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Determine the bond polarity, molecular polarity; electron geometry & molecular geometry and bond angles and Lewis structure:

1

BF3

2 CCI

3 PCI;

HCO

5_ HCI

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02:27

Determine the molecular shape, bond angle, and hybrid orbitals for each molecule.$$\mathrm{BF}_{3}$$

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Determine the molecular shape, bond angle, and hybrid orbitals for each molecule.$$\mathrm{BeF}_{2}$$

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Determine the molecular shape, bond angle, and hybrid orbitals for each molecule.$$\mathrm{OCl}_{2}$$

04:48

For each molecule listed below, predict its molecularshape and bond angle, and identify the hybrid orbitals.Drawing the Lewis structure might help you.$$\begin{array}{ll}{\text { a. } \mathrm{SCl}_{2}} & {\text { c. HOF }} \\ {\text { b. } \mathrm{NH}_{2} \mathrm{Cl}} & {\text { d. } \mathrm{BF}_{3}}\end{array}$$

Transcript

Hello everyone. So in the given question we have to find out the molecular popularity molecular polarity which means whether a molecule is polar or not. We have to find out the bond angle born angle also we have to determine the Levis structure. Okay of these molecules. So the molecules are BF three second is CCL four. We have PCL five also H 2, 3 and exhale. Let’s see BF three has molecular polarity or not. So the structure of BF three is like that tribunal cleaner. Okay. So in the levees and structural flooring will have three lone pairs of electrons. Each florian will have three lone pairs. Okay, so this is the levees the structure and boron has Total of six electrons. So the bond angle between them is 1 20°. So born angle is 1 20 degree. Also we have to find out the molecular popularity. So this molecule is not cooler at all because the disciple movement is moving in this direction. So they are cancel out each other. So a new value will be equal to zero. Right, so the molecule is not a bullet, let’s see the second structure which is the CCL four. So CCL four has tetra federal structure, carbon cl cl Okay, this is the structure of CSL four. Each chlorine will have three lone pair of electrons. So this is the latest structure of CCL four. Now we have to find out whether you will be equal to zero or not. So we can see here that disciple movement is not equal to zero.…

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