General Chemistry Model Of Quiz 2 By Dr.Loay Ehab 2017

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  • 1/34 Questions

    1. For the two solvents cyclohexane and benzene, the electronic transition that exists in benzene but absent in cyclohexane is:
     (π à π*) ]  transitionThere fore,the solvent that absorbs at longer wavelength is 

    • Cyclohexane
    • Benzene
    • Option 3
    • Option 4
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About This Quiz

This quiz, crafted by Dr. Loay Ehab in 2017, explores molecular interactions and electronic transitions in organic chemistry. It assesses understanding of molecular geometry, electronic transitions in solvents like benzene and ammonia, and bonding in iodine water. Key for learners aiming to deepen their knowledge in organic chemistry.

General Chemistry Model Of Quiz 2 By Dr.Loay Ehab 2017 - Quiz

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  • 2. 

    1. The coordination bond between ammonia and boron triflouride , ammonia undergoes ( oxidation  ) . according to the VSEPR theory BCl3 (  trignoal )in ammonia molecule NH3 , the molecular geometry is 

    • Tetrahedral

    • Trigonal

    • Option 3

    • Option 4

    Correct Answer
    A. Tetrahedral
    Explanation
    The correct answer is tetrahedral. According to the VSEPR theory, ammonia (NH3) has a tetrahedral molecular geometry. In this geometry, the central nitrogen atom is surrounded by three hydrogen atoms and one lone pair of electrons. The coordination bond between ammonia and boron trifluoride does not affect the molecular geometry of ammonia. Therefore, the correct answer is tetrahedral.

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  • 3. 

    Electronic transition available in ammonia are       [(α à α*)   /   ( n à π*)    /   (  n à α*)    /  (π à π*)  /  1st and 4th answers]In ammonia ion NH4+, the molecular geometry is ( tetrahedral )The electronic transition are

    • Option 1

    • [(α  α*)

    • ( n  Ï€*)

    • ( n  α*)

    • (Ï€  Ï€*)

    Correct Answer
    A. [(α  α*)
  • 4. 

    Iodine water is important it`s formed of the triIodide ( I3-) of structure ( I à I 2 )
    • I2 molecule acts as [ Lewis acid ]
    It undergoes [  / reduction ] in this bonding .
    • I -   ion acts as [/ lewis base ] it undergoes        
     ( oxidation ) in this bonding.
    • The empty molecular orbital in  I 2 molecule is  .....................ØŸ

    • Sigma star 5pz

    • Sigma 5pz

    • Option 3

    • Option 4

    Correct Answer
    A. Sigma star 5pz
    Explanation
    The correct answer is sigma star 5pz. In the given explanation, it is mentioned that I2 molecule acts as a Lewis acid and undergoes reduction in this bonding. This indicates that I2 molecule accepts an electron pair, which means it has an empty molecular orbital. The sigma star 5pz orbital is the empty molecular orbital in I2 molecule, making it the correct answer.

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  • 5. 

    Iodine water is important it`s formed of the triIodide ( I3-) of structure ( I à I 2 )It undergoes  >>>>>>>>>>>> in this bonding 

    • Reduction

    • Oxidation

    • Option 3

    • Option 4

    Correct Answer
    A. Reduction
    Explanation
    Iodine water is formed by the triiodide (I3-) ion, which has a structure of I à I 2. In this bonding, the triiodide ion is undergoing reduction.

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  • 6. 

    The (π) molecular orbital has >>>>>>>> compared with the ( sigma ) molecular orbital.

    • Higher

    • Lower

    • Option 3

    • Option 4

    Correct Answer
    A. Higher
    Explanation
    The π molecular orbital has a higher energy compared to the σ molecular orbital.

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  • 7. 

