<span>Ans- bond with the highest bond energy is for ( C=C \rightarrow 620 (kJ/mol)
N=N \rightarrow 418 (kJ/mol)
Si=O \rightarrow 452 (kJ/mol)
O=O \rightarrow 498.7 (kJ/mol)
C=N \rightarrow 615 (kJ/mol)
C=C \rightarrow 620 (kJ/mol)</span>
The uneven distribution of charge between the hydrogen and oxygen atoms primarily leads to its <span>ability to dissolve ionic substances. This is because the unequal sharing leads to the polarity of water. Its polarity makes it dissolve to ionic substances.</span>
Answer:
ΔH > O and ΔH > TΔS
Explanation:
When a reaction is non-spontaneous even though the standard entropy change of the reaction is large and positive the reason is the reaction is highly endothermic and dominates the term TΔS in the standard Gibbs free energy expression:
ΔG = ΔH-TΔS
Remember that for an endothermic reaction to be spontaneous ΔG < 0 and this can occur only if the term TΔS is greater than ΔH. This is not the case for quartz which requires a good quantity of energy to overcome its crystaline 3D lattice in which the oxygen are shared between 2 silicon atoms.
The hybridization for the Br in BrO4⁻ is 
. So, the correct option is (e).
In chemistry, the idea of combining atomic orbitals to create new hybrid orbitals (with different energies, shapes, etc., than the component atomic orbitals) is known as orbital hybridisation (or hybridization). These new hybrid orbitals are suitable for the pairing of electrons to form chemical bonds in valence bond theory.
Because more directional hybridised orbitals result in higher overlap when creating bonds, stronger bonds are formed, which favours the hybridization of orbitals. When hybridization takes place, this leads to more stable molecules.
One s orbital and three p orbitals combine to form four orbitals, each of which has a 25% s character and a 75% p character. This process is known as hybridization. Anytime an atom is surrounded by four groups of electrons, this kind of hybridization is necessary.
Learn more about hybridization here:
brainly.com/question/12207339
#SPJ4
Im not sure I understand the question at hand..
