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Chemistry NEET Questions — Free MCQ Practice

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The primary valency of Fe in K₃[Fe(Cn)₆] is:

A.

3

B.

2

C.

1

D.

zero

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Explanation

(a) According to Werner's theory, the primary valency of a metal is equal to the no. of charge on complex ion,i.e, 3 on [Fe(Cn)₆]^3-

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The IUPAC name for the complex [Co(NO₂)(NH₃)₅]Cl₂ is:

A.

nitro-N-pentaamminecobalt(III)chloride

B.

nitro-N-pentaamminecobalt(II)chloride

C.

pentaamminenitro-N-cobalt(II)chloride

D.

pentaamminenitro-N-cobalt(III)chloride

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Explanation

(d)

n nitro , becoz N is donating electron to metal

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The total number of possible isomers for the complex compound[Cu^II(NH₃)₄][Pt^IICl₄] are

A.

5

B.

6

C.

3

D.

4

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Explanation

(d) Total isomers of the given complex are four.

these are

(i) [Cu(NH₃)₄][PtCl₄]

(ii) [CU(NH₃)₃Cl][Pt(NH₃)Cl₃]

(iii)[Pt(NH₃)₃Cl] [Cu(NH₃)Cl₃]

(iv)[Pt(NH₃)₄][CuCl₄]

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Out of TiF₆^2-,CoF₆^3-,Cu₂Cl₂and NiCl₄^2-(At. no. Z of Ti=22, Co=27, Cu=29, Ni=28), the colourless species are

A.

TiF₆^2- and CoF₆^3-

B.

Cu₂Cl₂ and NiCl₄^2-

C.

TiF₆^2- and Cu₂Cl₂

D.

CoF₆^3- and NiCl₄^2-

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Ferrocene is :

A.

Fe( $η^{5}$ -C₅H₅)₂

B.

Fe( $η^{2}$ -C₅H₅)₂

C.

Cr( $η^{5}$ -C₅H₅)₅

D.

Os( $η^{5}$ -C₅H₅)₂

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Explanation

(a) Ferrocene is a $π$ complex Fe( $η^{5}$ -C₅H₅)₂

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The ionization isomer of [Cr(H₂O)₄Cl(NO₂)]Cl is :

A.

[Cr(H₂O)₄(O₂N)]Cl₂

B.

[Cr(H₂O)₄Cl₂](NO₂)

C.

[Cr(H₂O)₄Cl(ONO)]Cl

D.

[Cr(H₂O)₄Cl₂(NO₂)].H₂O

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Explanation

(b) [Cr(H₂O)₄Cl(NO₂)]Cl $\to$ [Cr(H₂O)₄Cl₂(NO₂)]⁺ + Cl^-

[Cr(H₂O)₄Cl₂](NO₂) $\to$ [Cr(H₂O)₄Cl₂]⁺ + NO₂^-

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Among [Ni(CO)₄],[Ni(CN)₄]^2-, [NiCl₄]^2- species, the hybridisation states of the Ni atom are, respectively (At. no. of Ni=28)

A.

sp³,dsp².dsp²

B.

sp³,dsp²,sp³

C.

sp³,sp³,dsp²

D.

dsp²,sp³,sp³

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Explanation

In [Ni(CO)â‚„], the Ni atom is spÂ³ hybridized. In [Ni(CN)â‚„]Â²â», the Ni atom is dspÂ² hybridized. In [NiClâ‚„]Â²â», the Ni atom is spÂ³ hybridized, as it forms a tetrahedral complex with the four chloride ligands.

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The correct order of magnetic moments (only spin value in BM)among is:

A.

[Fe(Cn)₆]^4->[CoCl₄]^2->[MnCl₄]^2-

B.

[MnCl₄]^2->[Fe(Cn)₆]⁴>[CoCl⁴]^2-

C.

[Fe(Cn)₆]⁴>[MnCl₄]^2->[CoCl₄]^2-

D.

[MnCl₄]^2->[CoCl₄]^2->[Fe(Cn)₆]^4-

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Explanation

(d) Mn²⁺,Fe²⁺ and Co²⁺ have 3d⁵,3d⁶, and 3d⁷ configuration respectively having 5,0 and 1 unpaired electrons. Pairing occurs in Fe²⁺ and Co²⁺

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Among the following complexes(K,P),K₃[Fe(CN)₆(k),[Co(NH₃)₆Cl₃(L),Na₃[Co(oxalate)₃](M),[Ni(H₂O)6Cl₂(N),K₂[Pt(CN)₄(O), and [Zn(H₂O)₆(NO₃)₂(P) the diamagnetic complexes are:

A.

K,L,M,N

B.

K,M,O,P

C.

L,M,O,P

D.

L,M,N,O

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Explanation

(c) [K]:K₃[Fe(CN)₆]:Fe³⁺ has 3d⁵ configuration; CN^- is strong field ligand and thus four electrons are paired leaving one unpaired electron with d²sp³ hybridization and thus paramagnetic.

[L]:[Co(NH₃)₆Cl₃: Co³⁺ has 3d⁶ configuration;NH₃ is strong field ligand and thus all the six electrons are paired with d²sp³ hybridization and therefore dismagnetic.

[M]: Na₃[Co(ox)₃]: Co³⁺ has 3d⁶ configuration; C₂O₄^2- is strong field ligand and thus all the six electrons are paired with d²sp³ hybridization and therefore diamagnetic.

[N]: [Ni(H₂O)6Cl₂]: Ni²⁺ has 3d⁸ configuration with two unpaired electrons. H₂O is weak field ligand and thus sp³d² hybridization and paramagnetic.

[O]: K₂[Pt(CN)₄]: Pt2+ has 3d⁸ configuration;CN^- is strong field ligand and thus all the eight electrons are paired with dsp² hybridzation and therefore diamagnetic.

[P]: [Zn(H₂O)₆(NO₃)_2]:Zn²⁺ has 3d¹⁰ configuration and thus all are paired showing sp³d² hybridization and so diamagnetic.

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A solution of CuCl in NH₄OH is used to measure the amount of which gas is a sample by simply measuring change in volume?

A.

CO₂

B.

H₂

C.

CO

D.

All of these

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Explanation

CuCl in NH₄OH absorbs CO

Solutions of CuCl in NH₃ absorb carbon monoxide to form colourless complexes such as the crystalline halogen-bridged dimer [CuCl(CO)]₂.

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