Mock Test 4: MCQs with Explanations

1. The atomic number of potassium is 19 and that of manganese is 25. Although the colour of $MnO_{4}^{-}$ is dark violet, yet the $K^{+}$ is colourless. This is due to the fact that

• (a) Mn is a transition element; while $K^{+}$ is not
• (b) $[MnO_{4}]^{-}$ is negatively charged; while $K^{+}$ has a positive charge
• (c) the effective atomic number of Mn in $[MnO_{4}]^{-}$ is 26; while for $K^{+}$ the effective atomic number is 18
• (d) the Mn in a high positive oxidation state allows charge transfer transitions.

Correct Answer: (d)

Explanation: In $MnO_{4}^{-}$, Mn is in its highest oxidation state (+7) with a $d^0$ configuration. The intense color arises not from $d-d$ transitions, but from charge transfer from the oxygen ligands to the metal center.

2. How many $P-O$ bonds and how many lone pairs respectively are present in $P_{4}O_{6}$ molecule?

• (a) 12, 4
• (b) 8, 8
• (c) 12, 16
• (d) 12, 12

Correct Answer: (c)

Explanation: $P_{4}O_{6}$ has a cage-like structure where each of the 4 Phosphorus atoms is bonded to 3 Oxygen atoms, totaling 12 P-O bonds. There are 16 lone pairs: 1 on each P atom (4) and 2 on each O atom (12).

3. If the radii of $La^{3+}$ and $Lu^{3+}$ are x and y respectively, then the value of lanthanide contraction would be equal to

• (a) $x - y$
• (b) $y - x$
• (c) $x + y$
• (d) none of these.

Correct Answer: (a)

Explanation: Lanthanide contraction is the steady decrease in atomic and ionic radii from Lanthanum to Lutetium. Thus, the total contraction is the initial radius minus the final radius ($x - y$).

4. White vitriol has the formula

• (a) $CaSO_{4} \cdot 2H_{2}O$
• (b) $CuSO_{4} \cdot 5H_{2}O$
• (c) $ZnSO_{4} \cdot 7H_{2}O$
• (d) $FeSO_{4} \cdot 7H_{2}O$

Correct Answer: (c)

Explanation: Zinc sulphate heptahydrate ($ZnSO_{4} \cdot 7H_{2}O$) is commonly known as white vitriol.

5. The property of halogen acids, that indicated incorrect is

• (a) $HF > HCl > HBr > HI$ ..... acidic strength
• (b) $HI > HBr > HCl > HF$ ... reducing strength
• (c) $HI > HBr > HCl > HF$ ..... bond length
• (d) $HF > HCl > HBr > HI$ .... thermal stability

Correct Answer: (a)

Explanation: The acidic strength of halogen acids increases down the group ($HI > HBr > HCl > HF$) because the bond dissociation energy decreases. Thus, HF is the weakest acid.

6. Low spin complex of $d^{6}$ cation in an octahedral field will have the following energy

• (a) $\frac{-12}{5}\Delta_{o} + P$
• (b) $\frac{-12}{5}\Delta_{o} + 3P$
• (c) $\frac{-2}{5}\Delta_{o} + 2P$
• (d) $\frac{-2}{5}\Delta_{o} + P$

Correct Answer: (b)

Explanation: In a low spin $d^6$ octahedral complex, all 6 electrons occupy the $t_{2g}$ orbitals. Energy = $6 \times (-0.4\Delta_{o}) = -2.4\Delta_{o}$ or $-12/5\Delta_{o}$. Since all electrons are paired in 3 orbitals, it includes 3 pairing energies (3P).

7. Which of the following will give a pair of enantiomers?

• (a) $[Pt(NH_{3})_{4}][PtCl_{6}]$
• (b) $[Co(NH_{3})_{4}Cl_{2}]NO_{2}$
• (c) $[Cr(NH_{3})_{6}][Co(CN)_{6}]$
• (d) $[Co(en)_{2}Cl_{2}]Cl$

Correct Answer: (d)

Explanation: The cis-isomer of $[Co(en)_{2}Cl_{2}]Cl$ is chiral and exists as a pair of enantiomers, whereas the trans-isomer is achiral.

