Chemistry M.C.Q (1 Marks)

Electronic configuration of $\mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 d^4$

Electronic configuration of $\mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 d^5$

Electronic configuration of $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 d^\beta$

Electronic configuration of $\mathrm{Cr}^{2+}=[\mathrm{Ar}] 3 d^4$

As $\mathrm{Mn}^{3+}$ from $d^4$ configuration goes to more stable $d^5$ configuration (Half filled), due to more exchange energy in $d^5$ configuration.

Spin only magnetic moment is given by $\sqrt{n(n+2)} \mathrm{BM}$

$\therefore \mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{2+}$ will have same spin only magnetic moment.

$\mathrm{n} \rightarrow$ number of unpaired electron

$Image$

Hence $\mathrm{Ce}^{4+}$ and $\mathrm{Yb}^{2+}$ are only diamagnetic.

The correct order of group number of ions is $\underset{\text { (B) }}{\mathrm{Cu}^{2+}} < \underset{\text { (A) }}{\mathrm{Al}^{3+}} < \underset{\text { (D) }}{\mathrm{Co}^{2+}} < \underset{\text { (C) }}{\mathrm{Ba}^{2+}} < \underset{\text { (E) }}{\mathrm{Mg}^{2+}}$

$\therefore$ The correct order is $\mathrm{B}, \mathrm{A}, \mathrm{D}, \mathrm{C}, \mathrm{E}$

$\Delta \text { hyd } H ^{\circ} \text { of } Cu ^{2+}( aq )=-2121\,kJ\,mol ^{-1}$

$\Delta_{ i } H _1^0$ of $Cu =+745\,kJ\,mol ^{-1}$

$\Delta_{ i } H_2^{\circ} \text { of } Cu =+1960\,kJ\,mol ^{-1}$

- The highest oxidation number corresponding to the group number in transition metal oxides in attained in $Sc _2 O _3$ to $Mn _2 O _7$.

- Acidic character increases from $V _2 O _3$ to $V _2 O _4$ to $V _2 O _5$.

- $V _2 O _4$ dissolves in acids to give $VO ^{2+}$.

- $CrO$ is basic but $Cr _2 O _3$ is amphoteric.

$Gd ^{+2}=[ Xe ] 4 f ^{7} 5 d ^{1}$

After losing $5 d$ electron $4 f$ has maximum exchange energy so Gd has low value of Third Ionisation energy

Due to lanthanoid contration size of $4 \mathrm{~d}$ $\&$ $5 \mathrm{~d}$ series elements of same group are approx same.

Scandium shows only one oxidation state
i.e., +3 .
Cu $^{+}$ undergoes disproportionation reaction in aqueous solution
$2 Cu ^{*}( aq ) \longrightarrow Cu ^{2+}( aq )+ Cu ( s )$

dichromate $\left( Cr _{2} O _{7}\,^{-2}\right) \Rightarrow$ oxidation state $=+6$

oxidation state are same.

$SnO _{2}:$ amphoteric

$SiO _{2}:$ acidic

$GeO _{2}:$ acidic

Fluorine $\rightarrow$ Non-metal

Silicon $\rightarrow$ Metalloid

Cerium $\rightarrow$ Lanthanoid

Where $MnSO_4$ is colourless solution

$5 f, \,6 d$ and  $7s$ subshell. Thats why  exitation will be easier.

The appearance of green colour is due to the reduction of chromium metal.

$\Rightarrow n=2$

$N i^{+2}$

Here, $n=$ number of unpaired electrons

$\Rightarrow 2.83=\sqrt{n(n+2)}$

$\Rightarrow(2.83)^{2}=n(n+2)$

$0=8.00+n^{2}+2 n$

$n^{2}+ 2n-8=0$

$(n+4)(n-2)=0$

$n=2$      $n=-\,4$

$-\,4$ not be consider

Hence, $Ni^{2+}$ possesses a magnetic moment of $2.83\,\mathrm{B} . \mathrm{M}$

$5 \mathrm{O}_{2}+2 \mathrm{Mn} \mathrm{SO}_{4}+8 \mathrm{H}_{2} \mathrm{O}+\mathrm{K}_{2} \mathrm{SO}_{4}$

In the above reaction, $K M n O_{4}$ oxidises $H_{2} O_{2}$ to $O_{2}$ and itself $\left[M n O_{4}^{-}\right]$ gets reduced to $M n^{2+}$ ion as $M n S O_{4}$

Hence, aqueous solution of $K M n O_{4} .$ with $H_{2} O_{2}$ yields $M n^{2+}$ and $O_{2}$ in acidic conditions.

In $S{c}$, the $e^{-}$ configuration is $3 d^{1} 4 s^{2}$

wheareas in $Z{n}$ configuration is $3 d^{10} 4 s^{2}$

$\therefore \,\ln \,Z{n}$, the $3 d$ leaving completely filled, its not a transtion element.

$\mathrm{Ce}_{58}=[\mathrm{Xe}] 4 \mathrm{f}^{2} 5 \mathrm{d}^{0} 6 \mathrm{s}^{2}$

$\mathrm{Ce}^{2+}=[\mathrm{Xe}] 4 \mathrm{f}^{2}$ (two unpaired electrons)

$\mathrm{Sm}_{62}=[\mathrm{Xe}] 4 \mathrm{f}^{6} 5 \mathrm{d}^{0} 6 \mathrm{s}^{2}$

$\mathrm{Sm}^{2+}=[\mathrm{Xe}] 4 \mathrm{f}^{6}(\mathrm{six} \text { unpaired electron })$

$\mathrm{Eu}_{63}=[\mathrm{Xe}] 4 \mathrm{f}^{7} 5 \mathrm{d}^{0} 6 \mathrm{s}^{2}$

$\mathrm{Eu}^{2+}=[\mathrm{Xel}] 4 \mathrm{f}^{7}$ (seven unpaired electron)

$\mathrm{Yb}_{70}=[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{d}^{0} 6 \mathrm{s}^{2}$

$\mathrm{Yb}^{2+}=[\mathrm{Xe}] 4 \mathrm{f}^{14}$ (No unpaired electron)

Because of the absence of unpaired electrons, $Yb ^{2+}$ is diamagnetic.

$N{{i}^{4+}}=3{{d}^{6}}4{{s}^{0}}$  $(2)$

Maximum no. of ${e^ - }$ in last shell $= 6$

Group is $ VI-B.$

$Z{n^{ + 2 }} - 3{d^{10}}4{s^0}$

                                                                       $2$      $+$     $2$      $=4$

$2 + x - 8 = 0$

$x = 6$

$O.N.$ of $KMn{O_4}$

$1 + x - 8 = 0$

$x = 7$

Hence Options $\mathrm{B} \& \mathrm{C}$ are correct.