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Chemistry GATE - 7

Description: GATE
Number of Questions: 15
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Tags: GATE M.O. and V.B. Approaches for Diatomic Molecules VSEPR Theory and Shape of Molecules The Respiratory System Thermodynamics The Immune Response Proteins Ionic and Covalent Bonding s.p. and d Block Elements Enzyme Catalysis Aromatic Electrophilic Substitution
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In molecular ion ICl2+, the geometrical arrangement of orbital and shape are

  1. tetrahedral geometry, bent shape

  2. tetrahedral geometry, pyramidal shape

  3. trigonal bipyramidal geometry, linear shape

  4. trigonal bipyramidal geometry, bent shape

  5. tetrahedral geometry, T-shape

Show answer
Correct Option: A
Explanation:

ICl2+ has sp3 hybridisation and its geometrical arrangement of orbitals is tetrahedral but the shape (according to VESPR theory) is bent due to the presence of two lone pairs. Thus, this option is correct.

Directions: In the question given below, there are two statements labeled as Assertion (A) and Reason (R). Mark your answer as per the codes provided below: Assertion: CO and NO both combine with haemoglobin. Reason: Both have equal affinity for haemoglobin.

  1. Both A and R are true and R is the correct explanation of A.

  2. Both A and R are true but R is not the correct explanation of A.

  3. A is true but R is false.

  4. A is false but R is true.

  5. Both A and R are false.

Show answer
Correct Option: C
Explanation:

This option is correct because CO and NO both combine with haemoglobin, but NO has greater affinity for haemoglobin than CO and it is unable to enter the blood from atmosphere.

The spectroscopic term symbol for the ground state of a fluorine atom is

  1. 5P0

  2. 2P0

  3. 2P1/2

  4. 2P3/2

  5. 3P0

Show answer
Correct Option: D
Explanation:

This option is correct because the number of electrons in the ‘p’ orbital is greater than the half the capacity of this orbital. So, the greater the J value, the lower the energy. Thus, the spectroscopic term symbol for the ground state of a fluorine atom is 2P3/2.

If E is the energy of the combining atomic orbitals, E1 and E2 are the energies of the antibonding and bonding molecular orbitals formed, then

  1. E-E2 < E1-E

  2. E-E2 = E1-E

  3. E-E2 > E-E1

  4. E-E2 > E1-E

  5. Any one of these is possible

Show answer
Correct Option: A
Explanation:

The antibonding orbital is raised more in energy than the energy by which bonding molecular orbital is lowered. Thus, “E-E2 < E1-E” is the correct answer.

The ratio of σ- and π-bonds in carbon suboxide is

  1. 2 : 3

  2. 2 : 1

  3. 1 : 2

  4. 3 : 2

  5. 1 : 1

Show answer
Correct Option: E
Explanation:

O=C=C=C=O (carbon suboxide) Number of σ-bonds = 4 Number of π-bonds = 4 So that the ratio of σ- and π-bonds in carbon suboxide is 4 : 4 or 1 : 1. Thus, this option is correct.

Which of the following is not an example of β-lactam antibiotic?

  1. Terramycin

  2. Amoxicillin

  3. Ampicillin

  4. Penicillin

  5. Cephalosporin

Show answer
Correct Option: A
Explanation:

This option is correct because β-lactam antibiotics (beta-lactam antibiotics) is a broad class of antibiotics, which contains a β-lactam ring in their molecular structures. But terramycin has no β-lactam ring in their molecular structures so this antibiotic is not a β-lactam antibiotic.

Which of the statements is/are wrong?

  1. Halogens are all coloured because absorption of visible light by their molecules results in the excitation of the outer electrons to higher energy levels.

  2. Halogens are all coloured because the small F2 molecules absorb high energy violet radiation and appear yellow.

  3. Halogens are all coloured because larger I2 molecules absorb low energy yellow and green radiations and appear violet in colour.

  4. Options 2 and 3

  5. Halogens are all coloured because the excitation energy required by the small fluorine atoms is smaller than that required by the large iodine atoms.

Show answer
Correct Option: E
Explanation:

This option is correct because excitation energy of F (2p electron) is more than the excitation energy of iodine (5p electron).

Which of the following enzymes helps in hydrolysis of the protein into α-amino acids? A. Pepsin B. Carbonic anhydrase C. Trypsin D. Chymotrypsin

  1. Only A

  2. Only B

  3. Only B and D

  4. Only D

  5. Only A, C and D

Show answer
Correct Option: E
Explanation:

This option is correct because carbonic anhydrase does not hydrolyse protein to α-amino acids, but chymotripsin, trypsin and pepsin all hydrolyse protein into α-amino acids.

