Open Question
Draw the most likely shape for the following compound:XeF4
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12. Molecular Shapes & Valence Bond Theory
Equatorial and Axial Positions
Multiple Choice
How many lone pairs reside in the equatorial position of the KrCl5+ ion.
A
0
B
2
C
1
D
3
E
4
Verified step by step guidance1
Step 1: Begin by determining the total number of valence electrons available for the KrCl5+ ion. Krypton (Kr) has 8 valence electrons, and each chlorine (Cl) atom has 7 valence electrons. Since there are 5 chlorine atoms, calculate the total valence electrons from chlorine as 5 * 7 = 35. The positive charge on the ion indicates the loss of one electron, so subtract 1 from the total.
Step 2: Add the valence electrons from krypton and chlorine, and then subtract the electron lost due to the positive charge: 8 (Kr) + 35 (Cl) - 1 = 42 valence electrons.
Step 3: Determine the molecular geometry of the KrCl5+ ion. With 42 valence electrons, the central atom (Kr) will form bonds with the 5 chlorine atoms. The geometry that accommodates 5 bonds and lone pairs is typically a trigonal bipyramidal shape.
Step 4: In a trigonal bipyramidal geometry, there are two types of positions: axial and equatorial. The equatorial positions are typically occupied by lone pairs if present. Calculate the number of lone pairs by considering the electron count and the geometry.
Step 5: Since the KrCl5+ ion has 5 bonding pairs and no additional lone pairs needed to complete the electron count, there are no lone pairs in the equatorial position. Therefore, the number of lone pairs in the equatorial position is 0.
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