(B)

            Electronic configuration of Nd can be written by the following Aufbau rule as: ${}_{60}\mathrm{Nd}={}_{54}\left[\mathrm{Xe}\right]4f^{4}6s^2$.

(B)

           The common oxidation state of Lanthanoids is +3. Due to stable half-filled and fully-filled configurations other oxidation states of few elements such as ${\mathrm{Ce}}^{4+}$ $\left(4{\mathrm{f}}^0\right),$ ${\mathrm{Tb}}^{4+}\left(4{\mathrm{f}}^7\right),$ ${\mathrm{Eu}}^{2+}$ $\left(4{\mathrm{f}}^7\right)$ and ${\mathrm{Yb}}^{2+}$ $\left(4{\mathrm{f}}^{14}\right)$ also exist. However, these have tendency to revert to common oxidation state of +3. Hence, ${\mathrm{Ce}}^{4+}$ and ${\mathrm{Tb}}^{4+}$ are good oxidizing agents, ${\mathrm{Eu}}^{2+}$ and ${\mathrm{Yb}}^{2+}$ are good reducing agents.

(B) 

(D)

      (A)(D) In transition series elements, melting point and enthalpy of atomization first increase, reaches a maxima and then decrease. This happens because unpaired $d$ electrons have a strong tendency to participate in metallic bonding. Since Zn, Cd and Hg have comparatively low melting points in respective transition series, these are called soft metals.

(B) Mercury shows +1 and +2 oxidation states.

(C) Oxidation state of Zn and Cd is +2 in their compounds i.e. they have $d^{10}$ configuration in their compounds, which makes them diamagnetic. ${\mathrm{Hg}}^{2+}$ is also $d^{10}$ and diamagnetic. In +1 oxidation state, mercury exists as a dimer $\left({\mathrm{Hg}}_2^{2+}\right)$, hence in +1 oxidation state also mercury is diamagnetic.

(B)

(A)

Ionization energy increases from left to right across a period. But third ionization energy of Mn is more than that of Fe because third electron from Mn is to be removed from stable $d^5$ configuration.

Hence among given options Mn has highest third ionization energy.

(B)

 Assertion:

As ${\mathrm{Cr}}^{3+}/{\mathrm{Cr}}^{2+}$ potential is negative, ${\mathrm{Cr}}^{2+}/{\mathrm{Cr}}^{3+}$ potential is positive, i.e. ${\mathrm{Cr}}^{2+}$ has a tendency to get oxidized to ${\mathrm{Cr}}^{3+}$. Thus aqueous solution of ${\mathrm{Cr}}^{2+}$ is reducing.

As ${\mathrm{Mn}}^{3+}/{\mathrm{Mn}}^{2+}$ potential is positive, ${\mathrm{Mn}}^{3+}$ has a tendency to get reduced to ${\mathrm{Mn}}^{2+}$. Thus aqueous solution of ${\mathrm{Mn}}^{3+}$ is oxidizing. 

Reason:  

${\mathrm{Mn}}^{2+}$ $\left({}_{18}\left[\mathrm{Ar}\right]3{\mathrm{d}}^5\right)$ has stable half filled subshell. Thus ${\mathrm{Mn}}^{2+}$ has a tendency to be formed. This makes ${\mathrm{Mn}}^{3+}/{\mathrm{Mn}}^{2+}$ positive. ${\mathrm{Cr}}^{3+}$ has half filled ${\mathrm{t}}_{2\mathrm{g}}$ orbital. Thus ${\mathrm{Cr}}^{3+}$ has a tendency to be formed. This makes ${\mathrm{Cr}}^{2+}/{\mathrm{Cr}}^{3+}$ potential positive.

(C)

            Sc shows only +3 oxidation state.

(D)

        Spin only magnetic moment ($\mu$) is related to number of unpaired electrons $n$ by:

         $\mu =\sqrt{n(n+2)} \text{BM}$

         $3.86=\sqrt{n(n+2)}$

         $\mathrm{n}=3$

(C)

Oxidation states exhibited by various first transition series $d$ block elements are as follows: