JEE MAIN - Chemistry (2023 - 10th April Morning Shift - No. 6)
Which of the following statements are correct?
(A) The M$$^{3+}$$/M$$^{2+}$$ reduction potential for iron is greater than manganese.
(B) The higher oxidation states of first row d-block elements get stabilized by oxide ion.
(C) Aqueous solution of Cr$$^{2+}$$ can liberate hydrogen from dilute acid.
(D) Magnetic moment of V$$^{2+}$$ is observed between 4.4 - 5.2 BM.
Choose the correct answer from the options given below :
(B), (C) only
(A), (B) only
(A), (B), (D) only
(C), (D) only
Explanation
The reduction electrode potential of $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$ is $+1.57 \mathrm{~V}$ while that of $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ is $+0.77 \mathrm{~V}$, hence $\mathrm{A}$ is wrong. Higher oxidation state of smaller $d$-block elements is stabilized (or say form compounds) with smaller anion oxide that can be explained by stearic reason hence $B$ is correct.
The oxidation electrode potential of $\mathrm{Cr}^{2+} / \mathrm{Cr}^{3+}$ is $+0.41 \mathrm{~V}$ hence it can reduce $\mathrm{H}^{+}$and so liberate $\mathrm{H}_2$. The unpaired electrons in $\mathrm{V}^{2+}$ are 3 hence the magnetic moment of $\mathrm{V}^{2+}$ will be lesser than 4.4 BM.
Hence, only $B$ and $C$ are correct.
The oxidation electrode potential of $\mathrm{Cr}^{2+} / \mathrm{Cr}^{3+}$ is $+0.41 \mathrm{~V}$ hence it can reduce $\mathrm{H}^{+}$and so liberate $\mathrm{H}_2$. The unpaired electrons in $\mathrm{V}^{2+}$ are 3 hence the magnetic moment of $\mathrm{V}^{2+}$ will be lesser than 4.4 BM.
Hence, only $B$ and $C$ are correct.
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