Question

what is the electron geometry of kro₄? note that you must draw the correct lewis - dot structure to answer this question. trigonal pyramidal tetrahedral octohedral trigonal bipyramidal see - saw question 4 1 pts what is the electron geometry of the xecl₃⁺ ion? note that you must draw the correct lewis - dot structure to answer this question. t - shaped tetrahedral trigonal planar octohedral trigonal bipyramidal

Explanation:

Step1: Determine valence - electrons for $KrO_4$

Kr has 8 valence electrons and each O has 6 valence electrons. So total valence electrons for $KrO_4$ is $8 + 4\times6=32$. In the Lewis - dot structure of $KrO_4$, Kr is the central atom and is double - bonded to each of the 4 O atoms. There are 4 bonding pairs and 0 lone pairs around the Kr atom. According to VSEPR theory, the electron geometry is determined by the number of electron groups (bonding + lone pairs) around the central atom. Here, the number of electron groups is 4.

Step2: Identify electron geometry for $KrO_4$

When the number of electron groups around the central atom is 4, the electron geometry is tetrahedral.

Step3: Determine valence - electrons for $XeCl_3^+$

Xe has 8 valence electrons, each Cl has 7 valence electrons, and we subtract 1 for the positive charge. So total valence electrons for $XeCl_3^+$ is $8+3\times7 - 1=28$. In the Lewis - dot structure of $XeCl_3^+$, Xe is the central atom. Xe has 3 bonding pairs with Cl atoms and 2 lone pairs. The total number of electron groups around Xe is $3 + 2=5$.

Step4: Identify electron geometry for $XeCl_3^+$

When the number of electron groups around the central atom is 5, according to VSEPR theory, the electron geometry is trigonal bipyramidal.

Answer:

Question 3: B. Tetrahedral
Question 4: E. Trigonal bipyramidal