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d-Block Elements (Hard)

Description: GAte
Number of Questions: 15
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Tags: GAte Transition Elements Important Compounds of Transition Elements General Characteristics of 3d Elements
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Match the following:

 
Group - I (element) Group - II (electronic configuration)
P. Cu 1. 4d7, 5s1
Q. Mo 2. 4d8, 5s1
R. Ru 3. 3d10, 4s1
S. Rh 4. 4d5, 5s1
  1. P - 3, Q - 4, R - 1, S - 2

  2. P - 2, Q - 4, R - 1, S - 3

  3. P - 3, Q - 1, R - 4, S - 2

  4. P - 2, Q - 3, R - 1, S - 4

  5. P - 2, Q - 3, R - 4, S - 1


Correct Option: A
Explanation:

29Cu has the electronic configuration of 3d10, 4s1. 42Mo has the electronic configuration of 4d5, 5s1. 44Ru has the electronic configuration of 4d7, 5s1. 45Rh has the electronic configuration of 4d8, 5s1. Hence, the correct answer using the codes is P - 3, Q - 4, R - 1, S - 2.

Which of the following pairs of ‘element/ion - its electronic configuration’ is not correct?

  1. V - 3d34s2

  2. Cu - 3d104s1

  3. Fe2+ - 3d64s0

  4. Mn2+ - 3d74s0

  5. Sc+ - 3d14s1


Correct Option: D
Explanation:

Mn2+ has the electronic configuration of 3d54s0. 3d74s0 is the electronic configuration of Co2+.

Which metal ion represents zero value of CFSE if it associated with weak field ligand?

  1. Fe3+

  2. Fe2+

  3. Cr2+

  4. Mn+

  5. Mn3+


Correct Option: A
Explanation:

In weak field ligand, the filling of electrons are in accordance with Hund’s rule. The CFSE is given by the formula: CFSE = value of t2g (n) + value of eg (n) Dq, Fe3+ corresponds to the d5 configuration. Now, CFSE for Fe3+ (d5) = -4 (3) + 6 (2) = -12 + 12 = 0 Dq CFSE for Fe2+ (d6) = -4 (4) + 6 (2) = -16 + 12 = -4 Dq CFSE for Cr2+ (d4) = -4 (3) + 6 (1) = -12 + 6 = -6 Dq CFSE for Mn+ (d6) = -4 (4) + 6 (2) = -16 + 12 = -4 Dq CFSE for Mn3+ (d4) = -4 (3) + 6 (1) = -12 + 6 = -6 Dq Hence, the metal ion with zero CFSE is Fe3+ when it associated with weak field ligand.

Which of the following statements are incorrect?

  1. In floatation process, ZnO and SO3 are formed from sulphide ores.
  2. To prevent corrosion of Zn, Fe is used as a coating material.
  3. Cd metal is used in nuclear reactors as neutron excitation material.
  4. Mercury is used extensively in the production of sodium hydroxide and chlorine gas commercially.
  1. 1 and 2

  2. 2 and 3

  3. 3 and 4

  4. 1, 2 and 3

  5. 2, 3 and 4


Correct Option: B
Explanation:
  1. The floatation process is the concentration method of sulphide ores. Zinc ores, mainly ZnS, are concentrated by floatation method, then roasted in air to give ZnO and SO3.
    1. Zn is used extensively for coating for the appliances made up of corrosive metals like Fe to prevent them from rusting.
    2. Cd metal absorbs neutrons very well and is used to make control rods in nuclear reactors.
    3. The largest use of mercury is in electrolytic cells for the production of sodium hydroxide and chlorine gas at commercial level. Hence, statements 2 and 3 are incorrect.

Which of the following pairs of ‘transition metal and its ore’ is incorrectly matched?

  1. Ni - pentlandite

  2. Co - linnaeite

  3. Nb - scheelite

  4. Cu - malachite

  5. Hg - cinnabar


Correct Option: C
Explanation:

Scheelite is a calcium tungstate mineral with the chemical formula CaWO4, while pyrochlorite (CaNaNb2O6) is an important ore/mineral of niobium.