    In SF6 the molecular geometry is   

    • Octahedral

    • Linear

    • trignoal

    • Option 4

    Correct Answer
    A. Octahedral
    Explanation
    SF6 is a molecule composed of one sulfur atom bonded to six fluorine atoms. Each fluorine atom forms a single bond with the sulfur atom, resulting in a total of six bonds. The molecule has a central sulfur atom surrounded by six fluorine atoms, giving it a coordination number of six. This arrangement of atoms forms an octahedral geometry, where the sulfur atom is at the center and the fluorine atoms are positioned at the corners of an octahedron. Therefore, the correct answer is octahedral.

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  • 8. 

    In Ch4 the molecular geometry is  

    • Tetrahedral

    • Linear

    • Octahedral

    • Option 4

    Correct Answer
    A. Tetrahedral
    Explanation
    The correct answer is tetrahedral. In molecular geometry, tetrahedral refers to a shape where a central atom is surrounded by four other atoms, forming a three-dimensional pyramid-like structure. This geometry is commonly found in molecules with four bonding pairs of electrons and no lone pairs.

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  • 9. 

    1. In the electromagnetic spectrum , x-rays comes next to 
                  

    • UV

    • IR

    • Visible region

    • B-rays

    Correct Answer
    A. UV
    Explanation
    X-rays come next to UV in the electromagnetic spectrum because they have a higher frequency and shorter wavelength than UV rays. X-rays have wavelengths between 0.01 to 10 nanometers, while UV rays have wavelengths between 10 to 400 nanometers. This means that x-rays have higher energy and can penetrate matter more easily than UV rays. X-rays are commonly used in medical imaging and industrial applications due to their ability to pass through soft tissues and materials.

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  • 10. 

    1. The photo-toxicity of drugs is attributed to the generation of :
       (   /    /   /   )

    • O

    • O2

    • Singlet oxygen

    • 2nd and 3rd answers

    Correct Answer
    A. 2nd and 3rd answers
    Explanation
    The photo-toxicity of drugs is attributed to the generation of both O2 and singlet oxygen. O2 can react with other molecules in the body, leading to oxidative damage and toxicity. Singlet oxygen is a highly reactive form of oxygen that can also cause damage to cells and tissues. Therefore, the generation of both O2 and singlet oxygen contributes to the photo-toxicity of drugs.

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  • 11. 

    The oxygen-oxygen bond in ozone is ( Shorter / longer / equal) compared with the oxygen-oxygen bond in H2O2 .

    • Shorter

    • Longer

    • Equal

    • Option 4

    Correct Answer
    A. Shorter
    Explanation
    The oxygen-oxygen bond in ozone is shorter compared to the oxygen-oxygen bond in H2O2. This is because ozone (O3) has a bent molecular shape, with the oxygen atoms bonded at an angle of 117 degrees. This bent shape creates a greater electron density between the oxygen atoms, resulting in a stronger bond and a shorter bond length. In contrast, H2O2 has a linear molecular shape, with the oxygen atoms bonded at a 180-degree angle. This linear shape leads to a lower electron density between the oxygen atoms, resulting in a weaker bond and a longer bond length.

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  • 12. 

    1. In photodynamic therapy ( PDT )
     

    • Hema to prophyrin derivatives are common sensitizers

    • These sensitizers molecules accumulate in healthy cell

    • Ozone molecules are generated that destroy malignant twis

    • All are true

    Correct Answer
    A. Hema to prophyrin derivatives are common sensitizers
    Explanation
    Hematoporphyrin derivatives are commonly used as sensitizers in photodynamic therapy (PDT). These sensitizers molecules have the ability to accumulate in healthy cells. During PDT, when these sensitizers are exposed to light, they can generate ozone molecules that can destroy malignant tissues. Therefore, all the statements given are true.

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  • 13. 

