Question

1. an atom with the electron configuration, $1s^2 2s^2 2p^4$, forms chemical bonds with two hydrogen atoms. draw electron dot structures to determine the number of lone pairs of electrons on the central atom. what is the shape of the molecule as explained by vsepr theory?

Explanation:

Step1: Identify the central atom

The electron configuration \(1s^2 2s^2 2p^4\) corresponds to oxygen (O). So the central atom is O, and it bonds with two H atoms, forming \(H_2O\).

Step2: Draw electron dot structure

Oxygen has 6 valence electrons. Each H contributes 1 valence electron. In \(H_2O\), O forms single bonds with two H atoms (sharing 2 electrons, 1 from each H - O bond), so the bonding pairs are 2. The remaining valence electrons on O: \(6 - 2\times1 = 4\), which form 2 lone pairs (since a lone pair is 2 electrons).

Step3: Apply VSEPR theory

The central atom (O) has 2 bonding pairs (BP) and 2 lone pairs (LP). The electron - pair geometry is tetrahedral (because the total number of electron pairs around the central atom is \(BP + LP=2 + 2 = 4\)). But the molecular geometry (shape) is determined by the arrangement of the bonding pairs, considering the lone pairs. For a central atom with 2 BP and 2 LP, the molecular shape is bent (or angular).

Answer:

The number of lone pairs on the central (oxygen) atom is 2. The shape of the \(H_2O\) molecule (as per VSEPR theory) is bent (angular).