please explain why the answer is that answer How many electrons are involved in
ID: 1001711 • Letter: P
Question
please explain why the answer is that answer
How many electrons are involved in pi bonding in benzene, C_6H_6? 12 30 3 6 18 The C-C-H bond angles in ethylene, C_2H_4, are 120 degree. What is the hybridization of the carbon orbitals? Consider the molecular orbital energy level diagrams for O_2 and NO. Which of the following is true? Both molecules are paramagnetic. The bond strength of O_2 is greater than the bond strength of NO. NO is an example of a homonuclear diatomic molecule. The ionization energy of NO is smaller than the ionization energy of NO^+. I only I and II I and IV II and III I, II, and IV In the molecular orbital description of CO: The highest energy electrons occupy antibonding orbitals. Six molecular orbitals contain electrons. There are two impaired electrons. The bond order is 3. All of the above are false. The number of molecular orbitals formed is always the number of atomic orbitals combined.Explanation / Answer
16) In 1 bond formation between 2 C atoms each one contributes 1 unhybridized electron.
In Benzene there are 6 C atoms bonded through 3 bonds and hence,
Number of electrons involved in pi bonding 6 x 1 e from each C = 6 electrons.
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17)
C-C-H angles in ethylene (H2C=CH2) molecule are 120o. Hence its clear that there is sp2 hybridization of the Carbon orbitals.
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18) Molecular orbital electronic configuration for O2 and NO shown here,
O2 (16 e filling) : (KK) (2s)2 (*2s)2 (2Px)2 (2Py2, 2Pz2) (*2Py1, *2Pz1)
NO (15 e filling) : (KK) (2s)2 (*2s)2 (2Px)2 (2Py2, 2Pz2) (*2Py1, *2Pz0)
I. True. It clear that both O2 and NO have unpaired electrons in voutermost valence MO and hence both are paramagnetic in nature
II. FALSE. As bond order increases bond strength increases. Bond order of O2 = 2 and in NO its 2.5. It means Strength of NO is greater than O2.
III. FALSE. NO is not-homonuclear but is Heteronuclear.
IV. TRUE. There is just one unpaired electron in valence MO and on ionization of that we get a MO known to have vacant orbital stability.
Now in NO+ formed there are all paired electrons and hence it will be difficult to ionize further electrons.
Hence option-C) i.e. I and IV are true.
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19) MO electronic configuration of CO is
CO (14 e filling) : (KK) (2s)2 (*2s)2 (2Px)2 (2Py2, 2Pz2) (*2Py, *2Pz0)
Bond order = (Number of electrons in BMO – Number of electrons in ABMO )/2 = (10 – 4) / 2 = 6/2 = 3
Bond order is 3
Answer: Option-D) Bond order is 3.
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20) Number of MO formed is always equal to the number of AOs involved.
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