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Chemistry Mixed Test

Description: s block elements p block elements d block elements coordination complexes B, Al, Si, N, P, and SCoordination ComplexesGeneral Characteristics of 3d ElementsChemistry
Number of Questions: 17
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Tags: s block elements p block elements d block elements coordination complexes B, Al, Si, N, P, and S Coordination Complexes General Characteristics of 3d Elements Chemistry
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Which of the following statements is/ are false? P. In Raschig process hydrazine is prepared from the reduction of ammonia with sodium hypochlorite in dilute aqueous solution. Q. Borazine is known as inorganic benzene. R. Boric acid is a dibasic acid.

  1. Only P

  2. Only Q

  3. Only P and Q

  4. Only Q and R

  5. Only P and R

Show answer
Correct Option: E
Explanation:

P) Hydrazine (N2H4) is a nitrogen hydride. In Raschig process it is prepared by the reaction of ammonia with sodium hypochlorite in dilute aqueous solution. Q) Borazine, B3N3H6 known as inorganic benzene in view of its ring structure with alternate BH and NH groups. R) Boric acid is a weak monobasic acid. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion. Hence, statements P and R are false.

Which of the following groups shows the isoelectronic species?

  1. CO, N2, O2, NO+

  2. CO, N2, O2+, CN-

  3. CO, F2, NO+, CN-

  4. CO, N2, NO+, CN-

  5. CO, N2+, NO+, CN-

Show answer
Correct Option: D
Explanation:

Isoelectronic species have the same number of electrons: CO, N2, NO+ and CN- have 14 electrons each.

Identify the false statements. P. Hydrazine is a colourless gas and is used as a rocket fuel. Q. PCl3 has more Lewis acid strength than PI3. R. H2S has T-shaped geometry. S. The sulphur atom exhibits +3 oxidation state in sodium dithionate.

  1. P and Q

  2. Q and R

  3. R and S

  4. P and S

  5. P, R and S

Show answer
Correct Option: E
Explanation:

P. Hydrazine is an inorganic compound with the formula N2H4. It is a colourless flammable liquid with an ammonia-like odor and is used as rocket fuel. Q. In the case of PX3, Lewis acid strength decreases in the order of; PF3>PCl3>PBr3>PI3. So, PCl3 has more Lewis acid strength than PI3.
R. H2S has sp3 hybridisation and 2 lone pairs with V-shaped geometry. S. The oxidation state of sulphur atom in sodium dithionate is +5. Hence, P, Q and R are false.

Among the following, an Arachno-carborane is

  1. B3C2H11

  2. B3C2H7

  3. B5CH9

  4. B7C2H13

  5. B10C2H12

Show answer
Correct Option: D
Explanation:

B7C2H13 is an example of Arachno-carborane.

The complex with tetrahedral geometry is ____________.

  1. [NiCl4]2-

  2. [Ni(CN)4]2-

  3. Cis-platin

  4. Carboplatin

  5. Zeise’s salt

Show answer
Correct Option: A
Explanation:

In case of tetrahedral configuration, the compound would have a magnetic moment corresponding to two unpaired electrons. [NiCl4]2- has two unpaired electrons and a tetrahedral geometry.

Match the Group-I (oxy acids of phosphorus) with Group-II (oxidation state of P atom). ||| |---|---| |Group-I |Group-II| |A. Orthophosphoric acid|1. +1| |B. Hypophosphoric acid|2. +3| |C. Phosphorous acid|3. +4| |D. Metaphosphoric acid|4. +5|

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

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

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

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

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

Show answer
Correct Option: A
Explanation:

The oxidation state of P in orthophosphoric acid (H3PO4) is +5. The oxidation state of P in phosphorous acid (H3PO3) is +3. The oxidation state of P in hypophosphoric acid (H4P2O6) is +4. The oxidation state of P in metaphosphoric acid (HPO3) is +5. Hence, the correct representation of the codes is A - 4, B - 3, C - 2, D - 4.

Identify the unmatched pair.

  1. Wilkinson’s catalyst - Ph3RhCl/Et3Al

  2. Lindlar catalyst - PtO2

  3. Lazier’s catalyst - Cu2Cr2O5

  4. Vaska's complex - IrCl(CO)(PPh3)2

  5. Adkins catalyst - Cr/CuO

Show answer
Correct Option: B
Explanation:

Lindlar catalyst is the combination of Pd/CaCO3/PbO. PtO2 is the Adam’s catalyst.

Which of the following is not an example of Closo-carboranes?

  1. B3C2H11

  2. B4C2H6

  3. B6C2H8

  4. B9C2H11

  5. B8C2H10

Show answer
Correct Option: A
Explanation:

B3C2H11 is an example of Nido-carboranes.

Match the entries in Group-I with Group-II ||| |---|---| |Group-I (Ores) |Group-II (Corresponding metal)| |P. Patronite|1. V| |Q. Crocoisite|2. Ni| |R. Garnierite|3. Pt| |S. Cooperate|4. Cr| | |5. Cu|

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

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

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

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

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

Show answer
Correct Option: A
Explanation:

Patronite (V2S5.3CuS2) is the sulphide ore of vanadium. Crocoisite (PbCrO4) is the ore of chromium. Garnierite (NiMgSiO3.nH2O) is the ore of nickel. Cooperate (PtS) is the sulphide ore of platinum. Hence, the correct answer using the codes is P - 1, Q - 4, R - 2, S - 3.

