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Baby dies in India after man ‘cuts open pregnant wife’s belly’

Husband allegedly wanted to check whether child was male

Chiara Giordano
Monday 21 September 2020 17:01 BST
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Stock image. (Getty Images/iStockphoto)
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A baby has died in India after a man allegedly cut open his pregnant wife’s belly with a sickle.  

The man is accused of attacking his spouse with the hand tool because he wanted to find out the baby’s gender. 

The woman’s family told local media the man wanted a son but a priest had predicted the couple were expecting their sixth daughter. 

The mother was in a critical condition when she was taken to a hospital in the capital, Delhi, where doctors fought to save her life and that of her unborn child.

However the woman later gave birth to a stillborn son, the BBC reports.

It is not clear what stage in her pregnancy the woman, who is said to be in a stable condition, was at.

The incident is said to have taken place at the couple’s home in Budaun district, Uttar Pradesh, on Saturday night.  

Police arrested the man but he claimed he did not hurt his wife intentionally, according to the BBC.

He reportedly admitted he threw the sickle at her but told local media he had not intended to injure her so seriously.

An investigation into suspected sex-selective abortions was launched in a district of northern India last year after government data revealed none of the 216 children born across 132 villages over three months were girls.  

India outlawed the selective abortion of female foetuses in 1994 but the practice remains commonplace in the country, where parents often see boys as breadwinners and girls as costly liabilities.  

The last population census, conducted in 2011, found there were only 943 females per 1,000 males in India. 

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