Among H_2S, H_2O, NF_3, NH_3 and CHCl_3, iden | vedclass

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(C) The dipole moments of the given molecules are as follows: $H_2S = 0.95 \ D$ $H_2O = 1.85 \ D$ $NF_3 = 0.23 \ D$ $NH_3 = 1.47 \ D$ $CHCl_3 = 1.04 \

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(C) The dipole moments of the given molecules are as follows: $H_2S = 0.95 \ D$ $H_2O = 1.85 \ D$ $NF_3 = 0.23 \ D$ $NH_3 = 1.47 \ D$ $CHCl_3 = 1.04 \ D$ Comparing these values, $NF_3$ has the lowest dipole moment. Therefore, the molecule $(X)$ is $NF_3$. In $NF_3$, the central atom is Nitrogen $(N)$. Nitrogen has $5$ valence electrons. It forms $3$ single bonds with $3$ Fluorine atoms, leaving $1$ lone pair on the Nitrogen atom. Thus, the number of lone pairs on the central atom is $1$.

Among $H_2S$, $H_2O$, $NF_3$, $NH_3$ and $CHCl_3$, identify the molecule $(X)$ with the lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule $(X)$ is:

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