Which of the following orders is correct for | vedclass

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(C) The standard electrode potential $(E^{\circ})$ for $M^{3+}/M^{2+}$ follows the order $Mn^{3+}/Mn^{2+} > Fe^{3+}/Fe^{2+} > Cr^{3+}/Cr^{2+}$. Thus,o

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$VO_2^{+} < Cr_2O_7^{2-} < MnO_4^{-}$ - oxidizing power

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(C) The standard electrode potential $(E^{\circ})$ for $M^{3+}/M^{2+}$ follows the order $Mn^{3+}/Mn^{2+} > Fe^{3+}/Fe^{2+} > Cr^{3+}/Cr^{2+}$. Thus,option $A$ is incorrect.Magnetic moment is calculated as $\mu = \sqrt{n(n+2)}$,where $n$ is the number of unpaired electrons.For $Mn^{2+}$ $(3d^5)$,$n=5$. For $Cr^{2+}$ $(3d^4)$,$n=4$. For $Fe^{2+}$ $(3d^6)$,$n=4$. The order is $Mn^{2+} > Cr^{2+} \approx Fe^{2+}$. Thus,option $B$ is incorrect.Oxidizing power depends on the reduction potential and the stability of the lower oxidation state. $MnO_4^-$ $(Mn^{+7})$ is a stronger oxidizing agent than $Cr_2O_7^{2-}$ $(Cr^{+6})$ and $VO_2^+$ $(V^{+5})$. The correct order is $VO_2^+ First ionization enthalpy generally increases across the period but shows irregularities due to electronic configuration. The order is $Ti < V < Cr$ is incorrect; the correct order is $Ti < V < Cr$ is not followed due to stability of half-filled $d$-orbitals in $Cr$.

List-$I$	List-$II$
$A$. Oxidation state of $V$ in $VOCl_2$	$I$. $0$
$B$. Number of unpaired electrons in $MnO_4^{2-}$ ion	$II$. $1$
$C$. Number of unpaired electrons in $[NiCl_4]^{2-}$ ion	$III$. $5$
$D$. This oxidation state is exhibited by all lanthanide ions	$IV$. $3$
	$V$. $4$
	$VI$. $2$

Property	Metal
$a$. Element with highest second ionization enthalpy $(\Delta_{i} H_2)$	$i$. $Co$
$b$. Element with highest third ionization enthalpy $(\Delta_{i} H_3)$	$ii$. $Cr$
$c$. $M$ in $[M(CO)_6]$	$iii$. $Cu$
$d$. Element with highest heat of atomization $(\Delta_{a} H)$	$iv$. $Zn$
	$v$. $Ni$

Which of the following orders is correct for the property given?

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