$(b)\,\mathop {BaS{O_4} \downarrow }\limits_{(InSo\operatorname{lub} le)} + \mathop {2NaOH}\limits_{(So\operatorname{lub} le)} \xrightarrow{{Aqueous}}\mathop {Ba{{(OH)}_2}}\limits_{(So\operatorname{lub} le)} + \mathop {N{a_2}S{O_4}}\limits_{(So\operatorname{lub} le)} $
$(c)\,\mathop {Pb{{(N{O_3})}_2}}\limits_{(So\operatorname{lub} le)} + \mathop {2C{H_3}COONa}\limits_{(So\operatorname{lub} le)} \xrightarrow{{Aqueous}}\mathop {Pb{{(OAc)}_2}}\limits_{(So\operatorname{lub} le)} + \mathop {2NaN{O_3}}\limits_{(So\operatorname{lub} le)} $
$(d)\,\mathop {N{a_2}Cr{O_7}}\limits_{(So\operatorname{lub} le)} + \mathop {BaC{l_2}}\limits_{(So\operatorname{lub} le)} \xrightarrow{{Aqueous}}\mathop {BaCr{O_4}}\limits_{(InSo\operatorname{lub} le)} \downarrow + \mathop {2NaCl}\limits_{(So\operatorname{lub} le)} $
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$1.$ Among the following, the correct statement is
$(A)$ Phosphates have no biological significance in humans
$(B)$ Between nitrates and phosphates, phosphates are less abundant in earth's crust
$(C)$ Between nitrates and phosphates, nitrates are less abundant in earth's crust
$(D)$ Oxidation of nitrates is possible in soil
$2.$ Among the following, the correct statement is
$(A)$ Between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{NH}_3$ is better electron donor because the lone pair of electrons occupies spherical ' $\mathrm{s}$ ' orbital and is less directional
$(B)$ Between $\mathrm{NH}_5$ and $\mathrm{PH}_5, \mathrm{PH}_5$ is better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$ orbital and is more directional
$(C)$ Between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{NH}_3$ is a better electron donor because the lone pair of electrons occupies $\mathrm{sp}^3$ orbital and is more directional
$(D)$ Between $\mathrm{NH}_3$ and $\mathrm{PH}_3, \mathrm{PH}_3$ is better electron donor because the lone pair of electrons occupies spherical ' $\mathrm{s}$ ' orbital and is less directional
$3.$ White phosphorus on reaction with $\mathrm{NaOH}$ gives $\mathrm{PH}_3$ as one of the products. This is a
$(A)$ dimerization reaction
$(B)$ disproportionation reaction
$(C)$ condensation reaction
$(D)$ precipitation reaction
Give the answer question $1,2$ and $3.$
| $Pb ^{2+} / Pb$ | $-0.13 V$ |
| $Ni ^{2+} / Ni$ | $-0.24 V$ |
| $Cd ^{2+} / Cd$ | $-0.40 V$ |
| $Fe ^{2+} / Fe$ | $V-0.44 V$ |
To a solution containing $0.001 M$ of $X ^{2+}$ and $0.1 M$ of $Y ^{2+}$, the metal rods $X$ and $Y$ are inserted (at $298 K$ ) and connected by a conducting wire. This resulted in dissolution of $X$. The correct combination(s) of $X$ and $Y$, respectively, is (are)
(Given: Gas constant, $R =8.314 J K ^{-1} mol ^{-1}$,
Faraday constant, $F =96500 C mol ^{-1}$ )
$(A)$ $Cd$ and $Ni$ $(B)$ $Cd$ and $Fe$ $(C)$ $Ni$ and $Pb$ $(D)$ $Ni$ and $Fe$
