CrCl3⋅6H2O dissolves in water to form a complex. If a solution is prepared by dissolving 0.1 mole of CrCl3⋅6H2O in water is treated with excess AgNO3 solution, 0.2 mole of AgCl is precipitated. Select correct option for complex: (Mass number of Cr=52,O=16,Cl=35.5)

CrCl3⋅6H2O→CrH2O5ClCl2⋅H2OCr′=[Ar]3d54s1 Cr3+=[Ar]Formation of complex Hybridisation ⇒d2sp3Magnetism ⇒ paramagneticIt will exist in hydrated isomerism CrH2O5ClCl2⋅H2O,CrH2O6Cl3

CrCl3⋅6H2O dissolves in water to form a complex. If a solution is prepared by dissolving 0.1 mole of CrCl3⋅6H2O in water is treated with excess AgNO3 solution, 0.2 mole of AgCl is precipitated. Select correct option for complex: (Mass number of Cr=52,O=16,Cl=35.5)

CrCl3⋅6H2O→CrH2O5ClCl2⋅H2OCr′=[Ar]3d54s1 Cr3+=[Ar]Formation of complex Hybridisation ⇒d2sp3Magnetism ⇒ paramagneticIt will exist in hydrated isomerism CrH2O5ClCl2⋅H2O,CrH2O6Cl3

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${CrCl}_{3} \cdot 6 H_{2} O$ dissolves in water to form a complex. If a solution is prepared by dissolving 0.1 mole of ${CrCl}_{3} \cdot 6 H_{2} O$ in water is treated with excess ${AgNO}_{3}$ solution, 0.2 mole of AgCl is precipitated. Select correct option for complex: (Mass number of $Cr = 52, O = 16, Cl = 35.5)$

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It will exist in hydrate isomerism

It loses almost 6.75% by mass of its original mass on treatment with $Conc. H_{2} {SO}_{4}$

It is a paramagnetic complex in which central metal is in $d^{2} {sp}^{3}$ hybridisation

It will exist in optical isomeric form

answer is A, B, C.

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Detailed Solution

${CrCl}_{3} \cdot 6 H_{2} O \to [Cr {(H_{2} O)}_{5} Cl] {Cl}_{2} \cdot H_{2} O$
${Cr}^{'} = [Ar] 3 d^{5} 4 s^{1}$ Question Image
${Cr}^{3 +} = [Ar]$
Formation of complex
Hybridisation $\Rightarrow d^{2} s p^{3}$
Magnetism $\Rightarrow$ paramagnetic
It will exist in hydrated isomerism $[Cr {(H_{2} O)}_{5} Cl] {Cl}_{2} \cdot H_{2} O, [Cr {(H_{2} O)}_{6}] {Cl}_{3}$