Answer:
SF2 > H2O > PBr3 > NCl3
Explanation:
Compare the electronegativity values for the atoms and classify the nature of the bonding based on the electronegativity difference.
P has an electronegativity of 2.1, while Br has an electronegativity of 2.96. The difference is 0.86, indicating that these atoms will form covalent bonds.
S has an electronegativity of 2.58 while F has an electronegativity of 4.0. The difference is 1.42, indicating that these atoms will form polar covalent bonds.
O has an electronegativity of 3.5 while H has an electronegativity of 2.1. The difference is 1.4, indicating that these atoms will form polar covalent bonds.
N has an electronegativity of 3.04, whereas Cl has an electronegativity of 3.5. This difference of 0.46 indicates that these atoms will form covalent bonds.
We know that the greater the electronegativity, the higher the polarity. In decreasing order of polarity, we have:
SF2 > H2O > PBr3 > NCl3
The last answer is correct
The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.
The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).
The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).
The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.
Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.