8. The bonds present in the structure of dichromate ion are

• (a) six equivalent Cr-O bonds and one Cr-O-Cr bond
• (b) six equivalent Cr-O bonds and one Cr-Cr bond
• (c) six equivalent Cr-O bonds and one O-O bond
• (d) none of these.

Correct Answer: (a)

Explanation: The dichromate ion ($Cr_{2}O_{7}^{2-}$) consists of two tetrahedra sharing one oxygen atom. There are 6 equivalent terminal Cr-O bonds and a bridging Cr-O-Cr linkage.

9. Which of the following represents the correct order of increasing $pK_{a}$ values of the given acids?

• (a) $HClO_{4} < HNO_{3} < H_{2}CO_{3} < B(OH)_{3}$
• (b) $HNO_{3} < HClO_{4} < B(OH)_{3} < H_{2}CO_{3}$
• (c) $B(OH)_{3} < H_{2}CO_{3} < HClO_{4} < HNO_{3}$
• (d) $HClO_{4} < HNO_{3} < B(OH)_{3} < H_{2}CO_{3}$

Correct Answer: (a)

Explanation: Lower $pK_a$ signifies a stronger acid. $HClO_{4}$ is a very strong acid, followed by $HNO_{3}$, then the weak $H_{2}CO_{3}$, and finally $B(OH)_{3}$ which is a very weak Lewis acid.

10. Which of the following compounds is explosive in nature?

• (a) Phosphorus trichloride
• (b) Nitrogen trichloride
• (c) Hyponitrous acid
• (d) Nitrosyl chloride

Correct Answer: (b)

Explanation: Nitrogen trichloride ($NCl_{3}$) is a highly unstable and explosive yellow oily liquid.

11. When an inorganic compound reacts with $SO_{2}$ in aqueous medium produces (A). (A) on reaction with $Na_{2}CO_{3}$ gives the compound (B) which with sulphur gives a substance (C) used in photography. The compound (C) is

• (a) $Na_{2}S_{2}O_{3}$
• (b) $Na_{2}SO_{4}$
• (c) $Na_{2}S$
• (d) $Na_{2}S_{2}O_{7}$

Correct Answer: (a)

Explanation: Compound (C) is Sodium thiosulphate ($Na_{2}S_{2}O_{3}$), also known as "hypo," widely used in photography for fixing films.

12. Which of the following reactions will not evolve phosphine?

• (a) White $P + Ca(OH)_{2} \rightarrow$
• (b) $AlP + H_{2}SO_{4} \rightarrow$
• (c) $H_{3}PO_{3} \xrightarrow{\text{Heat}}$
• (d) $PH_{4}I + NaOH \rightarrow$

Correct Answer: (a)

Explanation: While the heating of white phosphorus with $Ca(OH)_{2}$ is a standard method to prepare phosphine, the key indicates (a), suggesting it might produce something else under specific conditions or is the least efficient. (Note: Scientifically, (b), (c), and (d) definitely produce PH3; (a) also produces it, so check source nuance).

13. Match the List I (Catalyst) with the List II (Process):

P. Ni ($H_2$); Q. $Cu_2Cl_2$; R. $V_2O_5$; S. Fe

• (a) 4, 2, 3, 1
• (b) 3, 4, 2, 1
• (c) 2, 3, 1, 4
• (d) 2, 1, 4, 3

Correct Answer: (c)

Explanation: Ni catalyzes vegetable oil to ghee (P-2); $Cu_2Cl_2$ is used in Sandmeyer reaction (Q-3); $V_2O_5$ is for Contact process (R-1); and Fe is for Haber's process (S-4).

14. Effective magnetic moment of $Sc^{3+}$ ion is

• (a) 0
• (b) 1
• (c) 2
• (d) 3

Correct Answer: (a)

Explanation: $Sc^{3+}$ has an electronic configuration of $[Ar]3d^0$. Since there are zero unpaired electrons, the magnetic moment is zero.

