1890 Belgian general election

1890 Belgian general election

← 1888 10 June 1890 (1890-06-10) 1892 →
69 of the 138 seats in the Chamber of Representatives 70 seats needed for a majority
First party Second party   Leader Auguste Beernaert Party Catholic Liberal Leader's seat Candidate for PM Seats before 98 seats 40 seats Seats won 29 40 Seats after 94 44 Seat change 4 4 Popular vote 17,253 20,829 Percentage 45.15% 54.51%
Government before election Beernaert Catholic Government after election Beernaert Catholic

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Partial general elections were held in Belgium on 10 June 1890.^[1][2] In the elections for the Chamber of Representatives the result was a victory for the Catholic Party, which won 94 of the 138 seats.^[2]

Under the alternating system, elections were held in only four out of the nine provinces: Hainaut, Limburg, Liège and East Flanders. Thus, only 69 seats out of the 138 were up for election. Additionally, a special election was held in Neufchâteau on 3 June 1890.^[3]

Party		Votes	%	Seats
				Won	Total	+/–
	Liberal Party	20,829	54.51	40	44	+4
	Catholic Party	17,253	45.15	29	94	–4
	Belgian Labour Party	98	0.26	0	0	New
	Others	30	0.08	0	0	0
Total		38,210	100.00	69	138	0
Total votes		38,210	–
Registered voters/turnout		59,452	64.27
Source: Mackie & Rose,[4] Sternberger et al.

The distribution of seats among the electoral districts was as follows:^[5]