Question

Electronic Structure of Atoms 1.20 Write the ground-state electron configuration for each atom. After each atom is its atomic number in parentheses. (a) Sodium (11) (b) Magnesium (12) (c) Oxygen (8) (d) Nitrogen (7) 1.21 Identify the atom that has each ground-state electron configuration. (a) 1s 2s 2p 3:23p (b) 1:2 2:22 1.23 How many electrons are in the valence shell of each atom? (a) Carbon (b) Nitrogen (c) Chlorine (d) Aluminum Lewis Structures and Formal Charge 1.25 Which compounds have nonpolar covalent bonds, which have polar covalent bonds, and which have ions? (a) LiF (b) CHF (c) MgCl, (d) HCI 1.26 Using the symbols 8- and 8+, indicate the direction of polarity, if any, in each covalent bond. (a) C-CI (b) S-H (c) C-S (a) P-H

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Answer 1

1.20) a)Atomic number of sodium (Na), Z = 11 No of electrons = 12 Electronic configuration of Na= 152 2s 2p 3s" =[ Ne]3s" - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - b) Atomic number of magnesium (Mg), Z = 12 No of electrons = 12 Electronic configuration of Mg = 152 252 2p 3s2 =[ Ne]3s2 - - - - - - - - - - - - - - - - - - - - - - c) Atomic number of oxygen (O), Z = 8 No of electrons = 8 Electronic configuration of O= 1s22s22p* = [He]2s22p* - - - - - - - - - - - - - - - - d) Atomic number of nitrogen (N), Z = 7 No of electrons = 7 Electronic configuration of N = 1s22s22p* = [He]2s22p

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1.21) a) Electronic configuration of the element = 1s22s 2p 3sº3p No of electrons present in the element = 2+2+6 +2+4 = 16 Atomic number = no of electrons = 16 According to periodic table, element with atomic number 16 is Sulfur (S) Thus, the identity of the element is Sulfur (S) b) Electronic configuration of the element = 1822s22p* No of electrons present in the element = 2+2+4 = 8 Atomic number = no of electrons = 8 According to periodic table, element with atomic number 8 is Oxygen (O) Thus, the identity of the element is Oxygen (O) - - - - - - - - - - - - - - - - - - - - 1.23)a) Atomic number of Carbon(C), Z = 6 Electronic configuration of C= 152s22p2 No of valence electrons = 2+2 = 4 b) Atomic number of nitrogen (N), Z = 7 Electronic configuration of N = 1822s22p3 No of valence electrons = 2+3 = 5 c) Atomic number of chlorine(a),Z = 17 Electronic configuration of a = 1522s22p 3s23p3 No of valence electrons = 2+5 = 7 d) Atomic number of Aluminum (Al), Z =13 Electronic configuration of Al = 1522s 2p 3s 3p1 No of valence electrons = 2+1 = 3