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Determine the molecular geometry of each molecule.(a)

Chapter 10, Problem 65P

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QUESTION:

Determine the molecular geometry of each molecule.

(a) \(C B r_{4}\)

(b) \(\mathrm{H}_{2} \mathrm{CO}\)

(c) \(C S_{2}\)

(d) \(\mathrm{BH}_{3}\)

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QUESTION:

Determine the molecular geometry of each molecule.

(a) \(C B r_{4}\)

(b) \(\mathrm{H}_{2} \mathrm{CO}\)

(c) \(C S_{2}\)

(d) \(\mathrm{BH}_{3}\)

ANSWER:

In order to predict the molecular geometry, we have to draw the Lewis structure of the given molecule and then molecular geometry can be predicted by VSEPR theory (steric number)

Steric number(SN) = (number of atoms bonded to central atom) +(number of lone pair on central atom)

VSEPR.png

 

Step 1 of 4

(a) \(\mathrm{CBr}_{4}\)

\({ }_{6} \mathrm{C} 1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{2}\)    \({ }_{35} \mathrm{Br}[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{5}\)

Total number of valence electrons in \(\mathrm{CBr}_{4}=4+7(4)=32\)

Thus the Lewis structure is,

CBr4.jpg

Now we will calculate the steric number (SN) = (number of atoms bonded to central atom) + (number of lone pair on central atom)

SN = 4 + 0 = 4

SN is 4 i.e the molecule has tetrahedral shape with bond angle \(109.5^{\circ}\), \(\mathrm{sp}^{3}\) hybridization.

 

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