Kolkata: A pregnant woman was allegedly poisoned to death by her husband and in-laws in East Midnapore district in Bengal after she refused to abort a girl child.
The victim has been identified as Rubina Bibi, a resident of Kotbar village. Last year, she married Sheikh Rezabul and was seven months pregnant at the time of the incident.
Rezabul and his parents are absconding and no arrest has been made so far in this connection.
The incident took place on August 6 when Rubina was asked to abort the girl child but she refused. Her husband and in-laws allegedly poisoned her to get rid of both (the mother and her girl child).
Later, they admitted her (in an unconscious state) to Tamluk Hospital claiming that she accidently consumed pesticide at home. On Tuesday (August 8), her condition started deteriorating and she died in the evening.
Speaking to News18, victim’s mother Sakina Bibi, said, “They used to harass my daughter in the name of dowry. They demanded Rs 1 lakh from us but we told them that it is not possible to arrange such a huge amount of money. Somehow we managed to arrange Rs 30,000 and gave it to her husband but he demanded more. My daughter used to tell me that they will kill her one day but I told her that things will be fine. I never imagined that they will actually kill her.”
“They poisoned my daughter after they came to know that she is carrying a girl child in her womb. We have already lodged a complaint at Chandipur police station and demand stern action against them,” she added.
The Superintendent of Police, Alok Rajoria, said, “It is a case of poisoning and we are looking for the accused in this case. We are also looking for the doctors who did the ultrasonography and revealed the gender of the baby.”
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