# Which of the following angles corresponds to $s{{p}^{2}}$ hybridization?

(A) $90{}^\circ$

(B) $120{}^\circ$

(C) $180{}^\circ$

(D) $109{}^\circ$

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**Hint:**To answer this question we should know about the geometries of the various hybridizations. The geometries give us the angles corresponding to the hybridization. $s{{p}^{2}}$ hybridization corresponds to triangular planar geometry.

**Complete answer:**

Let’s look at the answer to this question.

> The angles corresponding to a hybridization depends on the geometry of the molecule. Each hybridization has a unique geometry.

> The bond angles of a molecule are measured between the bonds present in the molecule. Greater the bonds greater is the stability of the molecule.

Let’s look at some hybridizations and the corresponding geometries:

> In sp hybridization the geometry is linear. This means that the angle between the bonds present in sp hybridization is $180{}^\circ$. So, option (C) is incorrect.

> In $s{{p}^{3}}$ hybridization the geometry is tetrahedral. This means that the angle between the bonds present is $109.5{}^\circ$.

Now, let’s analyse $s{{p}^{2}}$ hybridization.

> In $s{{p}^{2}}$ hybridization the geometry is triangular planar. We all know that in triangular planar geometry if the central atom is placed at the center and three equidistant bonds are made by it, then each bond angle will come out to be $120{}^\circ$. So, option (B) is the correct answer.

**Hence, the answer to the given question is option (B).****Note:**The presence of a lone pair in a molecule changes its bond angle. This happens due to increased repulsions in the molecule. For example the hybridization and geometry in ammonia molecules is tetrahedral but the bond angle is $107{}^\circ$ and not $109.5{}^\circ$.