All Free Chemistry MCQs with Answers
Every Chemistry question in the bank, across all chapters, each with the correct answer and a written explanation. Free and unlimited, with no account needed.
13,720 questions · page 542 of 686
- A. [Co(NH3)4NO2Cl] [(NH3)Cl]
- B. [Co(NH3)5Cl] [Cl(NO2)]
- C. [Co(NH3)5NO2]Cl2
- D. [Co(NH3)5] [(NO2)2Cl2]
Explanation: Because only (C) gives two moles of AgCl with AgNO3 and yields three ions (including two ionizable Cl- ions) in an aqueous…
Correct answer: [Co(NH3)5NO2]Cl2- A. -6
- B. +3
- C. -3
- D. +6
Explanation: The algebraic sum of oxidation no. of all atoms in a complex ion is equal to the charge on that ion,i.e., a + 6 × (-1) = - 3. a = + 3.
Correct answer: +3- A. FeSO4
- B. KCl.MgCl26H2O
- C. Al2(CO3)3
- D. NH4Br.
Explanation: A simple salt is a compound which has been prepared by the combination of two simple ions like Na+ and Cl- ions to form NaCl.
Correct answer: KCl.MgCl26H2O- A. Solution of iron salt does not give blue colour with indicator.
- B. Solution of iron salt gives blue colour with indicator.
- C. No colour before and after with indicator.
- D. Always blue colour before and after
Explanation: At the end point, there is no Fe2+. All iron in Fe3+ hence no blue colour with K3[Fe(CN)6] hence (A).
Correct answer: Solution of iron salt does not give blue colour with indicator.- A. +1.
- B. +3.
- C. -1.
- D. -3.
Explanation: We know that CO is a neutral ligand and K carries a charge of +1. Therefore, the complex can be written as K+ [Co(CO)4]-.
Correct answer: -1.- A. a central metal ion with d8 configuration has two unpaired electrons.
- B. an ion with d6 configuration is diamagnetic in a strong field.
- C. an ion with d5 configuration has one unpaired electron both in weak and strong fields.
- D. in d4, d5, d6 and d7 configurations, weak and strong field complexes have different numbers of unpaired electrons.
Explanation: This statement is fundamentally incorrect because the number of unpaired electrons changes:Weak field: All five electrons are unpaired…
Correct answer: an ion with d5 configuration has one unpaired electron both in weak and strong fields.- A. Geometrical isomers.
- B. Linkage isomers
- C. Ligand isomers.
- D. Ionisation isomers.
Explanation: The ligand NO2 has two types of linkage with the central atom. In NO2, N donates an electron pair, while in O NO the O atom donates the…
Correct answer: Linkage isomers- A. [Pt(NH3)Cl3]Br2 and [Pt(NH3)4Br2]Cl2
- B. [Co(NH3)5NO3]SO4 and [Co(NH3)5SO4]NO3
- C. [Cr(H2O)5SCN]2+ and [Cr(H2O)5NCS]2+
- D. [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl2]Cl2.H2O
Explanation: [Cr(H2O)5SCN]2+ and [Cr(H2O)5NCS]2+ exhibits linkage isomerism.
Correct answer: [Cr(H2O)5SCN]2+ and [Cr(H2O)5NCS]2+- A. 1
- B. 2
- C. 4
- D. 5
Explanation: Hexafluoroferrate (III) is [FeF6]3-. The O.N. of Fe in it is x + (-6) = - 3 or x - 6 = -3 or x = + 3.Thus in it, Fe is present as Fe3+…
Correct answer: 5- A. 1
- B. 2
- C. 5
- D. 3
Explanation: The configuration of Mn is [Ar]d5. Thus it is expected to have 5 unpaired electrons.
Correct answer: 1- A. CrCl3.6H2O
- B. [Cr(H2O)3Cl3].(H2O)3
- C. [CrCl2(H2O)4]Cl.2H2O
- D. [CrCl(H2O)5]Cl2.H2O
Explanation: [CrCl2(H2O)4]Cl.2H2O has only one ionisable Cl out of three Cl atoms present in the compound.
Correct answer: [CrCl2(H2O)4]Cl.2H2O- A. 0
- B. +1
- C. +2
- D. +1.5
Explanation: Ag in Tollens reagent exists as Ag2O2 × a + 1 × (-2) = 0a = + 1.
Correct answer: +1- A. d3, d5, d6, d9
- B. d4, d5, d6, d7
- C. d2, d5, d7, d9
- D. d4, d6, d7, d10
Explanation: Metal ions with d4, d5, d6 and d7 configuration can form high spin and low spin complexes.
Correct answer: d4, d5, d6, d7- A. [Ar](n-1)d1-10ns1-2
- B. [Ar](n-1)d1-10ns2
- C. [Ar](n-1)d1-2ns1-2
- D. [Ar](n-2)d1-2ns1-2
Explanation: d-block (outer) transition elements are those in which d-subshell of penultimate shell (inner to valence shell) is in process of…
Correct answer: [Ar](n-1)d1-10ns1-2- A. [Ar] 3d34s2
- B. [Ar] 3d44s2
- C. [Ar] 3d4s2
- D. [Ar] 3d54s2
Explanation: Electronic configuration of Mn is 25Mn 18[Ar] 4s23d5.
Correct answer: [Ar] 3d54s2- A. Sc
- B. Mn
- C. Ti
- D. Zn
Explanation: (A) 21Sc 18[Ar] 4s2 3d1Sc+3 18[Ar] 4sº 3dº (B) 21Ti 18[Ar] 4s2 3d2Ti+3 18[Ar] 4sº 3d1 (C) 25Mn 18[Ar] 4s2 3d5Mn+2 18[Ar] 4sº 3d5 (D) 30Zn…
Correct answer: Zn- A. V
- B. Ti
- C. Sc
- D. Cr
Explanation: 21Sc =18[Ar] 4s2 3d121Sc+3 = 18[Ar] 4s° 3d°Scandium in +3 state has same electronic configuration as 18Ar.Ar = 1s22s22p63s23p6
Correct answer: Sc- A. 0
- B. +2
- C. +1
- D. +3
Explanation: Cu+118[Ar] 4s° 3d10Zn+2 18[Ar] 4s° 3d10 As per the electronic configuration, Cu in +1 oxidation state, achieves electronic configuration…
Correct answer: +1- A. Its d-subshell is incomplete.
- B. Its d-subshell is complete
- C. It is relatively soft metal
- D. It has two electrons in outermost shell
Explanation: 30Zn 18[Ar] 4s2 3d10 Zn+2 18[Ar] 4s° 3d10 Zinc (Zn) has a completely filled d-subshell (3d10) in its electronic configuration.
Correct answer: Its d-subshell is complete- A. +7
- B. -7
- C. +6
- D. +4
Explanation: Mn belongs to group (VIIB)of periodic table. So, its maximum oxidation state should be +7.
Correct answer: +7