahedral compound is with the help of VSEPR theory. VSEPR stands for Valence Shell Electron Pair Repulsion theory. It assumes the number of valence electrons around the central atom plus the number of bonded atoms. In tetrahedral geometry the bond pair- bond pair repulsion leads to the bond angle of $${109}^{\circ }{28}^{\prime }$$. If a lone pair is present in spite of a bonded atom the bond angle changes along with the geometry due to greater repulsion between lone pair and bond pair.