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The complex ion which has no ‘d’ electrons in the central metal atom is
$[MnO_4]^-$ has Mn in the +7 oxidation state. Manganese in this state has an electronic configuration of $[Ar]3d^04s^0$, meaning it has no 'd' electrons.
The strongest ligand in the following is
According to the spectrochemical series, $CN^-$ (cyanide) is a strong field ligand compared to $Br^-$, $HO^-$, and $F^-$. This means it can cause a larger splitting of the d-orbitals, making it the strongest ligand among the given options.
The most stable ion is
The most stable ion among the given options is $[Fe(CN)_6]^{3-}$. This is because cyanide (CNâ») is a strong field ligand, which leads to a low-spin complex and greater stability due to the pairing energy being less than the crystal field splitting energy. In contrast, the other ligands such as oxalate (OX) and water (Hâ‚‚O) are weaker field ligands, resulting in less stable complexes. Chloride (Clâ») is also a weaker field ligand compared to cyanide.
Wilkinson’s catalyst is used in
Mixture X=0.02 mole of $ [Co(NH_3)_5SO_4]Br and 0.02 mole of [Co(NH_3)_5Br]SO_4$ was prepared in 2 litre of solution. 1 Litre of mixture $ X + excess AgNO_3 \rightarrow Y$ 1 Liter of mixture $ X+ excess BaCl_2 \rightarrow Z $ Number of moles of Y and Z are respectively
In $ [Ni(NH_3)_4]SO_4 $ the valency and coordinate number of Ni will be respectivly ?
In the complex $[Ni(NH_3)_4]SO_4$, the nickel ion is coordinated to 4 ammonia (NH3) molecules. Ammonia is a neutral ligand, so it does not affect the oxidation state of nickel. The sulfate ion (SO4) has a charge of -2. Therefore, for the complex to be neutral, the nickel ion must have a charge of +2. Thus, the valency (oxidation state) of Ni is +2 and the coordination number (number of ligands attached to Ni) is 4.
Which of the following compounds shows optical isomerism?
Optical isomerism occurs when a compound can exist in two non-superimposable mirror images. In the given options, $[Cr(C_2O_4)_3]^{3-}$ can show optical isomerism because it has a chiral center due to the arrangement of the oxalate ligands around the chromium ion. The other compounds do not have such an arrangement that leads to optical isomerism.
In the process of extraction of gold, Roasted gold ore $ +CN^– + H_2O \rightarrow X +OH^– [X] +Zn \rightarrow Y+ Au $. Identify the complexes [X] and [Y]
In the extraction of gold, the roasted gold ore reacts with cyanide ions to form the complex $[Au(CN)_2]^-$ and hydroxide ions (OH^-). This complex then reacts with zinc to form $[Zn(CN)_4]^{2-}$ and releases gold (Au). Hence, the complexes are $X = [Au(CN)_2]^-$ and $Y = [Zn(CN)_4]^{2-}$.
Which of the following statement is incorrect ?
Assertion and Reas on Read the assertion and reason carefully to mark the correct option out of the option given below : Assertion : Potassium ferrocyanide and potassium ferricyanide both are diamagnetic. Reason : Both have unpaired election.
Potassium ferrocyanide ($K_4[Fe(CN)_6]$) is diamagnetic because it has no unpaired electrons in its low-spin complex. However, potassium ferricyanide ($K_3[Fe(CN)_6]$) is paramagnetic because it contains unpaired electrons in its high-spin complex. The assertion is false because potassium ferricyanide is not diamagnetic, and the reason is also false as it states that both have unpaired electrons. Thus, the correct answer is o4: If the assertion and reason both are false.
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