Question

question 20 (1 point)
which of the following has the greatest number of unshared electrons around the central atom?
nh₄⁺
ch₄
clf₃
bcl₃
icl₅

Explanation:

• For \(NH_{4}^{+}\): The central \(N\) atom has \(0\) unshared electron pairs. The \(N\) atom forms \(4\) bonding pairs with \(H\) atoms and has a positive charge (loses one electron), so no unshared electrons.
• For \(CH_{4}\): The central \(C\) atom has \(0\) unshared electron pairs. \(C\) forms \(4\) bonding pairs with \(H\) atoms.
• For \(ClF_{3}\): The central \(Cl\) atom has \(2\) unshared electron pairs. \(Cl\) has \(7\) valence electrons. It forms \(3\) bonding pairs with \(F\) atoms (\(3\) electrons used in bonding), so \(7 - 3= 4\) electrons left, which is \(2\) unshared pairs (\(4\div2 = 2\)).
• For \(BCl_{3}\): The central \(B\) atom has \(0\) unshared electron pairs. \(B\) has \(3\) valence electrons and forms \(3\) bonding pairs with \(Cl\) atoms.
• For \(ICl_{5}\): The central \(I\) atom has \(1\) unshared electron pair. \(I\) has \(7\) valence electrons. It forms \(5\) bonding pairs with \(Cl\) atoms (\(5\) electrons used in bonding), so \(7 - 5 = 2\) electrons left, which is \(1\) unshared pair (\(2\div2=1\)).

Answer:

C. \(ClF_{3}\)