15. The number of chloride ion(s) produced by complex tetraamminedichloroplatinum(IV) chloride in an aqueous solution is/are

• (a) 1
• (b) 2
• (c) 8
• (d) 6

Correct Answer: (b)

Explanation: The complex is $[Pt(NH_{3})_{4}Cl_{2}]Cl_{2}$. Upon ionization, it produces one complex cation and two chloride ions from the outer coordination sphere.

16. The magnetic moment of transition metal ion is $\sqrt{15}$ B.M. The number of unpaired electrons present in it are

• (a) 4
• (b) 1
• (c) 2
• (d) 3

Correct Answer: (d)

Explanation: Using $\mu = \sqrt{n(n+2)}$, $\sqrt{15} = \sqrt{3(3+2)}$. Thus, there are 3 unpaired electrons.

17. The structure of paramagnetic nickel complex, $[NiCl_{4}]^{2-}$ is

• (a) tetrahedral
• (b) square planar
• (c) trigonal bipyramidal
• (d) distorted octahedral

Correct Answer: (a)

Explanation: In $[NiCl_{4}]^{2-}$, Cl is a weak field ligand that doesn't cause pairing. $Ni^{2+}$ undergoes $sp^3$ hybridization, resulting in a tetrahedral geometry.

18. The elements Re, Os and Ir belong to

• (a) first transition series
• (b) second transition series
• (c) third transition series
• (d) fourth transition series.

Correct Answer: (c)

Explanation: Rhenium, Osmium, and Iridium are part of the 5d or third transition series.

19. $[Co(NH_{3})_{5}Br]SO_{4}$ and $[Co(NH_{3})_{5}SO_{4}]Br$ are the examples of which type of isomerism?

• (a) Linkage
• (b) Geometrical
• (c) Ionization
• (d) Optical

Correct Answer: (c)

Explanation: These complexes give different ions in solution ($SO_4^{2-}$ vs $Br^{-}$) due to the exchange of ions between the coordination sphere and counter ions, signifying ionization isomerism.

20. Which of the following ions will exhibit colour in aqueous solution?

• (a) $Lu^{3+} (Z = 71)$
• (b) $Sc^{3+} (Z = 21)$
• (c) $La^{3+} (Z = 57)$
• (d) $Ti^{3+} (Z = 22)$

Correct Answer: (d)

Explanation: $Ti^{3+}$ ($3d^1$) has an unpaired electron allowing for $d-d$ transitions, which produce color. The others have $d^0$ or $f^{14}$ configurations and are colorless.

21. The correct order of absorption of wavelength of light in the visible region for the complexes, $[Co(NH_{3})_{6}]^{3+}$, $[Co(CN)_{6}]^{3-}$, $[Co(H_{2}O)_{6}]^{3+}$ is

• (a) $[Co(CN)_{6}]^{3-} > [Co(NH_{3})_{6}]^{3+} > [Co(H_{2}O)_{6}]^{3+}$
• (b) $[Co(NH_{3})_{6}]^{3+} > [Co(H_{2}O)_{6}]^{3+} > [Co(CN)_{6}]^{3-}$
• (c) $[Co(H_{2}O)_{6}]^{3+} > [Co(NH_{3})_{6}]^{3+} > [Co(CN)_{6}]^{3-}$
• (d) $[Co(CN)_{6}]^{3-} > [Co(H_{2}O)_{6}]^{3+} > [Co(NH_{3})_{6}]^{3+}$

Correct Answer: (c)

Explanation: Absorption wavelength is inversely proportional to crystal field splitting ($\Delta_{o}$). Since field strength is $CN^{-} > NH_{3} > H_{2}O$, the $\Delta_{o}$ follows the same order, and wavelength follows the reverse.

22. The value of 'spin only' magnetic moment for one of the following configuration is 2.84 B.M. The correct one is

• (a) $d^{8}$ (in weak field ligand)
• (b) $d^{4}$ (in weak field ligand)
• (c) $d^{3}$ (in weak as well as strong field ligands)
• (d) $d^{5}$ (in strong field ligand)

Correct Answer: (a)

Explanation: A magnetic moment of 2.84 BM indicates 2 unpaired electrons. In a $d^8$ configuration (octahedral), there are always 2 unpaired electrons in the $e_g$ level regardless of field strength.

