Question

question 21 (1 point)
which of the following compounds demonstrates an exception to the octet rule?
h₂o
c₂h₆
kcl
sf₆
question 22 (1 point)
which of the following has the greatest number of unshared electrons around the central atom?
i₃⁻
sf₄
so₂
nh₃
h₂o

Explanation:

Question 21

The octet rule states that atoms tend to gain, lose, or share electrons to have 8 electrons in their valence shell. In \(H_2O\), \(O\) has 8 valence electrons (2 from each \(H - O\) bond and 4 as lone pairs). In \(C_2H_6\), each \(C\) has 8 valence electrons (3 from \(C - H\) bonds and 1 from \(C - C\) bond). In \(KCl\), \(K^+\) has 8 valence electrons (after losing 1 electron) and \(Cl^-\) has 8 valence electrons (after gaining 1 electron). In \(SF_6\), \(S\) has 12 valence electrons (6 from each \(S - F\) bond), which is an exception to the octet rule.

• For \(I_3^-\): The central \(I\) atom has 9 valence electrons (7 + 1 from the negative charge). Using the formula for lone pairs (\(L=\frac{V - N}{2}\), where \(V\) is the valence electrons of the central atom and \(N\) is the number of bonding pairs). The central \(I\) forms 2 bonds (with the other two \(I\) atoms). \(V = 7+1=8\), \(N = 2\). Lone pairs \(L=\frac{8 - 2}{2}=3\) (6 unshared electrons).
• For \(SF_4\): \(S\) has 6 valence electrons. It forms 4 bonds with \(F\) atoms. Lone pairs \(L=\frac{6 - 4}{2}=1\) (2 unshared electrons).
• For \(SO_2\): \(S\) has 6 valence electrons. It forms 2 double - bonds (equivalent to 2 bonding pairs in terms of electron - pair counting for lone - pair formula). Lone pairs \(L=\frac{6 - 4}{2}=1\) (2 unshared electrons).
• For \(NH_3\): \(N\) has 5 valence electrons. It forms 3 bonds with \(H\) atoms. Lone pairs \(L=\frac{5 - 3}{2}=1\) (2 unshared electrons).
• For \(H_2O\): \(O\) has 6 valence electrons. It forms 2 bonds with \(H\) atoms. Lone pairs \(L=\frac{6 - 2}{2}=2\) (4 unshared electrons).

Answer:

\(SF_6\)

Question 22