Three is the answer expected. Higher valences of phosphorus, in PCl5 for example can be explained by hybridisation although this method is not the only explanation.
To find the hybridization of an atom, you can use the formula: hybridization number of sigma bonds number of lone pairs on the atom. Count the sigma bonds and lone pairs, then determine the hybridization based on the total.
A Ca atom in its 4s^2 electron configuration can make up to 2 bonds without hybridization. This is because it has two unpaired electrons in its 4s orbital available for bonding.
To determine the hybridization of the central atom in a molecule, you can use the formula: hybridization number of sigma bonds number of lone pairs on the central atom. Count the number of sigma bonds and lone pairs around the central atom, then use this formula to find the hybridization.
To determine the hybridization of a central atom in a molecule, you can use the formula: hybridization number of sigma bonds number of lone pairs on the central atom. Count the sigma bonds and lone pairs, then use this formula to find the hybridization.
To determine the hybridization of the central atom in a molecule, you can use the formula: hybridization number of sigma bonds number of lone pairs on the central atom. Count the number of sigma bonds and lone pairs around the central atom, then use this formula to find the hybridization.
To find the hybridization of an atom, you can use the formula: hybridization number of sigma bonds number of lone pairs on the atom. Count the sigma bonds and lone pairs, then determine the hybridization based on the total.
A Ca atom in its 4s^2 electron configuration can make up to 2 bonds without hybridization. This is because it has two unpaired electrons in its 4s orbital available for bonding.
To determine the hybridization of the central atom in a molecule, you can use the formula: hybridization number of sigma bonds number of lone pairs on the central atom. Count the number of sigma bonds and lone pairs around the central atom, then use this formula to find the hybridization.
To determine the hybridization of a central atom in a molecule, you can use the formula: hybridization number of sigma bonds number of lone pairs on the central atom. Count the sigma bonds and lone pairs, then use this formula to find the hybridization.
To determine the hybridization of the central atom in a molecule, you can use the formula: hybridization number of sigma bonds number of lone pairs on the central atom. Count the number of sigma bonds and lone pairs around the central atom, then use this formula to find the hybridization.
Mg in the 3s^2 configuration can form up to two bonds without hybridization. This is because it has two valence electrons in its 3s orbital, allowing it to form two bonds by losing or sharing these electrons.
The hybridization of MnO4- is sp3. Each oxygen atom contributes one electron to form single bonds with manganese, leading to the sp3 hybridization of the central manganese atom.
The hybridization of CH3 is sp3. Each carbon atom forms four sigma bonds with hydrogen atoms, resulting in a tetrahedral geometry and sp3 hybridization.
To determine the sp hybridization of a molecule, you can look at the number of sigma bonds and lone pairs around the central atom. If there are two sigma bonds and no lone pairs, the central atom is sp hybridized.
The central atom in BCl3 is boron, which has only three valence electrons. Since it forms three bonds with the chlorine atoms, the hybridization of the central boron atom is sp2.
The central atom Xe in XeCl2 is in a hybridization of sp3d. Xenon has 8 valence electrons, and to form two Xe-Cl bonds, it undergoes hybridization to utilize its 5d orbital along with the 2s and 3p orbitals, resulting in sp3d hybridization.
The central oxygen atom in H3O+ has sp3 hybridization. This means that the oxygen atom in H3O+ forms four equivalent bonds with the three hydrogen atoms and the lone pair, resulting in a tetrahedral geometry.