A transition metal M among Sc, Ti, V, Cr, Mn | vedclass

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(B) The second ionization enthalpy $(IE_2)$ involves removing an electron from the $M^{+}$ ion. For $Cr$ $(Z=24)$,the electronic configuration is $[Ar

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$6$

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(B) The second ionization enthalpy $(IE_2)$ involves removing an electron from the $M^{+}$ ion. For $Cr$ $(Z=24)$,the electronic configuration is $[Ar] 3d^5 4s^1$. The $Cr^{+}$ ion has the configuration $[Ar] 3d^5$. Removing the second electron from this stable half-filled $d^5$ subshell requires a very high amount of energy.Thus,$M = Cr$.The $Cr^{+}$ ion has the configuration $[Ar] 3d^5$,which means it has $n = 5$ unpaired electrons.The spin-only magnetic moment is calculated as $\mu = \sqrt{n(n+2)} \ BM$.$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \ BM$.Rounding to the nearest integer,we get $6 \ BM$.

$A$ transition metal $M$ among $Sc, Ti, V, Cr, Mn$ and $Fe$ has the highest second ionisation enthalpy. The spin only magnetic moment value of $M^{+}$ ion is $. . . . . . . BM$ (Near integer)
(Given atomic number $Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26$)

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$A$ transition metal $M$ among $Sc, Ti, V, Cr, Mn$ and $Fe$ has the highest second ionisation enthalpy. The spin only magnetic moment value of $M^{+}$ ion is $. . . . . . . BM$ (Near integer)(Given atomic number $Sc: 21, Ti: 22, V: 23, Cr: 24, Mn: 25, Fe: 26$)