For a polyatomic ion having the general formula AB₃ⁿ^±, where A is an atom from Group 6A, B is an atom from Group 7A, and the ionic charge is n±, what charge must the ion have in order to yield a molecular geometry that is trigonal pyramidal?

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In order to have a trigonal pyramidal molecular geometry, the polyatomic ion must have a central atom with four electron groups, three of which are bonding and one of which is a lone pair. This means that the central atom must have a tetrahedral electron geometry. Group 6A elements have 6 valence electrons, and Group 7A elements have 7 valence electrons. Therefore, if we use one atom from each group in the ion, we have a total of 26 valence electrons (6 from A, 7 from B, and 1 from the single negative charge). To achieve a tetrahedral electron geometry with this number of electrons, the central atom must have one lone pair. Looking at the formula given, there is only one atom that could be the central atom with one lone pair, and that is the A atom. Therefore, the ion must be in the form A1B3S1 with one negative charge, giving a total of 26 valence electrons. In order to achieve a trigonal pyramidal molecular geometry, the ion must have a total of four electron groups around the central A atom, which means that there must be three bonding groups (corresponding to the three B atoms) and one lone pair. This requires that the ion have an overall charge of 1+, so that the central A atom can have a lone pair of electrons. Therefore, the correct answer is C) 1+.

For a Polyatomic Ion Having the General Formula AB3n±, Where

For a Polyatomic Ion Having the General Formula AB₃ⁿ^±, Where