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Give the bond angle, how many lone pairs, electron geometry,

molecular geometry for CH3Br Divide your answer with a coma.

(Example: 103 degrees, 3 LP, tetrahedral, trigonal

planar)

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02:20

Draw the Lewis dot structure for CH2Cl2 . Determine the electrongeometry of CH2Cl2 . trigonal planar tetrahedral linear Determinethe molecular geometry of CH2Cl2 . trigonal planar tetrahedrallinear bent trigonal pyramidal Identify the bond angle in CH2Cl2 .90° 120° 180° 109.5°

00:30

Give the approximate bond angle for a molecule with trigonalbipyramidal electron geometry and linear molecular geometry.Give the approximate bond angle for a molecule with trigonalbipyramidal electron geometry and linear molecular geometry.180°<120°<109.5°<180°>180°

01:29

Give the approximate bond angle for a molecule with octahedralelectron geometry and square pyramidal molecular geometry.Give the approximate bond angle for a molecule with octahedralelectron geometry and square pyramidal molecular geometry.>90°>120°<90°90°180°

01:20

Part AGive the approximate bond angle for a molecule with tetrahedralelectron geometry and trigonal pyramidal molecular geometry.Give the approximate bond angle for a molecule with tetrahedralelectron geometry and trigonal pyramidal molecular geometry.<120°<180°<109.5°180°109.5°

Transcript

We’Ve got methyl bromine or c h, 3 b r and we’ve got to determine bond in how many lone pairs, electron geometry and molecular geometry. For this molecule so recall that we should typically start by putting the carbon or the least electro negative or the most electro neutral atom in the center of this molecule, hydrogen is not as neutral and neither is bromine bromine in fact rather electronegative. So we should expect that we have a carbon and that it should have a bond to each the remaining atoms- hydrogen, hydrogen, hydrogen and bromine. We can also double check now that we have the correct number of electrons used in a sign lone pairs where needed. For example, we know that carbon should bring 4 electrons to the party that 3 hydrogen should bring 3 electrons to the party and that bromine. This is again in terms of vaillance electrons should bring 7 electrons to the party to give us a total of 14 electrons. Now we’ve used 2468 of them, so we should have the remaining 6 electrons. Well note that everything has a full octet or duet in the case of hydrogen, except for bromine, which at this point needs 6 more electrons. So it would be rather nice to just put…

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