23. In the reaction $P_{4} + 3KOH + 3H_{2}O \rightarrow PH_{3} + 3KH_{2}PO_{2}$, phosphorus is

• (a) reduced
• (b) oxidised
• (c) both oxidised and reduced
• (d) neither oxidised nor reduced.

Correct Answer: (c)

Explanation: Phosphorus in $P_4$ (0) is reduced to $PH_3$ (-3) and oxidized to $KH_2PO_2$ (+1). This is a disproportionation reaction.

24. The correct order of thermal stability of hydrides of oxygen family is

• (a) $H_{2}O > H_{2}S > H_{2}Te > H_{2}Se > H_{2}Po$
• (b) $H_{2}Po < H_{2}Te < H_{2}Se < H_{2}S < H_{2}O$
• (c) $H_{2}O < H_{2}S < H_{2}Se < H_{2}Te < H_{2}Po$
• (d) $H_{2}S > H_{2}O < H_{2}Se < H_{2}Te < H_{2}Po$

Correct Answer: (b)

Explanation: Thermal stability of Group 16 hydrides decreases down the group as the H-M bond becomes weaker due to increasing size of the central atom.

25. $Cr_2O_7^{2-} \xrightarrow{pH_1} CrO_4^{2-} \xrightarrow{pH_2} Cr_2O_7^{2-}$. Values of $pH_{1}$ and $pH_{2}$ are

• (a) 3 and 5
• (b) 4 and 8
• (c) 8 and 4
• (d) 9 and 10

Correct Answer: (c)

Explanation: Dichromate converts to chromate in alkaline medium (higher pH), and chromate converts back to dichromate in acidic medium (lower pH).

26. When a chromite ore (P) is fused with sodium carbonate in excess air, product Q is obtained... P, Q, R and S respectively are

• (a) $FeCr_{2}O_{4}$, $Na_{2}CrO_{4}$, $Na_{2}Cr_{2}O_{7}$, $K_{2}Cr_{2}O_{7}$
• (b) $FeCr_{2}O_{7}$, $Na_{2}Cr_{2}O_{7}$, $Na_{2}CrO_{4}$, $K_{2}Cr_{2}O_{7}$
• (c) $Na_{2}Cr_{2}O_{7}$, $Na_{2}CrO_{4}$, $K_{2}Cr_{2}O_{7}$, $FeCr_{2}O_{4}$
• (d) $Cr_{2}O_{3}$, $FeCr_{2}O_{4}$, $Na_{2}Cr_{2}O_{7}$, $K_{2}Cr_{2}O_{7}$

Correct Answer: (a)

Explanation: Chromite ore (P) $\rightarrow$ Sodium chromate (Q, yellow) $\rightarrow$ Sodium dichromate (R) $\rightarrow$ Potassium dichromate (S, orange).

27. Name of the structure of silicates in which three oxygen atoms of $[SiO_{4}]^{4-}$ are shared is

• (a) pyrosilicate
• (b) sheet silicate
• (c) linear chain silicate
• (d) three-dimensional silicate.

Correct Answer: (b)

Explanation: In sheet silicates, three oxygen atoms per tetrahedron are shared with adjacent tetrahedra to form a 2D network.

28. $H_{3}BO_{3}$ is

• (a) monobasic and weak Lewis acid
• (b) monobasic and weak Bronsted acid
• (c) monobasic and strong Lewis acid
• (d) tribasic and weak Bronsted acid.

Correct Answer: (a)

Explanation: Boric acid ($H_{3}BO_{3}$) acts as a weak monobasic Lewis acid by accepting $OH^{-}$ from water rather than donating a proton.

29. Match List I with list II (Properties):

P. $SiO_4^{4-}$; Q. $SiO_2$; R. $PbO_2$; S. CO

• (a) 1, 2, 3, 4
• (b) 4, 3, 2, 1
• (c) 2, 1, 4, 3
• (d) 2, 4, 1, 3

Correct Answer: (c)

Explanation: $SiO_4^{4-}$ is tetrahedral (P-2); $SiO_2$ is acidic (Q-1); $PbO_2$ is a strong oxidising agent (R-4); and CO is neutral (S-3).

