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Chemistry MCQs for NEET — Practice Questions with Answers

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IUPAC name of $H_{2} [{PtCl}_{6}]$ is:

hydrogen hexachloroplatinate (IV)

dihydrogen hexachloroplatinate (IV)

hydrogen hexachloroplatinic (IV) acid

hexachloroplatinic (IV) acid

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Explanation

$H_{2} [{PtCl}_{6}]$ is an acid, not salt hence its name : hexachloroplatinic (IV) acid

whenever HYDROGEN is present as cation , it is called as acid rather than salt.

KNO2 is known as POT nitrite

but HNO2 is not known as HYDROGEN NITRITE, its known as NITROUS ACID.

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Which of the following ligand does not as $π$ -acid ligand?

$N_{2}$

$C_{2} H_{4}$

$O_{2}^{2 -}$

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Explanation

(d) Due to absence of vacant atomic orbital as well as $π^{*}$ molecular orbital $O_{2}^{2 -}$ does not as $π$ acid ligand .

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$[Mn (CO)_{4} NO]$ is diamagnetic because:

Mn metal is diamagnetic in free state

Mn is in +1 oxidation state in this complex

NO is present as positive ligand

All of the above

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Explanation

$[\overset{+ 1}{Mn} (CO)_{4} (NO^{-})^{}]$ : Paramagnetic due to presence of unpaired $e^{-^{}}$ in ${NO}^{-}$ $[\overset{- 1}{Mn} (CO)_{4} ({NO}^{+})]$ : No. of unpaired $e^{-}$ either on ligands or on $\overset{- 1}{Mn}$ hence it is diamagnetic

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If CO ligands are substituted by NO in respective neutral carbonyl compounds then which of the following will not be correct formula ?

$Cr (CO)_{3} (NO)_{2}$

$Fe (CO)_{2} (NO)_{2}$

$Cr (NO)_{4}$

$Ni (CO)_{2} (NO)_{2}$

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Explanation

(d) Ligand NO is $3 e^{-}$ donar hence three CO ligands can be substituted by two NO ligands.

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Which of the following species can act as reducing agent ?

$[Co (CO)_{4}]^{-}$

$Mn (CO)_{6}$

$Mn (CO)_{5}$

$Cr (CO)_{6}$

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Explanation

(b) $Mn (CO)_{6}$ can act as reducing agent because the metal carbonyl is stable when EAN is equal to nearest noble gas configuration.

$[Mn (CO)_{6}] \overset{- e^{-}}{\to} [Mn (CO)_{6}]^{+} EAN = 37 EAN = 36 (less stable) (more stable)$

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What is electronic arranegment of metal atom/ionin octahedral complex with $d^{4}$ configuration , if $∆_{0}$ < pairing energy ?

$t_{2 g}^{4} e_{g}^{0}$

$e_{g}^{4} t_{2 g}^{0}$

$t_{2 g}^{3} e_{g}^{1}$

$e_{g}^{2} t_{2 g}^{2}$

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Which of the following statement is not correct ?

Bis(glycinato)Zinc(II) is optically active

$[{NiCl}_{4}]^{2 -} and [{PtCl}_{4}]^{2 -}$ have different shape

$[Ni (CN)_{4}]^{4 -}$ is square planar complex

$[Ni (CN)_{4}]^{2 -} and [Ni (CO)_{4}]$ have the same magnetic moment

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Explanation

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Give the correct of initial T or F for following statements. Use T is statement is true and F if it is false

(I) Co(III) is stabilised in presence of weak ligands , while Co(II) is stabilised in presence of strong field ligand.

(II) Four coordinated complexes of Pd(II) and Pt(II) are diamagnetic and square planar.

(III) $[Ni (CN)_{4}]^{4 -}$ ion and $[Ni (CO)_{4}]$ are diamagnetic tetrahedral and square planar.

(IV) ${Ni}^{2 +}$ ion does not inner orbital octahedral complexes.

TFTF

TTTF

TTFT

FTFT

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Match List-I with List-II and select the correct answer using the codes given below :

List-I List-II

(I) $[{FeF}_{6}]^{3 -}$ (A) 1.73 BM

(II) $[Ti (H_{2} O)_{6}]^{3 +}$ (B) 5.93 BM

(III) $[Cr ({NH}_{3})_{6}]^{3 +}$ (C) 0.00 BM

(IV) $[Ni (H_{2} O)_{6}]^{2 +}$ (D) 2.83 BM

(V) $[Fe (CN)_{6}]^{4 -}$ (E) 3.88 BM

(I) (II) (III) (IV) (V) (I) (II) (III) (IV) (V)

B A C D E

B A E D C

B C D E A

D E A B C

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Explanation

The correct matching is: (I) [FeFâ‚†]Â³â» - (B) 5.93 BM (high spin dâµ configuration), (II) [Ti(Hâ‚‚O)â‚†]Â³â» - (A) 1.73 BM (dÂ¹ configuration), (III) [Cr(NHâ‚ƒ)â‚†]Â³âº - (E) 3.88 BM (dÂ³ configuration), (IV) [Ni(Hâ‚‚O)â‚†]Â²âº - (D) 2.83 BM (dâ¸ configuration), (V) [Fe(CN)â‚†]â´â» - (C) 0.00 BM (diamagnetic, low spin dâ¶ configuration).

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Set of d-orbitals which is used by central metal during formation of ${MnO}_{4}^{-}$ ?

$d_{x^{2} - y^{2}}, d_{z^{2}}, d_{xy}$

$d_{xy}, d_{yz}, d_{xz}$

$d_{x^{2} - y^{2}}, d_{xy}, d_{xz}$

$d_{x^{2} - y^{2}}, d_{z^{2}}, d_{xz}$

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Explanation

The central metal atom manganese (Mn) in the MnO4- ion has an oxidation state of +7. According to the aufbau principle, the five 3d orbitals (dxy, dyz, dxz, dx2-y2, dz2) are progressively filled. In the +7 oxidation state, Mn has a d0 configuration, meaning all 3d orbitals are vacant. Therefore, the set of d-orbitals used by the central metal during the formation of MnO4- is dxy, dyz, and dxz.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.