Question

question 3 (2 points)
the lewis structure for oxygen dichloride, ocl₂, has 1 reds, the shape is 2, and the bond angles are 3°.
a. 0 b. 1 c. 2 d. 3 e. 4 f. 5 g. 6 h. 7 i. 8
j. no shape, atoms or ions k. diatomic l. linear m. trigonal planar
n. bent o. tetrahedral p. pyramidal q. 109.5 r. 120 s. 180
t. no bond angles, no central atom

Explanation:

Step1: Calculate the number of REDs (Region of Electron Density)

The central atom is \(O\). The number of valence electrons of \(O\) is \(6\). Each \(Cl\) atom has \(7\) valence electrons. In \(OCl_{2}\), \(O\) forms two single bonds with \(Cl\) atoms. The number of lone pairs on \(O\) is \(\frac{6 - 2\times1}{2}=2\). The number of REDs (bonds + lone pairs) is \(2 + 2=4\).

Step2: Determine the shape

Using VSEPR (Valence - Shell Electron - Pair Repulsion) theory, for \(4\) REDs with \(2\) bonding pairs and \(2\) lone pairs, the shape is bent.

Step3: Estimate the bond angle

For a molecule with \(4\) REDs (tetrahedral electron - pair geometry) and \(2\) lone pairs, the bond angle is less than \(109.5^{\circ}\) (due to lone - pair repulsion). The bond angle in \(OCl_{2}\) is approximately \(109.5^{\circ}\) (closer to the tetrahedral angle considering the VSEPR model for \(AX_{2}E_{2}\) type molecules).

Answer:

1. E. 4
2. N. bent
3. Q. 109.5