Which of the following enzymes is used to check heart attacks due to blood clot formation in the coronary artery?

  1. RNA polymerase

  2. DNA polymerase

  3. Phenylalanine hydroxylase

  4. Tyrosinase

  5. Streptokinase

Show answer
Correct Option: E
Explanation:

This option is correct because streptokinase is an enzyme secreted by several species of streptococci that can bind and activate human plasminogen. Streptokinase is used as an effective and inexpensive thrombolysis medication in some cases of myocardial infarction (heart attack) and pulmonary embolism.

In enzymatic fermentation, C2H5OH and CO2 are produced from which of the following sugars? I. Glucose II. Invert sugar III. Fructose

  1. Only I and II

  2. Only I and III

  3. Only II and III

  4. All of these

  5. None of these

Show answer
Correct Option: D
Explanation:

This option is correct because zymase converts glucose, fructose and invert sugar (equimolar ratio of glucose and fructose) into ethyl alcohol.

Match the List I with List II and select the correct answer using the codes given below ||| |---|---| |List I |List II| |A. (dG / dP)T|1. V| |B. (dG / dT)P|2. T| |C. (dH / dS)P|3. –S| |D. (dT / dP)H|4. P| | |5. μJT|

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

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

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

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

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

Show answer
Correct Option: B
Explanation:

From thermodynamics, dG = VdP – SdT At constant T, dT = 0 so that (dG / dP)T =V At constant P, dP = 0 so that (dG / dT)P = −S At constant P, dP = 0 so that (dH / dS)P = T and also (dT / dP)H= μJT. Thus, this option is correct.

Directions: In the question given below, there are two statements labeled as Assertion (A) and Reason (R). Mark your answer as per the codes provided below: Assertion: Chlorination of ethylbenzene with Cl2 in presence of heat and light mainly yields 1-chloro-2-phenylethane. Reason: The reaction occurs through intermediate formation of the radical, C6H5CH2C*H2.

  1. Both A and R are true and R is the correct explanation of A.

  2. Both A and R are true but R is NOT the correct explanation of A.

  3. A is true, but R is false.

  4. A is false, but R is true.

  5. Both A and R are false.

Show answer
Correct Option: E
Explanation:

This option is correct because chlorination of ethylbenzene with Cl2 in presence of heat and light gives 1-chloro-1-phenylethane, and the reaction occurs through intermediate formation of the benzylic radical, C6H5C.HCH3.

Directions: In the question given below, there are two statements labeled as Assertion (A) and Reason (R). Mark your answer as per the codes provided below: Assertion: Patient suffering from CO poisoning is kept in CO2 at a pressure of 2 atmospheres. Reason: CO of carboxyhaemoglobin is replaced by CO2.

  1. Both A and R are true and R is the correct explanation of A.

  2. Both A and R are true, but R is NOT the correct explanation of A.

  3. A is true, but R is false.

  4. A is false. but R is true.

  5. Both A and R are false.

Show answer
Correct Option: E
Explanation:

This option is correct because the patient suffering from CO poisoning is kept in O2 at a pressure of 2 to 2.5 atmospheres and CO of carboxyhaemoglobin is replaced by O2, and thus transport of O2 to different parts of the body starts.

Hybridizations present in central atom of ICl5, BrCl2-, ClF2+, ClF4+ are respectively:

  1. sp3d2, sp3, sp3, sp3d3

  2. sp3d3, sp3d, sp3, sp3d2

  3. sp3d, sp2, sp, sp3d2

  4. sp3d3, sp3d, sp3d, dsp3

  5. d3sp3, sp3d, sp3, sp3d2

Show answer
Correct Option: B
Explanation:

This option is correct because the hybridization of atomic orbital's of iodine in ICl5, BrCl2-, ClF2+ and ClF4+ are sp3d2, sp3, sp3, sp3d3 respectively.

Match the column I (components) with column II (functions) ||| |---|---| |Column I |Column II| |A. RBC|1. Blood clotting| |B. Fibrinogen|2. Maintenance of osmotic pressure of blood plasma| |C. Saccharide|3. Transports of O2 from lungs to the tissues| |D. Globulins|4. Transport of lipids as lipoproteins| |E. Albumins|5. Source of energy| |F. Blood platelets| |

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

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

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

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

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

Show answer
Correct Option: D
Explanation:

All are correctly matched. Thus, this option is correct.

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