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1886 Belgian general election

1886 Belgian general election

← 1884 8 June 1886 (1886-06-08) 1888 →

69 of the 138 seats in the Chamber of Representatives
70 seats needed for a majority
  First party Second party
  Auguste Beernaert(03).jpg No image.png
Leader Auguste Beernaert
Party Catholic Liberal
Leader since Candidate for PM
Seats before 86 seats 52 seats
Seats won 32 37
Seats after 98 40
Seat change Increase 12 Decrease 12
Popular vote 17,979 18,965
Percentage 48.67% 51.33%

Government before election

Beernaert
Catholic

Elected Government

Beernaert
Catholic

State Coat of Arms of Belgium.svg
This article is part of a series on the
politics and government of
Belgium
Constitution
Foreign relations

Partial general elections were held in Belgium on 8 June 1886.[1][2] In the elections for the Chamber of Representatives the result was a victory for the Catholic Party, which won 98 of the 138 seats.[2]

Under the alternating system, elections were only held in four out of the nine provinces: Hainaut, Limburg, Liège and East Flanders.

The Catholics continued their gains, after the 1884 elections that gave them a victory over the Liberals. With the defeat of the Liberals by the Catholics in the eight-member arrondissement of Ghent, the Catholics now had all "Flemish" seats. Catholics also gained both seats of the arrondissement of Waremme and two out of seven in the arrondissement of Charleroi.

Results

Chamber of Representatives

Party Votes % Seats
Won Total +/–
Liberal Party 18,965 51.3 37 40 –12
Catholic Party 17,979 48.7 32 98 +12
Invalid/blank votes
Total 36,944 100 69 138 0
Registered voters/turnout 57,962
Source: Mackie & Rose,[3] Sternberger et al.

References

  1. ^ Codebook Constituency-level Elections Archive, 2003
  2. ^ a b Sternberger, D, Vogel, B & Nohlen, D (1969) Die Wahl der Parlamente: Band I: Europa - Erster Halbband, p105
  3. ^ Thomas T Mackie & Richard Rose (1991) The International Almanac of Electoral History, Macmillan, pp50–51

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