On treatment of 100 mL of 0-1 M solution of CoCl 3 -6 H 2 O with excess AgNO 3 - 1-2 - 1022 ions are precipitated- The complex isA- - Co H 2 O 6- Cl 3B- - Co H 2 O 5 Cl - Cl 2- H 2 OC- - Co H 2 O 4 Cl 2- Cl -2 H 2 OD- - Co H 2 O 3 Cl 3- -3 H 2 O

The correct option is B [Co(H_2O)_5Cl]Cl_2.H_2OMillimoles of AgNO_3= 1.2 × 10^22/6 × 10^23× 1000=20 Millimoles of CoCl_3.6H_2O=0.1 × 100=10 ∴ Each mole of CoCl ...

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Byju's Answer

Standard XII

Chemistry

Coordination and Oxidation Number

On treatment ...

Question

On treatment of 100 mL of 0.1M solution of $C o C l_{3} .6 H_{2} O$ with excess $A g N O_{3}; 1.2 \times 10^{22}$ ions are precipitated. The complex is

[C o (H_{2} O)_{6}] C l_{3}

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[C o (H_{2} O)_{5} C l] C l_{2} . H_{2} O

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[C o (H_{2} O)_{4} C l_{2}] C l .2 H_{2} O

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[C o (H_{2} O)_{3} C l_{3}] .3 H_{2} O

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Solution

The correct option is B $[C o (H_{2} O)_{5} C l] C l_{2} . H_{2} O$
Millimoles of $A g N O_{3} = \frac{1.2 \times 10^{22}}{6 \times 10^{23}} \times 1000 = 20$
Millimoles of $C o C l_{3} .6 H_{2} O = 0.1 \times 100 = 10$
$∴$ Each mole of $C o C l_{3} .6 H_{2} O$ gives two chloride ions.
$∴ [C o (H_{2} O)_{5} C l] C l_{2} . H_{2} O$

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On treatment of 100 mL of 0.1M solution of CoCl3.6H2O with excess AgNO3; 1.2×1022 ions are precipitated. The complex is

The correct option is B [Co(H2O)5Cl]Cl2.H2OMillimoles of AgNO3=1.2×10226×1023×1000=20 Millimoles of CoCl3.6H2O=0.1×100=10 ∴ Each mole of CoCl3.6H2O gives two chloride ions. ∴[Co(H2O)5Cl]Cl2.H2O

On treatment of 100 mL of 0.1M solution of $C o C l_{3} .6 H_{2} O$ with excess $A g N O_{3}; 1.2 \times 10^{22}$ ions are precipitated. The complex is