The Styx number for pentaborane-11 is __________.

  1. 2002

  2. 3203

  3. 4012

  4. 4120

  5. 4220

Show answer
Correct Option: B
Explanation:

Styx number is the code system representing the overall valence bonding in boranes by four digit number. s = number of B-H-B bonds (I digit) t = number of B-B-B bonds (II digit) y = number of B-B bonds (III digit) x = number of BH2 groups (IV digit) The Styx number for pentaborane-11 (B5H11) is 3203.

Identify the incorrect statements. P. Pentaborane-11shows symmetrical square pyramidal geometry. Q. CsBr crystal has face centred cubic structure. R. MgAl2O4 and Fe3O4 are identical in crystal structure.

  1. P only

  2. Q only

  3. R only

  4. P, Q

  5. Q, R

Show answer
Correct Option: D
Explanation:

P. Pentaborane-11 (B5H11) has an unsymmetrical square pyramidal geometry, in which five boron atoms occupy the five corners of a square pyramid. Q. The crystal of CsBr has body centred cubic structure. R. MgAl2O4 and Fe3O4 have spinel structures. Hence, statement P and Q both are incorrect.

Match the entries in Column-I with Column-II ||| |---|---| |Column - I (Crystals) |Column - II (Corresponding structures)| |A. TiO2|1. perovskite| |B. Zn2TiO4|2. rutile| |C. CaTiO3|3. spinel| |D. CaTiO3|4. inverse spinel|

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

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

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

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

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

Show answer
Correct Option: B
Explanation:

TiO2 has rutile structure. Zn2TiO4 has inverse spinel structure. CaTiO3 has perovskite structure. MgAl2O4 belongs to spinel structure. Hence, the sequence of codes A - 2, B - 1, C - 4, D - 3 is correct.

Match the Column-I with Column-II and select the correct answer using the codes given below. ||| |---|---| |Column-I (metal salts) |Column-II (their common defects)| |P. NaCl|1. metal excess defect| |Q. FeO|2. metal deficiency defect| |R. FeS| | |S. NiO| | |T. AgBr| |

  1. P - 1, Q - 2, R - 2, S - 2, T - 1

  2. P - 1, Q - 1, R - 2, S - 1, T - 1

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

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

  5. P - 2, Q - 1, R - 2, S - 2, T - 1

Show answer
Correct Option: A
Explanation:

Generally, metal salts are impregnated with some defects like metal excess defects and metal deficiency defects. NaCl and AgBr exhibit metal excess defects. FeO, FeS and NiO exhibit metal deficiency defects. Hence, the correct answer using the codes is P - 1, Q - 2, R - 2, S - 2, T - 1.

An intrinsic semiconductor has a band gap of 1.5 eV. The wavelength of the electromagnetic radiation required to cause the material photoconducting, is ________.

  1. 3525 Å

  2. 7827 Å

  3. 8272 Å

  4. 8573 Å

  5. 9368 Å

Show answer
Correct Option: C
Explanation:

Given, Band gap (E) = 1.5 eV = 1.5 × 1.602 × 10-12 erg, h = 6.627 × 10-27 erg.s, c = 3×1010 cm/s Now, energy of material, E = hc/λ and λ = 6.627 × 10-27 × 3 × 1010/ 1.5 × 1.602 × 10-12 = 8.272 × 10-5 cm = 8272 Å.

The IUPAC name for the heteropoly complex, Na3[PVMo12O40] is ___________.

  1. Tris sodium 12-molybdophosphate

  2. Tris sodium 12-molybdophosphorus(V)

  3. Sodium 12-molybdophosphate

  4. Sodiumphospho(V) molybdenum-12

  5. Tris sodiumphosphate molybdenum-12

Show answer
Correct Option: C
Explanation:

The IUPAC name of Na3[PVMo12O40 is Sodium 12-molybdophosphate.

Select the correct match using the List-A (complex/ion) and List-B (coordination number) ||| |---|---| |List-A |List-B| |P. Mn(NO)3CO|1. 3| |Q. Mn2(CO)10|2. 2| |R. Mn[C(SiMe3)2]2|3. 6| |S. MnO3+|4. 4|

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

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

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

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

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

Show answer
Correct Option: C
Explanation:

Yes, it is correct. Mn(NO)3CO has coordination number of 4 with tetrahedral geometry. Mn2(CO)10 has coordination number of 6 with octahedral geometry. Mn[C(SiMe3)2]2 has the coordination number of 2 with linear geometry. MnO3+ has the coordination number of 3 with triangular planar shape.

Identify the false statement(s).
P. B-F bond in BF3 is larger than in BF4- molecule. Q. Borax is used as an indicator. R. Borazine is isoelectronic with benzene and exhibits D3h symmetry.

  1. P only

  2. Q only

  3. R only

  4. P and Q

  5. Q and R

Show answer
Correct Option: D
Explanation:

P. B-F bond is larger in BF4- than in BF3 molecule. This is due to the fact that in BF4-, the boron atom is sp3 hybridised while in BF3, the boron atom is sp2 hybridised and bond length decreases with increase in s-character as s-orbital is smaller than the p-orbital.

Q. Borax is not used as an indicator. Borax is used as a buffer because its aqueous solution contains equal amounts of weak acid and its salt.

R. Borazine is isoelectronic with benzene and the symmetry point group in borazine is D3h.

Hence, statement P and Q both are false.

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