QQuestionChemistry
QuestionChemistry
"Draw the Lewis Dot Structure for BH^−2.
What is the electron geometry?
What is the molecular geometry?"
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Answer
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Step 1:: Draw the Lewis structure for BH${}^{- 2}$.
H${}^{+}$ H${}^{-}$ H${}^{+}$ H${}^{-}$ H${}^{+}$
First, let's distribute the negative 2 charge to the hydrogen atoms, since boron cannot have a formal charge of + 1. This gives us three B-H bonds and two additional negative charges on two hydrogen atoms.
Step 2:: Determine the total number of valence electrons.
3(B) + 6(H) + 2(extra) = 11
Boron has 3 valence electrons, and each hydrogen atom has 1 valence electron. In addition, there are 2 extra electrons from the charge. Therefore, the total number of valence electrons is:
Step 3:: Place the electrons around the atoms using Lewis dot structure rules.
Step 4:
Start by placing one electron between each bonded atom. This accounts for 6 electrons (3 bonds): B : H : H : H : H : H
Step 5:
Place the remaining electrons as lone pairs on the atoms, following the octet rule. Distribute the lone pairs such that the central atom (boron) has the least possible formal charge. In this case, distribute the lone pairs on the hydrogen atoms: B : H : H : H : H : H : : : : : : : : : : : : : :
Step 6:: Calculate the formal charge for each atom.
For hydrogen (H): $FC = 1 - 2 - \frac{1}{2} = 0$
Formal charge is calculated as:
Step 7:: Determine the electron geometry and molecular geometry.
Electron geometry is determined by the total number of electron pairs (bonds and lone pairs) around the central atom. Here, there are 3 bonding pairs and 2 lone pairs, making the electron geometry trigonal bipyramidal. Molecular geometry is determined by the arrangement of the bonding pairs only. Here, there are 3 bonding pairs, making the molecular geometry trigonal planar.
Final Answer
The Lewis structure for BH${}^{- 2}$ is: B : H : H : H : H : H : : : : : : : : : : : : : : The electron geometry is trigonal bipyramidal, and the molecular geometry is trigonal planar.
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