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Hint: The answer of the assertion lies in the strength of O-H and S-H bond whereas the answer of reason lies in the electronegativity of oxygen and sulfur.
Complete step by step solution:
As mentioned in the hint, here, we need to predict the strength of the O-H and S-H bond. O-H bond is stronger than S-H bond. Also, O is more electronegative than S. Due to this, bond dissociation enthalpy of H-S bond is lower than that of H-O bond.
Therefore, ${{\text{H}}_{2}}\text{S}$ is more acidic than ${{\text{H}}_{2}}\text{O}$.
The polarity of a covalent bond depends on the difference in the electronegativity of the bonding atoms. The higher the difference in the electronegativities of the bonding atoms, the greater is the bond polarity. On the periodic table, (i) in each period: the electronegativity increases from left to right and (ii) in each group, the electronegativity decreases down the group from top to bottom. Using these trends on the periodic table, it is possible to predict which bond is polar. Therefore, the order of the polarity of the bonds of O and S with hydrogen is as follows: $\text{O-H >S-H}$
Hence, we can reach to the conclusion from the above discussion that
${{\text{H}}_{2}}\text{S}$ is more acidic than ${{\text{H}}_{2}}\text{O}$.
$\text{H-S}$ Bond is more polar than $\text{H-O}$bond.
Thus, the correct answer would be option (c) assertion is correct but reason is incorrect.
Note: Apart from that, we can also predict the nature of acidity on the basis of dissociation constant. Higher the dissociation constant, the stronger is the acid. Acids and bases are measured using the pH scale. Also, electronegativity can be explained as the tendency of an atom participating in a covalent bond to attract the bonding electrons. The most frequently used to measure electronegativity is the Pauling scale. Fluorine is the most electronegative element and has been assigned a value of 4.0 on Pauling scale.
Complete step by step solution:
As mentioned in the hint, here, we need to predict the strength of the O-H and S-H bond. O-H bond is stronger than S-H bond. Also, O is more electronegative than S. Due to this, bond dissociation enthalpy of H-S bond is lower than that of H-O bond.
Therefore, ${{\text{H}}_{2}}\text{S}$ is more acidic than ${{\text{H}}_{2}}\text{O}$.
The polarity of a covalent bond depends on the difference in the electronegativity of the bonding atoms. The higher the difference in the electronegativities of the bonding atoms, the greater is the bond polarity. On the periodic table, (i) in each period: the electronegativity increases from left to right and (ii) in each group, the electronegativity decreases down the group from top to bottom. Using these trends on the periodic table, it is possible to predict which bond is polar. Therefore, the order of the polarity of the bonds of O and S with hydrogen is as follows: $\text{O-H >S-H}$
Hence, we can reach to the conclusion from the above discussion that
${{\text{H}}_{2}}\text{S}$ is more acidic than ${{\text{H}}_{2}}\text{O}$.
$\text{H-S}$ Bond is more polar than $\text{H-O}$bond.
Thus, the correct answer would be option (c) assertion is correct but reason is incorrect.
Note: Apart from that, we can also predict the nature of acidity on the basis of dissociation constant. Higher the dissociation constant, the stronger is the acid. Acids and bases are measured using the pH scale. Also, electronegativity can be explained as the tendency of an atom participating in a covalent bond to attract the bonding electrons. The most frequently used to measure electronegativity is the Pauling scale. Fluorine is the most electronegative element and has been assigned a value of 4.0 on Pauling scale.
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