11th Standard Syllabus & Materials
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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Chemistry Test1.
What are chemical bonds ?
2.
How does the effect of the process C2 \(\rightarrow\) C2+ + e- affect the bond order ?
3.
Bond order increases with loss of electron in bonding molecular orbital - State True or False and give reason.
4.
How is the sequence of energy levels of molecular orbitals written in case of heavier diatomic molecules ?
5.
Identify the magnetic nature of the anion of Na2O2.
1.
The interatomic attractive forces which holds the constituent atoms/ions together in a molecule are called chemical bonds.
2.
\(C^+_2:(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2 (\pi_2p_x)^2=(\pi_2p_y)^1\)
Bond order = \({1\over2}(8-4)=2\)
\(C^+_2:(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2 (\pi_2p_x)^2=(\pi_2p_y)^1\)
Bond order = \({1\over2}(7-4)=1.5\)
\(\therefore\)The Bond order decreases by 1.5 during the process.
3.
1. False
2. Bond order increases only when the electron is last from the anti bonding molecular orbital.
4.
For heavier diatomic molecules, the order of energy levels of \(\sigma_2p_z,\pi_2p_x,\pi_2p_y\)is reversed, (i.e.) \(\sigma_2p_z\)has lesser energy than\(\pi_2p_x\) and \(\pi_2p_y\)
Hence, the order of increasing energies of Mos is,
\(\sigma_{1s}<\sigma^*_{1s}<\sigma_{2s}<\sigma^*_{2s}<\sigma_{2}p_z<\pi_2p_x\)=\(\pi_2p_y<\pi^*_2p_x=\pi^*_2p_y<\sigma_2p_z\).
5.
The anion of Na2O2 isO2-2 (peroxide ion).
No. of electrons in O2-2: 8 + 8 + 2 = 18 e-
Electronic configuration: \((\sigma_{1s})^2 (\sigma^*_{1s})^2(\sigma^s_{2})^2(\sigma^{s*}_{2})\)\((\sigma_2p_z)^2(\pi_2p_x)^2(\pi_2p_y)^2(\pi^*_2p_x)^2(\pi^*_2p_y)^2\)
Bond order: \({1\over2}(N_b-N_a)={1\over2}(10-8)=1\)
Magnetic Nature: Has no unpaired electrons. Hence diamagnetic.
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
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