Which of the following has a yellow colour?

  • A
    Potassium cobaltinitrite
  • B
    Potassium hexanitrocobaltate $(III)$
  • C
    Fischer's salt
  • D
    All of the above

Explore More

Similar Questions

The correct option(s) about entropy $(S)$ is(are)
$[R =$ gas constant, $F =$ Faraday constant, $T =$ Temperature $]$
$(A)$ For the reaction, $M_{(s)} + 2H^{+}_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)}$, if $\frac{dE_{cell}}{dT} = \frac{R}{F}$, then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes in entropy and internal energy are temperature independent).
$(B)$ The cell reaction, $Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 0.01 \ M) \parallel H^{+}(aq, 0.1 \ M) \mid H_2(g, 1 \ bar) \mid Pt_{(s)}$, is an entropy driven process.
$(C)$ For racemization of an optically active compound, $\Delta S > 0$.
$(D)$ $\Delta S > 0$, for $[Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O$ (where $en =$ ethylenediamine).

Given below are two statements:
Statement-$I$: Hybridisation,shape,and spin-only magnetic moment of $K_{3}[Co(CO_{3})_{3}]$ are $sp^{3}d^{2}$,octahedral,and $4.9 \ BM$ respectively.
Statement-$II$: Geometry,hybridisation,and spin-only magnetic moment values $(BM)$ of the ions $[Ni(CN)_{4}]^{2-}$,$[MnBr_{4}]^{2-}$,and $[CoF_{6}]^{3-}$ respectively are square planar,tetrahedral,octahedral; $dsp^{2}$,$sp^{3}$,$sp^{3}d^{2}$ and $0, 5.9, 4.9$.

Three arrangements are shown for the complex $[CoBr_2(NH_3)_2(en)]^{\oplus}$. Which one is the wrong statement?

The number of ions present in $2.0 \ L$ of a solution of $0.8 \ M \ K_4[Fe(CN)_6]$ is

Difficult
View Solution

The reaction of $K_3[Fe(CN)_6]$ with freshly prepared $FeSO_4$ solution produces a dark blue precipitate called Turnbull's blue. The reaction of $K_4[Fe(CN)_6]$ with $FeSO_4$ solution in the complete absence of air produces a white precipitate $X$,which turns blue in air. Mixing the $FeSO_4$ solution with $NaNO_3$,followed by a slow addition of concentrated $H_2SO_4$ through the side of the test tube,produces a brown ring.
Precipitate $X$ is:
$(A)$ $Fe_4[Fe(CN)_6]_3$
$(B)$ $Fe[Fe(CN)_6]$
$(C)$ $K_2Fe[Fe(CN)_6]$
$(D)$ $KFe[Fe(CN)_6]$
Among the following,the brown ring is due to the formation of:
$(A)$ $[Fe(NO)_2(SO_4)_2]^{2-}$
$(B)$ $[Fe(NO)_2(H_2O)_4]^{3+}$
$(C)$ $[Fe(NO)_4(SO_4)_2]$
$(D)$ $[Fe(H_2O)_5(NO)]SO_4$

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