The correct order of decreasing third ionisat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The third ionisation enthalpy involves the removal of an electron from the $M^{2+}$ ion.The electronic configurations of the $M^{2+}$ ions are:$Ti

Vedclass · Accepted Answer

$Mn > Cr > V > Ti$

Vedclass · Answer

(A) The third ionisation enthalpy involves the removal of an electron from the $M^{2+}$ ion.The electronic configurations of the $M^{2+}$ ions are:$Ti^{2+}: [Ar] 3d^2$$V^{2+}: [Ar] 3d^3$$Cr^{2+}: [Ar] 3d^4$$Mn^{2+}: [Ar] 3d^5$For $Mn^{2+}$,the $3d^5$ configuration is half-filled and highly stable,making the removal of the third electron very difficult (highest third ionisation enthalpy).For $Cr^{2+}$,the removal of the third electron leads to the stable $3d^3$ configuration ($t_{2g}^3$ in octahedral field),which is also quite stable.Comparing the values,the order of decreasing third ionisation enthalpy is $Mn > Cr > V > Ti$.

The correct order of decreasing third ionisation enthalpy of $Ti, V, Cr, Mn$ is

Similar Questions

The displacement-time equation of a particle executing $SHM$ is $x = A \sin \left( \omega t + \frac{\pi}{6} \right)$. The time taken by the particle to go directly from $x = -\frac{A}{2}$ to $x = +\frac{A}{2}$ is

In leaves of $C_4$ plants,malic acid synthesis during $CO_2$ fixation occurs in :-

$A$ group of atoms can function as a ligand only when

In $\triangle ABC$,$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

Vedclass Test Series

Exam Paper Generator

Online Exam Module