$$Ce \to \left[ {Xe} \right]4{f^1}5{d^1}6{s^2};C{e^{4 + }}$$       $$ \to \left[ {Xe} \right]$$
$$Yb \to \left[ {Xe} \right]4{f^{14}}6{s^2};Y{b^{2 + }}$$      $$ \to \left[ {Xe} \right]4{f^{14}}$$
$$Lu \to \left[ {Xe} \right]4{f^{14}}5{d^1}6{s^2};L{u^{3 + }}$$       $$ \to \left[ {Xe} \right]4{f^{14}}$$
$$Eu \to \left[ {Xe} \right]4{f^7}6{s^2};E{u^{2 + }}$$       $$ \to \left[ {Xe} \right]4{f^7}$$

In acid medium
$$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$$
( $$O.S.$$  of $$Mn$$  changes form +7 to +2 ).

More number of oxidation states are exhibited by the actinides than by the corresponding lanthanides due to lesser energy difference between $$5f$$ and $$6d$$  orbitals than that between $$4f$$ and $$5d$$ - orbitals.

No explanation is given for this question. Let's discuss the answer together.

Solder is an alloy containing $$Sn - 67\% \,\,{\text{and}}\,\,Pb - 33\% .$$

$$\eqalign{ & F{e^{3 + }}\left( {Z = 26} \right) \to 3{d^5} \cr & M{n^{2 + }}\left( {Z = 25} \right) \to 3{d^5} \cr} $$

$${\text{Grey tin}} \rightleftharpoons {\text{white tin }}$$
Grey tin is brittle and crumbles down to powder in very cold climate
The conversion of grey tin to white tin is acompained by increase in volume. This is known as tin plaque or tin disease.

$$\underset{\left(\text{Yellow}\right)}{2Cr{O}_4^{2-}}\underset{O{H}^{-}}{\overset{2H^{+}}{\rightleftharpoons }}\underset{\left(\text{Orange}\right)}{C{r}_2{O}_7^{2-}}+H_{2}O$$

$$\eqalign{ & \mu = \sqrt {n\left( {n + 2} \right)} \cr & 1.73 = \sqrt {n\left( {n + 2} \right)} \cr} $$
On calculating the value of $$n$$ we find $$n = 1$$
No. of unpaired electrons $$= 1$$
hence its configuration will be
$$\eqalign{ & V\left( {23} \right) = \left[ {Ar} \right]3{d^3}4{s^2} \cr & {V^{4 + }} = \left[ {Ar} \right]3{d^1} \cr} $$
∴ Its chloride has the formula $$VC{l_4}$$