1890 United States Gubernatorial Elections
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United States gubernatorial elections were held in 1890, in 27 states, concurrent with the House and Senate elections, on November 4, 1890 (except in Alabama, Arkansas, Georgia, Idaho, Maine, Oregon, Rhode Island, Vermont and Wyoming, which held early elections). In
New Hampshire New Hampshire is a state in the New England region of the northeastern United States. It is bordered by Massachusetts to the south, Vermont to the west, Maine and the Gulf of Maine to the east, and the Canadian province of Quebec to the nor ...
, the newly elected Governor's term began in the January following the election for the first time, rather than in the following June as previously.
Idaho Idaho ( ) is a state in the Pacific Northwest region of the Western United States. To the north, it shares a small portion of the Canada–United States border with the province of British Columbia. It borders the states of Montana and Wyomi ...
and
Wyoming Wyoming () is a state in the Mountain West subregion of the Western United States. It is bordered by Montana to the north and northwest, South Dakota and Nebraska to the east, Idaho to the west, Utah to the southwest, and Colorado to the s ...
held their first gubernatorial elections on achieving statehood.


Results


See also

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1890 United States elections The 1890 United States elections occurred in the middle of Republican President Benjamin Harrison's term. Members of the 52nd United States Congress were chosen in this election. The Republicans suffered major losses due to the Panic of 1890 an ...


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* * * * * {{USGovElections November 1890 events