Question

Predict the geometry of the following species: (a) \(\mathrm{SeOF}_{2}\) (b) \(\mathrm{TeF}_{5}\) (c) \(\mathrm{ICl}_{4}^{-}\) (d) \(\mathrm{RnF}_{2}\)

Short answer

Question: Predict the molecular geometry of the following species using the VSEPR theory: (a) SeOF2, (b) TeF5, (c) ICl4^-, and(d) RnF2. Answer: (a) The molecular geometry of SeOF2 is trigonal pyramidal. (b) The molecular geometry of TeF5 is square pyramidal. (c) The molecular geometry of ICl4^- is square planar. (d) The molecular geometry of RnF2 is linear.

Step-by-step solution

(a) SeOF2 geometry prediction

1. Count the total number of valence electrons: Selenium has 6, while oxygen has 6, and each fluorine has 7. Thus, SeOF2 has a total of 6 + 6 + 7(2) = 26 valence electrons. 2. Form electron pairs: Since SeOF2 has 26 valence electrons, we can form 13 electron pairs. 3. Arrange electron pairs around the central atom: Selenium is the central atom, and it is surrounded by 2 fluorine atoms, 1 oxygen atom, and 1 lone pair. Hence, Se has a steric number of 4, resulting in an sp3 hybridization. 4. Determine the molecular geometry: Since there are 3 bonding pairs and 1 lone pair, the geometry of SeOF2 is trigonal pyramidal.

(b) TeF5 geometry prediction

1. Count the total number of valence electrons: Tellurium has 6, and each fluorine has 7. Thus, TeF5 has a total of 6 + 7(5) = 41 valence electrons. 2. Form electron pairs: Since TeF5 has 41 valence electrons, we can form 20.5 electron pairs, but the number of pairs should be an integer. This indicates the presence of one unpaired electron, and the correct electron pair number is 20. 3. Arrange electron pairs around the central atom: Tellurium is the central atom, surrounded by 5 fluorine atoms and a free radical, resulting in a steric number of 6. Therefore, Te is sp3d2 hybridized. 4. Determine the molecular geometry: Since there are 5 bonding pairs and 1 free radical, the geometry of TeF5 is square pyramidal.

(c) ICl4^- geometry prediction

1. Count the total number of valence electrons: Iodine has 7, each chlorine has 7, and there is an extra electron due to the negative charge. Thus, ICl4^- has a total of 7 + 7(4) + 1 = 36 valence electrons. 2. Form electron pairs: Since ICl4^- has 36 valence electrons, we can form 18 electron pairs. 3. Arrange electron pairs around the central atom: Iodine is the central atom, surrounded by 4 chlorine atoms and 2 lone pairs. Hence, I has a steric number of 6, resulting in an sp3d2 hybridization. 4. Determine the molecular geometry: Since there are 4 bonding pairs and 2 lone pairs, the geometry of ICl4^- is square planar.

(d) RnF2 geometry prediction

1. Count the total number of valence electrons: Radon has 8, and each fluorine has 7. Thus, RnF2 has a total of 8 + 7(2) = 22 valence electrons. 2. Form electron pairs: Since RnF2 has 22 valence electrons, we can form 11 electron pairs. 3. Arrange electron pairs around the central atom: Radon is the central atom, surrounded by 2 fluorine atoms and 3 lone pairs. Hence, Rn has a steric number of 5, resulting in an sp3d hybridization. 4. Determine the molecular geometry: Since there are 2 bonding pairs and 3 lone pairs, the geometry of RnF2 is linear.