Based on electronegativity differences, which of the following is most likely to

Question

Based on electronegativity differences, which of the following is most likely to be ionic? BaF_2 Br_2 PH_3 NO_2 Cl_4 How many electrons can be contained in all of the orbitals with n = 4? 2 8 10 18 32 Which of the following molecules has no dipole moment? CO_2 NH_3 H_2O all none How many electrons can be described by the quantum numbers n = 2, l = 2, m_1 = 1? 0 2 6 10 14 Which of the following combinations of quantum numbers is not allowed? n = 1, l = 1, m_1 = 0, m_2 = 1/2 n = 3, l = 0, m_1 = 0, m_2 = -1/2 n = 2, l = 1, m_1 = -1, m_2 = 1/2 n = 4, l = 3, m_1 = -2, m_2 = -1/2 n = 4, l = 2, m_1 = 0, m_2 = 1/2

Explanation / Answer

5. The electronegativity of element is given in the bracket:

H (2.1), C (2.5), N (3.0), O (3.5), F (4), P (2.1), Cl (3.0), Br (2.8) and Ba (0.9), I (2.4)

Electronegativity difference in BaF2 = 4 – 0.9 = 3.1

Electronegativity difference in B2 = 2.8 – 2.8 = 0

Electronegativity difference in PH3 = 2.1 – 2.1 = ~0

Electronegativity difference in NO2 = 3.5 – 3.0 = 0.5

Electronegativity difference in CI4 = 2.5-2.4 = 0.1

For a molecule to be ionic the electronegativity difference should be >2. Therefore, only BaF2 will be ionic. (option A is correct)

6. The maximum number of electrons in a shell = 2n2 = 2 x (4)2 = 32 (option e)

7. CO2 due to linear structure does not have dipole moment (Option a)

8. One orbital can have two electrons, therefor, by the quantum number n = 2, l = 2, ml = 1, two electrons with +1/2 and -1/2 spins can be described (option b)

9. n = 1, l = 1, mI = 0, ms =1/2 is not allowed (option a)

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