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.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)