General, Organic, & Biological Chemistry 3rd Edition

Published by McGraw-Hill Education
ISBN 10: 0-07351-124-2
ISBN 13: 978-0-07351-124-5

Chapter 3 - Ionic Compounds - Problems - Page 97: 3.36

Answer

a) no. of protons = 19. No. of electrons = 18. b) no. of protons = 16. No. of electrons = 18 c) no. of protons = 25 No. of electrons = 23 d) no. of protons = 26 No. of electrons = 24 e) no. of protons = 55. No. of electrons = 54 f) no. of protons = 53. No. of electrons = 54

Work Step by Step

a) The element is potassium (K). Hence, no of protons = 19. Now, for neutral atom no. of electrons = 19. Here, there is +1 charge on the atom. Hence, no. of electrons = 19-1 = 18. b) The element is sulfur (S). hence, no. of protons = 16. Now, for neutral atom no. of electrons = 16. here, there is -2 charge on the atom i.e. two extra electrons are present. Hence, no. of electrons = 16+2 = 18. c) the element is Manganese (Mn). hence, no. of protons = 25. for neutral atom no. of electrons = 25. Here, there is +2 charge on the atom i.e. two electron is less than in the neutral atom. hence, no. of electrons = 23. d) the element is iron (Fe). Hence, no. of protons = 26. for neutral atom no. of electrons = 26. Here, there is +2 charge on the atom i.e. two electron is less than in the neutral atom. hence, no. of electrons = 24. e) the element is Cesium (Cs). Hence, no. of protons = 55. for neutral atom no. of electrons = 55. here, there is +1 charge on the atom i.e. one electron is less than in the neutral atom. hence, no. of electrons = 54. f) the element gained 1 electron
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