Analyzing the lone pairs of electrons in the given molecules.
\(H_2O\): Oxygen has 2 lone pairs, and each hydrogen atom has 0 lone pairs. Total = 2 lone pairs.
\(N_2\): Nitrogen in \(N_2\) has no lone pairs in the molecular structure. Total = 0 lone pairs.
\(CO\): Carbon in CO has no lone pairs, but oxygen has 2 lone pairs. Total = 2 lone pairs.
\(XeF_4\): Xenon in \(XeF_4\) has 2 lone pairs, and each fluorine atom has 3 lone pairs. Total = 2 lone pairs.
\(NH\_3\): Nitrogen in \(NH_3\) has 1 lone pair, and each hydrogen atom has 0 lone pairs. Total = 1 lone pair.
\(NO\): Nitrogen in NO has 1 lone pair, and oxygen has 2 lone pairs. Total = 3 lone pairs.
\(CO_2\): Carbon in \(CO_2\) has no lone pairs, and oxygen has 2 lone pairs on each oxygen atom. Total = 4 lone pairs.
\(F_2\): Each fluorine atom in \(F_2\) has 3 lone pairs. Total = 3 lone pairs. From the analysis, the molecules that contain only 2 lone pairs are \(H_2O\), \(CO\), and \(XeF_4\).
Therefore, the correct answer is 4 molecules.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 