Interpretation: The elements corresponding to the given electronic configuration are to be identified; the identified elements are to be arranged in the increasing order of their electronegativity value.
Concept introduction: The number of electrons can be obtained from the electronic configuration of an element. The number of electrons is equal to the
Electronegativity can be defined as the measure of the tendency of an atom to attract the shared pair of electrons towards itself.
To determine: The element corresponding to the given electron configurations and the correct arrangement of these elements in the increasing order of electronegativity.
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Chemistry: An Atoms First Approach
- Use the following data to estimate AH; for sodium bromide. Na(s) + Br2 (9) → NaBr(s) Lattice energy -742 kJ/mol Ionization energy for Na 495 kJ/mol -325 kJ/mol Electron affinity of Br Bond energy of Br2 Enthalpy of sublimation for Na 193 kJ/mol 109 kJ/mol AH; = | kJ/molarrow_forward(c) Draw the orbital diagrams and Lewis symbols to depict the formation of Na* and CI ions from the atoms. Give the formula of the compound formed. (d) The predicted bond length for HF is 109 pm (the sum of the covalent radii of H, 37 pm and F. 72 pm), however the actual bond length for HF is shorter (92 pm). It was observed that the difference between predicted and actual bond lengths becomes smalleor going down the halogen group from HF to HI Describe these observationsarrow_forwardComplete the blanks in each column, as in the first example: Name of the Magnesium Lithium ion Sulfide ion particle atom charge 3+ condensed e [He]3s? [He]2s?2p? [Kr]5s?4d10 configuration atomic Lewis • Mg. :Cl: structurearrow_forward
- The structure of sodium hyaluronate, a sodium salt of hyaluronic acid used in skincare products for its hydrating properties, is shown below. What is the shape of the bonds at (ii) ? Hint: atom (iii) has been completed for you as an example. (iii) bond angle: 109.5°, geometry of the electron pairs: tetrahedral, shape of the bonds: tetrahedral ОН, OH ·•·•·||||| N O H H Ol.. OH |||| iv + Naarrow_forwardAn important starting material for the manufacture ofpolyphosphazenes is the cyclic molecule (NPCl₂)₃. The mol-ecule has a symmetrical six-membered ring of alternating N and P atoms, with the Cl atoms bonded to the P atoms. The nitrogen-phosphorus bond length is significantly less than that expectedfor an N−P single bond.(a) Draw a likely Lewis structure for the molecule.(b) How many lone pairs of electrons do the ring atoms have?(c) What is the order of the nitrogen-phosphorus bond?arrow_forward1) Answer the following questions using principles of chemical bonding and molecular structure. (a) Consider the carbon dioxide molecule, CO2, and the carbonate ion, CO32– (i) Draw the complete Lewis electron-dot structure for each species. (ii) Account for the fact at the carbon-oxygen bond length in CO32– is greater than the carbon-oxygen bond length in CO2. 2) Explain the following in terms of the electronic structure and bonding of the compounds considered. (a) Liquid oxygen is attracted to a strong magnet, whereas liquid nitrogen is not. (b) The SO2 molecule has a dipole moment, whereas the CO2 molecule has no dipole moment. Include the Lewis (electron-dot) structures in your explanation. 3) Answer the following questions using principles of chemical bonding and molecular structure: (a) Consider the molecules CF4 and SF4. (i) Draw the complete Lewis electron-dot structure for each molecule. (ii) In terms of molecular geometry, account for the fact that the…arrow_forward
- Common exceptions to the octet rule are compounds and polyatomic ions with central atoms having more than 8 electrons around them. Phosphorus pentafluoride, PF5; sulfur tetrafluoride, SF4; xenon tetrafluoride, XeF4; and tri-iodide ion, I3, are all examples of exceptions to the octet rule. (a) Draw the Lewis structures of these substances.(b) For which elements in these substances can theatoms have more than 8 electrons around them?(c) How can the atoms of the elements youidentified in Part (b) be surrounded by morethan 8 electrons?arrow_forwardAmmonia reacts with boron trifluoride to form a stablecompound, as we saw in Section 8.7. (a) Draw the Lewisstructure of the ammonia–boron trifluoride reaction product.(b) The B—N bond is obviously more polar than the C—C bond. Draw the charge distribution you expect on theB—N bond within the molecule (using the delta plus anddelta minus symbols mentioned in Section 8.4). (c) Borontrichloride also reacts with ammonia in a similar way tothe trifluoride. Predict whether the B—N bond in the trichloridereaction product would be more or less polar thanthe B—N bond in the trifluoride product, and justify yourreasoning.arrow_forwardIdentify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom: (a) 1s 22s 22p 5 (b) 1s 22s 22p 63s 2 (c) 1s 22s 22p 63s 23p 64s 23d 10arrow_forward
- Describe the location of electrons; describe how electron placement determines chemical bonding, stability, and becoming an ionarrow_forwardThe HF bond length is 92 pm, 16% shorter than the sum of the covalent radii of H (37 pm) and F (72 pm). Suggest a reason for this difference. Similar data show that the difference becomes smaller down the group, from HF to HI. Explain.arrow_forward4, 134. Although both Br3 and Iz ions are known, the F3 ion has not been observed. Explain. 101 ond 102 Would vou make thearrow_forward
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning