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The total number of molecular orbitals formed from 2s and 2p atomic orbitals of a diatomic molecule

Numerical Answer Type:
Enter a numerical value Answer: 8 to 8 +4 marks

Solution & Explanation

### Core Logic According to Molecular Orbital Theory (MOT), the number of molecular orbitals (MOs) formed is equal to the total number of atomic orbitals (AOs) combined. ### Step 1: Counting atomic orbitals For a single atom in the 2nd period, the valence shell has: One 2s orbital Three 2p orbitals (2p_x, 2p_y, 2p_z) Total = 4 atomic orbitals per atom. For a diatomic molecule, two such atoms combine. Total atomic orbitals = 4 times 2 = 8. ### Step 2: Forming molecular orbitals Combining these 8 atomic orbitals yields 8 molecular orbitals: - From 2s: sigma_2s and sigma^*_2s (2 MOs) - From 2p: sigma_2p_z, pi_2p_x, pi_2p_y, pi^*_2p_x, pi^*_2p_y, sigma^*_2p_z (6 MOs) Total MOs = 2 + 6 = 8. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Reference Study Guides

Chemical Bonding And Molecular Structure

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 6

Q74 jee_main_2024_31_jan_evening Ionic Bond and Lattice Energy
Which of the following is least ionic?
  • A. text(1) BaCl_2
  • B. text(2) AgCl
  • C. text(3) KCl
  • D. text(4) CoCl_2

Solution

### Core Logic According to Fajan's rules, covalent character is favored by high charge and small size of the cation, and by cations with a pseudo-noble gas configuration. Ag^+ has a pseudo-noble gas configuration (ns^2np^6nd^10), which results in high polarizing power compared to s-block and typical transition elements. Therefore, AgCl has the maximum covalent character and is the least ionic among the given options. Ionic character order: AgCl < CoCl_2 < BaCl_2 < KCl ### Step 1: Final Selection Because AgCl is the most covalent, it is the least ionic. Hence, option (2) is correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q81 jee_main_2024_31_jan_evening Dipole Moment and Fractional Charge
A diatomic molecule has a dipole moment of 1.2text D. If the bond distance is 1mathrmAA, then fractional charge on each atom is _________ times 10^-10text esu. (Given: 1text D = 10^-18text esu cm)
Numerical Answer. Answer: 1.2 to 1.2

Solution

### Related Formula mu = q times d ### Core Logic Given dipole moment, mu = 1.2text D = 1.2 times 10^-18text esu cm. Bond distance, d = 1mathrmAA = 10^-8text cm. We need to find the fractional charge q. ### Step 1: Calculation q = fracmud q = frac1.2 times 10^-18text esu cm10^-8text cm q = 1.2 times 10^-10text esu ### Step 2: Final Formatting The question asks for the fractional charge in the form x times 10^-10text esu. Therefore, the value is 1.2. *Note: Based on NTA officially accepting 12 (if asked for x times 10^-11) or 1.2. We will format it exactly as calculated.* ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q77 jee_main_2024_31_jan_morning Molecular Orbital Theory
The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis Choose the most appropriate from the options given below:
  • A. textA, B, C only
  • B. textA and C only
  • C. textB, C, D only
  • D. textB and D only

Solution

### Core Logic Conditions for the linear combination of atomic orbitals (LCAO) to form molecular orbitals: 1. The combining atomic orbitals must have the same or nearly the same energy. 2. The combining atomic orbitals must have the same symmetry about the molecular axis. 3. The combining atomic orbitals must overlap to the maximum extent (not minimum). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q85 jee_main_2024_31_jan_morning Hybridization
The number of species from the following in which the central atom uses sp^3 hybrid orbitals in its bonding is NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3
Numerical Answer. Answer: 4 to 4

Solution

### Core Logic Analyzing the hybridization of the central atom in each species: - NH_3: 3 bp + 1 lp = 4 electron domains rightarrow sp^3 - SO_2: 2 bp + 1 lp = 3 electron domains rightarrow sp^2 - SiO_2: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally rightarrow sp^3 - BeCl_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - CO_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - H_2O: 2 bp + 2 lp = 4 electron domains rightarrow sp^3 - CH_4: 4 bp + 0 lp = 4 electron domains rightarrow sp^3 - BF_3: 3 bp + 0 lp = 3 electron domains rightarrow sp^2 Total species with sp^3 hybridization: NH_3, SiO_2, H_2O, CH_4. Total count = 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

More Chemical Bonding and Molecular Structure Questions — jee_main_2024_30_jan_morning

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