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(C) Transition metals exhibit variable oxidation states because the energy difference between the $(n - 1) \, d$ orbitals and the $ns$ orbitals is ver

Vedclass · Accepted Answer

$ns$ and $(n - 1) \, d$ only but not $np$

Vedclass · Answer

(C) Transition metals exhibit variable oxidation states because the energy difference between the $(n - 1) \, d$ orbitals and the $ns$ orbitals is very small. Due to this small energy gap,electrons from both the $(n - 1) \, d$ subshell and the $ns$ subshell can participate in bond formation. Therefore,both $ns$ and $(n - 1) \, d$ electrons are involved in exhibiting variable oxidation states.

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 electrons which take part in order to exhibit variable oxidation states by transition metals are

Similar Questions

The number of unpaired electrons in $Zn^{2+}$ is

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

Which of the following metal ions has the maximum number of unpaired electrons?

The atomic radii from $Cr$ to $Cu$ are almost identical because of:

Which element from the following is $NOT$ considered a transition element on the basis of electronic configuration?

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