    According to fajan`s rules

    • Covalence increases with increasing size of anion

    • Covalence increases with increasing size of cation

    • Covalence increases with decreasing size of anion

    • Both (a) and (b)

    Correct Answer
    A. Covalence increases with increasing size of anion
    Explanation
    According to Fajan's rules, covalence increases with increasing size of anion. This is because a larger anion can more easily polarize the electron cloud of the cation, leading to a greater degree of covalent character in the bond. As the size of the anion increases, the electron cloud becomes more easily distorted, allowing for stronger covalent interactions between the atoms. Therefore, the correct answer is that covalence increases with increasing size of anion.

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  • 14. 

    Given the radius of Ba is much larger than that of Be , then BeCo3 is (more  /  less  / equally ) covalent than BeCo3 . therefore BeCo3 decomposes at  lower temperature compared with BeCo3.

    • More

    • less

    • Equally

    • Option 4

    Correct Answer
    A. More
    Explanation
    BeCo3 is more covalent than BaCo3. This is because as the radius of an atom increases, the electronegativity decreases, resulting in weaker bonding between the atoms. Therefore, BeCo3, with a smaller radius for Be, has stronger covalent bonding compared to BaCo3, which has a larger radius for Ba. As a result, BeCo3 decomposes at a lower temperature compared to BaCo3.

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  • 15. 

    AlCl3 acts as lewis acid . 13Al 
    • The number of valence electrons is ( 2  /  3  /  5 )

    • 2

    • 3

    • 5

    • 10

    Correct Answer
    A. 3
    Explanation
    AlCl3 acts as a Lewis acid because it can accept an electron pair from a Lewis base to form a coordinate covalent bond. In AlCl3, aluminum (Al) has three valence electrons. Since it needs a total of eight electrons to achieve a stable electron configuration, it can accept an electron pair from a Lewis base to complete its octet. Therefore, the correct answer is 3, which represents the number of valence electrons in aluminum.

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  • 16. 

    The S character in HC =  CH is (larger  /  smaller ) than in H3C-CH3

    • Larger

    • Smaller

    • Option 3

    • Option 4

    Correct Answer
    A. Larger
    Explanation
    The S character in HC = CH is larger than in H3C-CH3 because in HC = CH, the carbon atom is bonded to two hydrogen atoms and one carbon atom, resulting in a higher s-character in the hybridized orbital. In H3C-CH3, the carbon atom is bonded to three hydrogen atoms and one carbon atom, resulting in a lower s-character in the hybridized orbital. This difference in s-character affects the bond length and bond strength, making the S character in HC = CH larger than in H3C-CH3.

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  • 17. 

    The S character in R-C = N is............ than in R-NH2

    • More

    • Less

    • Option 3

    • Option 4

    Correct Answer
    A. More
    Explanation
    The S character in R-C = N is more than in R-NH2. This means that the carbon atom in the R-C = N group has a higher degree of s-character in its hybridization compared to the carbon atom in the R-NH2 group. The higher the s-character, the more electronegative the carbon atom becomes, leading to a stronger electron-withdrawing effect. This can affect the reactivity and stability of the molecule.

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  • 18. 

    H2S is a gas but H2O is a liquid due to :

    • H2S has a higher molecular weight

    • Hydrogen bonding in H2O is stronger than H2S.

    • The polarity in O-H bond is higher than S-H bond

    • Both (b) and (b)

    Correct Answer
    A. Both (b) and (b)
    Explanation
    H2S is a gas but H2O is a liquid due to the stronger hydrogen bonding in H2O compared to H2S. Additionally, the polarity in the O-H bond in H2O is higher than the S-H bond in H2S, which contributes to the difference in their states of matter.

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  • 19. 

    A mixture of etanol and water has  less hydrogen bonds compared with pure water. It therefore has more (open –close ) structure compared with pure water.

    • Open

    • close

    • Option 3

    • Option 4

    Correct Answer
    A. close
    Explanation
    A mixture of ethanol and water has fewer hydrogen bonds compared to pure water because ethanol molecules do not form as many hydrogen bonds as water molecules. This results in a more open structure in the mixture compared to pure water, which has a more close-packed structure due to the extensive hydrogen bonding between water molecules.

