QQuestionChemistry
QuestionChemistry
What is the molecular geometry and Lewis structure of IF^−2(1)?
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Answer
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Step 1:**Step 1:** Determine the total number of valence electrons in an IF$^-$ ion.
- Iodine (I) is in the 7th group of the periodic table, so it has 7 valence electrons. - Fluorine (F) is in the 6th group of the periodic table, so it has 7 valence electrons. - There is one negative charge, which contributes an additional electron. - Therefore, the total number of valence electrons is: 7(I) + 7(F) + 1(-) = 15 electrons.
Step 2:**Step 2:** Draw the Lewis structure for the IF$^-$ ion.
- Place I as the central atom, surrounded by F atoms. - Distribute the 15 valence electrons around the I and F atoms using single bonds (2 electrons) and lone pairs (2 electrons). - The structure should have a single bond between I and each F atom, with one lone pair on I and three lone pairs on each F atom.
Step 3:**Step 3:** Determine the molecular geometry of the IF$^-$ ion using the Valence Shell Electron Pair Repulsion (VSEPR) theory.
- The central I atom has 5 electron domains (4 single bonds and 1 lone pair). - According to VSEPR theory, the molecular geometry is determined by the arrangement of these electron domains. - For 5 electron domains, the geometry is trigonal bipyramidal.
Step 4:**Step 4:** Describe the shape of the IF$^-$ ion.
- Since the lone pair occupies more space than the bonding pairs, the shape of the IF$^-$ ion is distorted from perfect trigonal bipyramidal.
- The trigonal bipyramidal geometry has two types of positions: equatorial and axial. - The four F atoms occupy the equatorial positions, while the lone pair is in the axial position. - The shape is better described as a seesaw, with the I atom at the pivot point, the lone pair at the top, and the four F atoms forming the two arms of the seesaw. **
Final Answer
** The molecular geometry of IF$^-$ is trigonal bipyramidal, and the shape is a distorted seesaw.
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