Questions: Draw the Lewis dot diagram for a Be+ cation.

Transcript text: Draw the Lewis dot diagram for a $\mathrm{Be}^{+}$cation.

Solution

Solution Steps

Step 1: Determine the Electron Configuration of Neutral Beryllium

Neutral beryllium (Be) has an atomic number of 4, which means it has 4 electrons. The electron configuration for neutral beryllium is:

\[ 1s^2 \, 2s^2 \]

Step 2: Account for the Positive Charge in $\mathrm{Be}^{+}$

The $\mathrm{Be}^{+}$ cation indicates that beryllium has lost one electron. Therefore, we need to remove one electron from the electron configuration of neutral beryllium. The electron is removed from the outermost shell, which is the 2s orbital:

\[ 1s^2 \, 2s^1 \]

Step 3: Draw the Lewis Dot Diagram

The Lewis dot diagram represents the valence electrons around the element symbol. For $\mathrm{Be}^{+}$, there is one valence electron in the 2s orbital. Therefore, the Lewis dot diagram for $\mathrm{Be}^{+}$ will have one dot around the symbol for beryllium (Be).

Final Answer

The Lewis dot diagram for a $\mathrm{Be}^{+}$ cation is:

\[ \boxed{\text{Be} \cdot} \]

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