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  • 20. 

    In the kinetic isotope effect, the rate of C-H breaking is (  Faster -slower) than the rate of C-D breaking.

    • faster

    • Slower

    • Option 3

    • Option 4

    Correct Answer
    A. faster
    Explanation
    The kinetic isotope effect refers to the difference in reaction rates between isotopes of the same element. In this case, the question is asking about the rate of breaking C-H bonds compared to the rate of breaking C-D bonds. The correct answer, "faster," suggests that the rate of C-H breaking is higher than the rate of C-D breaking. This is because the hydrogen isotope (H) is lighter than the deuterium isotope (D), resulting in stronger bonds and a slower rate of breaking for C-D bonds.

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  • 21. 

    Kinetic isotope is commonly obserred in ( Sn1  - Sn2 ) reactions.

    • Sn2

    • Sn1

    • Option 3

    • Option 4

    Correct Answer
    A. Sn2
    Explanation
    Kinetic isotope effect is commonly observed in Sn2 reactions. In Sn2 reactions, the rate-determining step involves a concerted attack of the nucleophile on the substrate, resulting in a transition state where the leaving group is partially bonded to the nucleophile. This transition state is highly sensitive to changes in mass, and the presence of a heavier isotope can significantly affect the rate of the reaction. Therefore, the correct answer is Sn2.

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  • 22. 

    For CO the valence electrons on C is (  4  ) the number of bonds is (  3 ) the number of electrons as lone pair is (  2  ) the formal charge on carbon is ( +1 / -1 ).

    • -1

    • +1

    • Option 3

    • Option 4

    Correct Answer
    A. -1
    Explanation
    The given information states that for CO, the number of valence electrons on carbon is 4, the number of bonds is 3, and the number of electrons as a lone pair is 2. To calculate the formal charge on carbon, we use the formula: Formal charge = (number of valence electrons) - (number of bonds) - (number of lone pair electrons). Plugging in the given values, we get: Formal charge = 4 - 3 - 2 = -1. Therefore, the correct answer is -1.

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  • 23. 

    CO poisoning gas, CO binds to hemoglobin through ( carbon / oxygen ) atoms.

    • Carbon

    • Oxygen

    • Option 3

    • Option 4

    Correct Answer
    A. Carbon
    Explanation
    CO poisoning occurs when carbon monoxide (CO) binds to hemoglobin in the blood, forming carboxyhemoglobin. This prevents oxygen from binding to hemoglobin, leading to tissue hypoxia and potentially fatal consequences. The statement in the question is correct, as CO binds to hemoglobin through carbon atoms. Oxygen, on the other hand, binds to hemoglobin through oxygen atoms.

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  • 24. 

    HCL + NaOH à NaCl + H2O ( redox – nonredox – disproprtho ]

    • Nonredox

    • Redox

    • Disproprtho

    • Option 4

    Correct Answer
    A. Nonredox
    Explanation
    The given chemical equation represents a neutralization reaction between hydrochloric acid (HCL) and sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water (H2O). This reaction does not involve any change in oxidation states of the elements involved, indicating that it is a nonredox reaction.

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  • 25. 

    CuSO4 is a ( hygroscopic ) CuSO4.5H2O is ( Hydroscopic / efflorescent / deliquescent / none of these ) .

    • None of these

    • Hydroscopic

    • Efflorescent

    • Deliquescent

    Correct Answer
    A. None of these
  • 26. 

    226Ra88 à 222Rn86 + 4He2This reaction ( increase / decrease ) N/p ratio.

    • increase

    • decrease

    • Option 3

    • Option 4

    Correct Answer
    A. increase
    Explanation
    The given reaction involves the decay of 226Ra into 222Rn and 4He. In this process, the number of protons (N) remains the same, but the number of neutrons (p) decreases by 4. Therefore, the N/p ratio increases, as the number of protons remains constant while the number of neutrons decreases.

