According to molecular orbital theory of benzene, each carbon atom is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridized with a bond angle of \({\rm{120^\circ }}{\rm{.}}\) There are six \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) atomic orbitals (one on each carbon atom) which are parallel to each other but are perpendicular to the plane of the ring and project above and below the plane. Each \({{\rm{p}}_{\rm{z}}}\) orbital interact equally with its two neighbours producing a circular double-doughnut shaped molecular orbital embracing all six carbon
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313593
Considering the state of hybridisation of carbon atoms, the linear molecule is
4 \({\mathrm{\mathrm{H}_{2} \mathrm{O}}}\) and \({\mathrm{\mathrm{NO}_{2}}}\)
Explanation:
In \({\mathrm{\mathrm{NH}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{BF}_{3}, \mathrm{~B}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{H}_{2} \mathrm{O}, \mathrm{O}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}, \mathrm{~N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. So, the correct option is (2).
According to molecular orbital theory of benzene, each carbon atom is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridized with a bond angle of \({\rm{120^\circ }}{\rm{.}}\) There are six \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) atomic orbitals (one on each carbon atom) which are parallel to each other but are perpendicular to the plane of the ring and project above and below the plane. Each \({{\rm{p}}_{\rm{z}}}\) orbital interact equally with its two neighbours producing a circular double-doughnut shaped molecular orbital embracing all six carbon
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313593
Considering the state of hybridisation of carbon atoms, the linear molecule is
4 \({\mathrm{\mathrm{H}_{2} \mathrm{O}}}\) and \({\mathrm{\mathrm{NO}_{2}}}\)
Explanation:
In \({\mathrm{\mathrm{NH}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{BF}_{3}, \mathrm{~B}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{H}_{2} \mathrm{O}, \mathrm{O}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}, \mathrm{~N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. So, the correct option is (2).
According to molecular orbital theory of benzene, each carbon atom is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridized with a bond angle of \({\rm{120^\circ }}{\rm{.}}\) There are six \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) atomic orbitals (one on each carbon atom) which are parallel to each other but are perpendicular to the plane of the ring and project above and below the plane. Each \({{\rm{p}}_{\rm{z}}}\) orbital interact equally with its two neighbours producing a circular double-doughnut shaped molecular orbital embracing all six carbon
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313593
Considering the state of hybridisation of carbon atoms, the linear molecule is
4 \({\mathrm{\mathrm{H}_{2} \mathrm{O}}}\) and \({\mathrm{\mathrm{NO}_{2}}}\)
Explanation:
In \({\mathrm{\mathrm{NH}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{BF}_{3}, \mathrm{~B}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{H}_{2} \mathrm{O}, \mathrm{O}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}, \mathrm{~N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. So, the correct option is (2).
According to molecular orbital theory of benzene, each carbon atom is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridized with a bond angle of \({\rm{120^\circ }}{\rm{.}}\) There are six \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) atomic orbitals (one on each carbon atom) which are parallel to each other but are perpendicular to the plane of the ring and project above and below the plane. Each \({{\rm{p}}_{\rm{z}}}\) orbital interact equally with its two neighbours producing a circular double-doughnut shaped molecular orbital embracing all six carbon
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313593
Considering the state of hybridisation of carbon atoms, the linear molecule is
4 \({\mathrm{\mathrm{H}_{2} \mathrm{O}}}\) and \({\mathrm{\mathrm{NO}_{2}}}\)
Explanation:
In \({\mathrm{\mathrm{NH}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{BF}_{3}, \mathrm{~B}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{H}_{2} \mathrm{O}, \mathrm{O}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}, \mathrm{~N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. So, the correct option is (2).
According to molecular orbital theory of benzene, each carbon atom is \({\rm{s}}{{\rm{p}}^{\rm{2}}}\) hybridized with a bond angle of \({\rm{120^\circ }}{\rm{.}}\) There are six \({\rm{2}}{{\rm{p}}_{\rm{z}}}\) atomic orbitals (one on each carbon atom) which are parallel to each other but are perpendicular to the plane of the ring and project above and below the plane. Each \({{\rm{p}}_{\rm{z}}}\) orbital interact equally with its two neighbours producing a circular double-doughnut shaped molecular orbital embracing all six carbon
CHXI04:CHEMICAL BONDING AND MOLECULAR STRUCTURE
313593
Considering the state of hybridisation of carbon atoms, the linear molecule is
4 \({\mathrm{\mathrm{H}_{2} \mathrm{O}}}\) and \({\mathrm{\mathrm{NO}_{2}}}\)
Explanation:
In \({\mathrm{\mathrm{NH}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{BF}_{3}, \mathrm{~B}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}^{-}, \mathrm{N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. In \({\mathrm{\mathrm{H}_{2} \mathrm{O}, \mathrm{O}}}\) is \({\mathrm{\mathrm{sp}^{3}}}\) hybridised. In \({\mathrm{\mathrm{NO}_{2}, \mathrm{~N}}}\) is \({\mathrm{\mathrm{sp}^{2}}}\) hybridised. So, the correct option is (2).