In the extraction of iron, which of the following constitutes the ‘charge’ in the blast furnace besides coke and oxide ore?

  1. Quicklime

  2. Limestone

  3. Dolomite

  4. Silica

  5. Soda lime


Correct Option: B
Explanation:

This option is correct because the mixture of iron ore, coke and limestone is called ‘charge’. Iron ore is not pure iron oxide; it also contains an assortment of rocky material. This wouldn't melt at the temperature of the furnace and would eventually clog it up. The limestone is added to convert this into slag which melts and runs to the bottom. The heat of the furnace decomposes the limestone to give calcium oxide.

Which of the following statements is/are not correct?

P. Manganese metal is used as scavenger in making steel. Q. All known isotopes of technetium are radioactive, except 98Tc. R. Rhenium is used in the making of electric furnace windings and spectrophotometer filaments.

  1. Only P

  2. Only Q

  3. Only R

  4. P and Q

  5. Q and R


Correct Option: B
Explanation:

This option is correct because technetium (99Tc) was the first man-made element with radioactive properties. All known isotopes of technetium (includes 97Tc, 98Tc) have radioactive properties. It is one of the fission products of uranium. It is obtained by spent fuel rods in nuclear reactors.

Match the lists:

   
List - A List - B List - C
P. Cu 1. Smaltite i. sulphide
Q. Zn 2. Chalcocite ii. arsenide
R. Fe 3. Franklinite iii. carbonate
S. Co 4. Siderite iv. oxide
  1. P - 3 - i, Q - 2 - iv, R - 4 - iii, S - 1 - ii

  2. P - 2 - i, Q - 3 - iv, R - 1 - iii, S - 4 - ii

  3. P - 2 - i, Q - 3 - ii, R - 1 - iii, S - 4 - iv

  4. P - 1 - i, Q - 2 - iv, R - 1 - ii, S - 4 - iii

  5. P - 2 - i, Q - 3 - iv, R - 4 - iii, S - 1 - ii


Correct Option: E
Explanation:

The correct answer using the codes is P - 2 - i, Q - 3 - iv, R - 4 - iii, S - 1 - ii.

The ores of titanium is/are

  1. ilmenite
  2. hausmannite
  3. linnaeite
  1. Only 1

  2. Only 2

  3. Only 3

  4. 1 and 2

  5. 2 and 3


Correct Option: A
Explanation:

Ilmenite (Fe.TiO3) is the oxide ore of titanium. Hausmannite (Mn3O4) is the oxide ore of manganese. Linnaeite (Co3S4) is the sulphide ore of cobalt. Hence, only ilmenite (Fe.TiO3) is the oxide ore of titanium.

Which of the following d-electronic configurations in an octahedral complex will give the maximum CFSE?

  1. Low spin-d4

  2. High spin-d4

  3. Low spin-d5

  4. High spin-d5

  5. High spin-d7


Correct Option: C
Explanation:

In octahedral complexes, CFSE is given by the formula:  CFSE = value of t2g (n) + value of eg (n) Dq Now, CFSE for low spin d4 = -4 (4) + 6 (0) = -16 + 0 = -16 Dq CFSE for high spin d4 = -4 (3) + 6 (1) = -12 + 6 = -6 Dq CFSE for low spin d5 = -4 (5) + 6 (0) = -20 + 0 = -20 Dq CFSE for high spin d5 = -4 (3) + 6 (2) = -12 + 12 = 0 Dq CFSE for low spin d7 = -4 (5) + 6 (2) = -20 + 12 = -8 Dq Hence, low spin d5 will give the maximum CFSE in an octahedral complex.

The clusters of H3Ru3NiCp(CO)9 are used to catalyse

  1. ammonia synthesis

  2. CO

  3. CO2 methanation

  4. hydrodesulfurisation

  5. butane hydrogenolysis


Correct Option: A
Explanation:

H3Ru3NiCp(CO)9 clusters contain Ru-Ni as core metals and catalyse the ammonia synthesis.

Which of the following statements is/are incorrect?

