An aqueous pink solution of Cobalt $(II)$ chloride changes to deep blue on addition of excess of $HCl$. This is because:

  • A
    $[Co(H_2O)_6]^{2+}$ is transformed into $[CoCl_6]^{4-}$
  • B
    $[Co(H_2O)_6]^{2+}$ is transformed into $[CoCl_4]^{2-}$
  • C
    Tetrahedral complexes have larger crystal field splitting than octahedral complexes
  • D
    Octahedral complexes have larger crystal field splitting than square planar complexes

Explore More

Similar Questions

Which of the following is the most stable complex?

Difficult
View Solution

$A$ first row transition metal $(M)$ does not liberate $H_{2}$ gas from dilute $HCl$. $1 \ mol$ of aqueous solution of $MSO_{4}$ is treated with excess of aqueous $KCN$ and then $H_{2}S_{(g)}$ is passed through the solution. The amount of $MS$ (metal sulphide) formed from the above reaction is . . . . . . $mol$.

The brown ring test for $NO_2^-$ and $NO_3^-$ is due to the formation of a complex ion with the formula:

On treatment of $100 \ mL$ of $0.1 \ M$ solution of $CoCl_3 \cdot 6H_2O$ with excess $AgNO_3$,$1.2 \times 10^{22}$ ions are precipitated. The complex is:

The reaction of $CuSO_4$ with excess $KCN$ solution leads to the formation of:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo