61. Each of the following is either linear , angular (bent), planar tetrahedral, or octahedral . Indicate the correct shape of (a) H_(2)S (b) N_(2)O_(4) ; (c) HCN; (d) SbCl_(6)^- (e) BF_(4)^-

To determine the shape of each molecule, we can use the VSEPR (Valence Shell Electron Pair Repulsion) theory. According to this theory, the shape of a molecule is determined by the repulsion between the electron pairs in the valence shell of the central atom.<br /><br />(a) $H_{2}S$: In $H_{2}S$, sulfur is the central atom. It has two bonded atoms (hydrogen) and two lone pairs of electrons. According to VSEPR theory, the shape of $H_{2}S$ is bent or angular.<br /><br />(b) $N_{2}O_{4}$: In $N_{2}O_{4}$, nitrogen is the central atom. It has two bonded atoms (nitrogen) and two lone pairs of electrons. According to VSEPR theory, the shape of $N_{2}O_{4}$ is planar tetrahedral.<br /><br />(c) HCN: In HCN, carbon is the central atom. It has one bonded atom (hydrogen) and one triple bond with nitrogen. According to VSEPR theory, the shape of HCN is linear.<br /><br />(d) $SbCl_{6}^{-}$: In $SbCl_{6}^{-}$, antimony is the central atom. It has six bonded atoms (chlorine) and no lone pairs of electrons. According to VSEPR theory, the shape of $SbCl_{6}^{-}$ is octahedral.<br /><br />(e) $BF_{4}^{-}$: In $BF_{4}^{-}$, boron is the central atom. It has four bonded atoms (fluorine) and no lone pairs of electrons. According to VSEPR theory, the shape of $BF_{4}^{-}$ is tetrahedral.<br /><br />Therefore, the correct shapes for each molecule are:<br />(a) $H_{2}S$: Bent or angular<br />(b) $N_{2}O_{4}$: Planar tetrahedral<br />(c) HCN: Linear<br />(d) $SbCl_{6}^{-}$: Octahedral<br />(e) $BF_{4}^{-}$: Tetrahedral