P. CuCl2 is used as a catalyst in the manufacture of ammonia by Haber process. Q. CuCl exhibits blue colour in aqueous solution.
R. Fractional crystallisation of metals to get maximum pure form is known as zone refining.

  1. Only P

  2. Only Q

  3. Only R

  4. P and Q

  5. Q and R


Correct Option: D
Explanation:

This option is correct because iron (Fe) is used as a catalyst in the manufacture of ammonia by Haber process, while CuCl2 is used as a catalyst in Deacon process for the preparation of Cl2 from HCl. CuCl in aqueous solution form Cu+ ions, which possess no colour due to the presence of completely filled d orbitals.

Which of the following statements is/are incorrect?

P. The central metal atom shows +4 oxidation state in potassium chromate and potassium dichromate. Q. Limonite and willemenite are ores of zinc. R. Zinc oxide is called as white lead with the formula (PbCO3)2•Pb(OH)2 and is used as a whitening material.

  1. Only P

  2. Only Q

  3. Only R

  4. P and Q

  5. Q and R


Correct Option: E
Explanation:

P. The central metal atom in potassium chromate and potassium dichromate is Cr. Chromium has +4 oxidation state in K2CrO4 and K2Cr2O7. Q. Limonite (Fe2O3.3H2O) is the ore of iron and willemite is the ore of zinc. R. Zinc oxide is an inorganic compound with the formula ZnO. ZnO is a white powder that is insoluble in water and it is widely used as an additive in numerous materials and products including rubbers, plastics, ceramics, glass, cement, etc. White lead (PbCO3)2•Pb(OH)2 is a complex chemical compound, containing both a carbonate and a hydroxide portion and occurs naturally as a mineral for lead. White lead compounds were used as pigmenting materials like cement, lubricants, etc, but due to its toxicity, it is not used frequently.

Hence, statements Q and R are incorrect.

Match the following:

 
Group - I Group - II
A. Wilkinson’s catalyst 1. Pt/PtO
B. Adams catalyst 2. [PH3P]3.Rh.Cl
C. Lazier catalyst 3. TiCl4/Al(C2H5)3
D. Ziegler-Natta catalyst 4. 2CuO.Cr2O3
  1. A - 2, B - 1, C - 4, D - 3

  2. A - 2, B - 4, C - 1, D - 3

  3. A - 1, B - 2, C - 4, D - 3

  4. A - 4, B - 3, C - 2, D - 1

  5. A - 4, B - 2, C - 1, D - 3


Correct Option: A
Explanation:

Wilkinson’s catalyst ([PH3P]3.Rh.Cl) is used in the hydrogenation of alkenes. Adams catalyst (Pt/PtO) is used in various reduction reactions. Lazier catalyst (2CuO.Cr2O3) is an effective catalyst for the reduction of aldehydes, ketones and esters to the corresponding alcohols and aliphatic double bonds. Ziegler-Natta catalyst (TiCl4/Al(C2H5)3) is used in the synthesis of polymers of alpha-olefins. Hence, the representation of codes A - 2, B - 1, C - 4, D - 3 is correct.

Match the lists:

 
List - A List - B
A. Hf 1. Polymerisation
B. AlEt3/TiCl4 2. Making of lead-free petrol
C. VC2 3. Neutron moderator
D. Pt-Re alloys 4. Ethyne production
  1. A - 2, B - 1, C - 4, D - 3

  2. A - 3, B - 4, C - 1, D - 2

  3. A - 3, B - 1, C - 4, D - 2

  4. A - 4, B - 2, C - 1, D - 2

  5. A - 4, B - 3, C - 2, D - 1


Correct Option: C
Explanation:

Hafnium is used to make control rods for regulating the free neutron levels in the nuclear reactors employed in submarines. The mixture of AlEt3 and TiCl4 is called Zeigler-Natta catalyst and is used extensively in polymerising alkenes (like ethylene to polyethylene). VC2 (vanadium carbide), when treated with water, produces ethyne and thus, is used in the production of ethyne. Pt-Re alloys are used as catalysts for making low-lead or lead-free petrol.

Hence, the correct sequence using the codes is A - 3, B - 1, C - 4, D - 2.  

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