30. Silver chloride dissolves in a solution of ammonia but not in water because

• (a) ammonia is a better solvent than water
• (b) silver ion forms a complex ion with ammonia
• (c) ammonia is a stronger base than water
• (d) the dipole moment of water molecule is higher than that of ammonia molecule.

Correct Answer: (b)

Explanation: AgCl reacts with ammonia to form the soluble complex $[Ag(NH_{3})_{2}]Cl$.

31. The chemical processes in the production of steel from haematite ore involve

• (a) reduction
• (b) oxidation
• (c) reduction followed by oxidation
• (d) oxidation followed by reduction.

Correct Answer: (c)

Explanation: Iron is first obtained by reduction of haematite in a blast furnace, and then impurities (like carbon) are removed by oxidation to convert it into steel.

32. A metal complex having composition $Cr(NH_{3})_{4}Cl_{2}Br$ has forms X and Y... X and Y are respectively

• (a) $[Cr(NH_{3})_{4}Br.Cl]Cl$ and $[Cr(NH_{3})_{4}Cl_{2}]Br$
• (b) $[Cr(NH_{3})_{4}Cl_{2}]Br$ and $[Cr(NH_{3})_{4}BrCl]Cl$
• (c) $[Cr(NH_{3})_{3}Cl_{2}.Br]NH_{3}$ and $[Cr(NH_{3})_{4}Cl_{2}]Br$
• (d) $[Cr(NH_{3})_{3}Cl_{2}.Br]NH_{3}$ and $[Cr(NH_{3})_{4}Cl.Br]Cl$

Correct Answer: (a)

Explanation: Form X gives white ppt (AgCl) which means Cl is outside the sphere. Form Y gives pale yellow ppt (AgBr) which means Br is outside. Thus, X is $[Cr(NH_{3})_{4}BrCl]Cl$ and Y is $[Cr(NH_{3})_{4}Cl_{2}]Br$.

33. Coordination compounds have great importance in biological systems... which statement is incorrect?

• (a) Chlorophylls are green pigments in plants and contain calcium.
• (b) Haemoglobin is the red pigment of blood and contains iron.
• (c) Cyanocobalamine is $B_{12}$ and contains cobalt.
• (d) Carboxypeptidase-A is an enzyme and contains zinc.

Correct Answer: (a)

Explanation: Chlorophyll contains Magnesium as its central metal atom, not Calcium.

34. The coordination number and the oxidation state of the element E in the complex $[E(en)_{2}(C_{2}O_{4})]NO_{2}$ are, respectively

• (a) 6 and 3
• (b) 6 and 2
• (c) 4 and 2
• (d) 4 and 3

Correct Answer: (a)

Explanation: Ethylenediamine (en) and oxalate ($C_{2}O_{4}^{2-}$) are both bidentate. Coordination number = $2 \times 2 + 2 = 6$. For oxidation state: $x + 2(0) + (-2) = +1 \Rightarrow x = +3$.

35. Titanium shows magnetic moment of 1.73 BM in its compound. What is the oxidation number of Ti?

• (a) +1
• (b) +4
• (c) +3
• (d) +2

Correct Answer: (c)

Explanation: $\sqrt{n(n+2)} = 1.73$ BM means $n=1$. Titanium ($3d^2 4s^2$) must lose 3 electrons to have 1 unpaired electron ($3d^1$), so the oxidation state is +3.

36. For the square planar complex $[Pt(NH_{3})(NH_{2}OH)py(NO_{2})]^{+}$, how many geometrical isomers are possible?

• (a) 3
• (b) 4
• (c) 5
• (d) 6

Correct Answer: (a)

Explanation: A square planar complex of the type [Mabcd] can exist as 3 geometrical isomers.

37. The most suitable method for extraction of copper from low grade sulphide ore is

• (a) smelting process
• (b) leaching process
• (c) electrolytic process
• (d) hydrometallurgical process.

Correct Answer: (d)

Explanation: Hydrometallurgy is used for recovering copper from low-grade ores using leaching and subsequent precipitation with scrap iron or $H_2$.

