- A$C{H_2}FCOOH$
- B$C{H_2}ClCOOH$
- C$CHC{l_2}COOH$
- ✓$CH{F_2}COOH$
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| Observation | $[A]$ | $[B]$ | Rate of reaction |
| $1$ | $0.1$ | $0.1$ | $2\times10^{-3}\, mol\, L^{-1}\,sec^{-1}$ |
| $2$ | $0.2$ | $0.1$ | $0.4\times10^{-2}\, mol\, L^{-1}\,sec^{-1}$ |
| $3$ | $0.1$ | $0.2$ | $1.4\times10^{-2}\, mol\, L^{-1}\,sec^{-1}$ |

Precipitate $X$ is
$(A)$ $Fe _4\left[ Fe ( CN )_6\right]_3$
$(B)$ $Fe \left[ Fe ( CN )_6\right]$
$(C)$ $K _2 Fe \left[ Fe ( CN )_6\right]$
$(D)$ $KFe \left[ Fe ( CN )_6\right]$
Among the following, the brown ring is due to the formation of
$(A)$ $\left[ Fe ( NO )_2\left( SO _4\right)_2\right]^{2-}$
$(B)$ $\left[ Fe ( NO )_2\left( H _2 O \right)_4\right]^{3+}$
$(C)$ $\left[ Fe ( NO )_4\left( SO _4\right)_2\right]$
$(D)$ $\left[ Fe ( NO )\left( H _2 O \right)_5\right]^{2+}$
$P =\left[ FeF _6\right]^{3-}, Q =\left[ V \left( H _2 O \right)_6\right]^{2+} \text { and } R =\left[ Fe \left( H _2 O \right)_6\right]^{2+} \text {. }$
The correct order of the complex ions, according to their spin-only magnetic moment values (in $B.M.$) is