Question

question 3 (1 point) which of the following has the greatest number of unshared electrons around the central atom? i3- sf4 so2 nh3 h2o

Explanation:

Step1: Determine valence - electrons and bonding for \(I_3^-\)

The central \(I\) atom in \(I_3^-\) has 7 valence electrons, plus 1 for the negative charge, total 8. It forms 2 bonds with the other \(I\) atoms. So the number of unshared electron - pairs is \(\frac{8 - 2}{2}=3\) unshared electron - pairs (6 unshared electrons).

Step2: Determine valence - electrons and bonding for \(SF_4\)

The central \(S\) atom has 6 valence electrons. It forms 4 bonds with \(F\) atoms. So the number of unshared electron - pairs is \(\frac{6 - 4}{2}=1\) unshared electron - pair (2 unshared electrons).

Step3: Determine valence - electrons and bonding for \(SO_2\)

The central \(S\) atom has 6 valence electrons. It forms 2 double - bonds with \(O\) atoms. So the number of unshared electron - pairs is \(\frac{6 - 4}{2}=1\) unshared electron - pair (2 unshared electrons).

Step4: Determine valence - electrons and bonding for \(NH_3\)

The central \(N\) atom has 5 valence electrons. It forms 3 bonds with \(H\) atoms. So the number of unshared electron - pairs is \(\frac{5 - 3}{2}=1\) unshared electron - pair (2 unshared electrons).

Step5: Determine valence - electrons and bonding for \(H_2O\)

The central \(O\) atom has 6 valence electrons. It forms 2 bonds with \(H\) atoms. So the number of unshared electron - pairs is \(\frac{6 - 2}{2}=2\) unshared electron - pairs (4 unshared electrons).

Answer:

\(I_3^-\)