Chemistry MCQs for NEET — Practice Questions with Answers

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The numbers of unpair electrone in $ Ni(CO)_4 $ is ?

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Explanation

In $Ni(CO)_4$, nickel is in the zero oxidation state and has the electronic configuration 3d^8 4s^2. Carbon monoxide (CO) is a strong field ligand and causes pairing of electrons. Thus, $Ni(CO)_4$ has no unpaired electrons and is diamagnetic.

Which of the following does not have optical isomer ?

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Explanation

$[Co(NH_3)_3Cl_3]$ does not show optical isomerism because it is a symmetrical molecule and lacks chirality. The other complexes listed can form non-superimposable mirror images, hence they can have optical isomers.

The colour of tetrammine copper (II) sulphate is ?

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Explanation

The tetrammine copper (II) sulphate complex, [Cu(NH3)4]SO4, typically exhibits a blue color. This is due to the d-d electronic transitions that occur in the copper ion when it is complexed with ammonia ligands.

Cuprammonium ion $ [Cu(NH_3)_4]^ {2+} $ is ?

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Explanation

The cuprammonium ion, [Cu(NH3)4]^{2+}, adopts a square planar geometry. This is because copper (II) has a d^9 electronic configuration which leads to a preference for a square planar coordination to minimize repulsions and stabilize the complex.

The type of hybridisation involved in the metal ion of $ [Ni(H_2O)_6]^{2+} $ complex is ?

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Explanation

In the complex [Ni(H2O)6]^{2+}, the nickel ion is in an octahedral environment formed by six water molecules. The hybridization involved in this geometry is sp^3d^2, which allows the formation of six equivalent bonds oriented at 90 degrees to each other.

Which one of the following has square planar geometry?

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The complex ion which has no ‘d’ electrons in the central metal atom is

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Explanation

$[MnO_4]^-$ 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

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Explanation

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

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Explanation

The most stable ion among the given options is $[Fe(CN)_6]^{3-}$. This is because cyanide (CN⁻) 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 (H₂O) are weaker field ligands, resulting in less stable complexes. Chloride (Cl⁻) is also a weaker field ligand compared to cyanide.

Wilkinson’s catalyst is used in

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