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  • 27. 

    Every redox reaction is a disproportionation    

    • True

    • False

    Correct Answer
    A. False
    Explanation
    Every redox reaction is not a disproportionation. Disproportionation is a specific type of redox reaction in which an element in a single compound is simultaneously oxidized and reduced. In a disproportionation reaction, the same element is both oxidized and reduced. However, not all redox reactions involve the same element being oxidized and reduced. Therefore, the statement is false.

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  • 28. 

    1. B-rays come next to r-rays in electro magnetic spectrum          

    • True

    • False

    Correct Answer
    A. False
    Explanation
    B-rays do not exist in the electromagnetic spectrum. The electromagnetic spectrum consists of various types of waves, including radio waves, microwaves, infrared waves, visible light, ultraviolet waves, X-rays, and gamma rays. B-rays are not a recognized category of waves in the electromagnetic spectrum. Therefore, the statement is false.

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  • 29. 

    One rad of Alpha has higher rem them one rad of        B-partials  

    • True

    • False

    Correct Answer
    A. True
    Explanation
    Alpha particles have a higher rem (roentgen equivalent man) value than B-particles. The rem is a unit used to measure the biological effect of radiation on living tissue. It takes into account the type of radiation and its energy. Alpha particles are larger and more massive than B-particles, and they have a higher ionizing power. This means that they can cause more damage to living tissue, resulting in a higher rem value. Therefore, it is true that one rad of Alpha has a higher rem than one rad of B-particles.

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  • 30. 

    The reaction H2O2 à H2O + 1/2 O2 is disproportion redox reaction.                                                                      

    • True

    • False

    Correct Answer
    A. True
    Explanation
    The given reaction, H2O2 → H2O + 1/2 O2, is a disproportionation reaction because the same element, oxygen, is both reduced and oxidized. In this reaction, hydrogen peroxide (H2O2) is being both reduced to water (H2O) and oxidized to oxygen gas (1/2 O2). Therefore, the statement "The reaction H2O2 → H2O + 1/2 O2 is a disproportion redox reaction" is true.

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  • 31. 

    Neutron rich Isotopes are usually B-emitters.    

    • True

    • False

    Correct Answer
    A. True
    Explanation
    Neutron-rich isotopes have an excess of neutrons compared to protons in their nucleus. This imbalance makes the nucleus unstable, and in order to regain stability, it undergoes beta decay. During beta decay, a neutron is converted into a proton, and a beta particle (electron or positron) is emitted. Therefore, neutron-rich isotopes are usually beta emitters, making the statement "Neutron-rich isotopes are usually B-emitters" true.

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  • 32. 

    59mTC is bone seeking Isotope.

    • True

    • False

    Correct Answer
    A. False
    Explanation
    59mTC is not a bone-seeking isotope. It is actually a technetium isotope that is commonly used in nuclear medicine imaging procedures, such as myocardial perfusion imaging and lung ventilation/perfusion scans. It has a short half-life of about 6 hours, which allows for timely imaging and reduces radiation exposure to the patient.

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  • 33. 

    13N7 , 17O8 are isotones

    • True

    • False

    Correct Answer
    A. True
    Explanation
    Isotones are atoms that have the same number of neutrons but different numbers of protons. In this case, 13N7 and 17O8 have the same number of neutrons (8) but different numbers of protons (7 and 8 respectively), making them isotones. Therefore, the correct answer is true.

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  • 34. 

    13C6,15O8 are Isobars. 

    • True

    • False

    Correct Answer
    A. False
    Explanation
    Isobars are atoms or ions that have the same mass number but different atomic numbers. In the given statement, 13C6 and 15O8 are mentioned as isobars. However, this is incorrect because the atomic numbers of carbon and oxygen are different (6 and 8 respectively), indicating that they are not isobars. Therefore, the correct answer is False.

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  • Mar 21, 2023
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