Select the correct statement.

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(C) Let us analyze each option:$A$: Acidic strength depends on bond dissociation energy. $HCl$ has a stronger bond than $HBr$,so $HBr$ is more acidic.

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Dipole moment of $CH_3Cl > CH_3F$,but the reverse is true for their $H-C-H$ bond angle.

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(C) Let us analyze each option:$A$: Acidic strength depends on bond dissociation energy. $HCl$ has a stronger bond than $HBr$,so $HBr$ is more acidic. Reducing property also follows the order $HI > HBr > HCl > HF$. Thus,both properties follow the same order. This is incorrect.$B$: Basic strength depends on the availability of the lone pair. As we go down the group,the lone pair is in a larger orbital,making it less available. Thus,$PH_3 > AsH_3$. Bond angle also follows the same order due to the decrease in electronegativity of the central atom. This is incorrect.$C$: The dipole moment of $CH_3Cl$ $(1.86 \ D)$ is greater than $CH_3F$ $(1.85 \ D)$ due to the larger bond length in $C-Cl$. The $H-C-H$ bond angle in $CH_3F$ is larger than in $CH_3Cl$ due to the higher electronegativity of $F$ causing more $s$-character in $C-H$ bonds. This is correct.$D$: Maleic acid (cis-isomer) has a higher $Ka_1$ due to intramolecular hydrogen bonding in the mono-anion,but its $Ka_2$ is lower than fumaric acid due to strong electrostatic repulsion between the two carboxylate groups. This is correct.Note: Both $C$ and $D$ are scientifically correct statements.

Similar Questions

Calculate the osmotic pressure in $atm$ at $0\,^oC$ of a $0.18\, m$ aqueous solution of $KCl$ which has a freezing point of $-0.68\,^oC$. Assume the volume of the solution is equal to the volume of pure water. $(K_f = 1.86\,^oC\, m^{-1})$

Which of the following has the lowest freezing point?

The elevation in boiling point for $1 \ m$ solution of non-volatile solute $A$ is $3 \ K$. The depression in freezing point for $2 \ m$ solution of $A$ in the same solvent is $6 \ K$. The ratio of $K_{b}$ and $K_{f}$ i.e.,$K_{b} / K_{f}$ is $1 : X$. The value of $X$ is [nearest integer].

If the mole fraction of the solvent in a solution decreases,then:

The freezing point of an aqueous solution is $-0.186 \ ^oC$. The boiling point of the same solution is ........ $^oC$. (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$ and $K_b = 0.512 \ K \ kg \ mol^{-1}$)

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