Given Molecules and Their Hybridization:
Correct Answer: The correct answer is 4.
• \(NH_3\): The nitrogen atom is surrounded by three bonded atoms and one lone pair, resulting in an sp3 hybridization due to the tetrahedral arrangement.
• \(SO_2\): Sulfur in \(SO_2\) is sp2 hybridized because it forms two sigma bonds and has one lone pair, giving a bent structure.
• \(SiO_2\): Each silicon atom forms four sigma bonds with oxygen atoms. However, due to its extended lattice structure, we consider the local bonding, indicating sp3 hybridization for the central Si atom.
• \(BeCl_2\): The beryllium atom is sp hybridized as it forms two sigma bonds with chlorine atoms, leading to a linear geometry.
• \(CO_2\): Carbon in \(CO_2\) is sp hybridized since it forms two sigma bonds with oxygen atoms, resulting in a linear structure.
• \(H_2O\): The oxygen atom has two sigma bonds and two lone pairs, leading to sp3 hybridization, resulting in a bent structure.
• \(CH_4\): Carbon in \(CH_4\) is sp3 hybridized as it forms four sigma bonds, resulting in a tetrahedral geometry.
• \(BF_3\): Boron in \(BF_3\) is sp2 hybridized, as it forms three sigma bonds with fluorine atoms, resulting in a planar triangular structure.
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 : \(A_2 \rightleftharpoons 2A\)