Question

question 18 (1 point)
for which of the following compounds is hydrogen bonding the main intermolecular force?
bh₃
ch₄
hcn
ch₂f₂
ch₃nh₃
question 19 (1 point)
which of the following is not a valid set of four quantum numbers? (n, l, ml, ms )
4, 0, 0, +1/2
2, 1, 0, -1/2
3, 2, -1, -1/2
1, 0, 0, +1/2
1, 1, 0, +1/2

Explanation:

Question 18

Hydrogen bonding occurs when \(H\) is bonded to \(N\), \(O\), or \(F\).

• For \(BH_3\): \(B - H\) bonds. No \(N\), \(O\), or \(F\), so no hydrogen bonding.
• For \(CH_4\): \(C - H\) bonds. No \(N\), \(O\), or \(F\), so no hydrogen bonding.
• For \(HCN\): \(C - H\) and \(C\equiv N\). \(H\) is not directly bonded to \(N\), so no hydrogen bonding as main force.
• For \(CH_2F_2\): \(C - H\) and \(C - F\). \(H\) is not directly bonded to \(F\), so no hydrogen bonding as main force.
• For \(CH_3NH_3\): \(N - H\) bonds. \(H\) is bonded to \(N\), so hydrogen bonding is the main intermolecular force.

The rules for quantum numbers:

• \(n\) (principal quantum number) \(\geq1\)
• \(l\) (azimuthal quantum number): \(0\leq l\leq n - 1\)
• \(m_l\) (magnetic quantum number): \(-l\leq m_l\leq l\)
• \(m_s\) (spin quantum number) \(=\pm\frac{1}{2}\)

For the set \(1,1,0,+\frac{1}{2}\):

• \(n = 1\), then \(l\) can only be \(0\) (\(l\leq n-1=0\)). But here \(l = 1\), so this set is not valid.

For \(4,0,0,+\frac{1}{2}\): \(n = 4\), \(l=0\) (\(0\leq4 - 1\)), \(m_l = 0\) (\(-0\leq0\leq0\)), \(m_s=+\frac{1}{2}\) (valid).
For \(2,1,0,-\frac{1}{2}\): \(n = 2\), \(l = 1\) (\(1\leq2 - 1\)), \(m_l=0\) (\(-1\leq0\leq1\)), \(m_s=-\frac{1}{2}\) (valid).
For \(3,2,-1,-\frac{1}{2}\): \(n = 3\), \(l = 2\) (\(2\leq3 - 1\)), \(m_l=-1\) (\(-2\leq-1\leq2\)), \(m_s=-\frac{1}{2}\) (valid).
For \(1,0,0,+\frac{1}{2}\): \(n = 1\), \(l = 0\) (\(0\leq1 - 1\)), \(m_l=0\) (\(-0\leq0\leq0\)), \(m_s=+\frac{1}{2}\) (valid).

Answer:

\(CH_3NH_3\)

Question 19