The number of valence electrons present in th | vedclass

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(C) The enthalpy of atomisation depends on the number of unpaired electrons available for metallic bonding. Among the given transition metals ($Cr$,$C

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The enthalpy of atomisation depends on the number of unpaired electrons available for metallic bonding. Among the given transition metals ($Cr$,$Co$,$Fe$,$Ni$),$Cr$ has the lowest enthalpy of atomisation because it has a stable $d^5$ configuration,which limits the number of electrons participating in metallic bonding compared to others.For $Cr$ $(Z=24)$,the electronic configuration is $[Ar] 3d^5 4s^1$.The total number of valence electrons is the sum of electrons in the outermost $s$ and $d$ orbitals.Valence electrons = $5 (d) + 1 (s) = 6$.

List-$I$ (Transition metal,$M$)	List-$II$ ($E_{M^{2+}/ M}^{\ominus}$ in $V$)
$A$. $Ni$	$I$. $-1.18$
$B$. $Mn$	$II$. $-0.91$
$C$. $Fe$	$III$. $-0.25$
$D$. $Cr$	$IV$. $-0.44$

The number of valence electrons present in the metal among $Cr$,$Co$,$Fe$,and $Ni$ which has the lowest enthalpy of atomisation is

Similar Questions

Match the following:

List-$I$ (Transition metal,$M$) List-$II$ ($E_{M^{2+}/ M}^{\ominus}$ in $V$)

$A$. $Ni$ $I$. $-1.18$

$B$. $Mn$ $II$. $-0.91$

$C$. $Fe$ $III$. $-0.25$

$D$. $Cr$ $IV$. $-0.44$

The correct answer is

Discuss the position of $d$-block elements in the periodic table.

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The pair of ions with paramagnetic nature and same number of electrons is

For the four consecutive transition elements $Cr, Mn, Fe,$ and $Co$,the correct order of stability of their $+2$ oxidation state is:

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