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Chapter 10: Problem 2

Write Lewis symbols for the following ions. (a) \(\mathrm{H}^{-}\) (b) \(\operatorname{Sn}^{2+} ;\) (c) \(\mathrm{K}^{+} ;\) (d) \(\mathrm{Br}^{-} ;\) (e) \(\mathrm{Se}^{2-} ;\) (f) \(\mathrm{Sc}^{3+}\).

Short Answer

Expert verified
The Lewis symbols for \( \mathrm{H}^{-},\operatorname{Sn}^{2+},\mathrm{K}^{+}, \mathrm{Br}^{-},\mathrm{Se}^{2-}, \mathrm{Sc}^{3+} \) are \( \stackrel{..}{\mathrm{H}}^{-},\stackrel{: :}{\mathrm{Sn}}^{2+}, \mathrm{K}^{+}, \stackrel{.. : .. : .. :}{\mathrm{Br}}^{-},\stackrel{.. : .. : .. :}{\mathrm{Se}}^{2-}, \mathrm{Sc}^{3+} \) respectively.

Step by step solution

01

Draw the Lewis symbol for \( \mathrm{H}^{-} \)

The hydrogen atom \( \mathrm{H} \) has one valence electron. Since the ion \( \mathrm{H}^{-} \) is negatively charged, it means it has gained an electron. Therefore, \( \mathrm{H}^{-} \) has two electrons. The Lewis symbol is : \( \stackrel{..}{\mathrm{H}}^{-} \)
02

Draw the Lewis symbol for \( \operatorname{Sn}^{2+} \)

The tin atom \( \operatorname{Sn} \) has four valence electrons. Since the ion \( \operatorname{Sn}^{2+} \) is positively charged, it means it has lost two electrons. Therefore, \( \operatorname{Sn}^{2+} \) has two electrons. The Lewis symbol is \( \stackrel{: :}{\mathrm{Sn}}^{2+} \)
03

Draw the Lewis symbol for \( \mathrm{K}^{+} \)

The potassium atom \( \mathrm{K} \) has one valence electron. Since the ion \( \mathrm{K}^{+} \) is positively charged, it means it has lost an electron. Therefore, \( \mathrm{K}^{+} \) has no electrons. The Lewis symbol is \( \mathrm{K}^{+} \)
04

Draw the Lewis symbol for \( \mathrm{Br}^{-} \)

The bromine atom \( \mathrm{Br} \) has seven valence electrons. Since the ion \( \mathrm{Br}^{-} \) is negatively charged, it means it has gained an electron. Therefore, \( \mathrm{Br}^{-} \) has eight electrons. The Lewis symbol is \( \stackrel{.. : .. : .. :}{\mathrm{Br}}^{-} \)
05

Draw the Lewis symbol for \( \mathrm{Se}^{2-} \)

The selenium atom \( \mathrm{Se} \) has six valence electrons. Since the ion \( \mathrm{Se}^{2-} \) is negatively charged, it means it has gained two electrons. Therefore, \( \mathrm{Se}^{2-} \) has eight electrons. The Lewis symbol is \( \stackrel{.. : .. : .. :}{\mathrm{Se}}^{2-} \)
06

Draw the Lewis symbol for \( \mathrm{Sc}^{3+} \)

The scandium atom \( \mathrm{Sc} \) has three valence electrons. Since the ion \( \mathrm{Sc}^{3+} \) is positively charged, it means it has lost three electrons. Therefore, \( \mathrm{Sc}^{3+} \) has no electrons. The Lewis symbol is \( \mathrm{Sc}^{3+} \)

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Most popular questions from this chapter

One possibility for the electron-group geometry for seven electron groups is pentagonal-bipyramidal, as found in the IF \(_{7}\) molecule. Write the VSEPR notation for this molecule. Sketch the structure of the molecule, labeling all the bond angles.

Explain the important distinctions between (a) ionic and covalent bonds; (b) lone-pair and bond-pair electrons; (c) molecular geometry and electron-group geometry; (d) bond dipole and resultant dipole moment; (e) polar molecule and nonpolar molecule.

Which molecule is nonlinear? (a) \(\mathrm{SO}_{2} ;\) (b) \(\mathrm{CO}_{2}\); (c) HCN; (d) NO.

Predict the shapes of the following sulfur-containing species. (a) \(\mathrm{SO}_{2} ;\) (b) \(\mathrm{SO}_{3}^{2-} ;\) (c) \(\mathrm{SO}_{4}^{2-}\).

The following polyatomic anions involve covalent bonds between O atoms and the central nonmetal atom. Propose an acceptable Lewis structure for each. (a) \(\mathrm{SO}_{3}^{2-} ;\) (b) \(\mathrm{NO}_{2}^{-} ;\) (c) \(\mathrm{CO}_{3}^{2-} ;\) (d) \(\mathrm{HO}_{2}^{-}\).
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