38. PbS can be separated from ZnS by electrostatic separation method. The property utilized in this method is

• (a) PbS is a good conductor and ZnS is a poor conductor of electricity
• (b) PbS is a bad conductor and ZnS is a good conductor of electricity
• (c) both PbS and ZnS are good conductors
• (d) both PbS and ZnS are bad conductors

Correct Answer: (a)

Explanation: Electrostatic separation relies on differences in electrical conductivity; PbS is a conductor while ZnS is a non-conductor.

39. Which of the following is not correct?

• (a) Atomic volume of transition elements are low compared to groups 1 and 2.
• (b) Atomic volume of transition elements are high compared to groups 1 and 2.
• (c) Melting point of transition elements are high.
• (d) Melting point of s and d-block elements are almost same.

Correct Answer: (b)

Explanation: Transition elements have lower atomic volumes and higher melting points than s-block elements due to stronger metallic bonding involving d-electrons.

40. Which is incorrect statement?

• (a) Lanthanoids do not form oxo ions
• (b) Actinoids form oxo anions like $UO_{2}^{2+}$
• (c) Lanthanoids show oxidation states of +3, +4, +5, +6 and +7
• (d) Actinoids have more tendency to form complexes than lanthanoids

Correct Answer: (c)

Explanation: Lanthanoids primarily show +3 oxidation states (with some +2 and +4). The higher oxidation states (+5 to +7) are characteristic of actinoids, not lanthanoids.

41. Which of the following ores can be concentrated by magnetic separation method?

I. $Fe_2O_3$; II. $FeCO_3$; III. ZnS; IV. ZnO; V. $CuFeS_2$

• (a) II, IV and V
• (b) I, II and V
• (c) III and IV
• (d) III only

Correct Answer: (b)

Explanation: Iron-containing ores like Haematite ($Fe_2O_3$), Siderite ($FeCO_3$), and Copper Pyrites ($CuFeS_2$) have magnetic properties allowing for magnetic separation.

42. Which property is not involved in making fluorine the strongest oxidising halogen?

• (a) Bond dissociation energy
• (b) Ionisation enthalpy
• (c) Hydration enthalpy
• (d) Electron affinity

Correct Answer: (b)

Explanation: Fluorine's oxidizing power is due to low bond dissociation energy, high hydration enthalpy, and electron affinity. Ionisation enthalpy is the energy required to remove an electron, which is not a step in forming an anion.

43. Ozone when reacts with potassium iodide solution liberates certain product, which turns starch paper blue. The liberated substance is

• (a) oxygen
• (b) iodine
• (c) hydrogen iodide
• (d) potassium hydroxide.

Correct Answer: (b)

Explanation: Ozone oxidizes iodide ions to iodine ($I_2$), which then reacts with starch to give a classic blue color.

44. Lanthanoid (Ln) reaction sequence products:

• (a) $A-LnO_{3}; B-LnS_{3}; C-LnC_{2}; D-LnX_{3}; E-LnN; F-H_{2}$
• (b) $A-LnO_{3}; B-LnX_{3}; C-LnC_{2}; D-LnS_{3}; E-H_{2}; F-LnN$
• (c) $A-H_{2}; B-LnX_{3}; C-LnC_{2}; D-LnN; E-Ln_{2}S_{3}; F-Ln_{2}O_{3}$
• (d) $A-H_{2}; B-LnX_{3}; C-Ln_{2}C_{3}; D-LnN; E-LnS_{2}; F-LnO_{2}$

Correct Answer: (c)

Explanation: Ln reacts with acids to give $H_2$ (A), halogens to give $LnX_3$ (B), carbon to give $LnC_2$ (C), nitrogen to give $LnN$ (D), sulfur to give $Ln_2S_3$ (E), and oxygen to give $Ln_2O_3$ (F).

45. $K_{3}[CoF_{6}]$ is high spin complex. What is the hybrid state of Co atom in this complex?

• (a) $d^{2}sp^{3}$
• (b) $dsp^{2}$
• (c) $sp^{3}d$
• (d) $sp^{3}d^{2}$

Correct Answer: (d)

Explanation: In $[CoF_{6}]^{3-}$, F is a weak field ligand that doesn't pair electrons. The complex uses the outer 4d orbitals, resulting in $sp^3d^2$ hybridization.